Polymath Wiki - User contributions [en]

"Low Dimensions" grant acknowledgments

2010-08-23T20:46:33Z

Jbd: /* Participants and contact information */

Participants should be arranged in alphabetical order of surname.

== Participants and contact information ==

(Note: this list is incomplete and unofficial. Inclusion or omission from this list should not be construed as any formal declaration of level of contribution to this project.)

* Kristal Cantwell[http://kristalcantwell.wordpress.com/]
* Kareem Carr, NYU [http://twofoldgaze.wordpress.com/]
* Jason Dyer, University of Arizona / Tucson Unified School District [http://numberwarrior.wordpress.com]
* Christian Elsholtz
* Kevin O'Bryant, CUNY (Staten Island and the Graduate Center), [http://www.math.csi.cuny.edu/obryant]
* Klas Markström, Umeå universitet, Sweden. [http://abel.math.umu.se/~klasm/]
* Michael Peake
* Terence Tao, UCLA, [http://www.math.ucla.edu/~tao]

== Grant information ==

* Kevin O'Bryant is supported by a grant from The City University of New York PSC-CUNY Research Award Program.
* Terence Tao is supported by a grant from the MacArthur Foundation, by NSF grant DMS-0649473, and by the NSF Waterman award.

== Other acknowledgments ==

Miscellaneous contributors to the project include KS Chua, Sune Kristian Jakobsen, Tyler Neylon (bounds and related quantities), Thomas Sauvaget.

Thanks to Michael Nielsen for hosting the polymath wiki for this project.

This project was a spinoff from the larger "Polymath1" project, initiated by Timothy Gowers.

Generalize to a graph-theoretic formulation

2010-05-25T16:39:20Z

Jbd:

This may turn out to be not as much a "proof strategy" as "another interesting way of thinking about the problem", but the hope is to generate examples much more tractable by hand and by computer which reveal graph theoretic properties that force certain discrepancies; and translate those into number theoretic properties we can search for (perhaps very rare ones, but ones we can prove exist nonetheless).

---

Define a set of nodes <math>a_1, a_2, ...</math> with sets of directed edges <math>e_j = \left \{ e_{j}^1, e_{j}^2, ... \right \}</math> such that each set of edges (with corresponding nodes) forms an acyclic path.

Give each node a value of 1 or -1. Consider any traversal corresponding to a set <math>e_j</math> that starts at the root node; define the absolute value of the sum of the values of the nodes visited to be <math>t_j</math>.

Possible conditions:

No subsets (NS): No set of edges is a subset of another set.

Omega set (OS): One set of edges connects <math>a_i</math> to <math>a_i+1</math> for all i.

Always increasing (AI): For any edge from <math>a_i</math> to <math>a_j</math>, i is less than j.

Isolated roots (IR): Each node can be the root node of only one set of edges.

---

Define an infinite set of nodes <math>a_1, a_2, ...</math> such that given a node <math>a_r</math> there is a edge in the set <math>e_r</math> from <math>a_r(n)</math> to <math>a_r(n+1)</math> for all <math>n \in \mathbb{N}</math>.

Asking if there is some fixed C that <math>t_j</math> is always bounded by is equivalent to the EDP.

---

Here is a possible meaning for “completely multiplicative”, although it is extremely messy and could likely be simplified.

Define everything as above, including the 5 extra conditions.

Call the set of edges from condition #1 the omega set, and the root node of that set to be the alpha node.

A node is called prime if the in-degree is 1; that is, the only edge incoming is from the omega set.

A node is a power if the in-degree is 2.

Consider all incoming edges to a particular node; trace the edges backwards to their root nodes. These are the divisors of the node.

Here is a prime factorization algorithm for the node n:

1. List the k divisors of n (excluding the alpha node) <math>d_1, d_2, ... d_k</math>. Call the path of a divisor to be the traversal going from the divisor to n. Given any divisor a that has divisor b in its path, remove a from the list.

2a. If n is already a power, jump to 2c.

2b. For any divisor that is not prime and is not a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisors of each of <math>d_1, d_2, ... d_k</math> as a branching tree (again excluding the alpha node); again, given any divisor a that has divisor b in its path, remove a from the list.

2c. For any divisor that is a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisor that is not the alpha node c a multiple number of times m+1, where m is the number of times powers occur in the traversal between the divisor of c and c.

3. Repeat #2a-c until all divisors on the bottom of the tree are prime.

4. The set of divisors at the bottom of the tree is the prime factorization of n.

So, a graph is completely multiplicative if that multiplying the values of all the nodes in the prime factorization of a node n (including repeated nodes as ncessary) gives the value of the node n.

--

Potential questions: Given a discrepancy d, what is a minimal graph (in terms of nodes and/or edges) where the discrepancy is greater than d? What conditions cause the discrepancy to “break”?

Using conditions 1-4, the minimal set of nodes where the discrepancy must be greater than 1 is 4.

Proof: Consider every possible set of edges using 3 nodes.

set A: 1->2->3
set B: 1->3

Then <math>a_1 = 1</math>, <math>a_2= -1</math>, <math>a_3= -1</math> has a discrepancy of 1.

However, consider these sets with 4 nodes:

set A: 1->2->3->4
set B: 1->3
set C: 1->4

Then no combinations of +1 and -1 (set A constrains the possibilities to + – + -, – + + -, + – - +, and – + – +) avoids having a discrepancy of 2 in some set of edges.

Generalize to a graph-theoretic formulation

2010-05-25T15:12:09Z

Jbd:

This may turn out to be not as much a "proof strategy" as "another interesting way of thinking about the problem", but the hope is to generate examples much more tractable by hand and by computer which reveal graph theoretic properties that force certain discrepancies; and translate those into number theoretic properties we can search for (perhaps very rare ones, but ones we can prove exist nonetheless).

---

(General formulation)
Define a set of nodes <math>a_1, a_2, ...</math> with k sets of directed edges <math>e_j = \left \{ e_{j}^1, e_{j}^2, ... \right \}</math> (for <math> 1 \leq j \leq k</math>) such that each set of edges (with corresponding nodes) forms an acyclic path.

Give each node a value of 1 or -1. Consider any traversal corresponding to a set <math>e_j</math> that starts at the root node; define the absolute value of the sum of the values of the nodes visited to be <math>t_j</math>.

Possible conditions given a set of nodes <math>a_1, a_2, ... </math> (or the finitary case <math>a_1, a_2, ... a_n</math>):

No subsets (NS): No set of edges is a subset of another set.

Omega set (OS): One set of edges connects <math>a_i</math> to <math>a_i+1</math> for all i (or in the finitary case, all i from 1 to n-1).

Always increasing (AI): For any edge from <math>a_i</math> to <math>a_j</math>, i is less than j.

Isolated roots (IR): Each node can be the root node of only one set of edges. [This condition is needed for the definition of completely multiplicative.]

---

Define an infinite set of nodes <math>a_1, a_2, ...</math> such that given a node <math>a_r</math> there is a edge in the set <math>e_r</math> from <math>a_r(n)</math> to <math>a_r(n+1)</math> for all <math>n \in \mathbb{N}</math>.

Asking if there is some fixed C that <math>t_j</math> is always bounded by is equivalent to our problem.

---

Here is a possible meaning for “completely multiplicative”, although it is extremely messy and could likely be simplified.

Define everything as above, including the 5 extra conditions.

Call the set of edges from condition #1 the omega set, and the root node of that set to be the alpha node.

A node is called prime if the in-degree is 1; that is, the only edge incoming is from the omega set.

A node is a power if the in-degree is 2.

Consider all incoming edges to a particular node; trace the edges backwards to their root nodes. These are the divisors of the node.

Here is a prime factorization algorithm for the node n:

1. List the k divisors of n (excluding the alpha node) <math>d_1, d_2, ... d_k</math>. Call the path of a divisor to be the traversal going from the divisor to n. Given any divisor a that has divisor b in its path, remove a from the list.

2a. If n is already a power, jump to 2c.

2b. For any divisor that is not prime and is not a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisors of each of <math>d_1, d_2, ... d_k</math> as a branching tree (again excluding the alpha node); again, given any divisor a that has divisor b in its path, remove a from the list.

2c. For any divisor that is a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisor that is not the alpha node c a multiple number of times m+1, where m is the number of times powers occur in the traversal between the divisor of c and c.

3. Repeat #2a-c until all divisors on the bottom of the tree are prime.

4. The set of divisors at the bottom of the tree is the prime factorization of n.

So, a graph is completely multiplicative if that multiplying the values of all the nodes in the prime factorization of a node n (including repeated nodes as ncessary) gives the value of the node n.

--

Potential questions: Given a discrepancy d, what is a minimal graph (in terms of nodes and/or edges) where the discrepancy is greater than d? What conditions cause the discrepancy to “break”?

Using conditions 1-4, the minimal set of nodes where the discrepancy must be greater than 1 is 4.

Proof: Consider every possible set of edges using 3 nodes.

set A: 1->2->3
set B: 1->3

Then <math>a_1 = 1</math>, <math>a_2= -1</math>, <math>a_3= -1</math> has a discrepancy of 1.

However, consider these sets with 4 nodes:

set A: 1->2->3->4
set B: 1->3
set C: 1->4

Then no combinations of +1 and -1 (set A constrains the possibilities to + – + -, – + + -, + – - +, and – + – +) avoids having a discrepancy of 2 in some set of edges.

Generalize to a graph-theoretic formulation

2010-05-25T14:55:04Z

Jbd:

This may turn out to be not as much a "proof strategy" as "another interesting way of thinking about the problem", but the hope is to generate examples much more tractable by hand and by computer which reveal graph theoretic properties that force certain discrepancies; and translate those into number theoretic properties we can search for (perhaps very rare ones, but ones we can prove exist nonetheless).

---

(General formulation)
Define a set of nodes <math>a_1, a_2, ...</math> with k sets of directed edges <math>e_{1}^j, e_{2}^j, ...</math> (for <math> 1 \leq j \leq k</math>) such that each set of edges (with corresponding nodes) forms an acyclic path.

Give each node a value of 1 or -1. Consider any traversal corresponding to a set <math>e_{1}^j, e_{2}^j, ...</math> that starts at the root node; define the absolute value of the sum of the values of the nodes visited to be <math>t_j</math>.

Possible conditions given a set of nodes <math>a_1, a_2, ..., a_n</math>:

No subsets (NS): No set of edges is a subset of another set.

Omega set (OS): One set of edges connects <math>a_i</math> to <math>a_i+1</math> for all i from 1 to n-1.

Always increasing (AI): For any edge from <math>a_i</math> to <math>a_j</math>, i is less than j.

Isolated roots (IR): Each node can be the root node of only one labelled set of directed edges. [This condition is needed for the definition of completely multiplicative.]

---

Define an infinite set of nodes <math>a_1, a_2, ...</math> such that given a node <math>a_i</math> there is a directed edge labelled <math>i</math> from <math>a_i(n)</math> to <math>a_i(n+1)</math> for all <math>n \in \mathbb{N}</math>.

Give each node a value 1 or -1. Consider a traversal starting at a node moving along edges with the same label a finite number of times; then the discrepancy of a given traversal is the absolute value of the sum of all nodes visited. If we only allow traversals to start at the root node of a particular set of labelled edges, asking if it is possible for the discrepancy to be bounded is equivalent to our problem.

The graph doesn’t have to follow the number system. Have a set of nodes (finite or infinite) such that there are labelled directed edges, but allow the edges to be arbitrary.

---

Here is a possible meaning for “completely multiplicative”, although it is extremely messy and could likely be simplified.

Define everything as above, including the 5 extra conditions.

Call the set of edges from condition #1 the omega set, and the root node of that set to be the alpha node.

A node is called prime if the in-degree is 1; that is, the only edge incoming is from the omega set.

A node is a power if the in-degree is 2.

Consider all incoming edges to a particular node; trace the edges backwards to their root nodes. These are the divisors of the node.

Here is a prime factorization algorithm for the node n:

1. List the k divisors of n (excluding the alpha node) <math>d_1, d_2, ... d_k</math>. Call the path of a divisor to be the traversal going from the divisor to n. Given any divisor a that has divisor b in its path, remove a from the list.

2a. If n is already a power, jump to 2c.

2b. For any divisor that is not prime and is not a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisors of each of <math>d_1, d_2, ... d_k</math> as a branching tree (again excluding the alpha node); again, given any divisor a that has divisor b in its path, remove a from the list.

2c. For any divisor that is a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisor that is not the alpha node c a multiple number of times m+1, where m is the number of times powers occur in the traversal between the divisor of c and c.

3. Repeat #2a-c until all divisors on the bottom of the tree are prime.

4. The set of divisors at the bottom of the tree is the prime factorization of n.

So, a graph is completely multiplicative if that multiplying the values of all the nodes in the prime factorization of a node n (including repeated nodes as ncessary) gives the value of the node n.

--

Potential questions: Given a discrepancy d, what is a minimal graph (in terms of nodes and/or edges) where the discrepancy is greater than d? What conditions cause the discrepancy to “break”?

Using conditions 1-4, the minimal set of nodes where the discrepancy must be greater than 1 is 4.

Proof: Consider every possible set of edges using 3 nodes.

set A: 1->2->3
set B: 1->3

Then <math>a_1 = 1</math>, <math>a_2= -1</math>, <math>a_3= -1</math> has a discrepancy of 1.

However, consider these sets with 4 nodes:

set A: 1->2->3->4
set B: 1->3
set C: 1->4

Then no combinations of +1 and -1 (set A constrains the possibilities to + – + -, – + + -, + – - +, and – + – +) avoids having a discrepancy of 2 in some set of edges.

Generalize to a graph-theoretic formulation

2010-05-25T14:23:48Z

Jbd:

This may turn out to be not as much a "proof strategy" as "another interesting way of thinking about the problem", but the hope is to generate examples much more tractable by hand and by computer which reveal graph theoretic properties that force certain discrepancies; and translate those into number theoretic properties we can search for (perhaps very rare ones, but ones we can prove exist nonetheless).

---

(General formulation)
Define a set of nodes <math>a_1, a_2, ...</math> with k sets of directed edges <math>e_{1}^j, e_{2}^j, ...</math> (for <math> 1 \leq j \leq k</math>) such that each set of edges (with corresponding nodes) forms an acyclic path.

Give each node a value of 1 or -1. Consider any traversal corresponding to a set <math>e_{1}^j, e_{2}^j, ...</math> that starts at the root node; define the absolute value of the sum of the values of the nodes visited to be <math>t_j</math>.

Possible conditions given a set of nodes <math>a_1, a_2, ..., a_n</math>:

No subsets (NS): No set of edges is a subset of another set.

Omega set (OS): One labelled set of edges connects <math>a_i</math> to <math>a_i+1</math> for all i from 1 to n-1.

Always increasing (AI): For any edge from <math>a_i</math> to <math>a_j</math>, i is less than j.

Isolated roots (IR): Each node can be the root node of only one labelled set of directed edges. [This condition is needed for the definition of completely multiplicative.]

---

Define an infinite set of nodes <math>a_1, a_2, ...</math> such that given a node <math>a_i</math> there is a directed edge labelled <math>i</math> from <math>a_i(n)</math> to <math>a_i(n+1)</math> for all <math>n \in \mathbb{N}</math>.

Give each node a value 1 or -1. Consider a traversal starting at a node moving along edges with the same label a finite number of times; then the discrepancy of a given traversal is the absolute value of the sum of all nodes visited. If we only allow traversals to start at the root node of a particular set of labelled edges, asking if it is possible for the discrepancy to be bounded is equivalent to our problem.

The graph doesn’t have to follow the number system. Have a set of nodes (finite or infinite) such that there are labelled directed edges, but allow the edges to be arbitrary.

---

Here is a possible meaning for “completely multiplicative”, although it is extremely messy and could likely be simplified.

Define everything as above, including the 5 extra conditions.

Call the set of edges from condition #1 the omega set, and the root node of that set to be the alpha node.

A node is called prime if the in-degree is 1; that is, the only edge incoming is from the omega set.

A node is a power if the in-degree is 2.

Consider all incoming edges to a particular node; trace the edges backwards to their root nodes. These are the divisors of the node.

Here is a prime factorization algorithm for the node n:

1. List the k divisors of n (excluding the alpha node) <math>d_1, d_2, ... d_k</math>. Call the path of a divisor to be the traversal going from the divisor to n. Given any divisor a that has divisor b in its path, remove a from the list.

2a. If n is already a power, jump to 2c.

2b. For any divisor that is not prime and is not a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisors of each of <math>d_1, d_2, ... d_k</math> as a branching tree (again excluding the alpha node); again, given any divisor a that has divisor b in its path, remove a from the list.

2c. For any divisor that is a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisor that is not the alpha node c a multiple number of times m+1, where m is the number of times powers occur in the traversal between the divisor of c and c.

3. Repeat #2a-c until all divisors on the bottom of the tree are prime.

4. The set of divisors at the bottom of the tree is the prime factorization of n.

So, a graph is completely multiplicative if that multiplying the values of all the nodes in the prime factorization of a node n (including repeated nodes as ncessary) gives the value of the node n.

--

Potential questions: Given a discrepancy d, what is a minimal graph (in terms of nodes and/or edges) where the discrepancy is greater than d? What conditions cause the discrepancy to “break”?

Using conditions 1-4, the minimal set of nodes where the discrepancy must be greater than 1 is 4.

Proof: Consider every possible set of edges using 3 nodes.

set A: 1->2->3
set B: 1->3

Then <math>a_1 = 1</math>, <math>a_2= -1</math>, <math>a_3= -1</math> has a discrepancy of 1.

However, consider these sets with 4 nodes:

set A: 1->2->3->4
set B: 1->3
set C: 1->4

Then no combinations of +1 and -1 (set A constrains the possibilities to + – + -, – + + -, + – - +, and – + – +) avoids having a discrepancy of 2 in some set of edges.

Generalize to a graph-theoretic formulation

2010-05-11T16:39:33Z

Jbd:

This may turn out to be not as much a "proof strategy" as "another interesting way of thinking about the problem", but the hope is to generate examples much more tractable by hand and by computer which reveal graph theoretic properties that force certain discrepancies; and translate those into number theoretic properties we can search for (perhaps very rare ones, but ones we can prove exist nonetheless).

---

(General formulation)
Define a set of nodes <math>a_1, a_2, ...</math> with k sets of directed edges <math>e_{1}^j, e_{2}^j, ...</math> (for <math> 1 \leq j \leq k</math>) such that each set of edges (with corresponding nodes) forms an acyclic path.

Give each node a value of 1 or -1. Consider any traversal corresponding to a set <math>e_{1}^j, e_{2}^j, ...</math> that starts at the root node; define the absolute value of the sum of the values of the nodes visited to be <math>t_j</math>.

Possible conditions given a set of nodes <math>a_1, a_2, ..., a_n</math>:

No subsets (NS): No set of edges is a subset of another set.

Omega set (OS): One labelled set of edges connects <math>a_i</math> to <math>a_i+1</math> for all i from 1 to n-1.

(Needs OS) Always increasing (AI): For any edge from <math>a_i</math> to <math>a_j</math>, i is less than j.

Isolated roots (IR): Each node can be the root node of only one labelled set of directed edges. [This condition is needed for the definition of completely multiplicative.]

---

Define an infinite set of nodes <math>a_1, a_2, ...</math> such that given a node <math>a_i</math> there is a directed edge labelled <math>i</math> from <math>a_i(n)</math> to <math>a_i(n+1)</math> for all <math>n \in \mathbb{N}</math>.

Give each node a value 1 or -1. Consider a traversal starting at a node moving along edges with the same label a finite number of times; then the discrepancy of a given traversal is the absolute value of the sum of all nodes visited. If we only allow traversals to start at the root node of a particular set of labelled edges, asking if it is possible for the discrepancy to be bounded is equivalent to our problem.

The graph doesn’t have to follow the number system. Have a set of nodes (finite or infinite) such that there are labelled directed edges, but allow the edges to be arbitrary.

---

Here is a possible meaning for “completely multiplicative”, although it is extremely messy and could likely be simplified.

Define everything as above, including the 5 extra conditions.

Call the set of edges from condition #1 the omega set, and the root node of that set to be the alpha node.

A node is called prime if the in-degree is 1; that is, the only edge incoming is from the omega set.

A node is a power if the in-degree is 2.

Consider all incoming edges to a particular node; trace the edges backwards to their root nodes. These are the divisors of the node.

Here is a prime factorization algorithm for the node n:

1. List the k divisors of n (excluding the alpha node) <math>d_1, d_2, ... d_k</math>. Call the path of a divisor to be the traversal going from the divisor to n. Given any divisor a that has divisor b in its path, remove a from the list.

2a. If n is already a power, jump to 2c.

2b. For any divisor that is not prime and is not a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisors of each of <math>d_1, d_2, ... d_k</math> as a branching tree (again excluding the alpha node); again, given any divisor a that has divisor b in its path, remove a from the list.

2c. For any divisor that is a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisor that is not the alpha node c a multiple number of times m+1, where m is the number of times powers occur in the traversal between the divisor of c and c.

3. Repeat #2a-c until all divisors on the bottom of the tree are prime.

4. The set of divisors at the bottom of the tree is the prime factorization of n.

So, a graph is completely multiplicative if that multiplying the values of all the nodes in the prime factorization of a node n (including repeated nodes as ncessary) gives the value of the node n.

--

Potential questions: Given a discrepancy d, what is a minimal graph (in terms of nodes and/or edges) where the discrepancy is greater than d? What conditions cause the discrepancy to “break”?

Using conditions 1-4, the minimal set of nodes where the discrepancy must be greater than 1 is 4.

Proof: Consider every possible set of edges using 3 nodes.

set A: 1->2->3
set B: 1->3

Then <math>a_1 = 1</math>, <math>a_2= -1</math>, <math>a_3= -1</math> has a discrepancy of 1.

However, consider these sets with 4 nodes:

set A: 1->2->3->4
set B: 1->3
set C: 1->4

Then no combinations of +1 and -1 (set A constrains the possibilities to + – + -, – + + -, + – - +, and – + – +) avoids having a discrepancy of 2 in some set of edges.

Generalize to a graph-theoretic formulation

2010-05-07T22:23:06Z

Jbd:

This may turn out to be not as much a "proof strategy" as "another interesting way of thinking about the problem", but the hope is to generate examples much more tractable by hand and by computer which reveal graph theoretic properties that force certain discrepancies; and translate those into number theoretic properties we can search for (perhaps very rare ones, but ones we can prove exist nonetheless).

---

Define an infinite set of nodes <math>a_1, a_2, ...</math> such that given a node <math>a_i</math> there is a directed edge labelled <math>i</math> from <math>a_i(n)</math> to <math>a_i(n+1)</math> for all <math>n \in \mathbb{N}</math>.

Give each node a value 1 or -1. Consider a traversal starting at a node moving along edges with the same label a finite number of times; then the discrepancy of a given traversal is the absolute value of the sum of all nodes visited. If we only allow traversals to start at the root node of a particular set of labelled edges, asking if it is possible for the discrepancy to be bounded is equivalent to our problem.

The graph doesn’t have to follow the number system. Have a set of nodes (finite or infinite) such that there are labelled directed edges, but allow the edges to be arbitrary.

(General formulation)
Define a set of nodes <math>a_1, a_2, ...</math> with k sets of directed edges <math>e_{1}^j, e_{2}^j, ...</math> (for <math> 1 \leq j \leq k</math>) such that each set of edges (with corresponding nodes) forms an acyclic path.

Give each node a value of 1 or -1. Consider any traversal corresponding to a set <math>e_{1}^j, e_{2}^j, ...</math> that starts at the root node; define the absolute value of the sum of the values of the nodes visited to be <math>t_j</math>.

Possible conditions given a set of nodes <math>a_1, a_2, ..., a_n</math>:

No arbitrary start points: No set of edges is a subset of another set.

Omega set: One labelled set of edges connects <math>a_i</math> to <math>a_i+1</math> for all i from 1 to n-1.

(Needs omega set) No reverses: For any edge from <math>a_i</math> to <math>a_j</math>, i is less than j.

One root per node: Each node can be the root node of only one labelled set of directed edges. [This condition is needed for the definition of completely multiplicative.]

Should any of the other conditions be built into the main definition?

---

Here is a possible meaning for “completely multiplicative”, although it is extremely messy and could likely be simplified.

Define everything as above, including the 5 extra conditions.

Call the set of edges from condition #1 the omega set, and the root node of that set to be the alpha node.

A node is called prime if the in-degree is 1; that is, the only edge incoming is from the omega set.

A node is a power if the in-degree is 2.

Consider all incoming edges to a particular node; trace the edges backwards to their root nodes. These are the divisors of the node.

Here is a prime factorization algorithm for the node n:

1. List the k divisors of n (excluding the alpha node) <math>d_1, d_2, ... d_k</math>. Call the path of a divisor to be the traversal going from the divisor to n. Given any divisor a that has divisor b in its path, remove a from the list.

2a. If n is already a power, jump to 2c.

2b. For any divisor that is not prime and is not a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisors of each of <math>d_1, d_2, ... d_k</math> as a branching tree (again excluding the alpha node); again, given any divisor a that has divisor b in its path, remove a from the list.

2c. For any divisor that is a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisor that is not the alpha node c a multiple number of times m+1, where m is the number of times powers occur in the traversal between the divisor of c and c.

3. Repeat #2a-c until all divisors on the bottom of the tree are prime.

4. The set of divisors at the bottom of the tree is the prime factorization of n.

So, a graph is completely multiplicative if that multiplying the values of all the nodes in the prime factorization of a node n (including repeated nodes as ncessary) gives the value of the node n.

--

Potential questions: Given a discrepancy d, what is a minimal graph (in terms of nodes and/or edges) where the discrepancy is greater than d? What conditions cause the discrepancy to “break”?

Using conditions 1-4, the minimal set of nodes where the discrepancy must be greater than 1 is 4.

Proof: Consider every possible set of edges using 3 nodes.

set A: 1->2->3
set B: 1->3

Then <math>a_1 = 1</math>, <math>a_2= -1</math>, <math>a_3= -1</math> has a discrepancy of 1.

However, consider these sets with 4 nodes:

set A: 1->2->3->4
set B: 1->3
set C: 1->4

Then no combinations of +1 and -1 (set A constrains the possibilities to + – + -, – + + -, + – - +, and – + – +) avoids having a discrepancy of 2 in some set of edges.

Generalize to a graph-theoretic formulation

2010-05-07T16:40:54Z

Jbd:

This may turn out to be not as much a "proof strategy" as "another interesting way of thinking about the problem", but the hope is to generate examples much more tractable by hand and by computer which reveal graph theoretic properties that force certain discrepancies; and translate those into number theoretic properties we can search for (perhaps very rare ones, but ones we can prove exist nonetheless).

---

Define an infinite set of nodes <math>a_1, a_2, ...</math> such that given a node <math>a_i</math> there is a directed edge labelled <math>i</math> from <math>a_i(n)</math> to <math>a_i(n+1)</math> for all <math>n \in \mathbb{N}</math>.

Give each node a value 1 or -1. Consider a traversal starting at a node moving along edges with the same label a finite number of times; then the discrepancy of a given traversal is the absolute value of the sum of all nodes visited. If we only allow traversals to start at the root node of a particular set of labelled edges, asking if it is possible for the discrepancy to be bounded is equivalent to our problem.

The graph doesn’t have to follow the number system. Have a set of nodes (finite or infinite) such that there are labelled directed edges, but allow the edges to be arbitrary.

(General formulation)
Define a set of nodes <math>a_1, a_2, ...</math> with k sets of directed edges <math>e_{1}^j, e_{2}^j, ...</math> (for <math> 1 \leq j \leq k</math>) such that each set of edges (with corresponding nodes) forms an acyclic path.

Give each node a value of 1 or -1. Consider any traversal corresponding to a set <math>e_{1}^j, e_{2}^j, ...</math> that starts at the root node; define the absolute value of the sum of the values of the nodes visited to be <math>t_j</math>.

Necessary condition?: No set of edges is a subset of another set.

---

While it’s fun to think about the completely general version of the structure (for example, cycles of odd degree cause unbounded discrepancy), for the sake of the original problem it might be useful to set some conditions.

Given a set of nodes <math>a_1, a_2, ..., a_n</math>:

1. One labelled set of edges connects <math>a_i</math> to <math>a_i+1</math> for all i from 1 to n-1.

2. For any edge from <math>a_i</math> to <math>a_j</math>, i is less than j.

3. Given a set consisting of all labelled edges of a given label, any traversal starting from the root node of the set can traverse the entire set of edges.

4. No set of edges is a subset of another set (this prevents having arbitrary start points).

5. (Optional, but required for a definition of completely multiplicative) Each node can be the root node of only one labelled set of directed edges.

---

Here is a possible meaning for “completely multiplicative”, although it is extremely messy and could likely be simplified.

Define everything as above, including the 5 extra conditions.

Call the set of edges from condition #1 the omega set, and the root node of that set to be the alpha node.

A node is called prime if the in-degree is 1; that is, the only edge incoming is from the omega set.

A node is a power if the in-degree is 2.

Consider all incoming edges to a particular node; trace the edges backwards to their root nodes. These are the divisors of the node.

Here is a prime factorization algorithm for the node n:

1. List the k divisors of n (excluding the alpha node) <math>d_1, d_2, ... d_k</math>. Call the path of a divisor to be the traversal going from the divisor to n. Given any divisor a that has divisor b in its path, remove a from the list.

2a. If n is already a power, jump to 2c.

2b. For any divisor that is not prime and is not a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisors of each of <math>d_1, d_2, ... d_k</math> as a branching tree (again excluding the alpha node); again, given any divisor a that has divisor b in its path, remove a from the list.

2c. For any divisor that is a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisor that is not the alpha node c a multiple number of times m+1, where m is the number of times powers occur in the traversal between the divisor of c and c.

3. Repeat #2a-c until all divisors on the bottom of the tree are prime.

4. The set of divisors at the bottom of the tree is the prime factorization of n.

So, a graph is completely multiplicative if that multiplying the values of all the nodes in the prime factorization of a node n (including repeated nodes as ncessary) gives the value of the node n.

--

Potential questions: Given a discrepancy d, what is a minimal graph (in terms of nodes and/or edges) where the discrepancy is greater than d? What conditions cause the discrepancy to “break”?

Using conditions 1-4, the minimal set of nodes where the discrepancy must be greater than 1 is 4.

Proof: Consider every possible set of edges using 3 nodes.

set A: 1->2->3
set B: 1->3

Then <math>a_1 = 1</math>, <math>a_2= -1</math>, <math>a_3= -1</math> has a discrepancy of 1.

However, consider these sets with 4 nodes:

set A: 1->2->3->4
set B: 1->3
set C: 1->4

Then no combinations of +1 and -1 (set A constrains the possibilities to + – + -, – + + -, + – - +, and – + – +) avoids having a discrepancy of 2 in some set of edges.

"Low Dimensions" grant acknowledgments

2010-04-26T21:34:23Z

Jbd: /* Participants and contact information */

Participants should be arranged in alphabetical order of surname.

== Participants and contact information ==

(Note: this list is incomplete and unofficial. Inclusion or omission from this list should not be construed as any formal declaration of level of contribution to this project.)

* Kristal Cantwell
* Kareem Carr, NYU [http://twofoldgaze.wordpress.com/]
* Jason Dyer, Tucson Unified School District (current to July 2010) [http://numberwarrior.wordpress.com]
* Kevin O'Bryant, CUNY (Staten Island and the Graduate Center), [http://www.math.csi.cuny.edu/obryant]
* Klas Markström, Umeå universitet, Sweden. [http://abel.math.umu.se/~klasm/]
* Michael Peake
* Terence Tao, UCLA, [http://www.math.ucla.edu/~tao]

== Grant information ==

* Kevin O'Bryant is supported by a grant from The City University of New York PSC-CUNY Research Award Program.
* Terence Tao is supported by a grant from the MacArthur Foundation, by NSF grant DMS-0649473, and by the NSF Waterman award.

== Other acknowledgments ==

Miscellaneous contributors to the project include KS Chua, Sune Kristian Jakobsen, Tyler Neylon (bounds and related quantities), Thomas Sauvaget.

Thanks to Michael Nielsen for hosting the polymath wiki for this project.

This project was a spinoff from the larger "Polymath1" project, initiated by Timothy Gowers.

Generalize to a graph-theoretic formulation

2010-04-23T16:40:28Z

Jbd:

This may turn out to be not as much a "proof strategy" as "another interesting way of thinking about the problem", but the hope is to generate examples much more tractable by hand and by computer which reveal graph theoretic properties that force certain discrepancies; and translate those into number theoretic properties we can search for (perhaps very rare ones, but ones we can prove exist nonetheless).

---

Define an infinite set of nodes <math>a_1, a_2, ...</math> such that given a node <math>a_i</math> there is a directed edge labelled <math>i</math> from <math>a_i(n)</math> to <math>a_i(n+1)</math> for all <math>n \in \mathbb{N}</math>.

Give each node a value 1 or -1. Consider a traversal starting at a node moving along edges with the same label a finite number of times; then the discrepancy of a given traversal is the absolute value of the sum of all nodes visited. If we only allow traversals to start at the root node of a particular set of labelled edges, asking if it is possible for the discrepancy to be bounded is equivalent to our problem.

The graph doesn’t have to follow the number system. Have a set of nodes (finite or infinite) such that there are labelled directed edges, but allow the edges to be arbitrary.

---

While it’s fun to think about the completely general version of the structure (for example, cycles of odd degree cause unbounded discrepancy), for the sake of the original problem it might be useful to set some conditions.

Given a set of nodes <math>a_1, a_2, ..., a_n</math>:

1. One labelled set of edges connects <math>a_i</math> to <math>a_i+1</math> for all i from 1 to n-1.

2. For any edge from <math>a_i</math> to <math>a_j</math>, i is less than j.

3. Given a set consisting of all labelled edges of a given label, any traversal starting from the root node of the set can traverse the entire set of edges.

4. No set of edges is a subset of another set (this prevents having arbitrary start points).

5. (Optional, but required for a definition of completely multiplicative) Each node can be the root node of only one labelled set of directed edges.

---

Here is a possible meaning for “completely multiplicative”, although it is extremely messy and could likely be simplified.

Define everything as above, including the 5 extra conditions.

Call the set of edges from condition #1 the omega set, and the root node of that set to be the alpha node.

A node is called prime if the in-degree is 1; that is, the only edge incoming is from the omega set.

A node is a power if the in-degree is 2.

Consider all incoming edges to a particular node; trace the edges backwards to their root nodes. These are the divisors of the node.

Here is a prime factorization algorithm for the node n:

1. List the k divisors of n (excluding the alpha node) <math>d_1, d_2, ... d_k</math>. Call the path of a divisor to be the traversal going from the divisor to n. Given any divisor a that has divisor b in its path, remove a from the list.

2a. If n is already a power, jump to 2c.

2b. For any divisor that is not prime and is not a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisors of each of <math>d_1, d_2, ... d_k</math> as a branching tree (again excluding the alpha node); again, given any divisor a that has divisor b in its path, remove a from the list.

2c. For any divisor that is a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisor that is not the alpha node c a multiple number of times m+1, where m is the number of times powers occur in the traversal between the divisor of c and c.

3. Repeat #2a-c until all divisors on the bottom of the tree are prime.

4. The set of divisors at the bottom of the tree is the prime factorization of n.

So, a graph is completely multiplicative if that multiplying the values of all the nodes in the prime factorization of a node n (including repeated nodes as ncessary) gives the value of the node n.

--

Potential questions: Given a discrepancy d, what is a minimal graph (in terms of nodes and/or edges) where the discrepancy is greater than d? What conditions cause the discrepancy to “break”?

Using conditions 1-4, the minimal set of nodes where the discrepancy must be greater than 1 is 4.

Proof: Consider every possible set of edges using 3 nodes.

set A: 1->2->3
set B: 1->3

Then <math>a_1 = 1</math>, <math>a_2= -1</math>, <math>a_3= -1</math> has a discrepancy of 1.

However, consider these sets with 4 nodes:

set A: 1->2->3->4
set B: 1->3
set C: 1->4

Then no combinations of +1 and -1 (set A constrains the possibilities to + – + -, – + + -, + – - +, and – + – +) avoids having a discrepancy of 2 in some set of edges.

Generalize to a graph-theoretic formulation

2010-04-23T16:35:02Z

Jbd:

This may turn out to be not as much a "proof strategy" as "another interesting way of thinking about the problem", but the hope is to generate examples much more tractable by hand and by computer which reveal graph theoretic properties that force certain discrepancies; and translate those into number theoretic properties we can search for (perhaps very rare ones, but ones we can prove exist nonetheless).

---

Define an infinite set of nodes <math>a_1, a_2, ...</math> such that given a node <math>a_i</math> there is a directed edge labelled <math>i</math> from <math>a_i(n)</math> to <math>a_i(n+1)</math> for all <math>n \in \mathbb{N}</math>.

Give each node a value 1 or -1. Consider a traversal starting at a node moving along edges with the same label a finite number of times; then the discrepancy of a given traversal is the absolute value of the sum of all nodes visited. If we only allow traversals to start at the root node of a particular set of labelled edges, asking if it is possible for the discrepancy to be bounded is equivalent to our problem.

The graph doesn’t have to follow the number system. Have a set of nodes (finite or infinite) such that there are labelled directed edges, but allow the edges to be arbitrary. Then there are interesting questions like;

---

While it’s fun to think about the completely general version of the structure (for example, cycles of odd degree cause unbounded discrepancy), for the sake of the original problem it might be useful to set some conditions.

Given a set of nodes <math>a_1, a_2, ..., a_n</math>:

1. One labelled set of edges connects <math>a_i</math> to <math>a_i+1</math> for all i from 1 to n-1.

2. For any edge from <math>a_i</math> to <math>a_j</math>, i is less than j.

3. Given a set consisting of all labelled edges of a given label, any traversal starting from the root node of the set can traverse the entire set of edges.

4. No set of edges is a subset of another set (this prevents having arbitrary start points).

5. (Optional, but required for a definition of completely multiplicative) Each node can be the root node of only one labelled set of directed edges.

---

Here is a possible meaning for “completely multiplicative”, although it is extremely messy and could likely be simplified.

Define everything as above, including the 5 extra conditions.

Call the set of edges from condition #1 the omega set, and the root node of that set to be the alpha node.

A node is called prime if the in-degree is 1; that is, the only edge incoming is from the omega set.

A node is a power if the in-degree is 2.

Consider all incoming edges to a particular node; trace the edges backwards to their root nodes. These are the divisors of the node.

Here is a prime factorization algorithm for the node n:

1. List the k divisors of n (excluding the alpha node) <math>d_1, d_2, ... d_k</math>. Call the path of a divisor to be the traversal going from the divisor to n. Given any divisor a that has divisor b in its path, remove a from the list.

2a. If n is already a power, jump to 2c.

2b. For any divisor that is not prime and is not a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisors of each of <math>d_1, d_2, ... d_k</math> as a branching tree (again excluding the alpha node); again, given any divisor a that has divisor b in its path, remove a from the list.

2c. For any divisor that is a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisor that is not the alpha node c a multiple number of times m+1, where m is the number of times powers occur in the traversal between the divisor of c and c.

3. Repeat #2a-c until all divisors on the bottom of the tree are prime.

4. The set of divisors at the bottom of the tree is the prime factorization of n.

So, a graph is completely multiplicative if that multiplying the values of all the nodes in the prime factorization of a node n (including repeated nodes as ncessary) gives the value of the node n.

--

Potential questions: Given a discrepancy d, what is a minimal graph (in terms of nodes and/or edges) where the discrepancy is greater than d? What conditions cause the discrepancy to “break”?

Using conditions 1-4, the minimal set of nodes where the discrepancy must be greater than 1 is 4.

Proof: Consider every possible set of edges using 3 nodes.

set A: 1->2->3
set B: 1->3

Then <math>a_1 = 1</math>, <math>a_2= -1</math>, <math>a_3= -1</math> has a discrepancy of 1.

However, consider these sets with 4 nodes:

set A: 1->2->3->4
set B: 1->3
set C: 1->4

Then no combinations of +1 and -1 (set A constrains the possibilities to + – + -, – + + -, + – - +, and – + – +) avoids having a discrepancy of 2 in some set of edges.

Generalize to a graph-theoretic formulation

2010-04-23T16:34:10Z

Jbd:

This may turn out to be not as much a "proof strategy" as "another interesting way of thinking about the problem", but the hope is to generate examples much more tractable by hand and by computer which reveal graph theoretic properties that force certain discrepancies; and translate those into number theoretic properties we can search for (perhaps very rare ones, but ones we can prove exist nonetheless).

---

Define an infinite set of nodes <math>a_1, a_2, ...</math> such that given a node <math>a_i</math> there is a directed edge labelled <math>i</math> from <math>a_i(n)</math> to <math>a_i(n+1)</math> for all <math>n \in \mathbb{N}</math>.

Give each node a value 1 or -1. Consider a traversal starting at a node moving along edges with the same label a finite number of times; then the discrepancy of a given traversal is the absolute value of the sum of all nodes visited. If we only allow traversals to start at the root node of a particular set of labelled edges, asking if it is possible for the discrepancy to be bounded is equivalent to our problem.

The graph doesn’t have to follow the number system. Have a set of nodes (finite or infinite) such that there are labelled directed edges, but allow the edges to be arbitrary. Then there are interesting questions like;

---

While it’s fun to think about the completely general version of the structure (for example, cycles of odd degree cause unbounded discrepancy), for the sake of the original problem it might be useful to set some conditions.

Given a set of nodes <math>a_1, a_2, ..., a_n</math>:

1. One labelled set of edges connects <math>a_i</math> to <math>a_i+1</math> for all i from 1 to n-1.

2. For any edge from <math>a_i</math> to <math>a_j</math>, i is less than j.

3. Given a set consisting of all labelled edges of a given label, any traversal starting from the root node of the set can traverse the entire set of edges.

4. No set of edges is a subset of another set (this prevents having arbitrary start points).

5. (Optional, but required for a definition of completely multiplicative) Each node can be the root node of only one labelled set of directed edges.

---

Here is a possible meaning for “completely multiplicative”, although it is extremely messy and could likely be simplified.

Define everything as above, including the 5 extra conditions.

Call the set of edges from condition #1 the omega set, and the root node of that set to be the alpha node.

A node is called prime if the in-degree is 1; that is, the only edge incoming is from the omega set.

A node is a power if the in-degree is 2.

Consider all incoming edges to a particular node; trace the edges backwards to their root nodes. These are the divisors of the node.

Here is a prime factorization algorithm for the node n:

1. List the k divisors of n (excluding the alpha node) <math>d_1, d_2, ... d_k</math>. Call the path of a divisor to be the traversal going from the divisor to n. Given any divisor a that has divisor b in its path, remove a from the list.

2a. If n is already a power, jump to 2c.

2b. For any divisor that is not prime and is not a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisors of each of <math>d_1, d_2, ... d_k</math> as a branching tree (again excluding the alpha node); again, given any divisor a that has divisor b in its path, remove a from the list.

2c. For any divisor that is a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisor that is not the alpha node c a multiple number of times m+1, where m is the number of times powers occur in the traversal between the divisor of c and c.

3. Repeat #2a-c until all divisors on the bottom of the tree are prime.

4. The set of divisors at the bottom of the tree is the prime factorization of n.

So, a graph is completely multiplicative if that multiplying the values of all the nodes in the prime factorization of a node n (including repeated nodes as ncessary) gives the value of the node n.

--

Potential questions: Given a discrepancy d, what is a minimal graph (in terms of nodes and/or edges) where the discrepancy is greater than d? What conditions cause the discrepancy to “break”?

Using all 5 conditions, the minimal set of nodes where the discrepancy must be greater than 1 is 4.

Proof: Consider every possible set of edges using 3 nodes.

set A: 1->2->3
set B: 1->3

Then <math>a_1 = 1</math>, <math>a_2= -1</math>, <math>a_3= -1</math> has a discrepancy of 1.

However, consider these sets with 4 nodes:

set A: 1->2->3->4
set B: 1->3
set C: 1->4

Then no combinations of +1 and -1 (set A constrains the possibilities to + – + -, – + + -, + – - +, and – + – +) avoids having a discrepancy of 2 in some set of edges.

Generalize to a graph-theoretic formulation

2010-04-23T16:29:24Z

Jbd:

This may turn out to be not as much a "proof strategy" as "another interesting way of thinking about the problem", but the hope is to generate examples much more tractable by hand and by computer which reveal graph theoretic properties that force certain discrepancies; and translate those into number theoretic properties we can search for (perhaps very rare ones, but ones we can prove exist nonetheless).

---

Define an infinite set of nodes <math>a_1, a_2, ...</math> such that given a node <math>a_i</math> there is a directed edge labelled <math>i</math> from <math>a_i(n)</math> to <math>a_i(n+1)</math> for all <math>n \in \mathbb{N}</math>.

Give each node a value 1 or -1. Consider a traversal starting at a node moving along edges with the same label a finite number of times; then the discrepancy of a given traversal is the absolute value of the sum of all nodes visited. If we only allow traversals to start at the root node of a particular set of labelled edges, asking if it is possible for the discrepancy to be bounded is equivalent to our problem.

The graph doesn’t have to follow the number system. Have a set of nodes (finite or infinite) such that there are labelled directed edges, but allow the edges to be arbitrary. Then there are interesting questions like;

---

While it’s fun to think about the completely general version of the structure (for example, cycles of odd degree cause unbounded discrepancy), for the sake of the original problem it might be useful to set some conditions.

Given a set of nodes <math>a_1, a_2, ..., a_n</math>:

1. One labelled set of edges connects <math>a_i</math> to <math>a_i+1</math> for all i from 1 to n-1.

2. For any edge from <math>a_i</math> to <math>a_j</math>, i is less than j.

3. Given a set consisting of all labelled edges of a given label, any traversal starting from the root node of the set can traverse the entire set of edges.

4. No set of edges is a subset of another set (this prevents having arbitrary start points).

5. Each node can be the root node of only one labelled set of directed edges.

---

Here is a possible meaning for “completely multiplicative”, although it is extremely messy and could likely be simplified.

Define everything as above, including the 5 extra conditions.

Call the set of edges from condition #1 the omega set, and the root node of that set to be the alpha node.

A node is called prime if the in-degree is 1; that is, the only edge incoming is from the omega set.

A node is a power if the in-degree is 2.

Consider all incoming edges to a particular node; trace the edges backwards to their root nodes. These are the divisors of the node.

Here is a prime factorization algorithm for the node n:

1. List the k divisors of n (excluding the alpha node) <math>d_1, d_2, ... d_k</math>. Call the path of a divisor to be the traversal going from the divisor to n. Given any divisor a that has divisor b in its path, remove a from the list.

2a. If n is already a power, jump to 2c.

2b. For any divisor that is not prime and is not a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisors of each of <math>d_1, d_2, ... d_k</math> as a branching tree (again excluding the alpha node); again, given any divisor a that has divisor b in its path, remove a from the list.

2c. For any divisor that is a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisor that is not the alpha node c a multiple number of times m+1, where m is the number of times powers occur in the traversal between the divisor of c and c.

3. Repeat #2a-c until all divisors on the bottom of the tree are prime.

4. The set of divisors at the bottom of the tree is the prime factorization of n.

So, a graph is completely multiplicative if that multiplying the values of all the nodes in the prime factorization of a node n (including repeated nodes as ncessary) gives the value of the node n.

--

Potential questions: Given a discrepancy d, what is a minimal graph (in terms of nodes and/or edges) where the discrepancy is greater than d? What conditions cause the discrepancy to “break”?

Using all 5 conditions, the minimal set of nodes where the discrepancy must be greater than 1 is 4.

Proof: Consider every possible set of edges using 3 nodes.

set A: 1->2->3
set B: 1->3

Then <math>a_1 = 1</math>, <math>a_2= -1</math>, <math>a_3= -1</math> has a discrepancy of 1.

However, consider these sets with 4 nodes:

set A: 1->2->3->4
set B: 1->3
set C: 1->4

Then no combinations of +1 and -1 (set A constrains the possibilities to + – + -, – + + -, + – - +, and – + – +) avoids having a discrepancy of 2 in some set of edges.

Generalize to a graph-theoretic formulation

2010-04-23T16:26:27Z

Jbd: New page: This may turn out to be not as much a "proof strategy" as "another interesting way of thinking about the problem", but the hope is to generate examples much more tractable by hand and by c...

This may turn out to be not as much a "proof strategy" as "another interesting way of thinking about the problem", but the hope is to generate examples much more tractable by hand and by computer which reveal graph theoretic properties that force certain discrepancies; then these properties might also appear in the original number theory version of the problem.

---

Define an infinite set of nodes <math>a_1, a_2, ...</math> such that given a node <math>a_i</math> there is a directed edge labelled <math>i</math> from <math>a_i(n)</math> to <math>a_i(n+1)</math> for all <math>n \in \mathbb{N}</math>.

Give each node a value 1 or -1. Consider a traversal starting at a node moving along edges with the same label a finite number of times; then the discrepancy of a given traversal is the absolute value of the sum of all nodes visited. If we only allow traversals to start at the root node of a particular set of labelled edges, asking if it is possible for the discrepancy to be bounded is equivalent to our problem.

The graph doesn’t have to follow the number system. Have a set of nodes (finite or infinite) such that there are labelled directed edges, but allow the edges to be arbitrary. Then there are interesting questions like; given a discrepancy d, what is a minimal graph (in terms of nodes and/or edges) where the discrepancy is greater than d? What conditions cause the discrepancy to “break”? My hope is we can isolate certain graph theoretic properties which cause the logic to break and translate those into number theoretic properties we can search for (perhaps very rare ones, but ones we can prove exist nonetheless).

---

While it’s fun to think about the completely general version of the structure (for example, cycles of odd degree cause unbounded discrepancy), for the sake of the original problem it might be useful to set some conditions.

Given a set of nodes <math>a_1, a_2, ..., a_n</math>:

1. One labelled set of edges connects <math>a_i</math> to <math>a_i+1</math> for all i from 1 to n-1.

2. For any edge from <math>a_i</math> to <math>a_j</math>, i is less than j.

3. Given a set consisting of all labelled edges of a given label, any traversal starting from the root node of the set can traverse the entire set of edges.

4. No set of edges is a subset of another set (this prevents having arbitrary start points).

5. Each node can be the root node of only one labelled set of directed edges.

Given these conditions, the minimal set of nodes where the discrepancy must be greater than 1 is 4.

Proof: Consider every possible set of edges using 3 nodes.

set A: 1->2->3
set B: 1->3

Then <math>a_1 = 1</math>, <math>a_2= -1</math>, <math>a_3= -1</math> has a discrepancy of 1.

However, consider these sets with 4 nodes:

set A: 1->2->3->4
set B: 1->3
set C: 1->4

Then no combinations of +1 and -1 (set A constrains the possibilities to + – + -, – + + -, + – - +, and – + – +) avoids having a discrepancy of 2 in some set of edges.

---

Here is a possible meaning for “completely multiplicative”, although it is extremely messy and could likely be simplified.

Define everything as above, including the 5 extra conditions.

Call the set of edges from condition #1 the omega set, and the root node of that set to be the alpha node.

A node is called prime if the in-degree is 1; that is, the only edge incoming is from the omega set.

A node is a power if the in-degree is 2.

Consider all incoming edges to a particular node; trace the edges backwards to their root nodes. These are the divisors of the node.

Here is a prime factorization algorithm for the node n:

1. List the k divisors of n (excluding the alpha node) <math>d_1, d_2, ... d_k</math>. Call the path of a divisor to be the traversal going from the divisor to n. Given any divisor a that has divisor b in its path, remove a from the list.

2a. If n is already a power, jump to 2c.

2b. For any divisor that is not prime and is not a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisors of each of <math>d_1, d_2, ... d_k</math> as a branching tree (again excluding the alpha node); again, given any divisor a that has divisor b in its path, remove a from the list.

2c. For any divisor that is a power on the list <math>d_1, d_2, ... d_k</math>, connect the divisor that is not the alpha node c a multiple number of times m+1, where m is the number of times powers occur in the traversal between the divisor of c and c.

3. Repeat #2a-c until all divisors on the bottom of the tree are prime.

4. The set of divisors at the bottom of the tree is the prime factorization of n.

So, a graph is completely multiplicative if that multiplying the values of all the nodes in the prime factorization of a node n (including repeated nodes as ncessary) gives the value of the node n.

The Erdős discrepancy problem

2010-04-23T16:10:56Z

Jbd: /* General proof strategies */

If you want to you can [http://michaelnielsen.org/polymath1/index.php?title=Experimental_results jump straight to the main experimental results page].

==Introduction and statement of problem==

Let <math>(x_n)</math> be a <math>\scriptstyle \pm 1</math> sequence and let <math>C</math> be a constant. Must there exist positive integers <math> d,k </math> such that

:<math> \left| \sum_{i=1}^k x_{id} \right| > C </math>?

Known colloquially as "The Erdős discrepancy problem", this question has remained unanswered since the 1930s (Erdős, 1957) and Erdős offered $500 for an answer. It was also asked by Chudakov (1956). It is extremely easy to state, and a very appealing problem, so perhaps Polymath can succeed where individual mathematicians have failed. If so, then, given the notoriety of the problem, it would be a significant coup for the multiply collaborative approach to mathematics. But even if a full solution is not reached, preliminary investigations have already thrown up some interesting ideas which could perhaps be developed into publishable results.

It seems likely that the answer to Erdős's question is yes. If it is, then an easy compactness argument tells us that for every <math>C</math> there exists <math>n</math> such that for every <math>\scriptstyle \pm 1</math> sequence of length <math>n</math> there exist <math>d,k</math> with <math>dk\leq n</math> such that <math> \left| \sum_{i=1}^k x_{id} \right| > C </math>. In view of this, we make some definitions that allow one to talk about the dependence between <math>C</math> and <math>n</math>. For any finite set <math>A</math> of integers, we define the error on <math>A</math> by
:<math> E(A) := \sum_{a\in A} x_a </math>,
and for a set <math>\mathcal{A}</math> of finite sets of integers, we define the discrepancy
:<math> \delta(\mathcal{A},x) := \sup_{A\in \mathcal{A}} |E(A)|. </math>
We can think of the values taken by the sequence <math>(x_n)</math> as a red/blue colouring of the integers that tries to make the number of reds and blues in each <math>\scriptstyle A\in\mathcal{A}</math> as equal as possible. The discrepancy measures the extent to which the sequence fails in this attempt. Taking <math>\scriptstyle \mathcal{HAP}(N)</math> to be the set of homogeneous arithmetic progressions <math>\{d, 2d, 3d, ..., nd\}</math> contained in <math>\{1, 2, ..., N\}</math>, we can restate the question as whether <math>\scriptstyle \delta(\mathcal{HAP}(N),x) \to \infty</math> for every sequence <math>x</math>.

Two related questions have already been solved in the literature. Letting <math>\scriptstyle \mathcal{AP}(N) </math> be the collection of all (not necessarily homogeneous) arithmetic progressions in <math>\{1, 2, ..., N\}</math>, Roth (1964) proved that <math> \scriptstyle \delta(\mathcal{AP}(N),x) \geq c n^{1/4}</math>, independent of the sequence <math>x</math>. Letting <math> \scriptstyle \mathcal{HQAP}(N) </math> be the collection of all homogeneous quasi-arithmetic progressions <math>\scriptstyle\{\lfloor \alpha \rfloor,\lfloor 2\alpha \rfloor,\dots,\lfloor k\alpha \rfloor\}</math> contained in <math>\{1, 2, ..., N\}</math>, Vijay (2008) proved that <math>\scriptstyle \delta(\mathcal{HQAP}(N),x) \geq 0.02 n^{1/6}</math>, independent of the sequence <math>x</math>.

==Some definitions and notational conventions==

A homogeneous arithmetic progression, or HAP, is an arithmetic progression of the form <math> \{d,2d,3d,...,nd\}</math>.

When <math> x</math> is clear from context, we write <math> \delta(N) </math> in place of <math> \delta(\mathcal{HAP}(N),x) </math>.

The discrepancy of a sequence <math>(x_n)</math> is the supremum of <math>|\sum_{n\in P}x_n|</math> over all homogeneous arithmetic progressions P. (Strictly speaking, we should call this something like the HAP-discrepancy, but since we will almost always be talking about HAPs, we adopt the convention that "discrepancy" always means "HAP-discrepancy" unless it is stated otherwise.)
<math>Insert formula here</math>
A sequence <math>(x_n)</math> has discrepancy at most φ(n) if for every natural number N the maximum value of <math>|\sum_{n\in P}x_n|</math> over all homogeneous arithmetic progressions P that are subsets of {1,2,...,N} is at most φ(N).

The EDP is the Erdős discrepancy problem. (This may conceivably be changed if enough people don't like it.)

A sequence <math>(x_n)</math> is completely multiplicative if <math>x_{mn}=x_mx_n</math> for any two positive integers m and n. We shall sometimes abbreviate this to "multiplicative", but the reader should be aware that the word "multiplicative" normally refers to the more general class of sequences such that <math>x_{mn}=x_mx_n</math> whenever m and n are coprime. A completely multiplicative function is determined by the values it takes at primes. The Liouville function λ is the unique completely multiplicative function that takes the value -1 at every prime: if the prime factorization of n is <math>\prod p_i^{a_i}</math> then λ(n) equals <math>(-1)^{\sum a_i}</math>. Another important multiplicative function is <math>\mu_3</math>, the multiplicative function that’s -1 at 3, 1 at numbers that are 1 mod 3 and -1 at numbers that are 2 mod 3. This function has mean-square partial sums which grow logarithmically, [http://gowers.wordpress.com/2010/01/26/edp3-a-very-brief-report-on-where-we-are/#comment-5585 see this calculation].

A HAP-subsequence of a sequence <math>(x_n)</math> is a subsequence of the form <math>x_d,x_{2d},x_{3d},...</math> for some d. If <math>(x_n)</math> is a multiplicative <math>\pm 1</math> sequence, then every HAP-subsequence is equal to either <math>(x_n)</math> or <math>(-x_n)</math>. We shall call a sequence weakly multiplicative if it has only finitely many distinct HAP-subsequences. Let us also call a <math>\pm 1</math> sequence quasi-multiplicative if it is a composition of a completely multiplicative function from <math>\mathbb{N}</math> to an Abelian group G with a function from G to {-1,1} (This definition is too general. See [http://gowers.wordpress.com/2010/01/09/erds-discrepancy-problem-continued/#comment-4843 this and the next comment]). [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4705 It can be shown] that if <math>(x_n)</math> is a weakly multiplicative sequence, then it has an HAP-subsequence that is quasi-multiplicative.

It is convenient to define the maps <math>T_k</math> for <math>k \geq 1</math> by <math>T_k(x)_n = x_{kn}</math>, where <math>x = (x_1, x_2, ...)</math> is a sequence.

==Simple observations==

*To answer the problem negatively, it is sufficient to find a completely multiplicative sequence (that is, a sequence <math>(x_n)</math> such that <math>x_{mn}=x_mx_n</math> for every m,n) such that its partial sums <math>x_1+\dots+x_n</math> are bounded. First mentioned in the proof of the theorem in [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/ this] post.

*The function that takes n to 1 if the last non-zero digit of n in its ternary representation is 1 and -1 if the last non-zero digit is 2 is completely multiplicative and the partial sum up to n is easily shown to be at most <math>\log_3n +1</math>. Therefore, the rate at which the worst discrepancy grows, as a function of the length of the homogeneous progression, can be as slow as logarithmic. [[Matryoshka_Sequences|More examples of this sort are here.]] [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/] <s>[http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4553 It turns out] that the base can be made significantly higher than 3, so this example is not best possible.</s> [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4757 Correction]

*To answer the problem negatively, it is also sufficient to find a completely multiplicative function f from <math>\mathbb{N}</math> to a finite Abelian group G (meaning that f(mn)=f(m)+f(n) for every m,n) and a function h from G to {-1,1} such that for each g in G there exists d with f(d)=g and with the partial sums hf(d)+hf(2d)+...+hf(nd) bounded. In that case, one can set <math>x_n=hf(n)</math>. (Though this observation is simple with the benefit of hindsight, it might not have been made had it not been for the fact that this sort of structure was identified in a long sequence that had been produced experimentally). First mentioned [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4696 here]. Se also [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/ this post] and [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4717 this comment].

*[http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4705 It can be shown] that if a <math>\pm 1</math> sequence <math>(x_n)</math> starts with 1 and has only finitely many distinct subsequences of the form <math>(x_d,x_{2d},x_{3d},\dots)</math>, then it must have a subsequence that arises in the above way. (This was mentioned just above in the terminology section.)

*The problem for the positive integers is equivalent to the problem for the positive rationals. [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4676](Sketch proof: if you have a function f from the positive integers to {-1,1} that works, then define <math>f_q(x)</math> to be f(q!x) whenever q!x is an integer. Then take a pointwise limit of a subsequence of the functions <math>f_q</math>. [[Limits with better properties|Click here for a more detailed discussion of this construction and what it is good for]].

*Define the [[drift]] of a sequence to be the maximal value of |f(md)+...+f(nd)|. The discrepancy problem is equivalent to showing that the drift is always infinite. It is obvious that it is at least 2 (because |f(2)+f(4)|, |f(2)+f(3)|, |f(3)+f(4)| cannot all be at most 1); one can show that [[drift|it is at least 3]] (which implies as a corollary that the discrepancy is at least 2). One can also show that the [[upper and lower discrepancy]] are each at least 2.

*One can [[topological dynamics formulation|formulate the problem using topological dynamics]].

*Using Fourier analysis, one can [[Fourier reduction|reduce the problem to one about completely multiplicative functions]].

*Here is an [[algorithm for finding multiplicative sequences with bounded discrepancy]].

*Here is a page whose aim is to record a [[human proof that completely multiplicative sequences have discrepancy at least 2]].

*Here is an argument that shows that [[bounded discrepancy multiplicative functions do not correlate with characters]].

* The answer to a [[function field version]] of the problem seems to be negative.

* The problem can be generalized to [[pseudointegers]]. Here we have found [[EDP on pseudointegers|problems similar to EDP]] with a negative answer.

==Experimental evidence==

''Main page: [[Experimental results]]''

A nice feature of the problem is that it lends itself well to investigation on a computer. Although the number of <math>\pm 1</math> sequences grows exponentially, the constraints on these sequences are such that depth-first searches can lead quite quickly to interesting examples. For instance, at the time of writing they have yielded several examples of sequences of length 1124 with discrepancy 2. (This is the current record.) See [[Experimental results|this page]] for more details and further links.

Another sort of evidence is to bound the discrepancy for specific sequences. For example, setting <math>x_1=-x_2=1</math> and <math> x_n=-x_{n/3}</math> or <math>x_n=x_{n-3}</math> according to whether <math>n</math> is a multiple of 3 or not, yields a sequence with

:<math> \delta(N) \leq \log_9(N)+1 </math>

for sufficiently large <math> N </math>. Currently, this is the record-holder for slow growing discrepancy.

The [[Thue-Morse-Hedlund Sequence|Thue-Morse sequence]] has discrepancy <math> \delta(N) \gg N^{\log_4(3)} </math>.

A random sequence (each <math> x_i </math> chosen independently) has discrepancy at least (asymptotically) <math> \sqrt{2N \log\log N} </math> by the law of the iterated logarithm.

Determining if the discrepancy of [http://en.wikipedia.org/wiki/Liouville_function Liouville's lambda function] is <math> O(n^{1/2+\epsilon})</math> is equivalent to solving the Riemann hypothesis. However, this growth cannot be bounded above by <math>n^{1/2-\epsilon}</math> for any positive ε.

==Interesting subquestions==

Given the length of time that the Erdős discrepancy problem has been open, the chances that it will be solved by Polymath5 are necessarily small. However, there are a number of interesting questions that we do not know the answers to, several of which have arisen naturally from the experimental evidence. Good answers to some of these would certainly constitute publishable results. Here is a partial list -- further additions are very welcome. (When the more theoretical part of the project starts, this section will probably grow substantially and become a separate page.)

*Is there an infinite sequence of discrepancy 2? (Given how long a finite sequence can be, it seems unlikely that we could answer this question just by a clever search of all possibilities on a computer.)

*The long sequences of low discrepancy discovered by computer all have some kind of approximate weak multiplicativity. If we take a hypothetical counterexample <math>(x_n)</math> (which we could even assume has discrepancy 2), can we prove that it, or some other counterexample derived from it by passing to HAP-subsequences and taking pointwise limits, has some kind of interesting multiplicative structure?

*Closely related to the previous question. If <math>(x_n)</math> is a hypothetical counterexample, must it satisfy a compactness property of the following kind: for every positive c there exists M such that if you take any M HAP-subsequences, then there must be two of them, <math>(y_n)</math> and <math>(z_n)</math>, such that <math>\lim\inf N^{-1}\sum_{n=1}^Ny_nz_n\geq 1-c</math>?

*An even weaker question. If <math>(x_n)</math> is a hypothetical counterexample, must it have two HAP-subsequences <math>(y_n)</math> and <math>(z_n)</math> such that the lim inf of <math>N^{-1}\sum_{n=1}^N y_nz_n</math> is greater than 0?

*A similar question, perhaps equivalent to the previous one (this should be fairly easy to check). Given a sequence <math>(x_n)</math> define f(N) to be the average of <math>x_ax_bx_cx_d</math> over all quadruples (a,b,c,d) such that ab=cd and <math>a,b,c,d\leq N</math>. If <math>(x_n)</math> is a counterexample, does that imply that <math>\lim\inf f(N)</math> is greater than 0? [http://thomas1111.wordpress.com/2010/01/15/average-over-quadruples-of-the-first-1124-sequence/ See here for a computation of this average for the first 1124 sequence].

*Is there any completely multiplicative counterexample? (This may turn out to be a very hard question. If so, then answering the previous questions could turn out to be the best we can hope for without making a substantial breakthrough in analytic number theory.)

*Does there exist a <math>\pm 1</math> sequence such that the corresponding discrepancy function grows at a rate that is slower than <math>c\log n</math> for any positive c?

*Does the number of sequences of length n and discrepancy at most C grow exponentially with n or more slowly than exponentially? (Obviously if the conjecture is true then it must be zero for large enough n, but the hope is that this question is a more realistic initial target.)

==General proof strategies==

This section contains links to other pages in which potential approaches to solving the problem are described.

[[First obtain multiplicative structure and then obtain a contradiction]]

[[Find a different parameter, show that it tends to infinity, and show that that implies that the discrepancy tends to infinity]]

[[Prove the result for shifted HAPs instead of HAPs]]

[[Find a good configuration of HAPs]]

[[Generalize to a graph-theoretic formulation]]

==Annotated Bibliography==

*Blog Discussion on Gowers's Weblog
**[http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/ Zeroth Post] (Dec 17 - Jan 6).
**[http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/ First Post] (Jan 6 - Jan 12).
**[http://gowers.wordpress.com/2010/01/09/erds-discrepancy-problem-continued/ Second Post] (Jan 9 - Jan 11).
**[http://gowers.wordpress.com/2010/01/11/the-erds-discrepancy-problem-iii/ Third Post] (Jan 11 - Jan 14).
**[http://gowers.wordpress.com/2010/01/14/the-erds-discrepancy-problem-iv/ Fourth Post] (Jan 14 - 16).
**[http://gowers.wordpress.com/2010/01/16/the-erds-discrepancy-problem-v/ Fifth Post] (Jan 16-19).
**[http://gowers.wordpress.com/2010/01/19/edp1-the-official-start-of-polymath5/ First Theoretical Post] (Jan 19-21)
**[http://gowers.wordpress.com/2010/01/21/edp2-a-few-lessons-from-edp1/ Second Theoretical Post] (Jan 21-26)
**[http://gowers.wordpress.com/2010/01/26/edp3-a-very-brief-report-on-where-we-are/ Third Theoretical Post] (Jan 26 -?)
**[http://gowers.wordpress.com/2010/01/30/edp4-focusing-on-multiplicative-functions/ Fourth Theoretical Post] (Jan 30- Feb 2)
**[http://gowers.wordpress.com/2010/02/02/edp5-another-very-brief-summary/ Fifth Theoretical Post] (Feb 2 - Feb 5)
**[http://gowers.wordpress.com/2010/02/05/edp6-what-are-the-chances-of-success/ Sixth Theoretical Post] (Feb 5- Feb 8)
**[http://gowers.wordpress.com/2010/02/08/edp7-emergency-post/ Seventh Theoretical Post] (Feb 8 - ?)

*[http://mathoverflow.net/questions/tagged/polymath5 Several questions on MathOverflow have been given the `polymath5' tag.]

*Mathias, A. R. D. [http://www.dpmms.cam.ac.uk/~ardm/erdoschu.pdf On a conjecture of Erdős and Čudakov]. Combinatorics, geometry and probability (Cambridge, 1993).

This one page paper establishes that the maximal length of sequence for the case where C=1 is 11, and is the starting point for our experimental studies.

*Tchudakoff, N. G. Theory of the characters of number semigroups. J. Indian Math. Soc. (N.S.) 20 (1956), 11--15. MR0083515 (18,719e)

The Mathias paper states that one of Erdős's problem lists states that this paper "studies related questions". The Math Review for this paper states that it summarizes results from seven other papers that are in Russian. The Math Review leaves the impression that the topic concerns characters of modulus 1.

*Borwein, Peter, and Choi, Stephen. [http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P204.pdf A variant of Liouville's lambda function: some surprizing formulae].

Explicit formulas for the discrepancy of some completely multiplicative functions whose discrepancy is logarithmic. A typical example is: Let <math>\lambda_3(n) = (-1)^{\omega_3(n)}</math> where <math>\omega_3(n)</math> is the number of distinct prime factors congruent to <math>-1 \bmod 3</math> in <math>n</math> (with multiple factors counted multiply). Then the discrepancy of the <math>\lambda_3</math> sequence up to <math>N</math> is exactly the number of 1's in the base three expansion of <math>N</math>. (This turns out not to be so surprising after all, since <math>\lambda_3(n)</math> is precisely the same as the ternary function defined in the second item of the Simple Observations section.)

*Borwein, Peter, Choi, Stephen and Coons, Michael. [http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P225.pdf Completely multiplicative functions taking values in {-1,1}]. The published version of the above paper, which explains the results and proofs more clearly than the preprint, and is more explicit about the relationship with Erdős's question.

*Granville and Soundararajan. [http://arxiv.org/abs/math/0702389 Multiplicative functions in arithmetic progressions]

In this [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4729 blog post] Terry Tao wrote: "Granville and Soundararajan have made some study of the discrepancy of bounded multiplicative functions. The situation is remarkably delicate and number-theoretical (and is closely tied with the Granville-Soundararajan theory of pretentious characters)."

*Wirsing, E. Das asymptotische Verhalten von Summen über multiplikative Funktionen. II. (German) [The asymptotic behavior of sums of multiplicative functions. II.] Acta Math. Acad. Sci. Hungar. 18 1967 411--467. MR0223318 (36 #6366) [http://michaelnielsen.org/polymath1/index.php?title=Wirsing_translation (partial) English translation]

*Newman, D. J. [http://www.jstor.org/stable/2036455 On the number of binary digits in a multiple of three]. Proc Amer Math Soc. 21(1969): 719--721.
Proves that the Thue-Morse sequence has discrepancy <math> \gg N^{\log_4(3)}</math>

*Halász, G. [http://gowers.wordpress.com/2010/02/02/edp5-another-very-brief-summary/#comment-5806 On random multiplicative functions]. Hubert Delange colloquium (Orsay, 1982), 74–96, Publ. Math. Orsay, 83-4, Univ. Paris XI, Orsay, 1983.

The discrepancy of a random multiplicative function is close to $\sqrt{N}$.

=== Problem lists on which this problem appears ===

*Erdős, P. and Graham, R. [http://www.math.ucsd.edu/~fan/ron/papers/79_09_combinatorial_number_theory.pdf Old and New Problems and Results in Combinatorial Number Theory: van der Waerden's Theorem and Related Topics], L'Enseignement Math. 25 (1979), 325-344.

Our problem is mentioned at the bottom of page 331, where they indicate knowledge of a coloring with logarithmic discrepancy. On pages 330-1, they review work on the non-homogeneous problem.

*Erdős, Paul, Some of my favourite unsolved problems. A tribute to Paul Erdős, 467--478, Cambridge Univ. Press, Cambridge, 1990. MR1117038 (92f:11003)

"Let <math>f(n)=\pm1</math>. Is it true that for every <math>c</math> there is a <math>d</math> and an <math>m</math> so that
:<math> \left| \sum_{k=1}^m f(kd) \right| > c</math>?
I will pay $500 for an answer. Let <math>f(n)=\pm1</math> and <math>f(ab)=f(a)f(b)</math>. Is it then true that <math>\left| \sum_{k=1}^m f(kd) \right|</math> cannot be bounded?"

*P. Erdős. [http://www.renyi.hu/~p_erdos/1957-13.pdf Some unsolved problems], Michigan Math. J. 4 (1957), 291--300 MR20 #5157; Zentralblatt 81,1.

Problem 9 of this paper is ours. Erdős dates the problem to around 1932, and notes what we know about Liouville's function (lower and upper bound).

*Finch, S. [http://algo.inria.fr/csolve/ec.pdf Two-colorings of positive integers]. Dated May 27, 2008.

Summarizes knowledge of discrepancy of two colorings when the discrepancy is restricted to homogeneous progressions, nonhomogeneous progressions, and homogeneous quasi-progressions. Contains bibliography with 17 entries, including most of those above.

=== non-homogeneous AP, quasi-AP, and other related discrepancy papers ===

*Hochberg, Robert. [http://www.springerlink.com/content/q02424284373qw45/ Large Discrepancy In Homogenous Quasi-Arithmetic Progressions]. Combinatorica, Volume 26 (2006), Number 1.

Roth's method for the <math> n^{1/4}</math> lower bound on nonhomogeneous AP discrepancy is adapted to give a lower bound for homogeneous quasi-AP discrepancy. This result is weaker than the Vijay result, but uses a different method.

*Vijay, Sujith. [http://www.combinatorics.org/Volume_15/PDF/v15i1r104.pdf On the discrepancy of quasi-progressions]. Electronic Journal of Combinatorics, R104 of Volume 15(1), 2008.

A quasi-progression is a sequence of the form <math>\lfloor k \alpha \rfloor </math>, with <math>1\leq k \leq K</math> for some K. The main theorem of interest is: If the integers from 0 to n are 2-coloured, there exists a quasi-progression that has discrepancy at least <math> (1/50)n^{1/6}</math>.

* Doerr, Benjamin; Srivastav, Anand; Wehr, Petra. [http://www.combinatorics.org/Volume_11/Abstracts/v11i1r5.html Discrepancy of Cartesian products of arithmetic progressions]. Electron. J. Combin. 11 (2004), no. 1, Research Paper 5, 16 pp. (electronic).

Abstract: We determine the combinatorial discrepancy of the hypergraph H of cartesian
products of d arithmetic progressions in the <math> [N]^d</math> –lattice.
The study of such higher dimensional arithmetic progressions is motivated by a
multi-dimensional version of van derWaerden’s theorem, namely the Gallai-theorem
(1933). We solve the discrepancy problem for d–dimensional arithmetic progressions
by proving <math>disc(H) = \Theta�(N^{d/4})</math> for every fixed positive integer d. This extends the famous
lower bound of <math> \Omega(N^{1/4})</math> of Roth (1964) and the matching upper bound <math> O(N^{1/4})</math>
of Matouˇsek and Spencer (1996) from d = 1 to arbitrary, fixed d. To establish
the lower bound we use harmonic analysis on locally compact abelian groups. For
the upper bound a product coloring arising from the theorem of Matouˇsek and
Spencer is sufficient. We also regard some special cases, e.g., symmetric arithmetic
progressions and infinite arithmetic progressions.

*Cilleruelo, Javier; Hebbinghaus, Nils [http://www.ams.org/mathscinet-getitem?mr=2547932 Discrepancy in generalized arithmetic progressions]. European J. Combin. 30 (2009), no. 7, 1607--1611. MR 2547932

*Valkó, Benedek. [Discrepancy of arithmetic progressions in higher dimensions]. (English summary)
J. Number Theory 92 (2002), no. 1, 117--130. MR1880588 (2003b:11071)

Abstract:
K. F. Roth (1964, Acta. Arith.9, 257–260) proved that the discrepancy of arithmetic progressions contained in [N] is at least <math> cN^{1/4} </math>, and later it was proved that this result is sharp. We consider the d-dimensional version of this problem. We give a lower estimate for the discrepancy of arithmetic progressions on <math> [N]^d </math> and prove that this result is nearly sharp. We use our results to give an upper estimate for the discrepancy of lines on an N×N lattice, and we also give an estimate for the discrepancy of a related random hypergraph.

*Roth, K. F., Remark concerning integer sequences. Acta Arith. 9 1964 257--260. MR0168545 (29 #5806)

Shows that the discrepancy on APs is at least <math>cN^{1/4}</math>.

Human proof that completely multiplicative sequences have discrepancy greater than 2

2010-02-04T03:25:28Z

Jbd:

It is [[multiplicative sequences|known from computer search]] that the longest completely multiplicative sequence <math>f: [N] \to \{-1,+1\}</math> of discrepancy 2 has length 246. Here we record the attempts to prove this by hand. Notation: we display our function in rows of 10, starting with 0. We use the notation that + and - denote values which are known to equal +1 or -1 respectively, and ? for unknown values. Thus, initially our array looks like this:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

since f(n^2) = 1 for all square numbers.

The notions of a ''cut'' and a ''block'' will be useful. A ''cut'' occurs between n and n+1 whenever it can be proved that f[1,n] sums to 0; these can only occur when n is even. We indicate a cut by a bar | between n and n+1. Thus for instance we begin with just one cut:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

A ''block'' is any interval between two cuts; thus inside each block one must sum to zero, and the discrepancy inside each block is at most 2. Thus we can "localise" the analysis to each block, though multiplicativity relates the blocks together.

A block is ''solved'' if all values are assigned in it. Solved blocks are uninteresting (other than for the values they provide by multiplicativity), so we shall adopt the convention of concatenating solved blocks with adjacent blocks to reduce clutter.

More notation: we use f[a,b] as short for f(a)+...+f(b). Note that

: f[a,b] = -4,-2,0,2,4 if a is odd and b is even;
: f[a,b] = -3,-1,1,3 if a and b are both odd or both even;
: f[a,b] = -2,0,2 if a is even and b is odd.

Also if [a,b] starts or ends at a cut, then f[a,b] has magnitude at most 2.

A key observation is that if f(2n)=f(2n+1) then there must be a cut between 2n and 2n+1.

== Case 1. f(2)=1 ==

Now, let us '''assume''' f(2)=1 and that the discrepancy is at most 2 and see what happens. From multiplicativity we obtain f(2n^2)=1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + ? + ? ? ? + + 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

As f(8)=f(9) we can now cut at 8=9. And then by using discrepancy we get f(3)=f(5)=-1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + - + - ? ? +|+ 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

which then by multiplicativity and adding cuts where obvious extends to

0 1 2 3 4 5 6 7 8 9

0 + + - + - -|? +|+ 0-9
-|? - ? ? + + ? + ? 10-19
- ? ? ? - + ? - ? ? 20-29
+ ? + ? ? ? + ? ? ? 30-39

Looking at the 7-8 block we see that f(7) is forced to be -1; from this and multiplicativity, and showing rows up to 100, we arrive at

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + ? 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? ? ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - ? ? ? 50-59
+ ? ? - + ? ? ? ? ? 60-69
+ ? + ? ? - ? ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? ? - ? + ? 90-99
+

Note that

: f[15,19] = 3 + f(17) + f(19)

cannot exceed 3, and

: f[168,171] = 2 - f(17) + f(19)

cannot exceed 2. Thus f(19) = -1. We use this information to update the board:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + - 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? - ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - + ? ? 50-59
+ ? ? - + ? ? ? ? ? 60-69
+ ? + ? ? - - ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? + - ? + ? 90-99
+

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
+

This seems to be about as far as one can get just from the assumption f(2)=+1.

== Case 1.a f(2)=f(37)=1 ==

The most profitable '''assumption''' here seems to be f(37)=+1. This creates a cut at 36-37, which then forces f(34)=-1 and then f(33)=-1 also. Filling in these values and adding in cuts, and extending a row gives

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+ ? +|- - + + + -|B 30-39
- ? + ? + - e ? - + 40-49
+|+ b ? -|- - + f ? 50-59
+ ? ? - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B e ? ? + - ? +|+ 90-99
+ ? + ? b - ? ? - ? 100-109

looking at f[30,32] we thus have f(31)=-1, while from f[99,103] we have f(101)=f(103)=-1,

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + + + -|B 30-39
- ? + ? + - e ? - + 40-49
+|+ b ? -|- - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B e + ? + - ? +|+ 90-99
+ - + - b - ? ? - ? 100-109

From the [51,54] block we have f(53) = B, which then forces f[90,106]=0, so there is a cut at 106-107:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + + + -|B 30-39
- ? + ? + - e ? - + 40-49
+|+ b B -|- - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B e + ? + - ? +|+ 90-99
+ - + - b - B|? - ? 100-109
-|- - ? + E f b ? + 110-119
+|+ ? ? -|- - ? + ? 120-129

From f[118-120] we have f(59)=-1; from f[111,113] we have f(113)=+1;

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + + + -|B 30-39
- ? + ? + - e ? - + 40-49
+|+ b B -|- - + f - 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B e + ? + - ? +|+ 90-99
+ - + - b - B|? - ? 100-109
-|- - + +|E f b - + 110-119
+|+ ? ? -|- - ? + ? 120-129

From f[13,30] we have 2b+e+f=0 while from f[115,120] we have -e+f+b=-1. This forces b=-1, e=f=1:

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? + - + ? - + 40-49
+|+ - + - - - + + - 50-59
+|? -|- +|+ - ? -|- 60-69
+|? + ? + - -|- +|? 70-79
- + ? ? +|+ ? - + ? 80-89
- + +|+ ? + - ? +|+ 90-99
+ - + - - - +|? - ? 100-109
-|- - + + - + - - + 110-119
+|+ ? ? -|- - ? + ? 120-129

From the 60-69 row we then get f(61)=+1, f(67)=+1; from f[85,92] we get f(43)=f(89)=-1; from f[93,98] we have f(47)=f(97)=-1; from f[125,127] one has f(127)=+1; from f[107,110] one gets f(107)=f(109)=+1:

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + - + - + - - + 40-49
+|+ - + - - - + + - 50-59
+ + - - + + - + -|- 60-69
+|? + ? + - -|- +|? 70-79
- + ? ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
+ - + - - - + + - + 100-109
-|- - + + - + - - + 110-119
+|+ + ? -|- - + + ? 120-129

From f[41,50] one has f(41)=-1; but then f[121,125]=2, a contradiction. Hence this case is not possible.

== Case 1.b f(2)=1, f(37)=-1 ==

This takes us back to case 1.b f(2)=1, f(37)=-1, which begins here:

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
+

So we assume f(37)=-1

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + - - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? -|- - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

From f[75,77] we have A=+1, then from f[11,13] we have b=-1, and then from f[75,79] we have f(79)=+1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|c + - 10-19
- + - e - + + - - f 20-29
+ ? +|+ c|+ + - -|- 30-39
- ? + ? - - e ? - + 40-49
+ C + ? - + - + f ? 50-59
+ ? ? - + - + ? c E 60-69
+ ? + + - - - + - + 70-79
- + ? ? + C ? F - ? 80-89
- - C ? ? + - ? + - 90-99
+

From f[33,34] we have c=-1, and from f[39,41] we have f(41)=+1;

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - e - + + - - f 20-29
+ ? + + -|+ + - - - 30-39
- + + ? - - e ? - + 40-49
+ + + ? - + - + f ? 50-59
+ ? ? - + - + ? + E 60-69
+ ? + + - - - + - + 70-79
- + ? ? + + ? F - ? 80-89
- - + ? ? + - ? + - 90-99
+

From f[19,23] we have e = +

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - f 20-29
+ ? + + -|+ + - - - 30-39
- + + ? - - + ? - + 40-49
+ + + ? - + - + f ? 50-59
+ ? ? - + - + ? + - 60-69
+ ? + + - - - + - + 70-79
- + ? ? + + ? F - ? 80-89
- - + ? ? + - ? + - 90-99
+

From f[27,29] we have f = +, so from f[27,34] we have f(31) = -, and sums and multiplicity force

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|? + - 60-69
+ ? + + - - - + - + 70-79
- + + ? + + + - - ? 80-89
- - + + - + - ? + - 90-99
+

The smallest sum before 70 is -1, so 71 must be -1. This forces 67 = -1, otherwise the sum at 73 is 3.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|? + + + - - ? 80-89
- - + + - + - ? + - 90-99
+

83 must be -, otherwise the sum is 3 at 85.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|- +|+ + - -|? 80-89
- - + + - + - ? + - 90-99
+

So 89 is + (otherwise the sum is -3 at 91)

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|- +|+ + - -|+ 80-89
- - +|+ -|+ -|? + - 90-99
-

(could use intermediate diagrams displayed here with text)

f(58) is positive so f(116) is positive

f(13) is positive and f(9) is a square so f(117) is positive

there is a cut between 116 and 117

f(118) is negative since f(59) is negative

f(7) and f(17) are negative so f(119) is positive

f(3) and f(5) are negative and f(2) is positive so 120 is positive

121 is a square so f(121) is positive

so at 116 the sum is zero at because there is a cut between
116 and 117 at 117 the sum is one since f(117) is positive
at 118 the sum is zero since f(118) is negative and since f(119) f(120) and f(121) are positive at 121 the sum is three and we have shown that if a completely multiplicative sequnce has f(2) equal to 1 and f(37) equal to -1 then it reaches discrepancy 3 before 246.

Since we have previously shown that if a completely multiplicative sequence that has f(2) equal to 1 and f(37)equal to 1 reaches discrepancy 3 before 246 we have any completely multiplicative sequence with f(2) equal to one reaches discrepancy 3 before 246 and that finishes case 1.

== Case 2. f(2) = -1 ==

Therefore f(2) = -1. Resetting the board:

0 1 2 3 4 5 6 7 8 9

0 + - ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39
? ? ? ? ? ? ? ? ? + 40-49
? ? ? ? ? ? ? ? ? ? 50-59
? ? ? ? + ? ? ? ? ? 60-69
? ? ? ? ? ? ? ? ? ? 70-79
? + ? ? ? ? ? ? ? ? 80-89
? ? ? ? ? ? ? ? ? ? 90-99

Adding in multiplicative values:

0 1 2 3 4 5 6 7 8 9

0|+ -|? + ? ? ? - + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? - ? ? ? + ? ? ? 30-39
? ? ? ? ? ? ? ? ? + 40-49
- ? ? ? ? ? ? ? ? ? 50-59
? ? ? ? + ? ? ? ? ? 60-69
? ? - ? ? ? ? ? ? ? 70-79
? + ? ? ? ? ? ? ? ? 80-89
? ? ? ? ? ? ? ? - ? 90-99

(Concatenation of various proofs, needs a lot of polish here)

''Assume'' f(5) = 1

Suppose f(11)= 1

Setting f(2)=-1... Setting f(5)=1... Setting f(11)=1... Deducing... Setting f(3)=-1... Deducing... Setting f(7)=-1... Setting f(241)=1... Deducing... Setting f(23)=-1... Setting f(17)=-1... Setting f(19)=1... Deducing... Setting f(13)=-1... Setting f(47)=-1... Setting f(59)=1... Setting f(67)=-1... Setting f(239)=1... Setting f(251)=1... Setting f(31)=1... Deducing... Setting f(29)=-1... Setting f(53)=-1... Contradiction! f(1) = 1 and f(1) = -1 (from f(58)=-1)

So f(11)=-1.

[program name] -2 5 -11
Setting f(2)=-1... Setting f(5)=1... Setting f(11)=-1... Deducing... Setting f(3)=-1... Deducing... Setting f(7)=-1... Setting f(13)=1... Setting f(23)=-1... Setting f(241)=1... Deducing... Contradiction! -1 >= f[1,23] >= 1

So f(5) = -1.

0 1 2 3 4 5 6 7 8 9

0|+ -|? + - ? ? - + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? - ? ? ? + ? ? ? 30-39
? ? ? ? ? ? ? ? ? + 40-49
- ? ? ? ? ? ? ? ? ? 50-59
? ? ? ? + ? ? ? ? ? 60-69
? ? - ? ? ? ? ? ? ? 70-79
? + ? ? ? ? ? ? ? ? 80-89
? ? ? ? ? ? ? ? - ? 90-99

If a completely multiplicative sequence with discrepancy 2 or less has length more than 246 then I think there is a human proof that f(2) and f(5) are -1. Given this I can show that if f(3) is 1 f(7) is -1.

Assume that f(3) is 1 and f(7) is 1. We have by the above that f(2) and f(5) are -1. Then this will give the partial sum at 10 the value 2 so f(11) must be -1.

Since f(12) is 1 the sum at 12 must be 2 and f(13) must be -1.

Then f(25), f(26),f(27),f(28) and f(30) must be 1 this forces f(31) to be -1.

Then f(62), f(63),f(64),f(65) and f(66) must be 1 and we have a contradiction.

So if a completely multiplicative sequence with discrepancy 2 or less has length more than 246 and f(3)=1 then f(7)=-1.

So supopse f(3) = 1 and f(7) = -1.

With these assumptions, the partial sum f[18,21]:=f(18)+f(19)+f(20)+f(21) equals -3+f(19). Since f[1,odd] is in {-1,0,1} it follows that f[18,21]>=-2 and thus f(19)=1.

Similarly we get:

f[168,171]=3+f(17) f(17)=-1,

f[34,37]=3+f(37) f(37)=-1 and

f[74,77]=3-f(11) f(11)=1.

Since f(9)=f(10)=f(11)=f(12)=1 it follows that f(13)=-1.

f[50,55]=-5+f(53)>=-2 which cannot be satisfied for f(53) in {-1,1}.

Hence f(7) cannot be -1 and thus f(3)=-1.

It remains to show (by hand) that there is no completely multiplicative sequence with discrepancy 2 and f(2)=f(3)=f(5)=-1.

0 1 2 3 4 5 6 7 8 9

0|+ -|- + - ? ? - + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? - ? ? ? + ? ? ? 30-39
? ? ? ? ? ? ? ? ? + 40-49
- ? ? ? ? ? ? ? ? ? 50-59
? ? ? ? + ? ? ? ? ? 60-69
? ? - ? ? ? ? ? ? ? 70-79
? + ? ? ? ? ? ? ? ? 80-89
? ? ? ? ? ? ? ? - ? 90-99

Putting in multiplicative values:

0 1 2 3 4 5 6 7 8 9

0|+ -|- +|- +|? - + 0-9
+ ? - ? ? + + ? - ? 10-19
- ? ? ? +|+ ? - ? ? 20-29
- ? - ? ? ? + ? ? ? 30-39
+ ? ? ? ? - ? ? - + 40-49
- ? ? ? - ? ? ? ? ? 50-59
+ ? ? ? + ? ? ? ? ? 60-69
(need to finish filling in)
? ? - ? ? ? ? ? ? ? 70-79
? + ? ? ? ? ? ? ? ? 80-89
? ? ? ? ? ? ? ? - ? 90-99

Here is a [[computer proof that completely multiplicative sequences have at least 2]], inspired by the above analysis.

Human proof that completely multiplicative sequences have discrepancy greater than 2

2010-02-03T17:29:41Z

Jbd:

It is [[multiplicative sequences|known from computer search]] that the longest completely multiplicative sequence <math>f: [N] \to \{-1,+1\}</math> of discrepancy 2 has length 246. Here we record the attempts to prove this by hand. Notation: we display our function in rows of 10, starting with 0. We use the notation that + and - denote values which are known to equal +1 or -1 respectively, and ? for unknown values. Thus, initially our array looks like this:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

since f(n^2) = 1 for all square numbers.

The notions of a ''cut'' and a ''block'' will be useful. A ''cut'' occurs between n and n+1 whenever it can be proved that f[1,n] sums to 0; these can only occur when n is even. We indicate a cut by a bar | between n and n+1. Thus for instance we begin with just one cut:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

A ''block'' is any interval between two cuts; thus inside each block one must sum to zero, and the discrepancy inside each block is at most 2. Thus we can "localise" the analysis to each block, though multiplicativity relates the blocks together.

A block is ''solved'' if all values are assigned in it. Solved blocks are uninteresting (other than for the values they provide by multiplicativity), so we shall adopt the convention of concatenating solved blocks with adjacent blocks to reduce clutter.

More notation: we use f[a,b] as short for f(a)+...+f(b). Note that

: f[a,b] = -4,-2,0,2,4 if a is odd and b is even;
: f[a,b] = -3,-1,1,3 if a and b are both odd or both even;
: f[a,b] = -2,0,2 if a is even and b is odd.

Also if [a,b] starts or ends at a cut, then f[a,b] has magnitude at most 2.

A key observation is that if f(2n)=f(2n+1) then there must be a cut between 2n and 2n+1.

== Case 1. f(2)=1 ==

Now, let us '''assume''' f(2)=1 and that the discrepancy is at most 2 and see what happens. From multiplicativity we obtain f(2n^2)=1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + ? + ? ? ? + + 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

As f(8)=f(9) we can now cut at 8=9. And then by using discrepancy we get f(3)=f(5)=-1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + - + - ? ? +|+ 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

which then by multiplicativity and adding cuts where obvious extends to

0 1 2 3 4 5 6 7 8 9

0 + + - + - -|? +|+ 0-9
-|? - ? ? + + ? + ? 10-19
- ? ? ? - + ? - ? ? 20-29
+ ? + ? ? ? + ? ? ? 30-39

Looking at the 7-8 block we see that f(7) is forced to be -1; from this and multiplicativity, and showing rows up to 100, we arrive at

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + ? 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? ? ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - ? ? ? 50-59
+ ? ? - + ? ? ? ? ? 60-69
+ ? + ? ? - ? ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? ? - ? + ? 90-99
+

Note that

: f[15,19] = 3 + f(17) + f(19)

cannot exceed 3, and

: f[168,171] = 2 - f(17) + f(19)

cannot exceed 2. Thus f(19) = -1. We use this information to update the board:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + - 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? - ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - + ? ? 50-59
+ ? ? - + ? ? ? ? ? 60-69
+ ? + ? ? - - ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? + - ? + ? 90-99
+

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
+

This seems to be about as far as one can get just from the assumption f(2)=+1.

== Case 1.a f(2)=f(37)=1 ==

The most profitable '''assumption''' here seems to be f(37)=+1. This creates a cut at 36-37, which then forces f(34)=-1 and then f(33)=-1 also. Filling in these values and adding in cuts, and extending a row gives

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+ ? +|- - + + + -|B 30-39
- ? + ? + - e ? - + 40-49
+|+ b ? -|- - + f ? 50-59
+ ? ? - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B e ? ? + - ? +|+ 90-99
+ ? + ? b - ? ? - ? 100-109

looking at f[30,32] we thus have f(31)=-1, while from f[99,103] we have f(101)=f(103)=-1,

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + + + -|B 30-39
- ? + ? + - e ? - + 40-49
+|+ b ? -|- - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B e + ? + - ? +|+ 90-99
+ - + - b - ? ? - ? 100-109

From the [51,54] block we have f(53) = B, which then forces f[90,106]=0, so there is a cut at 106-107:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + + + -|B 30-39
- ? + ? + - e ? - + 40-49
+|+ b B -|- - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B e + ? + - ? +|+ 90-99
+ - + - b - B|? - ? 100-109
-|- - ? + E f b ? + 110-119
+|+ ? ? -|- - ? + ? 120-129

From f[118-120] we have f(59)=-1; from f[111,113] we have f(113)=+1;

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + + + -|B 30-39
- ? + ? + - e ? - + 40-49
+|+ b B -|- - + f - 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B e + ? + - ? +|+ 90-99
+ - + - b - B|? - ? 100-109
-|- - + +|E f b - + 110-119
+|+ ? ? -|- - ? + ? 120-129

From f[13,30] we have 2b+e+f=0 while from f[115,120] we have -e+f+b=-1. This forces b=-1, e=f=1:

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? + - + ? - + 40-49
+|+ - + - - - + + - 50-59
+|? -|- +|+ - ? -|- 60-69
+|? + ? + - -|- +|? 70-79
- + ? ? +|+ ? - + ? 80-89
- + +|+ ? + - ? +|+ 90-99
+ - + - - - +|? - ? 100-109
-|- - + + - + - - + 110-119
+|+ ? ? -|- - ? + ? 120-129

From the 60-69 row we then get f(61)=+1, f(67)=+1; from f[85,92] we get f(43)=f(89)=-1; from f[93,98] we have f(47)=f(97)=-1; from f[125,127] one has f(127)=+1; from f[107,110] one gets f(107)=f(109)=+1:

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + - + - + - - + 40-49
+|+ - + - - - + + - 50-59
+ + - - + + - + -|- 60-69
+|? + ? + - -|- +|? 70-79
- + ? ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
+ - + - - - + + - + 100-109
-|- - + + - + - - + 110-119
+|+ + ? -|- - + + ? 120-129

From f[41,50] one has f(41)=-1; but then f[121,125]=2, a contradiction. Hence this case is not possible.

== Case 1.b f(2)=1, f(37)=-1 ==

This takes us back to case 1.b f(2)=1, f(37)=-1, which begins here:

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
+

So we assume f(37)=-1

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + - - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? -|- - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

From f[75,77] we have A=+1, then from f[11,13] we have b=-1, and then from f[75,79] we have f(79)=+1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|c + - 10-19
- + - e - + + - - f 20-29
+ ? +|+ c|+ + - -|- 30-39
- ? + ? - - e ? - + 40-49
+ C + ? - + - + f ? 50-59
+ ? ? - + - + ? c E 60-69
+ ? + + - - - + - + 70-79
- + ? ? + C ? F - ? 80-89
- - C ? ? + - ? + - 90-99
+

From f[33,34] we have c=-1, and from f[39,41] we have f(41)=+1;

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - e - + + - - f 20-29
+ ? + + -|+ + - - - 30-39
- + + ? - - e ? - + 40-49
+ + + ? - + - + f ? 50-59
+ ? ? - + - + ? + E 60-69
+ ? + + - - - + - + 70-79
- + ? ? + + ? F - ? 80-89
- - + ? ? + - ? + - 90-99
+

From f[19,23] we have e = +

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - f 20-29
+ ? + + -|+ + - - - 30-39
- + + ? - - + ? - + 40-49
+ + + ? - + - + f ? 50-59
+ ? ? - + - + ? + - 60-69
+ ? + + - - - + - + 70-79
- + ? ? + + ? F - ? 80-89
- - + ? ? + - ? + - 90-99
+

From f[27,29] we have f = +, so from f[27,34] we have f(31) = -, and sums and multiplicity force

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|? + - 60-69
+ ? + + - - - + - + 70-79
- + + ? + + + - - ? 80-89
- - + + - + - ? + - 90-99
+

The smallest sum before 70 is -1, so 71 must be -1. This forces 67 = -1, otherwise the sum at 73 is 3.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|? + + + - - ? 80-89
- - + + - + - ? + - 90-99
+

83 must be -, otherwise the sum is 3 at 85.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|- +|+ + - -|? 80-89
- - + + - + - ? + - 90-99
+

So 89 is + (otherwise the sum is -3 at 91)

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|- +|+ + - -|+ 80-89
- - +|+ -|+ -|? + - 90-99
-

(could use intermediate diagrams displayed here with text)

f(58) is positive so f(116) is positive

f(13) is positive and f(9) is a square so f(117) is positive

there is a cut between 116 and 117

f(118) is negative since f(59) is negative

f(7) and f(17) are negative so f(119) is positive

f(3) and f(5) are negative and f(2) is positive so 120 is positive

121 is a square so f(121) is positive

so at 116 the sum is zero at because there is a cut between
116 and 117 at 117 the sum is one since f(117) is positive
at 118 the sum is zero since f(118) is negative and since f(119) f(120) and f(121) are positive at 121 the sum is three and we have shown that if a completely multiplicative sequnce has f(2) equal to 1 and f(37) equal to -1 then it reaches discrepancy 3 before 246.

Since we have previously shown that if a completely multiplicative sequence that has f(2) equal to 1 and f(37)equal to 1 reaches discrepancy 3 before 246 we have any completely multiplicative sequence with f(2) equal to one reaches discrepancy 3 before 246 and that finishes case 1.

== Case 2. f(2) = -1 ==

Therefore f(2) = -1. Resetting the board:

0 1 2 3 4 5 6 7 8 9

0 + - ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39
? ? ? ? ? ? ? ? ? + 40-49
? ? ? ? ? ? ? ? ? ? 50-59
? ? ? ? + ? ? ? ? ? 60-69
? ? ? ? ? ? ? ? ? ? 70-79
? + ? ? ? ? ? ? ? ? 80-89
? ? ? ? ? ? ? ? ? ? 90-99

Adding in multiplicative values:

0 1 2 3 4 5 6 7 8 9

0|+ -|? + ? ? ? - + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? - ? ? ? + ? ? ? 30-39
? ? ? ? ? ? ? ? ? + 40-49
- ? ? ? ? ? ? ? ? ? 50-59
? ? ? ? + ? ? ? ? ? 60-69
? ? - ? ? ? ? ? ? ? 70-79
? + ? ? ? ? ? ? ? ? 80-89
? ? ? ? ? ? ? ? - ? 90-99

(Concatenation of various proofs, needs a lot of polish here)

''Assume'' f(5) = 1

Suppose f(11)= 1

Setting f(2)=-1... Setting f(5)=1... Setting f(11)=1... Deducing... Setting f(3)=-1... Deducing... Setting f(7)=-1... Setting f(241)=1... Deducing... Setting f(23)=-1... Setting f(17)=-1... Setting f(19)=1... Deducing... Setting f(13)=-1... Setting f(47)=-1... Setting f(59)=1... Setting f(67)=-1... Setting f(239)=1... Setting f(251)=1... Setting f(31)=1... Deducing... Setting f(29)=-1... Setting f(53)=-1... Contradiction! f(1) = 1 and f(1) = -1 (from f(58)=-1)

So f(11)=-1.

[program name] -2 5 -11
Setting f(2)=-1... Setting f(5)=1... Setting f(11)=-1... Deducing... Setting f(3)=-1... Deducing... Setting f(7)=-1... Setting f(13)=1... Setting f(23)=-1... Setting f(241)=1... Deducing... Contradiction! -1 >= f[1,23] >= 1

So f(5) = -1.

0 1 2 3 4 5 6 7 8 9

0|+ -|? + - ? ? - + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? - ? ? ? + ? ? ? 30-39
? ? ? ? ? ? ? ? ? + 40-49
- ? ? ? ? ? ? ? ? ? 50-59
? ? ? ? + ? ? ? ? ? 60-69
? ? - ? ? ? ? ? ? ? 70-79
? + ? ? ? ? ? ? ? ? 80-89
? ? ? ? ? ? ? ? - ? 90-99

If a completely multiplicative sequence with discrepancy 2 or less has length more than 246 then I think there is a human proof that f(2) and f(5) are -1. Given this I can show that if f(3) is 1 f(7) is -1.

Assume that f(3) is 1 and f(7) is 1. We have by the above that f(2) and f(5) are -1. Then this will give the partial sum at 10 the value 2 so f(11) must be -1.

Since f(12) is 1 the sum at 12 must be 2 and f(13) must be -1.

Then f(25), f(26),f(27),f(28) and f(30) must be 1 this forces f(31) to be -1.

Then f(62), f(63),f(64),f(65) and f(66) must be 1 and we have a contradiction.

So if a completely multiplicative sequence with discrepancy 2 or less has length more than 246 and f(3)=1 then f(7)=-1.

So supopse f(3) = 1 and f(7) = -1.

With these assumptions, the partial sum f[18,21]:=f(18)+f(19)+f(20)+f(21) equals -3+f(19). Since f[1,odd] is in {-1,0,1} it follows that f[18,21]>=-2 and thus f(19)=1.

Similarly we get:

f[168,171]=3+f(17) f(17)=-1,

f[34,37]=3+f(37) f(37)=-1 and

f[74,77]=3-f(11) f(11)=1.

Since f(9)=f(10)=f(11)=f(12)=1 it follows that f(13)=-1.

f[50,55]=-5+f(53)>=-2 which cannot be satisfied for f(53) in {-1,1}.

Hence f(7) cannot be -1 and thus f(3)=-1.

It remains to show (by hand) that there is no completely multiplicative sequence with discrepancy 2 and f(2)=f(3)=f(5)=-1.

0 1 2 3 4 5 6 7 8 9

0|+ -|- + - ? ? - + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? - ? ? ? + ? ? ? 30-39
? ? ? ? ? ? ? ? ? + 40-49
- ? ? ? ? ? ? ? ? ? 50-59
? ? ? ? + ? ? ? ? ? 60-69
? ? - ? ? ? ? ? ? ? 70-79
? + ? ? ? ? ? ? ? ? 80-89
? ? ? ? ? ? ? ? - ? 90-99

Putting in multiplicative values:

0 1 2 3 4 5 6 7 8 9

0|+ -|- +|- +|? - + 0-9
+ ? - ? ? + + ? - ? 10-19
- ? ? ? +|+ ? - ? ? 20-29
- ? - ? ? ? + ? ? ? 30-39

(need to finish filling in)
+ ? ? ? ? ? ? ? ? + 40-49
- ? ? ? ? ? ? ? ? ? 50-59
? ? ? ? + ? ? ? ? ? 60-69
? ? - ? ? ? ? ? ? ? 70-79
? + ? ? ? ? ? ? ? ? 80-89
? ? ? ? ? ? ? ? - ? 90-99

Here is a [[computer proof that completely multiplicative sequences have at least 2]], inspired by the above analysis.

Human proof that completely multiplicative sequences have discrepancy greater than 2

2010-02-03T17:23:53Z

Jbd:

It is [[multiplicative sequences|known from computer search]] that the longest completely multiplicative sequence <math>f: [N] \to \{-1,+1\}</math> of discrepancy 2 has length 246. Here we record the attempts to prove this by hand. Notation: we display our function in rows of 10, starting with 0. We use the notation that + and - denote values which are known to equal +1 or -1 respectively, and ? for unknown values. Thus, initially our array looks like this:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

since f(n^2) = 1 for all square numbers.

The notions of a ''cut'' and a ''block'' will be useful. A ''cut'' occurs between n and n+1 whenever it can be proved that f[1,n] sums to 0; these can only occur when n is even. We indicate a cut by a bar | between n and n+1. Thus for instance we begin with just one cut:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

A ''block'' is any interval between two cuts; thus inside each block one must sum to zero, and the discrepancy inside each block is at most 2. Thus we can "localise" the analysis to each block, though multiplicativity relates the blocks together.

A block is ''solved'' if all values are assigned in it. Solved blocks are uninteresting (other than for the values they provide by multiplicativity), so we shall adopt the convention of concatenating solved blocks with adjacent blocks to reduce clutter.

More notation: we use f[a,b] as short for f(a)+...+f(b). Note that

: f[a,b] = -4,-2,0,2,4 if a is odd and b is even;
: f[a,b] = -3,-1,1,3 if a and b are both odd or both even;
: f[a,b] = -2,0,2 if a is even and b is odd.

Also if [a,b] starts or ends at a cut, then f[a,b] has magnitude at most 2.

A key observation is that if f(2n)=f(2n+1) then there must be a cut between 2n and 2n+1.

== Case 1. f(2)=1 ==

Now, let us '''assume''' f(2)=1 and that the discrepancy is at most 2 and see what happens. From multiplicativity we obtain f(2n^2)=1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + ? + ? ? ? + + 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

As f(8)=f(9) we can now cut at 8=9. And then by using discrepancy we get f(3)=f(5)=-1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + - + - ? ? +|+ 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

which then by multiplicativity and adding cuts where obvious extends to

0 1 2 3 4 5 6 7 8 9

0 + + - + - -|? +|+ 0-9
-|? - ? ? + + ? + ? 10-19
- ? ? ? - + ? - ? ? 20-29
+ ? + ? ? ? + ? ? ? 30-39

Looking at the 7-8 block we see that f(7) is forced to be -1; from this and multiplicativity, and showing rows up to 100, we arrive at

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + ? 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? ? ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - ? ? ? 50-59
+ ? ? - + ? ? ? ? ? 60-69
+ ? + ? ? - ? ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? ? - ? + ? 90-99
+

Note that

: f[15,19] = 3 + f(17) + f(19)

cannot exceed 3, and

: f[168,171] = 2 - f(17) + f(19)

cannot exceed 2. Thus f(19) = -1. We use this information to update the board:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + - 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? - ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - + ? ? 50-59
+ ? ? - + ? ? ? ? ? 60-69
+ ? + ? ? - - ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? + - ? + ? 90-99
+

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
+

This seems to be about as far as one can get just from the assumption f(2)=+1.

== Case 1.a f(2)=f(37)=1 ==

The most profitable '''assumption''' here seems to be f(37)=+1. This creates a cut at 36-37, which then forces f(34)=-1 and then f(33)=-1 also. Filling in these values and adding in cuts, and extending a row gives

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+ ? +|- - + + + -|B 30-39
- ? + ? + - e ? - + 40-49
+|+ b ? -|- - + f ? 50-59
+ ? ? - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B e ? ? + - ? +|+ 90-99
+ ? + ? b - ? ? - ? 100-109

looking at f[30,32] we thus have f(31)=-1, while from f[99,103] we have f(101)=f(103)=-1,

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + + + -|B 30-39
- ? + ? + - e ? - + 40-49
+|+ b ? -|- - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B e + ? + - ? +|+ 90-99
+ - + - b - ? ? - ? 100-109

From the [51,54] block we have f(53) = B, which then forces f[90,106]=0, so there is a cut at 106-107:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + + + -|B 30-39
- ? + ? + - e ? - + 40-49
+|+ b B -|- - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B e + ? + - ? +|+ 90-99
+ - + - b - B|? - ? 100-109
-|- - ? + E f b ? + 110-119
+|+ ? ? -|- - ? + ? 120-129

From f[118-120] we have f(59)=-1; from f[111,113] we have f(113)=+1;

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + + + -|B 30-39
- ? + ? + - e ? - + 40-49
+|+ b B -|- - + f - 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B e + ? + - ? +|+ 90-99
+ - + - b - B|? - ? 100-109
-|- - + +|E f b - + 110-119
+|+ ? ? -|- - ? + ? 120-129

From f[13,30] we have 2b+e+f=0 while from f[115,120] we have -e+f+b=-1. This forces b=-1, e=f=1:

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? + - + ? - + 40-49
+|+ - + - - - + + - 50-59
+|? -|- +|+ - ? -|- 60-69
+|? + ? + - -|- +|? 70-79
- + ? ? +|+ ? - + ? 80-89
- + +|+ ? + - ? +|+ 90-99
+ - + - - - +|? - ? 100-109
-|- - + + - + - - + 110-119
+|+ ? ? -|- - ? + ? 120-129

From the 60-69 row we then get f(61)=+1, f(67)=+1; from f[85,92] we get f(43)=f(89)=-1; from f[93,98] we have f(47)=f(97)=-1; from f[125,127] one has f(127)=+1; from f[107,110] one gets f(107)=f(109)=+1:

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + - + - + - - + 40-49
+|+ - + - - - + + - 50-59
+ + - - + + - + -|- 60-69
+|? + ? + - -|- +|? 70-79
- + ? ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
+ - + - - - + + - + 100-109
-|- - + + - + - - + 110-119
+|+ + ? -|- - + + ? 120-129

From f[41,50] one has f(41)=-1; but then f[121,125]=2, a contradiction. Hence this case is not possible.

== Case 1.b f(2)=1, f(37)=-1 ==

This takes us back to case 1.b f(2)=1, f(37)=-1, which begins here:

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
+

So we assume f(37)=-1

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + - - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? -|- - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

From f[75,77] we have A=+1, then from f[11,13] we have b=-1, and then from f[75,79] we have f(79)=+1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|c + - 10-19
- + - e - + + - - f 20-29
+ ? +|+ c|+ + - -|- 30-39
- ? + ? - - e ? - + 40-49
+ C + ? - + - + f ? 50-59
+ ? ? - + - + ? c E 60-69
+ ? + + - - - + - + 70-79
- + ? ? + C ? F - ? 80-89
- - C ? ? + - ? + - 90-99
+

From f[33,34] we have c=-1, and from f[39,41] we have f(41)=+1;

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - e - + + - - f 20-29
+ ? + + -|+ + - - - 30-39
- + + ? - - e ? - + 40-49
+ + + ? - + - + f ? 50-59
+ ? ? - + - + ? + E 60-69
+ ? + + - - - + - + 70-79
- + ? ? + + ? F - ? 80-89
- - + ? ? + - ? + - 90-99
+

From f[19,23] we have e = +

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - f 20-29
+ ? + + -|+ + - - - 30-39
- + + ? - - + ? - + 40-49
+ + + ? - + - + f ? 50-59
+ ? ? - + - + ? + - 60-69
+ ? + + - - - + - + 70-79
- + ? ? + + ? F - ? 80-89
- - + ? ? + - ? + - 90-99
+

From f[27,29] we have f = +, so from f[27,34] we have f(31) = -, and sums and multiplicity force

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|? + - 60-69
+ ? + + - - - + - + 70-79
- + + ? + + + - - ? 80-89
- - + + - + - ? + - 90-99
+

The smallest sum before 70 is -1, so 71 must be -1. This forces 67 = -1, otherwise the sum at 73 is 3.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|? + + + - - ? 80-89
- - + + - + - ? + - 90-99
+

83 must be -, otherwise the sum is 3 at 85.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|- +|+ + - -|? 80-89
- - + + - + - ? + - 90-99
+

So 89 is + (otherwise the sum is -3 at 91)

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|- +|+ + - -|+ 80-89
- - +|+ -|+ -|? + - 90-99
-

(could use intermediate diagrams displayed here with text)

f(58) is positive so f(116) is positive

f(13) is positive and f(9) is a square so f(117) is positive

there is a cut between 116 and 117

f(118) is negative since f(59) is negative

f(7) and f(17) are negative so f(119) is positive

f(3) and f(5) are negative and f(2) is positive so 120 is positive

121 is a square so f(121) is positive

so at 116 the sum is zero at because there is a cut between
116 and 117 at 117 the sum is one since f(117) is positive
at 118 the sum is zero since f(118) is negative and since f(119) f(120) and f(121) are positive at 121 the sum is three and we have shown that if a completely multiplicative sequnce has f(2) equal to 1 and f(37) equal to -1 then it reaches discrepancy 3 before 246.

Since we have previously shown that if a completely multiplicative sequence that has f(2) equal to 1 and f(37)equal to 1 reaches discrepancy 3 before 246 we have any completely multiplicative sequence with f(2) equal to one reaches discrepancy 3 before 246 and that finishes case 1.

== Case 2. f(2) = -1 ==

Therefore f(2) = -1. Resetting the board:

0 1 2 3 4 5 6 7 8 9

0 + - ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39
? ? ? ? ? ? ? ? ? + 40-49
? ? ? ? ? ? ? ? ? ? 50-59
? ? ? ? + ? ? ? ? ? 60-69
? ? ? ? ? ? ? ? ? ? 70-79
? + ? ? ? ? ? ? ? ? 80-89
? ? ? ? ? ? ? ? ? ? 90-99

Adding in multiplicative values:

0 1 2 3 4 5 6 7 8 9

0|+ -|? + ? ? ? - + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? - ? ? ? + ? ? ? 30-39
? ? ? ? ? ? ? ? ? + 40-49
- ? ? ? ? ? ? ? ? ? 50-59
? ? ? ? + ? ? ? ? ? 60-69
? ? - ? ? ? ? ? ? ? 70-79
? + ? ? ? ? ? ? ? ? 80-89
? ? ? ? ? ? ? ? - ? 90-99

(Concatenation of various proofs, needs a lot of polish here)

''Assume'' f(5) = 1

Suppose f(11)= 1

Setting f(2)=-1... Setting f(5)=1... Setting f(11)=1... Deducing... Setting f(3)=-1... Deducing... Setting f(7)=-1... Setting f(241)=1... Deducing... Setting f(23)=-1... Setting f(17)=-1... Setting f(19)=1... Deducing... Setting f(13)=-1... Setting f(47)=-1... Setting f(59)=1... Setting f(67)=-1... Setting f(239)=1... Setting f(251)=1... Setting f(31)=1... Deducing... Setting f(29)=-1... Setting f(53)=-1... Contradiction! f(1) = 1 and f(1) = -1 (from f(58)=-1)

So f(11)=-1.

[program name] -2 5 -11
Setting f(2)=-1... Setting f(5)=1... Setting f(11)=-1... Deducing... Setting f(3)=-1... Deducing... Setting f(7)=-1... Setting f(13)=1... Setting f(23)=-1... Setting f(241)=1... Deducing... Contradiction! -1 >= f[1,23] >= 1

So f(5) = -1.

0 1 2 3 4 5 6 7 8 9

0|+ -|? + - ? ? - + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? - ? ? ? + ? ? ? 30-39
? ? ? ? ? ? ? ? ? + 40-49
- ? ? ? ? ? ? ? ? ? 50-59
? ? ? ? + ? ? ? ? ? 60-69
? ? - ? ? ? ? ? ? ? 70-79
? + ? ? ? ? ? ? ? ? 80-89
? ? ? ? ? ? ? ? - ? 90-99

If a completely multiplicative sequence with discrepancy 2 or less has length more than 246 then I think there is a human proof that f(2) and f(5) are -1. Given this I can show that if f(3) is 1 f(7) is -1.

Assume that f(3) is 1 and f(7) is 1. We have by the above that f(2) and f(5) are -1. Then this will give the partial sum at 10 the value 2 so f(11) must be -1.

Since f(12) is 1 the sum at 12 must be 2 and f(13) must be -1.

Then f(25), f(26),f(27),f(28) and f(30) must be 1 this forces f(31) to be -1.

Then f(62), f(63),f(64),f(65) and f(66) must be 1 and we have a contradiction.

So if a completely multiplicative sequence with discrepancy 2 or less has length more than 246 and f(3)=1 then f(7)=-1.

So supopse f(3) = 1 and f(7) = -1.

With these assumptions, the partial sum f[18,21]:=f(18)+f(19)+f(20)+f(21) equals -3+f(19). Since f[1,odd] is in {-1,0,1} it follows that f[18,21]>=-2 and thus f(19)=1.

Similarly we get:

f[168,171]=3+f(17) f(17)=-1,

f[34,37]=3+f(37) f(37)=-1 and

f[74,77]=3-f(11) f(11)=1.

Since f(9)=f(10)=f(11)=f(12)=1 it follows that f(13)=-1.

f[50,55]=-5+f(53)>=-2 which cannot be satisfied for f(53) in {-1,1}.

Hence f(7) cannot be -1 and thus f(3)=-1.

It remains to show (by hand) that there is no completely multiplicative sequence with discrepancy 2 and f(2)=f(3)=f(5)=-1.

0 1 2 3 4 5 6 7 8 9

0|+ -|- + - ? ? - + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? - ? ? ? + ? ? ? 30-39
? ? ? ? ? ? ? ? ? + 40-49
- ? ? ? ? ? ? ? ? ? 50-59
? ? ? ? + ? ? ? ? ? 60-69
? ? - ? ? ? ? ? ? ? 70-79
? + ? ? ? ? ? ? ? ? 80-89
? ? ? ? ? ? ? ? - ? 90-99

Putting in multiplicative values:

0 1 2 3 4 5 6 7 8 9

0|+ -|- +|- +|? - + 0-9
+ ? - ? ? + + ? - ? 10-19
- ? ? ? + + ? - ? ? 20-29

(need to finish filling in)
? ? - ? ? ? + ? ? ? 30-39
? ? ? ? ? ? ? ? ? + 40-49
- ? ? ? ? ? ? ? ? ? 50-59
? ? ? ? + ? ? ? ? ? 60-69
? ? - ? ? ? ? ? ? ? 70-79
? + ? ? ? ? ? ? ? ? 80-89
? ? ? ? ? ? ? ? - ? 90-99

Here is a [[computer proof that completely multiplicative sequences have at least 2]], inspired by the above analysis.

Human proof that completely multiplicative sequences have discrepancy greater than 2

2010-02-02T03:08:04Z

Jbd: /* Case 2. f(2) = -1 */

It is [[multiplicative sequences|known from computer search]] that the longest completely multiplicative sequence <math>f: [N] \to \{-1,+1\}</math> of discrepancy 2 has length 246. Here we record the attempts to prove this by hand. Notation: we display our function in rows of 10, starting with 0. We use the notation that + and - denote values which are known to equal +1 or -1 respectively, and ? for unknown values. Thus, initially our array looks like this:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

since f(n^2) = 1 for all square numbers.

The notions of a ''cut'' and a ''block'' will be useful. A ''cut'' occurs between n and n+1 whenever it can be proved that f[1,n] sums to 0; these can only occur when n is even. We indicate a cut by a bar | between n and n+1. Thus for instance we begin with just one cut:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

A ''block'' is any interval between two cuts; thus inside each block one must sum to zero, and the discrepancy inside each block is at most 2. Thus we can "localise" the analysis to each block, though multiplicativity relates the blocks together.

A block is ''solved'' if all values are assigned in it. Solved blocks are uninteresting (other than for the values they provide by multiplicativity), so we shall adopt the convention of concatenating solved blocks with adjacent blocks to reduce clutter.

More notation: we use f[a,b] as short for f(a)+...+f(b). Note that

: f[a,b] = -4,-2,0,2,4 if a is odd and b is even;
: f[a,b] = -3,-1,1,3 if a and b are both odd or both even;
: f[a,b] = -2,0,2 if a is even and b is odd.

Also if [a,b] starts or ends at a cut, then f[a,b] has magnitude at most 2.

A key observation is that if f(2n)=f(2n+1) then there must be a cut between 2n and 2n+1.

== Case 1. f(2)=1 ==

Now, let us '''assume''' f(2)=1 and that the discrepancy is at most 2 and see what happens. From multiplicativity we obtain f(2n^2)=1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + ? + ? ? ? + + 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

As f(8)=f(9) we can now cut at 8=9. And then by using discrepancy we get f(3)=f(5)=-1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + - + - ? ? +|+ 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

which then by multiplicativity and adding cuts where obvious extends to

0 1 2 3 4 5 6 7 8 9

0 + + - + - -|? +|+ 0-9
-|? - ? ? + + ? + ? 10-19
- ? ? ? - + ? - ? ? 20-29
+ ? + ? ? ? + ? ? ? 30-39

Looking at the 7-8 block we see that f(7) is forced to be -1; from this and multiplicativity, and showing rows up to 100, we arrive at

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + ? 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? ? ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - ? ? ? 50-59
+ ? ? - + ? ? ? ? ? 60-69
+ ? + ? ? - ? ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? ? - ? + ? 90-99
+

Note that

: f[15,19] = 3 + f(17) + f(19)

cannot exceed 3, and

: f[168,171] = 2 - f(17) + f(19)

cannot exceed 2. Thus f(19) = -1. We use this information to update the board:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + - 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? - ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - + ? ? 50-59
+ ? ? - + ? ? ? ? ? 60-69
+ ? + ? ? - - ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? + - ? + ? 90-99
+

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
+

This seems to be about as far as one can get just from the assumption f(2)=+1.

== Case 1.a f(2)=f(37)=1 ==

The most profitable '''assumption''' here seems to be f(37)=+1. This creates a cut at 36-37, which then forces f(34)=-1 and then f(33)=-1 also. Filling in these values and adding in cuts, and extending a row gives

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+ ? +|- - + + + -|B 30-39
- ? + ? + - e ? - + 40-49
+|+ b ? -|- - + f ? 50-59
+ ? ? - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B e ? ? + - ? +|+ 90-99
+ ? + ? b - ? ? - ? 100-109

looking at f[30,32] we thus have f(31)=-1, while from f[99,103] we have f(101)=f(103)=-1,

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + + + -|B 30-39
- ? + ? + - e ? - + 40-49
+|+ b ? -|- - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B e + ? + - ? +|+ 90-99
+ - + - b - ? ? - ? 100-109

From the [51,54] block we have f(53) = B, which then forces f[90,106]=0, so there is a cut at 106-107:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + + + -|B 30-39
- ? + ? + - e ? - + 40-49
+|+ b B -|- - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B e + ? + - ? +|+ 90-99
+ - + - b - B|? - ? 100-109
-|- - ? + E f b ? + 110-119
+|+ ? ? -|- - ? + ? 120-129

From f[118-120] we have f(59)=-1; from f[111,113] we have f(113)=+1;

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + + + -|B 30-39
- ? + ? + - e ? - + 40-49
+|+ b B -|- - + f - 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B e + ? + - ? +|+ 90-99
+ - + - b - B|? - ? 100-109
-|- - + +|E f b - + 110-119
+|+ ? ? -|- - ? + ? 120-129

From f[13,30] we have 2b+e+f=0 while from f[115,120] we have -e+f+b=-1. This forces b=-1, e=f=1:

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? + - + ? - + 40-49
+|+ - + - - - + + - 50-59
+|? -|- +|+ - ? -|- 60-69
+|? + ? + - -|- +|? 70-79
- + ? ? +|+ ? - + ? 80-89
- + +|+ ? + - ? +|+ 90-99
+ - + - - - +|? - ? 100-109
-|- - + + - + - - + 110-119
+|+ ? ? -|- - ? + ? 120-129

From the 60-69 row we then get f(61)=+1, f(67)=+1; from f[85,92] we get f(43)=f(89)=-1; from f[93,98] we have f(47)=f(97)=-1; from f[125,127] one has f(127)=+1; from f[107,110] one gets f(107)=f(109)=+1:

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + - + - + - - + 40-49
+|+ - + - - - + + - 50-59
+ + - - + + - + -|- 60-69
+|? + ? + - -|- +|? 70-79
- + ? ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
+ - + - - - + + - + 100-109
-|- - + + - + - - + 110-119
+|+ + ? -|- - + + ? 120-129

From f[41,50] one has f(41)=-1; but then f[121,125]=2, a contradiction. Hence this case is not possible.

== Case 1.b f(2)=1, f(37)=-1 ==

This takes us back to case 1.b f(2)=1, f(37)=-1, which begins here:

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
+

So we assume f(37)=-1

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + - - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? -|- - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

From f[75,77] we have A=+1, then from f[11,13] we have b=-1, and then from f[75,79] we have f(79)=+1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|c + - 10-19
- + - e - + + - - f 20-29
+ ? +|+ c|+ + - -|- 30-39
- ? + ? - - e ? - + 40-49
+ C + ? - + - + f ? 50-59
+ ? ? - + - + ? c E 60-69
+ ? + + - - - + - + 70-79
- + ? ? + C ? F - ? 80-89
- - C ? ? + - ? + - 90-99
+

From f[33,34] we have c=-1, and from f[39,41] we have f(41)=+1;

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - e - + + - - f 20-29
+ ? + + -|+ + - - - 30-39
- + + ? - - e ? - + 40-49
+ + + ? - + - + f ? 50-59
+ ? ? - + - + ? + E 60-69
+ ? + + - - - + - + 70-79
- + ? ? + + ? F - ? 80-89
- - + ? ? + - ? + - 90-99
+

From f[19,23] we have e = +

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - f 20-29
+ ? + + -|+ + - - - 30-39
- + + ? - - + ? - + 40-49
+ + + ? - + - + f ? 50-59
+ ? ? - + - + ? + - 60-69
+ ? + + - - - + - + 70-79
- + ? ? + + ? F - ? 80-89
- - + ? ? + - ? + - 90-99
+

From f[27,29] we have f = +, so from f[27,34] we have f(31) = -, and sums and multiplicity force

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|? + - 60-69
+ ? + + - - - + - + 70-79
- + + ? + + + - - ? 80-89
- - + + - + - ? + - 90-99
+

The smallest sum before 70 is -1, so 71 must be -1. This forces 67 = -1, otherwise the sum at 73 is 3.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|? + + + - - ? 80-89
- - + + - + - ? + - 90-99
+

83 must be -, otherwise the sum is 3 at 85.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|- +|+ + - -|? 80-89
- - + + - + - ? + - 90-99
+

So 89 is + (otherwise the sum is -3 at 91)

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|- +|+ + - -|+ 80-89
- - +|+ -|+ -|? + - 90-99
-

(could use intermediate diagrams displayed here with text)

f(58) is positive so f(116) is positive

f(13) is positive and f(9) is a square so f(117) is positive

there is a cut between 116 and 117

f(118) is negative since f(59) is negative

f(7) and f(17) are negative so f(119) is positive

f(3) and f(5) are negative and f(2) is positive so 120 is positive

121 is a square so f(121) is positive

so at 116 the sum is zero at because there is a cut between
116 and 117 at 117 the sum is one since f(117) is positive
at 118 the sum is zero since f(118) is negative and since f(119) f(120) and f(121) are positive at 121 the sum is three and we have shown that if a completely multiplicative sequnce has f(2) equal to 1 and f(37) equal to -1 then it reaches discrepancy 3 before 246.

Since we have previously shown that if a completely multiplicative sequence that has f(2) equal to 1 and f(37)equal to 1 reaches discrepancy 3 before 246 we have any completely multiplicative sequence with f(2) equal to one reaches discrepancy 3 before 246 and that finishes case 1.

== Case 2. f(2) = -1 ==

Therefore f(2) = -1. Resetting the board:

0 1 2 3 4 5 6 7 8 9

0 + - ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39
? ? ? ? ? ? ? ? ? + 40-49
? ? ? ? ? ? ? ? ? ? 50-59
? ? ? ? + ? ? ? ? ? 60-69
? ? ? ? ? ? ? ? ? ? 70-79
? + ? ? ? ? ? ? ? ? 80-89
? ? ? ? ? ? ? ? ? ? 90-99

Adding in multiplicative values:

0 1 2 3 4 5 6 7 8 9

0|+ -|? + ? ? ? - + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? - ? ? ? + ? ? ? 30-39
? ? ? ? ? ? ? ? ? + 40-49
- ? ? ? ? ? ? ? ? ? 50-59
? ? ? ? + ? ? ? ? ? 60-69
? ? - ? ? ? ? ? ? ? 70-79
? + ? ? ? ? ? ? ? ? 80-89
? ? ? ? ? ? ? ? - ? 90-99

Here is a [[computer proof that completely multiplicative sequences have at least 2]], inspired by the above analysis.

Human proof that completely multiplicative sequences have discrepancy greater than 2

2010-02-02T03:02:58Z

Jbd:

It is [[multiplicative sequences|known from computer search]] that the longest completely multiplicative sequence <math>f: [N] \to \{-1,+1\}</math> of discrepancy 2 has length 246. Here we record the attempts to prove this by hand. Notation: we display our function in rows of 10, starting with 0. We use the notation that + and - denote values which are known to equal +1 or -1 respectively, and ? for unknown values. Thus, initially our array looks like this:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

since f(n^2) = 1 for all square numbers.

The notions of a ''cut'' and a ''block'' will be useful. A ''cut'' occurs between n and n+1 whenever it can be proved that f[1,n] sums to 0; these can only occur when n is even. We indicate a cut by a bar | between n and n+1. Thus for instance we begin with just one cut:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

A ''block'' is any interval between two cuts; thus inside each block one must sum to zero, and the discrepancy inside each block is at most 2. Thus we can "localise" the analysis to each block, though multiplicativity relates the blocks together.

A block is ''solved'' if all values are assigned in it. Solved blocks are uninteresting (other than for the values they provide by multiplicativity), so we shall adopt the convention of concatenating solved blocks with adjacent blocks to reduce clutter.

More notation: we use f[a,b] as short for f(a)+...+f(b). Note that

: f[a,b] = -4,-2,0,2,4 if a is odd and b is even;
: f[a,b] = -3,-1,1,3 if a and b are both odd or both even;
: f[a,b] = -2,0,2 if a is even and b is odd.

Also if [a,b] starts or ends at a cut, then f[a,b] has magnitude at most 2.

A key observation is that if f(2n)=f(2n+1) then there must be a cut between 2n and 2n+1.

== Case 1. f(2)=1 ==

Now, let us '''assume''' f(2)=1 and that the discrepancy is at most 2 and see what happens. From multiplicativity we obtain f(2n^2)=1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + ? + ? ? ? + + 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

As f(8)=f(9) we can now cut at 8=9. And then by using discrepancy we get f(3)=f(5)=-1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + - + - ? ? +|+ 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

which then by multiplicativity and adding cuts where obvious extends to

0 1 2 3 4 5 6 7 8 9

0 + + - + - -|? +|+ 0-9
-|? - ? ? + + ? + ? 10-19
- ? ? ? - + ? - ? ? 20-29
+ ? + ? ? ? + ? ? ? 30-39

Looking at the 7-8 block we see that f(7) is forced to be -1; from this and multiplicativity, and showing rows up to 100, we arrive at

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + ? 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? ? ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - ? ? ? 50-59
+ ? ? - + ? ? ? ? ? 60-69
+ ? + ? ? - ? ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? ? - ? + ? 90-99
+

Note that

: f[15,19] = 3 + f(17) + f(19)

cannot exceed 3, and

: f[168,171] = 2 - f(17) + f(19)

cannot exceed 2. Thus f(19) = -1. We use this information to update the board:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + - 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? - ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - + ? ? 50-59
+ ? ? - + ? ? ? ? ? 60-69
+ ? + ? ? - - ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? + - ? + ? 90-99
+

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
+

This seems to be about as far as one can get just from the assumption f(2)=+1.

== Case 1.a f(2)=f(37)=1 ==

The most profitable '''assumption''' here seems to be f(37)=+1. This creates a cut at 36-37, which then forces f(34)=-1 and then f(33)=-1 also. Filling in these values and adding in cuts, and extending a row gives

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+ ? +|- - + + + -|B 30-39
- ? + ? + - e ? - + 40-49
+|+ b ? -|- - + f ? 50-59
+ ? ? - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B e ? ? + - ? +|+ 90-99
+ ? + ? b - ? ? - ? 100-109

looking at f[30,32] we thus have f(31)=-1, while from f[99,103] we have f(101)=f(103)=-1,

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + + + -|B 30-39
- ? + ? + - e ? - + 40-49
+|+ b ? -|- - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B e + ? + - ? +|+ 90-99
+ - + - b - ? ? - ? 100-109

From the [51,54] block we have f(53) = B, which then forces f[90,106]=0, so there is a cut at 106-107:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + + + -|B 30-39
- ? + ? + - e ? - + 40-49
+|+ b B -|- - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B e + ? + - ? +|+ 90-99
+ - + - b - B|? - ? 100-109
-|- - ? + E f b ? + 110-119
+|+ ? ? -|- - ? + ? 120-129

From f[118-120] we have f(59)=-1; from f[111,113] we have f(113)=+1;

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + + + -|B 30-39
- ? + ? + - e ? - + 40-49
+|+ b B -|- - + f - 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B e + ? + - ? +|+ 90-99
+ - + - b - B|? - ? 100-109
-|- - + +|E f b - + 110-119
+|+ ? ? -|- - ? + ? 120-129

From f[13,30] we have 2b+e+f=0 while from f[115,120] we have -e+f+b=-1. This forces b=-1, e=f=1:

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? + - + ? - + 40-49
+|+ - + - - - + + - 50-59
+|? -|- +|+ - ? -|- 60-69
+|? + ? + - -|- +|? 70-79
- + ? ? +|+ ? - + ? 80-89
- + +|+ ? + - ? +|+ 90-99
+ - + - - - +|? - ? 100-109
-|- - + + - + - - + 110-119
+|+ ? ? -|- - ? + ? 120-129

From the 60-69 row we then get f(61)=+1, f(67)=+1; from f[85,92] we get f(43)=f(89)=-1; from f[93,98] we have f(47)=f(97)=-1; from f[125,127] one has f(127)=+1; from f[107,110] one gets f(107)=f(109)=+1:

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + - + - + - - + 40-49
+|+ - + - - - + + - 50-59
+ + - - + + - + -|- 60-69
+|? + ? + - -|- +|? 70-79
- + ? ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
+ - + - - - + + - + 100-109
-|- - + + - + - - + 110-119
+|+ + ? -|- - + + ? 120-129

From f[41,50] one has f(41)=-1; but then f[121,125]=2, a contradiction. Hence this case is not possible.

== Case 1.b f(2)=1, f(37)=-1 ==

This takes us back to case 1.b f(2)=1, f(37)=-1, which begins here:

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
+

So we assume f(37)=-1

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + - - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? -|- - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

From f[75,77] we have A=+1, then from f[11,13] we have b=-1, and then from f[75,79] we have f(79)=+1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|c + - 10-19
- + - e - + + - - f 20-29
+ ? +|+ c|+ + - -|- 30-39
- ? + ? - - e ? - + 40-49
+ C + ? - + - + f ? 50-59
+ ? ? - + - + ? c E 60-69
+ ? + + - - - + - + 70-79
- + ? ? + C ? F - ? 80-89
- - C ? ? + - ? + - 90-99
+

From f[33,34] we have c=-1, and from f[39,41] we have f(41)=+1;

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - e - + + - - f 20-29
+ ? + + -|+ + - - - 30-39
- + + ? - - e ? - + 40-49
+ + + ? - + - + f ? 50-59
+ ? ? - + - + ? + E 60-69
+ ? + + - - - + - + 70-79
- + ? ? + + ? F - ? 80-89
- - + ? ? + - ? + - 90-99
+

From f[19,23] we have e = +

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - f 20-29
+ ? + + -|+ + - - - 30-39
- + + ? - - + ? - + 40-49
+ + + ? - + - + f ? 50-59
+ ? ? - + - + ? + - 60-69
+ ? + + - - - + - + 70-79
- + ? ? + + ? F - ? 80-89
- - + ? ? + - ? + - 90-99
+

From f[27,29] we have f = +, so from f[27,34] we have f(31) = -, and sums and multiplicity force

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|? + - 60-69
+ ? + + - - - + - + 70-79
- + + ? + + + - - ? 80-89
- - + + - + - ? + - 90-99
+

The smallest sum before 70 is -1, so 71 must be -1. This forces 67 = -1, otherwise the sum at 73 is 3.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|? + + + - - ? 80-89
- - + + - + - ? + - 90-99
+

83 must be -, otherwise the sum is 3 at 85.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|- +|+ + - -|? 80-89
- - + + - + - ? + - 90-99
+

So 89 is + (otherwise the sum is -3 at 91)

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|- +|+ + - -|+ 80-89
- - +|+ -|+ -|? + - 90-99
-

(could use intermediate diagrams displayed here with text)

f(58) is positive so f(116) is positive

f(13) is positive and f(9) is a square so f(117) is positive

there is a cut between 116 and 117

f(118) is negative since f(59) is negative

f(7) and f(17) are negative so f(119) is positive

f(3) and f(5) are negative and f(2) is positive so 120 is positive

121 is a square so f(121) is positive

so at 116 the sum is zero at because there is a cut between
116 and 117 at 117 the sum is one since f(117) is positive
at 118 the sum is zero since f(118) is negative and since f(119) f(120) and f(121) are positive at 121 the sum is three and we have shown that if a completely multiplicative sequnce has f(2) equal to 1 and f(37) equal to -1 then it reaches discrepancy 3 before 246.

Since we have previously shown that if a completely multiplicative sequence that has f(2) equal to 1 and f(37)equal to 1 reaches discrepancy 3 before 246 we have any completely multiplicative sequence with f(2) equal to one reaches discrepancy 3 before 246 and that finishes case 1.

== Case 2. f(2) = -1 ==

Therefore f(2) = -1. Resetting the board:

0 1 2 3 4 5 6 7 8 9

0 + - ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39
? ? ? ? ? ? ? ? ? + 40-49
? ? ? ? ? ? ? ? ? ? 50-59
? ? ? ? + ? ? ? ? ? 60-69
? ? ? ? ? ? ? ? ? ? 70-79
? + ? ? ? ? ? ? ? ? 80-89
? ? ? ? ? ? ? ? ? ? 90-99

Here is a [[computer proof that completely multiplicative sequences have at least 2]], inspired by the above analysis.

Human proof that completely multiplicative sequences have discrepancy greater than 2

2010-02-01T20:15:14Z

Jbd:

It is [[multiplicative sequences|known from computer search]] that the longest completely multiplicative sequence <math>f: [N] \to \{-1,+1\}</math> of discrepancy 2 has length 246. Here we record the attempts to prove this by hand. Notation: we display our function in rows of 10, starting with 0. We use the notation that + and - denote values which are known to equal +1 or -1 respectively, and ? for unknown values. Thus, initially our array looks like this:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

since f(n^2) = 1 for all square numbers.

The notions of a ''cut'' and a ''block'' will be useful. A ''cut'' occurs between n and n+1 whenever it can be proved that f[1,n] sums to 0; these can only occur when n is even. We indicate a cut by a bar | between n and n+1. Thus for instance we begin with just one cut:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

A ''block'' is any interval between two cuts; thus inside each block one must sum to zero, and the discrepancy inside each block is at most 2. Thus we can "localise" the analysis to each block, though multiplicativity relates the blocks together.

A block is ''solved'' if all values are assigned in it. Solved blocks are uninteresting (other than for the values they provide by multiplicativity), so we shall adopt the convention of concatenating solved blocks with adjacent blocks to reduce clutter.

More notation: we use f[a,b] as short for f(a)+...+f(b). Note that

: f[a,b] = -4,-2,0,2,4 if a is odd and b is even;
: f[a,b] = -3,-1,1,3 if a and b are both odd or both even;
: f[a,b] = -2,0,2 if a is even and b is odd.

Also if [a,b] starts or ends at a cut, then f[a,b] has magnitude at most 2.

A key observation is that if f(2n)=f(2n+1) then there must be a cut between 2n and 2n+1.

== Case 1. f(2)=1 ==

Now, let us '''assume''' f(2)=1 and that the discrepancy is at most 2 and see what happens. From multiplicativity we obtain f(2n^2)=1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + ? + ? ? ? + + 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

As f(8)=f(9) we can now cut at 8=9. And then by using discrepancy we get f(3)=f(5)=-1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + - + - ? ? +|+ 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

which then by multiplicativity and adding cuts where obvious extends to

0 1 2 3 4 5 6 7 8 9

0 + + - + - -|? +|+ 0-9
-|? - ? ? + + ? + ? 10-19
- ? ? ? - + ? - ? ? 20-29
+ ? + ? ? ? + ? ? ? 30-39

Looking at the 7-8 block we see that f(7) is forced to be -1; from this and multiplicativity, and showing rows up to 100, we arrive at

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + ? 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? ? ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - ? ? ? 50-59
+ ? ? ? + ? ? ? ? ? 60-69
+ ? + ? ? - ? ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? ? - ? + ? 90-99
-

Note that

: f[15,19] = 3 + f(17) + f(19)

cannot exceed 3, and

: f[168,171] = 2 - f(17) + f(19)

cannot exceed 2. Thus f(19) = -1. We use this information to update the board:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + - 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? - ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - + ? ? 50-59
+ ? ? ? + ? ? ? ? ? 60-69
+ ? + ? ? - - ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? + - ? + ? 90-99
-

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

This seems to be about as far as one can get just from the assumption f(2)=+1.

== Case 1.a f(2)=f(37)=1 ==

The most profitable '''assumption''' here seems to be f(37)=+1. This creates a cut at 36-37, which then forces f(34)=-1 and then f(33)=-1 also. Filling in these values and adding in cuts gives

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+ ? +|- - + +|+ - B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? -|- - + f ? 50-59
+ ? ? - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B + ? ? + - ? + + 90-99
-

looking at f[30,32] we thus have f(31)=-1,

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + +|+ - B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? -|- - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B +|+ ? + - ? + + 90-99
-

== Case 1.a.1 f(2)=f(37)=f(13)=1 ==

Now suppose we '''assume''' b=+1. Then by looking at f(1)+...+f(23) we have e=-1, and by looking at f(1)+...+f(37) we then have f=-1, leaving us with

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + -|- 30-39
- ? + ? + - - ? - + 40-49
+|+ + ? -|- - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - ? 70-79
- + ? ? +|+ ? + + ? 80-89
-|- +|+ ? + - ? + + 90-99
-

from f[39,41] we have f(41)=+1, and from f[77,79] we have f(79)=+1, from f[51,54] we have f(53)=-1, from f[47,50] we have f(47)=-1, and from f[85,88] we have f(86)=-1, hence f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + - - 30-39
- + + - +|- - - - + 40-49
+ +|+ - - - - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - + 70-79
- + +|? + + - + + ? 80-89
- - + + ? + - ? + + 90-99
-

But this creates a contradiction at the block [45,51]. Thus this case is impossible.

== Case 1.a f(2)=f(37)=1 ==

We have thus concluded in this case that f(13)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + +|e - + -|- - f 20-29
+|- + - - + +|+ - + 30-39
-|? + ? + - e ? - + 40-49
+|+ - ? -|- - + f ? 50-59
+ ? -|- + + -|? - E 60-69
+ ? + ? + - -|- + ? 70-79
- + ? ? +|+ ? F + ? 80-89
- + +|+ ? + - ? + + 90-99
-

From the [23,26] and [27,30] blocks we have e=+1 and f=+1; from the block [51,54] we have f(53)=+1. Looking at [93,95] we see that f(94)=-1 hence f(47)=+1, and we have a cut at 92-93, hence also at 96-97:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? +|- + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + -|? - - 60-69
+ ? + ? + - -|- +|? 70-79
- + ? ? +|+ ? - + ? 80-89
- + +|+ - + -|? + + 90-99
-

from f[97,99] we have f(97)=-1. From the [41,44] block we have f(41)=f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + -|? -|- 60-69
+ ? + ? + - -|- +|? 70-79
- + - ? +|+ - - +|? 80-89
- + +|+ - + - - + + 90-99
-|

Using the cuts we can fill in f(67)=+1 and f(89)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+|+ - ? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + +|+ - + - - + + 90-99
-|

The 41-50 block then forces f(41)=-1, hence f(82)=-1.

If 79 and 83 are both positive the sum at 85 is 3, and if 79 and 83 are both negative the sum at 82 is -3, so 79 and 83 must be alternate signs, so there is a cut before 85:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + -|? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

Looking at 53-56 we see that f(53)=+1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + + ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

Suppose f(59)=+1. Then f(118)=f(120)=f(121)=1 and f(119)=-f(17), so f(17)=+1. Also, f(57)=f(59)=f(60)=1 and f(58)=f(29), hence f(29)=-1.
Next, f(15)+…+f(22) = f(11)+2, hence f(11)=-1; and then f(1)+…+f(13) = f(13)-2, so f(13)=+1. But then f(1)+…+f(23)=f(23)+2, so f(23)=-1. But then f(1)+…+f(29)=-3, contradiction.

Thus we must also have f(59)=-1, so f(61)=+1.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

The remaining question marks are unconstrained other than requiring every block to sum to 0. Now consider the sequence up to 119, with the multiplicative values filled in:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? - - + ? - ? 100-109
- - - ? + - + - - + 110-119

109 is clearly + (otherwise there’s a sequence of 5 -s).

Consider 101 and 103.

Both can’t be + because the sum at 103 would reach 3.

Both can’t be – because then at 105 the sum would reach -3.

Therefore one has to be + and the other -, so the sum is 0 immediately before 105 and 0 immediately before 107.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|? - + 100-109
- - - ? + - + - - + 110-119

Since the sum is 0 before 107, 107 must be + (otherwise the sum is -3 at 111).

Also, the sum is -2 before 103, so 103 is +.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119

Continuing with known multiplicative values and values set by the sum being 2 or -2 up to 149, this gives deductions:
120 = +1, 121=+1, 122=+1, 123=+1 and together with 119=+1 this provides a contradiction based on earlier values at 2,3,5; 2,61; 3,41 and 7,17
This concludes case 1a.

== Case 1.b f(2)=1, f(37)=-1 ==

This takes us back to case 1.b f(2)=1, f(37)=-1, which begins here:

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-
This seems to be about as far as one can get just from the assumption f(2)=+1.

So we assume f(37)=-1

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + - - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? - - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

a and b can't both be - (five -s starting at 10).

A,B can't both be - (five -s starting at 74)

Also note that there must be a cut between 74 and 75 due to the rule that when f(2n)=f(2n+1) there is a cut between.

Therefore a and b can't both be +, so a and b are alternating, so there is a cut before 17.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + +|c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + - - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? -|- - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

A must be + (otherwise the sum reaches -3 at 76) and since A and B are also alternating, B must be - also forcing 79 to be +.

A = +1 and B = -1 so a = -1 and b = +1.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|c + - 10-19
- + - e - + + - - f 20-29
+ ? + + c|+ + - - - 30-39
- + + ? - - e ? - + 40-49
+ C + ? - + - + f ? 50-59
+ ? ? - + - + ? c E 60-69
+ ? + + - - - + - + 70-79
- + ? ? + C ? F - ? 80-89
- - C ? ? + - ? + - 90-99
-

+ + - - - - forces cut before the start of the sequence, so the only way for sum to be 0 with + + ? | is if ? = -1, so c = -1.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - e - + + - - f 20-29
+ ? + + -|+ + - - - 30-39
- + + ? - - e ? - + 40-49
+ + + ? - + - + f ? 50-59
+ ? ? - + - + ? + E 60-69
+ ? + + - - - + - + 70-79
- + ? ? + + ? F - ? 80-89
- - + ? ? + - ? + - 90-99
-

So sums force e = +

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - f 20-29
+ ? + + -|+ + - - - 30-39
- + + ? - - + ? - + 40-49
+ + + ? - + - + f ? 50-59
+ ? ? - + - + ? + - 60-69
+ ? + + - - - + - + 70-79
- + ? ? + + ? F - ? 80-89
- - + ? ? + - ? + - 90-99
-

So f = +, so 31 = -, and sums and multiplicity force

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|? + - 60-69
+ ? + + - - - + - + 70-79
- + + ? + + + - - ? 80-89
- - + + - + - ? + - 90-99
-

The smallest sum before 70 is -1, so 71 must be -1. This forces 67 = -1, otherwise the sum at 73 is 3.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|? + + + - - ? 80-89
- - + + - + - ? + - 90-99
-

83 must be -, otherwise the sum is 3 at 85.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|- +|+ + - -|? 80-89
- - + + - + - ? + - 90-99
-

So 89 is + (otherwise the sum is -3 at 91)

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|- +|+ + - -|+ 80-89
- - +|+ -|+ -|? + - 90-99
-

Human proof that completely multiplicative sequences have discrepancy greater than 2

2010-02-01T18:36:41Z

Jbd:

It is [[multiplicative sequences|known from computer search]] that the longest completely multiplicative sequence <math>f: [N] \to \{-1,+1\}</math> of discrepancy 2 has length 246. Here we record the attempts to prove this by hand. Notation: we display our function in rows of 10, starting with 0. We use the notation that + and - denote values which are known to equal +1 or -1 respectively, and ? for unknown values. Thus, initially our array looks like this:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

since f(n^2) = 1 for all square numbers.

The notions of a ''cut'' and a ''block'' will be useful. A ''cut'' occurs between n and n+1 whenever it can be proved that f[1,n] sums to 0; these can only occur when n is even. We indicate a cut by a bar | between n and n+1. Thus for instance we begin with just one cut:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

A ''block'' is any interval between two cuts; thus inside each block one must sum to zero, and the discrepancy inside each block is at most 2. Thus we can "localise" the analysis to each block, though multiplicativity relates the blocks together.

A block is ''solved'' if all values are assigned in it. Solved blocks are uninteresting (other than for the values they provide by multiplicativity), so we shall adopt the convention of concatenating solved blocks with adjacent blocks to reduce clutter.

More notation: we use f[a,b] as short for f(a)+...+f(b). Note that

: f[a,b] = -4,-2,0,2,4 if a is odd and b is even;
: f[a,b] = -3,-1,1,3 if a and b are both odd or both even;
: f[a,b] = -2,0,2 if a is even and b is odd.

Also if [a,b] starts or ends at a cut, then f[a,b] has magnitude at most 2.

A key observation is that if f(2n)=f(2n+1) then there must be a cut between 2n and 2n+1.

== Case 1. f(2)=1 ==

Now, let us '''assume''' f(2)=1 and that the discrepancy is at most 2 and see what happens. From multiplicativity we obtain f(2n^2)=1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + ? + ? ? ? + + 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

As f(8)=f(9) we can now cut at 8=9. And then by using discrepancy we get f(3)=f(5)=-1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + - + - ? ? +|+ 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

which then by multiplicativity and adding cuts where obvious extends to

0 1 2 3 4 5 6 7 8 9

0 + + - + - -|? +|+ 0-9
-|? - ? ? + + ? + ? 10-19
- ? ? ? - + ? - ? ? 20-29
+ ? + ? ? ? + ? ? ? 30-39

Looking at the 7-8 block we see that f(7) is forced to be -1; from this and multiplicativity, and showing rows up to 100, we arrive at

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + ? 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? ? ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - ? ? ? 50-59
+ ? ? ? + ? ? ? ? ? 60-69
+ ? + ? ? - ? ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? ? - ? + ? 90-99
-

Note that

: f[15,19] = 3 + f(17) + f(19)

cannot exceed 3, and

: f[168,171] = 2 - f(17) + f(19)

cannot exceed 2. Thus f(19) = -1. We use this information to update the board:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + - 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? - ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - + ? ? 50-59
+ ? ? ? + ? ? ? ? ? 60-69
+ ? + ? ? - - ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? + - ? + ? 90-99
-

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

This seems to be about as far as one can get just from the assumption f(2)=+1.

== Case 1.a f(2)=f(37)=1 ==

The most profitable '''assumption''' here seems to be f(37)=+1. This creates a cut at 36-37, which then forces f(34)=-1 and then f(33)=-1 also. Filling in these values and adding in cuts gives

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+ ? +|- - + +|+ - B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? -|- - + f ? 50-59
+ ? ? - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B + ? ? + - ? + + 90-99
-

looking at f[30,32] we thus have f(31)=-1,

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + +|+ - B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? -|- - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B +|+ ? + - ? + + 90-99
-

== Case 1.a.1 f(2)=f(37)=f(13)=1 ==

Now suppose we '''assume''' b=+1. Then by looking at f(1)+...+f(23) we have e=-1, and by looking at f(1)+...+f(37) we then have f=-1, leaving us with

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + -|- 30-39
- ? + ? + - - ? - + 40-49
+|+ + ? -|- - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - ? 70-79
- + ? ? +|+ ? + + ? 80-89
-|- +|+ ? + - ? + + 90-99
-

from f[39,41] we have f(41)=+1, and from f[77,79] we have f(79)=+1, from f[51,54] we have f(53)=-1, from f[47,50] we have f(47)=-1, and from f[85,88] we have f(86)=-1, hence f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + - - 30-39
- + + - +|- - - - + 40-49
+ +|+ - - - - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - + 70-79
- + +|? + + - + + ? 80-89
- - + + ? + - ? + + 90-99
-

But this creates a contradiction at the block [45,51]. Thus this case is impossible.

== Case 1.a f(2)=f(37)=1 ==

We have thus concluded in this case that f(13)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + +|e - + -|- - f 20-29
+|- + - - + +|+ - + 30-39
-|? + ? + - e ? - + 40-49
+|+ - ? -|- - + f ? 50-59
+ ? -|- + + -|? - E 60-69
+ ? + ? + - -|- + ? 70-79
- + ? ? +|+ ? F + ? 80-89
- + +|+ ? + - ? + + 90-99
-

From the [23,26] and [27,30] blocks we have e=+1 and f=+1; from the block [51,54] we have f(53)=+1. Looking at [93,95] we see that f(94)=-1 hence f(47)=+1, and we have a cut at 92-93, hence also at 96-97:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? +|- + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + -|? - - 60-69
+ ? + ? + - -|- +|? 70-79
- + ? ? +|+ ? - + ? 80-89
- + +|+ - + -|? + + 90-99
-

from f[97,99] we have f(97)=-1. From the [41,44] block we have f(41)=f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + -|? -|- 60-69
+ ? + ? + - -|- +|? 70-79
- + - ? +|+ - - +|? 80-89
- + +|+ - + - - + + 90-99
-|

Using the cuts we can fill in f(67)=+1 and f(89)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+|+ - ? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + +|+ - + - - + + 90-99
-|

The 41-50 block then forces f(41)=-1, hence f(82)=-1.

If 79 and 83 are both positive the sum at 85 is 3, and if 79 and 83 are both negative the sum at 82 is -3, so 79 and 83 must be alternate signs, so there is a cut before 85:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + -|? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

Looking at 53-56 we see that f(53)=+1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + + ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

Suppose f(59)=+1. Then f(118)=f(120)=f(121)=1 and f(119)=-f(17), so f(17)=+1. Also, f(57)=f(59)=f(60)=1 and f(58)=f(29), hence f(29)=-1.
Next, f(15)+…+f(22) = f(11)+2, hence f(11)=-1; and then f(1)+…+f(13) = f(13)-2, so f(13)=+1. But then f(1)+…+f(23)=f(23)+2, so f(23)=-1. But then f(1)+…+f(29)=-3, contradiction.

Thus we must also have f(59)=-1, so f(61)=+1.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

The remaining question marks are unconstrained other than requiring every block to sum to 0. Now consider the sequence up to 119, with the multiplicative values filled in:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? - - + ? - ? 100-109
- - - ? + - + - - + 110-119

109 is clearly + (otherwise there’s a sequence of 5 -s).

Consider 101 and 103.

Both can’t be + because the sum at 103 would reach 3.

Both can’t be – because then at 105 the sum would reach -3.

Therefore one has to be + and the other -, so the sum is 0 immediately before 105 and 0 immediately before 107.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|? - + 100-109
- - - ? + - + - - + 110-119

Since the sum is 0 before 107, 107 must be + (otherwise the sum is -3 at 111).

Also, the sum is -2 before 103, so 103 is +.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119

Continuing with known multiplicative values and values set by the sum being 2 or -2 up to 149, this gives deductions:
120 = +1, 121=+1, 122=+1, 123=+1 and together with 119=+1 this provides a contradiction based on earlier values at 2,3,5; 2,61; 3,41 and 7,17
This concludes case 1a.

== Case 1.b f(2)=1, f(37)=-1 ==

This takes us back to case 1.b f(2)=1, f(37)=-1, which begins here:

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-
This seems to be about as far as one can get just from the assumption f(2)=+1.

So we assume f(37)=-1

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + - - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? - - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

a and b can't both be - (five -s starting at 10).

A,B can't both be - (five -s starting at 74)

Therefore a and b can't both be +, so a and b are alternating, so there is a cut before 17.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + +|c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + - - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? - - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

Since A and B are also alternating, 79 must be + (otherwise the sum reaches -3 at 80).

Also note that there must be a cut between 74 and 75 due to the rule that when f(2n)=f(2n+1) there is a cut between.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + +|c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + - - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? -|- - A B + 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

Because the sum would reach -3 at 77, A = +1, so B = -1 and a = -1 and b = +1.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|c + - 10-19
- + - e - + + - - f 20-29
+ ? + + c|+ + - - - 30-39
- + + ? - - e ? - + 40-49
+ C + ? - + - + f ? 50-59
+ ? ? - + - + ? c E 60-69
+ ? + + - - - + - + 70-79
- + ? ? + C ? F - ? 80-89
- - C ? ? + - ? + - 90-99
-

+ + - - - - forces cut before the start of the sequence, so the only way for sum to be 0 with + + ? | is if ? = -1, so c = -1.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - e - + + - - f 20-29
+ ? + + -|+ + - - - 30-39
- + + ? - - e ? - + 40-49
+ + + ? - + - + f ? 50-59
+ ? ? - + - + ? + E 60-69
+ ? + + - - - + - + 70-79
- + ? ? + + ? F - ? 80-89
- - + ? ? + - ? + - 90-99
-

So sums force e = +

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - f 20-29
+ ? + + -|+ + - - - 30-39
- + + ? - - + ? - + 40-49
+ + + ? - + - + f ? 50-59
+ ? ? - + - + ? + - 60-69
+ ? + + - - - + - + 70-79
- + ? ? + + ? F - ? 80-89
- - + ? ? + - ? + - 90-99
-

So f = +, so 31 = -, and sums and multiplicity force

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|? + - 60-69
+ ? + + - - - + - + 70-79
- + + ? + + + - - ? 80-89
- - + + - + - ? + - 90-99
-

The smallest sum before 70 is -1, so 71 must be -1. This forces 67 = -1, otherwise the sum at 73 is 3.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|? + + + - - ? 80-89
- - + + - + - ? + - 90-99
-

83 must be -, otherwise the sum is 3 at 85.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|- +|+ + - -|? 80-89
- - + + - + - ? + - 90-99
-

So 89 is + (otherwise the sum is -3 at 91)

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|- +|+ + - -|+ 80-89
- - +|+ -|+ -|? + - 90-99
-

Human proof that completely multiplicative sequences have discrepancy greater than 2

2010-02-01T18:23:13Z

Jbd:

It is [[multiplicative sequences|known from computer search]] that the longest completely multiplicative sequence <math>f: [N] \to \{-1,+1\}</math> of discrepancy 2 has length 246. Here we record the attempts to prove this by hand. Notation: we display our function in rows of 10, starting with 0. We use the notation that + and - denote values which are known to equal +1 or -1 respectively, and ? for unknown values. Thus, initially our array looks like this:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

since f(n^2) = 1 for all square numbers.

The notions of a ''cut'' and a ''block'' will be useful. A ''cut'' occurs between n and n+1 whenever it can be proved that f[1,n] sums to 0; these can only occur when n is even. We indicate a cut by a bar | between n and n+1. Thus for instance we begin with just one cut:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

A ''block'' is any interval between two cuts; thus inside each block one must sum to zero, and the discrepancy inside each block is at most 2. Thus we can "localise" the analysis to each block, though multiplicativity relates the blocks together.

A block is ''solved'' if all values are assigned in it. Solved blocks are uninteresting (other than for the values they provide by multiplicativity), so we shall adopt the convention of concatenating solved blocks with adjacent blocks to reduce clutter.

More notation: we use f[a,b] as short for f(a)+...+f(b). Note that

: f[a,b] = -4,-2,0,2,4 if a is odd and b is even;
: f[a,b] = -3,-1,1,3 if a and b are both odd or both even;
: f[a,b] = -2,0,2 if a is even and b is odd.

Also if [a,b] starts or ends at a cut, then f[a,b] has magnitude at most 2.

A key observation is that if f(2n)=f(2n+1) then there must be a cut between 2n and 2n+1.

== Case 1. f(2)=1 ==

Now, let us '''assume''' f(2)=1 and that the discrepancy is at most 2 and see what happens. From multiplicativity we obtain f(2n^2)=1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + ? + ? ? ? + + 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

As f(8)=f(9) we can now cut at 8=9. And then by using discrepancy we get f(3)=f(5)=-1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + - + - ? ? +|+ 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

which then by multiplicativity and adding cuts where obvious extends to

0 1 2 3 4 5 6 7 8 9

0 + + - + - -|? +|+ 0-9
-|? - ? ? + + ? + ? 10-19
- ? ? ? - + ? - ? ? 20-29
+ ? + ? ? ? + ? ? ? 30-39

Looking at the 7-8 block we see that f(7) is forced to be -1; from this and multiplicativity, and showing rows up to 100, we arrive at

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + ? 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? ? ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - ? ? ? 50-59
+ ? ? ? + ? ? ? ? ? 60-69
+ ? + ? ? - ? ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? ? - ? + ? 90-99
-

Note that

: f[15,19] = 3 + f(17) + f(19)

cannot exceed 3, and

: f[168,171] = 2 - f(17) + f(19)

cannot exceed 2. Thus f(19) = -1. We use this information to update the board:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + - 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? - ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - + ? ? 50-59
+ ? ? ? + ? ? ? ? ? 60-69
+ ? + ? ? - - ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? + - ? + ? 90-99
-

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

This seems to be about as far as one can get just from the assumption f(2)=+1.

== Case 1.a f(2)=f(37)=1 ==

The most profitable '''assumption''' here seems to be f(37)=+1. This creates a cut at 36-37, which then forces f(34)=-1 and then f(33)=-1 also. Filling in these values and adding in cuts gives

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+ ? +|- - + +|+ - B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? -|- - + f ? 50-59
+ ? ? - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B + ? ? + - ? + + 90-99
-

looking at f[30,32] we thus have f(31)=-1,

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + +|+ - B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? -|- - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B +|+ ? + - ? + + 90-99
-

== Case 1.a.1 f(2)=f(37)=f(13)=1 ==

Now suppose we '''assume''' b=+1. Then by looking at f(1)+...+f(23) we have e=-1, and by looking at f(1)+...+f(37) we then have f=-1, leaving us with

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + -|- 30-39
- ? + ? + - - ? - + 40-49
+|+ + ? -|- - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - ? 70-79
- + ? ? +|+ ? + + ? 80-89
-|- +|+ ? + - ? + + 90-99
-

from f[39,41] we have f(41)=+1, and from f[77,79] we have f(79)=+1, from f[51,54] we have f(53)=-1, from f[47,50] we have f(47)=-1, and from f[85,88] we have f(86)=-1, hence f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + - - 30-39
- + + - +|- - - - + 40-49
+ +|+ - - - - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - + 70-79
- + +|? + + - + + ? 80-89
- - + + ? + - ? + + 90-99
-

But this creates a contradiction at the block [45,51]. Thus this case is impossible.

== Case 1.a f(2)=f(37)=1 ==

We have thus concluded in this case that f(13)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + +|e - + -|- - f 20-29
+|- + - - + +|+ - + 30-39
-|? + ? + - e ? - + 40-49
+|+ - ? -|- - + f ? 50-59
+ ? -|- + + -|? - E 60-69
+ ? + ? + - -|- + ? 70-79
- + ? ? +|+ ? F + ? 80-89
- + +|+ ? + - ? + + 90-99
-

From the [23,26] and [27,30] blocks we have e=+1 and f=+1; from the block [51,54] we have f(53)=+1. Looking at [93,95] we see that f(94)=-1 hence f(47)=+1, and we have a cut at 92-93, hence also at 96-97:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? +|- + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + -|? - - 60-69
+ ? + ? + - -|- +|? 70-79
- + ? ? +|+ ? - + ? 80-89
- + +|+ - + -|? + + 90-99
-

from f[97,99] we have f(97)=-1. From the [41,44] block we have f(41)=f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + -|? -|- 60-69
+ ? + ? + - -|- +|? 70-79
- + - ? +|+ - - +|? 80-89
- + +|+ - + - - + + 90-99
-|

Using the cuts we can fill in f(67)=+1 and f(89)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+|+ - ? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + +|+ - + - - + + 90-99
-|

The 41-50 block then forces f(41)=-1, hence f(82)=-1.

If 79 and 83 are both positive the sum at 85 is 3, and if 79 and 83 are both negative the sum at 82 is -3, so 79 and 83 must be alternate signs, so there is a cut before 85:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + -|? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

Looking at 53-56 we see that f(53)=+1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + + ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

Suppose f(59)=+1. Then f(118)=f(120)=f(121)=1 and f(119)=-f(17), so f(17)=+1. Also, f(57)=f(59)=f(60)=1 and f(58)=f(29), hence f(29)=-1.
Next, f(15)+…+f(22) = f(11)+2, hence f(11)=-1; and then f(1)+…+f(13) = f(13)-2, so f(13)=+1. But then f(1)+…+f(23)=f(23)+2, so f(23)=-1. But then f(1)+…+f(29)=-3, contradiction.

Thus we must also have f(59)=-1, so f(61)=+1.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

The remaining question marks are unconstrained other than requiring every block to sum to 0. Now consider the sequence up to 119, with the multiplicative values filled in:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? - - + ? - ? 100-109
- - - ? + - + - - + 110-119

109 is clearly + (otherwise there’s a sequence of 5 -s).

Consider 101 and 103.

Both can’t be + because the sum at 103 would reach 3.

Both can’t be – because then at 105 the sum would reach -3.

Therefore one has to be + and the other -, so the sum is 0 immediately before 105 and 0 immediately before 107.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|? - + 100-109
- - - ? + - + - - + 110-119

Since the sum is 0 before 107, 107 must be + (otherwise the sum is -3 at 111).

Also, the sum is -2 before 103, so 103 is +.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119

Continuing with known multiplicative values and values set by the sum being 2 or -2 up to 149, this gives deductions:
120 = +1, 121=+1, 122=+1, 123=+1 and together with 119=+1 this provides a contradiction based on earlier values at 2,3,5; 2,61; 3,41 and 7,17
This concludes case 1a.

== Case 1.b f(2)=1, f(37)=-1 ==

This takes us back to case 1.b f(2)=1, f(37)=-1, which begins here:

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-
This seems to be about as far as one can get just from the assumption f(2)=+1.

So we assume f(37)=-1

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + - - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? - - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

a and b can't both be - (five -s starting at 10).

A,B can't both be - (five -s starting at 74)

Therefore a and b can't both be +, so a and b are alternating, so there is a cut before 17.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + +|c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + - - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? - - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

Since A and B are also alternating, 79 must be + (otherwise the sum reaches -3 at 80).

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + +|c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + - - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? - - - A B + 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

'''Assume''' a = - and b = +.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|c + - 10-19
- + - e - + + - - f 20-29
+ ? + + c|+ + - - - 30-39
- + + ? - - e ? - + 40-49
+ C + ? - + - + f ? 50-59
+ ? ? - + - + ? c E 60-69
+ ? + + - - - + - + 70-79
- + ? ? + C ? F - ? 80-89
- - C ? ? + - ? + - 90-99
-

+ + - - - - forces cut before the start of the sequence, so the only way for sum to be 0 with + + ? | is if ? = -1, so c = -1.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - e - + + - - f 20-29
+ ? + + -|+ + - - - 30-39
- + + ? - - e ? - + 40-49
+ + + ? - + - + f ? 50-59
+ ? ? - + - + ? + E 60-69
+ ? + + - - - + - + 70-79
- + ? ? + + ? F - ? 80-89
- - + ? ? + - ? + - 90-99
-

So sums force e = +

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - f 20-29
+ ? + + -|+ + - - - 30-39
- + + ? - - + ? - + 40-49
+ + + ? - + - + f ? 50-59
+ ? ? - + - + ? + - 60-69
+ ? + + - - - + - + 70-79
- + ? ? + + ? F - ? 80-89
- - + ? ? + - ? + - 90-99
-

So f = +, so 31 = -, and sums and multiplicity force

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|? + - 60-69
+ ? + + - - - + - + 70-79
- + + ? + + + - - ? 80-89
- - + + - + - ? + - 90-99
-

The smallest sum before 70 is -1, so 71 must be -1. This forces 67 = -1, otherwise the sum at 73 is 3.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|? + + + - - ? 80-89
- - + + - + - ? + - 90-99
-

83 must be -, otherwise the sum is 3 at 85.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|- +|+ + - -|? 80-89
- - + + - + - ? + - 90-99
-

So 89 is + (otherwise the sum is -3 at 91)

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|- - + - + +|- +|- 10-19
- + - + - + +|- - + 20-29
+|- + + -|+ + - -|- 30-39
- + +|+ -|- +|- - + 40-49
+ + + - -|+ -|+ + - 50-59
+ - -|- + - +|- +|- 60-69
+|- +|+ -|- - + - + 70-79
- + +|- +|+ + - -|+ 80-89
- - +|+ -|+ -|? + - 90-99
-

Human proof that completely multiplicative sequences have discrepancy greater than 2

2010-02-01T17:57:11Z

Jbd:

It is [[multiplicative sequences|known from computer search]] that the longest completely multiplicative sequence <math>f: [N] \to \{-1,+1\}</math> of discrepancy 2 has length 246. Here we record the attempts to prove this by hand. Notation: we display our function in rows of 10, starting with 0. We use the notation that + and - denote values which are known to equal +1 or -1 respectively, and ? for unknown values. Thus, initially our array looks like this:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

since f(n^2) = 1 for all square numbers.

The notions of a ''cut'' and a ''block'' will be useful. A ''cut'' occurs between n and n+1 whenever it can be proved that f[1,n] sums to 0; these can only occur when n is even. We indicate a cut by a bar | between n and n+1. Thus for instance we begin with just one cut:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

A ''block'' is any interval between two cuts; thus inside each block one must sum to zero, and the discrepancy inside each block is at most 2. Thus we can "localise" the analysis to each block, though multiplicativity relates the blocks together.

A block is ''solved'' if all values are assigned in it. Solved blocks are uninteresting (other than for the values they provide by multiplicativity), so we shall adopt the convention of concatenating solved blocks with adjacent blocks to reduce clutter.

More notation: we use f[a,b] as short for f(a)+...+f(b). Note that

: f[a,b] = -4,-2,0,2,4 if a is odd and b is even;
: f[a,b] = -3,-1,1,3 if a and b are both odd or both even;
: f[a,b] = -2,0,2 if a is even and b is odd.

Also if [a,b] starts or ends at a cut, then f[a,b] has magnitude at most 2.

A key observation is that if f(2n)=f(2n+1) then there must be a cut between 2n and 2n+1.

== Case 1. f(2)=1 ==

Now, let us '''assume''' f(2)=1 and that the discrepancy is at most 2 and see what happens. From multiplicativity we obtain f(2n^2)=1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + ? + ? ? ? + + 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

As f(8)=f(9) we can now cut at 8=9. And then by using discrepancy we get f(3)=f(5)=-1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + - + - ? ? +|+ 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

which then by multiplicativity and adding cuts where obvious extends to

0 1 2 3 4 5 6 7 8 9

0 + + - + - -|? +|+ 0-9
-|? - ? ? + + ? + ? 10-19
- ? ? ? - + ? - ? ? 20-29
+ ? + ? ? ? + ? ? ? 30-39

Looking at the 7-8 block we see that f(7) is forced to be -1; from this and multiplicativity, and showing rows up to 100, we arrive at

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + ? 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? ? ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - ? ? ? 50-59
+ ? ? ? + ? ? ? ? ? 60-69
+ ? + ? ? - ? ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? ? - ? + ? 90-99
-

Note that

: f[15,19] = 3 + f(17) + f(19)

cannot exceed 3, and

: f[168,171] = 2 - f(17) + f(19)

cannot exceed 2. Thus f(19) = -1. We use this information to update the board:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + - 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? - ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - + ? ? 50-59
+ ? ? ? + ? ? ? ? ? 60-69
+ ? + ? ? - - ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? + - ? + ? 90-99
-

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

This seems to be about as far as one can get just from the assumption f(2)=+1.

== Case 1.a f(2)=f(37)=1 ==

The most profitable '''assumption''' here seems to be f(37)=+1. This creates a cut at 36-37, which then forces f(34)=-1 and then f(33)=-1 also. Filling in these values and adding in cuts gives

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+ ? +|- - + +|+ - B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? -|- - + f ? 50-59
+ ? ? - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B + ? ? + - ? + + 90-99
-

looking at f[30,32] we thus have f(31)=-1,

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + +|+ - B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? -|- - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B +|+ ? + - ? + + 90-99
-

== Case 1.a.1 f(2)=f(37)=f(13)=1 ==

Now suppose we '''assume''' b=+1. Then by looking at f(1)+...+f(23) we have e=-1, and by looking at f(1)+...+f(37) we then have f=-1, leaving us with

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + -|- 30-39
- ? + ? + - - ? - + 40-49
+|+ + ? -|- - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - ? 70-79
- + ? ? +|+ ? + + ? 80-89
-|- +|+ ? + - ? + + 90-99
-

from f[39,41] we have f(41)=+1, and from f[77,79] we have f(79)=+1, from f[51,54] we have f(53)=-1, from f[47,50] we have f(47)=-1, and from f[85,88] we have f(86)=-1, hence f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + - - 30-39
- + + - +|- - - - + 40-49
+ +|+ - - - - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - + 70-79
- + +|? + + - + + ? 80-89
- - + + ? + - ? + + 90-99
-

But this creates a contradiction at the block [45,51]. Thus this case is impossible.

== Case 1.a f(2)=f(37)=1 ==

We have thus concluded in this case that f(13)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + +|e - + -|- - f 20-29
+|- + - - + +|+ - + 30-39
-|? + ? + - e ? - + 40-49
+|+ - ? -|- - + f ? 50-59
+ ? -|- + + -|? - E 60-69
+ ? + ? + - -|- + ? 70-79
- + ? ? +|+ ? F + ? 80-89
- + +|+ ? + - ? + + 90-99
-

From the [23,26] and [27,30] blocks we have e=+1 and f=+1; from the block [51,54] we have f(53)=+1. Looking at [93,95] we see that f(94)=-1 hence f(47)=+1, and we have a cut at 92-93, hence also at 96-97:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? +|- + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + -|? - - 60-69
+ ? + ? + - -|- +|? 70-79
- + ? ? +|+ ? - + ? 80-89
- + +|+ - + -|? + + 90-99
-

from f[97,99] we have f(97)=-1. From the [41,44] block we have f(41)=f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + -|? -|- 60-69
+ ? + ? + - -|- +|? 70-79
- + - ? +|+ - - +|? 80-89
- + +|+ - + - - + + 90-99
-|

Using the cuts we can fill in f(67)=+1 and f(89)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+|+ - ? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + +|+ - + - - + + 90-99
-|

The 41-50 block then forces f(41)=-1, hence f(82)=-1.

If 79 and 83 are both positive the sum at 85 is 3, and if 79 and 83 are both negative the sum at 82 is -3, so 79 and 83 must be alternate signs, so there is a cut before 85:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + -|? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

Looking at 53-56 we see that f(53)=+1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + + ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

Suppose f(59)=+1. Then f(118)=f(120)=f(121)=1 and f(119)=-f(17), so f(17)=+1. Also, f(57)=f(59)=f(60)=1 and f(58)=f(29), hence f(29)=-1.
Next, f(15)+…+f(22) = f(11)+2, hence f(11)=-1; and then f(1)+…+f(13) = f(13)-2, so f(13)=+1. But then f(1)+…+f(23)=f(23)+2, so f(23)=-1. But then f(1)+…+f(29)=-3, contradiction.

Thus we must also have f(59)=-1, so f(61)=+1.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

The remaining question marks are unconstrained other than requiring every block to sum to 0. Now consider the sequence up to 119, with the multiplicative values filled in:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? - - + ? - ? 100-109
- - - ? + - + - - + 110-119

109 is clearly + (otherwise there’s a sequence of 5 -s).

Consider 101 and 103.

Both can’t be + because the sum at 103 would reach 3.

Both can’t be – because then at 105 the sum would reach -3.

Therefore one has to be + and the other -, so the sum is 0 immediately before 105 and 0 immediately before 107.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|? - + 100-109
- - - ? + - + - - + 110-119

Since the sum is 0 before 107, 107 must be + (otherwise the sum is -3 at 111).

Also, the sum is -2 before 103, so 103 is +.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119

Continuing with known multiplicative values and values set by the sum being 2 or -2 up to 149, this gives deductions:
120 = +1, 121=+1, 122=+1, 123=+1 and together with 119=+1 this provides a contradiction based on earlier values at 2,3,5; 2,61; 3,41 and 7,17
This concludes case 1a.

== Case 1.b f(2)=1, f(37)=-1 ==

This takes us back to case 1.b f(2)=1, f(37)=-1, which begins here:

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-
This seems to be about as far as one can get just from the assumption f(2)=+1.

So we assume f(37)=-1

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + - - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? - - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

a and b can't both be - (five -s starting at 10).

A,B can't both be - (five -s starting at 74)

Therefore a and b can't both be +, so a and b are alternating, so there is a cut before 17.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + +|c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + - - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? - - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

Since A and B are also alternating, 79 must be + (otherwise the sum reaches -3 at 80).

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + +|c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + - - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? - - - A B + 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

Human proof that completely multiplicative sequences have discrepancy greater than 2

2010-02-01T17:43:51Z

Jbd:

It is [[multiplicative sequences|known from computer search]] that the longest completely multiplicative sequence <math>f: [N] \to \{-1,+1\}</math> of discrepancy 2 has length 246. Here we record the attempts to prove this by hand. Notation: we display our function in rows of 10, starting with 0. We use the notation that + and - denote values which are known to equal +1 or -1 respectively, and ? for unknown values. Thus, initially our array looks like this:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

since f(n^2) = 1 for all square numbers.

The notions of a ''cut'' and a ''block'' will be useful. A ''cut'' occurs between n and n+1 whenever it can be proved that f[1,n] sums to 0; these can only occur when n is even. We indicate a cut by a bar | between n and n+1. Thus for instance we begin with just one cut:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

A ''block'' is any interval between two cuts; thus inside each block one must sum to zero, and the discrepancy inside each block is at most 2. Thus we can "localise" the analysis to each block, though multiplicativity relates the blocks together.

A block is ''solved'' if all values are assigned in it. Solved blocks are uninteresting (other than for the values they provide by multiplicativity), so we shall adopt the convention of concatenating solved blocks with adjacent blocks to reduce clutter.

More notation: we use f[a,b] as short for f(a)+...+f(b). Note that

: f[a,b] = -4,-2,0,2,4 if a is odd and b is even;
: f[a,b] = -3,-1,1,3 if a and b are both odd or both even;
: f[a,b] = -2,0,2 if a is even and b is odd.

Also if [a,b] starts or ends at a cut, then f[a,b] has magnitude at most 2.

A key observation is that if f(2n)=f(2n+1) then there must be a cut between 2n and 2n+1.

== Case 1. f(2)=1 ==

Now, let us '''assume''' f(2)=1 and that the discrepancy is at most 2 and see what happens. From multiplicativity we obtain f(2n^2)=1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + ? + ? ? ? + + 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

As f(8)=f(9) we can now cut at 8=9. And then by using discrepancy we get f(3)=f(5)=-1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + - + - ? ? +|+ 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

which then by multiplicativity and adding cuts where obvious extends to

0 1 2 3 4 5 6 7 8 9

0 + + - + - -|? +|+ 0-9
-|? - ? ? + + ? + ? 10-19
- ? ? ? - + ? - ? ? 20-29
+ ? + ? ? ? + ? ? ? 30-39

Looking at the 7-8 block we see that f(7) is forced to be -1; from this and multiplicativity, and showing rows up to 100, we arrive at

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + ? 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? ? ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - ? ? ? 50-59
+ ? ? ? + ? ? ? ? ? 60-69
+ ? + ? ? - ? ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? ? - ? + ? 90-99
-

Note that

: f[15,19] = 3 + f(17) + f(19)

cannot exceed 3, and

: f[168,171] = 2 - f(17) + f(19)

cannot exceed 2. Thus f(19) = -1. We use this information to update the board:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + - 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? - ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - + ? ? 50-59
+ ? ? ? + ? ? ? ? ? 60-69
+ ? + ? ? - - ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? + - ? + ? 90-99
-

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

This seems to be about as far as one can get just from the assumption f(2)=+1.

== Case 1.a f(2)=f(37)=1 ==

The most profitable '''assumption''' here seems to be f(37)=+1. This creates a cut at 36-37, which then forces f(34)=-1 and then f(33)=-1 also. Filling in these values and adding in cuts gives

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+ ? +|- - + +|+ - B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? -|- - + f ? 50-59
+ ? ? - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B + ? ? + - ? + + 90-99
-

looking at f[30,32] we thus have f(31)=-1,

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + +|+ - B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? -|- - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B +|+ ? + - ? + + 90-99
-

== Case 1.a.1 f(2)=f(37)=f(13)=1 ==

Now suppose we '''assume''' b=+1. Then by looking at f(1)+...+f(23) we have e=-1, and by looking at f(1)+...+f(37) we then have f=-1, leaving us with

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + -|- 30-39
- ? + ? + - - ? - + 40-49
+|+ + ? -|- - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - ? 70-79
- + ? ? +|+ ? + + ? 80-89
-|- +|+ ? + - ? + + 90-99
-

from f[39,41] we have f(41)=+1, and from f[77,79] we have f(79)=+1, from f[51,54] we have f(53)=-1, from f[47,50] we have f(47)=-1, and from f[85,88] we have f(86)=-1, hence f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + - - 30-39
- + + - +|- - - - + 40-49
+ +|+ - - - - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - + 70-79
- + +|? + + - + + ? 80-89
- - + + ? + - ? + + 90-99
-

But this creates a contradiction at the block [45,51]. Thus this case is impossible.

== Case 1.a f(2)=f(37)=1 ==

We have thus concluded in this case that f(13)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + +|e - + -|- - f 20-29
+|- + - - + +|+ - + 30-39
-|? + ? + - e ? - + 40-49
+|+ - ? -|- - + f ? 50-59
+ ? -|- + + -|? - E 60-69
+ ? + ? + - -|- + ? 70-79
- + ? ? +|+ ? F + ? 80-89
- + +|+ ? + - ? + + 90-99
-

From the [23,26] and [27,30] blocks we have e=+1 and f=+1; from the block [51,54] we have f(53)=+1. Looking at [93,95] we see that f(94)=-1 hence f(47)=+1, and we have a cut at 92-93, hence also at 96-97:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? +|- + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + -|? - - 60-69
+ ? + ? + - -|- +|? 70-79
- + ? ? +|+ ? - + ? 80-89
- + +|+ - + -|? + + 90-99
-

from f[97,99] we have f(97)=-1. From the [41,44] block we have f(41)=f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + -|? -|- 60-69
+ ? + ? + - -|- +|? 70-79
- + - ? +|+ - - +|? 80-89
- + +|+ - + - - + + 90-99
-|

Using the cuts we can fill in f(67)=+1 and f(89)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+|+ - ? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + +|+ - + - - + + 90-99
-|

The 41-50 block then forces f(41)=-1, hence f(82)=-1.

If 79 and 83 are both positive the sum at 85 is 3, and if 79 and 83 are both negative the sum at 82 is -3, so 79 and 83 must be alternate signs, so there is a cut before 85:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + -|? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

Looking at 53-56 we see that f(53)=+1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + + ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

Suppose f(59)=+1. Then f(118)=f(120)=f(121)=1 and f(119)=-f(17), so f(17)=+1. Also, f(57)=f(59)=f(60)=1 and f(58)=f(29), hence f(29)=-1.
Next, f(15)+…+f(22) = f(11)+2, hence f(11)=-1; and then f(1)+…+f(13) = f(13)-2, so f(13)=+1. But then f(1)+…+f(23)=f(23)+2, so f(23)=-1. But then f(1)+…+f(29)=-3, contradiction.

Thus we must also have f(59)=-1, so f(61)=+1.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

The remaining question marks are unconstrained other than requiring every block to sum to 0. Now consider the sequence up to 119, with the multiplicative values filled in:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? - - + ? - ? 100-109
- - - ? + - + - - + 110-119

109 is clearly + (otherwise there’s a sequence of 5 -s).

Consider 101 and 103.

Both can’t be + because the sum at 103 would reach 3.

Both can’t be – because then at 105 the sum would reach -3.

Therefore one has to be + and the other -, so the sum is 0 immediately before 105 and 0 immediately before 107.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|? - + 100-109
- - - ? + - + - - + 110-119

Since the sum is 0 before 107, 107 must be + (otherwise the sum is -3 at 111).

Also, the sum is -2 before 103, so 103 is +.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119

Continuing with known multiplicative values and values set by the sum being 2 or -2 up to 149, this gives deductions:
120 = +1, 121=+1, 122=+1, 123=+1 and together with 119=+1 this provides a contradiction based on earlier values at 2,3,5; 2,61; 3,41 and 7,17
This concludes case 1a.

== Case 1.b f(2)=1, f(37)=-1 ==

This takes us back to case 1.b f(2)=1, f(37)=-1, which begins here:

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-
This seems to be about as far as one can get just from the assumption f(2)=+1.

So we assume f(37)=-1

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + - - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? - - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

a and b can't both be - (five -s starting at 10).

A,B can't both be - (five -s starting at 74)

Therefore a and b can't both be +, so a and b are alternating, so there is a cut before 17.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + +|c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + - - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? - - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

Human proof that completely multiplicative sequences have discrepancy greater than 2

2010-01-30T00:43:39Z

Jbd:

It is [[multiplicative sequences|known from computer search]] that the longest completely multiplicative sequence <math>f: [N] \to \{-1,+1\}</math> of discrepancy 2 has length 246. Here we record the attempts to prove this by hand. Notation: we display our function in rows of 10, starting with 0. We use the notation that + and - denote values which are known to equal +1 or -1 respectively, and ? for unknown values. Thus, initially our array looks like this:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

since f(n^2) = 1 for all square numbers.

The notions of a ''cut'' and a ''block'' will be useful. A ''cut'' occurs between n and n+1 whenever it can be proved that f[1,n] sums to 0; these can only occur when n is even. We indicate a cut by a bar | between n and n+1. Thus for instance we begin with just one cut:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

A ''block'' is any interval between two cuts; thus inside each block one must sum to zero, and the discrepancy inside each block is at most 2. Thus we can "localise" the analysis to each block, though multiplicativity relates the blocks together.

A block is ''solved'' if all values are assigned in it. Solved blocks are uninteresting (other than for the values they provide by multiplicativity), so we shall adopt the convention of concatenating solved blocks with adjacent blocks to reduce clutter.

More notation: we use f[a,b] as short for f(a)+...+f(b). Note that

: f[a,b] = -4,-2,0,2,4 if a is odd and b is even;
: f[a,b] = -3,-1,1,3 if a and b are both odd or both even;
: f[a,b] = -2,0,2 if a is even and b is odd.

Also if [a,b] starts or ends at a cut, then f[a,b] has magnitude at most 2.

A key observation is that if f(2n)=f(2n+1) then there must be a cut between 2n and 2n+1.

== Case 1. f(2)=1 ==

Now, let us '''assume''' f(2)=1 and that the discrepancy is at most 2 and see what happens. From multiplicativity we obtain f(2n^2)=1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + ? + ? ? ? + + 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

As f(8)=f(9) we can now cut at 8=9. And then by using discrepancy we get f(3)=f(5)=-1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + - + - ? ? +|+ 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

which then by multiplicativity and adding cuts where obvious extends to

0 1 2 3 4 5 6 7 8 9

0 + + - + - -|? +|+ 0-9
-|? - ? ? + + ? + ? 10-19
- ? ? ? - + ? - ? ? 20-29
+ ? + ? ? ? + ? ? ? 30-39

Looking at the 7-8 block we see that f(7) is forced to be -1; from this and multiplicativity, and showing rows up to 100, we arrive at

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + ? 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? ? ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - ? ? ? 50-59
+ ? ? ? + ? ? ? ? ? 60-69
+ ? + ? ? - ? ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? ? - ? + ? 90-99
-

Note that

: f[15,19] = 3 + f(17) + f(19)

cannot exceed 3, and

: f[168,171] = 2 - f(17) + f(19)

cannot exceed 2. Thus f(19) = -1. We use this information to update the board:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + - 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? - ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - + ? ? 50-59
+ ? ? ? + ? ? ? ? ? 60-69
+ ? + ? ? - - ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? + - ? + ? 90-99
-

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

This seems to be about as far as one can get just from the assumption f(2)=+1.

== Case 1.a f(2)=f(37)=1 ==

The most profitable '''assumption''' here seems to be f(37)=+1. This creates a cut at 36-37, which then forces f(34)=-1 and then f(33)=-1 also. Filling in these values and adding in cuts gives

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+ ? +|- - + +|+ - B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? -|- - + f ? 50-59
+ ? ? - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B + ? ? + - ? + + 90-99
-

looking at f[30,32] we thus have f(31)=-1,

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + +|+ - B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? -|- - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B +|+ ? + - ? + + 90-99
-

== Case 1.a.1 f(2)=f(37)=f(13)=1 ==

Now suppose we '''assume''' b=+1. Then by looking at f(1)+...+f(23) we have e=-1, and by looking at f(1)+...+f(37) we then have f=-1, leaving us with

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + -|- 30-39
- ? + ? + - - ? - + 40-49
+|+ + ? -|- - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - ? 70-79
- + ? ? +|+ ? + + ? 80-89
-|- +|+ ? + - ? + + 90-99
-

from f[39,41] we have f(41)=+1, and from f[77,79] we have f(79)=+1, from f[51,54] we have f(53)=-1, from f[47,50] we have f(47)=-1, and from f[85,88] we have f(86)=-1, hence f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + - - 30-39
- + + - +|- - - - + 40-49
+ +|+ - - - - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - + 70-79
- + +|? + + - + + ? 80-89
- - + + ? + - ? + + 90-99
-

But this creates a contradiction at the block [45,51]. Thus this case is impossible.

== Case 1.a f(2)=f(37)=1 ==

We have thus concluded in this case that f(13)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + +|e - + -|- - f 20-29
+|- + - - + +|+ - + 30-39
-|? + ? + - e ? - + 40-49
+|+ - ? -|- - + f ? 50-59
+ ? -|- + + -|? - E 60-69
+ ? + ? + - -|- + ? 70-79
- + ? ? +|+ ? F + ? 80-89
- + +|+ ? + - ? + + 90-99
-

From the [23,26] and [27,30] blocks we have e=+1 and f=+1; from the block [51,54] we have f(53)=+1. Looking at [93,95] we see that f(94)=-1 hence f(47)=+1, and we have a cut at 92-93, hence also at 96-97:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? +|- + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + -|? - - 60-69
+ ? + ? + - -|- +|? 70-79
- + ? ? +|+ ? - + ? 80-89
- + +|+ - + -|? + + 90-99
-

from f[97,99] we have f(97)=-1. From the [41,44] block we have f(41)=f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + -|? -|- 60-69
+ ? + ? + - -|- +|? 70-79
- + - ? +|+ - - +|? 80-89
- + +|+ - + - - + + 90-99
-|

Using the cuts we can fill in f(67)=+1 and f(89)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+|+ - ? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + +|+ - + - - + + 90-99
-|

The 41-50 block then forces f(41)=-1, hence f(82)=-1.

If 79 and 83 are both positive the sum at 85 is 3, and if 79 and 83 are both negative the sum at 82 is -3, so 79 and 83 must be alternate signs, so there is a cut before 85:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + -|? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

Looking at 53-56 we see that f(53)=+1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + + ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

Suppose f(59)=+1. Then f(118)=f(120)=f(121)=1 and f(119)=-f(17), so f(17)=+1. Also, f(57)=f(59)=f(60)=1 and f(58)=f(29), hence f(29)=-1.
Next, f(15)+…+f(22) = f(11)+2, hence f(11)=-1; and then f(1)+…+f(13) = f(13)-2, so f(13)=+1. But then f(1)+…+f(23)=f(23)+2, so f(23)=-1. But then f(1)+…+f(29)=-3, contradiction.

Thus we must also have f(59)=-1, so f(61)=+1.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

The remaining question marks are unconstrained other than requiring every block to sum to 0. Now consider the sequence up to 119, with the multiplicative values filled in:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? - - + ? - ? 100-109
- - - ? + - + - - + 110-119

109 is clearly + (otherwise there’s a sequence of 5 -s).

Consider 101 and 103.

Both can’t be + because the sum at 103 would reach 3.

Both can’t be – because then at 105 the sum would reach -3.

Therefore one has to be + and the other -, so the sum is 0 immediately before 105 and 0 immediately before 107.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|? - + 100-109
- - - ? + - + - - + 110-119

Since the sum is 0 before 107, 107 must be + (otherwise the sum is -3 at 111).

Also, the sum is -2 before 103, so 103 is +.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119

Continuing with known multiplicative values and values set by the sum being 2 or -2 up to 149:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+ + ? - + - ? - + ? 140-149

107 and 109 can't both be + (sum reaches 3 at 141) and can't both be - (5 -s in a row) so before 141 the sum must be 0:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+|+ ? - + - ? - + ? 140-149

Extending through multiplicative knowledge ...

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+|+ ? - + - ? - + ? 140-149
- ? - - - + + ? - - 150-159

We know 71 and 73 must be alternate signs, so by multiplicative properties 142 (2 * 71) and 146 (2 * 73) must have alternate signs, so the sum is zero immediately before 147 and 149. Now extending the list to 159:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+|+ ? - + - ?|- +|? 140-149
- ? - - - + + ? - - 150-159

151 must be + because otherwise we have 5 -s in a row, and 149 must be + because otherwise the sum will be -3 at 153, so immediately before 157 the sum is 0:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+|+ ? - + - ?|- +|+ 140-149
- + - - - + +|? - - 150-159

Human proof that completely multiplicative sequences have discrepancy greater than 2

2010-01-30T00:40:48Z

Jbd: Took out the 79 and 83 deduction

It is [[multiplicative sequences|known from computer search]] that the longest completely multiplicative sequence <math>f: [N] \to \{-1,+1\}</math> of discrepancy 2 has length 246. Here we record the attempts to prove this by hand. Notation: we display our function in rows of 10, starting with 0. We use the notation that + and - denote values which are known to equal +1 or -1 respectively, and ? for unknown values. Thus, initially our array looks like this:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

since f(n^2) = 1 for all square numbers.

The notions of a ''cut'' and a ''block'' will be useful. A ''cut'' occurs between n and n+1 whenever it can be proved that f[1,n] sums to 0; these can only occur when n is even. We indicate a cut by a bar | between n and n+1. Thus for instance we begin with just one cut:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

A ''block'' is any interval between two cuts; thus inside each block one must sum to zero, and the discrepancy inside each block is at most 2. Thus we can "localise" the analysis to each block, though multiplicativity relates the blocks together.

A block is ''solved'' if all values are assigned in it. Solved blocks are uninteresting (other than for the values they provide by multiplicativity), so we shall adopt the convention of concatenating solved blocks with adjacent blocks to reduce clutter.

More notation: we use f[a,b] as short for f(a)+...+f(b). Note that

: f[a,b] = -4,-2,0,2,4 if a is odd and b is even;
: f[a,b] = -3,-1,1,3 if a and b are both odd or both even;
: f[a,b] = -2,0,2 if a is even and b is odd.

Also if [a,b] starts or ends at a cut, then f[a,b] has magnitude at most 2.

A key observation is that if f(2n)=f(2n+1) then there must be a cut between 2n and 2n+1.

== Case 1. f(2)=1 ==

Now, let us '''assume''' f(2)=1 and that the discrepancy is at most 2 and see what happens. From multiplicativity we obtain f(2n^2)=1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + ? + ? ? ? + + 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

As f(8)=f(9) we can now cut at 8=9. And then by using discrepancy we get f(3)=f(5)=-1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + - + - ? ? +|+ 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

which then by multiplicativity and adding cuts where obvious extends to

0 1 2 3 4 5 6 7 8 9

0 + + - + - -|? +|+ 0-9
-|? - ? ? + + ? + ? 10-19
- ? ? ? - + ? - ? ? 20-29
+ ? + ? ? ? + ? ? ? 30-39

Looking at the 7-8 block we see that f(7) is forced to be -1; from this and multiplicativity, and showing rows up to 100, we arrive at

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + ? 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? ? ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - ? ? ? 50-59
+ ? ? ? + ? ? ? ? ? 60-69
+ ? + ? ? - ? ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? ? - ? + ? 90-99
-

Note that

: f[15,19] = 3 + f(17) + f(19)

cannot exceed 3, and

: f[168,171] = 2 - f(17) + f(19)

cannot exceed 2. Thus f(19) = -1. We use this information to update the board:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + - 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? - ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - + ? ? 50-59
+ ? ? ? + ? ? ? ? ? 60-69
+ ? + ? ? - - ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? + - ? + ? 90-99
-

Let's write f(11)=a, f(13)=b, f(17)=c, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + - 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? - B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - + f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - - A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? + - ? + a 90-99
-

This seems to be about as far as one can get just from the assumption f(2)=+1.

== Case 1.a f(2)=f(37)=1 ==

The most profitable '''assumption''' here seems to be f(37)=+1. This creates a cut at 36-37, which then forces f(34)=-1 and then f(33)=-1 also. Filling in these values and adding in cuts gives

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+ ? +|- - + +|+ - B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? -|- - + f ? 50-59
+ ? ? - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B + ? ? + - ? + + 90-99
-

looking at f[30,32] we thus have f(31)=-1,

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+|- + - - + +|+ - B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? -|- - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - -|- B ? 70-79
- + ? ? +|+ ? F + ? 80-89
- B +|+ ? + - ? + + 90-99
-

== Case 1.a.1 f(2)=f(37)=f(13)=1 ==

Now suppose we '''assume''' b=+1. Then by looking at f(1)+...+f(23) we have e=-1, and by looking at f(1)+...+f(37) we then have f=-1, leaving us with

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + -|- 30-39
- ? + ? + - - ? - + 40-49
+|+ + ? -|- - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - ? 70-79
- + ? ? +|+ ? + + ? 80-89
-|- +|+ ? + - ? + + 90-99
-

from f[39,41] we have f(41)=+1, and from f[77,79] we have f(79)=+1, from f[51,54] we have f(53)=-1, from f[47,50] we have f(47)=-1, and from f[85,88] we have f(86)=-1, hence f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + - - 30-39
- + + - +|- - - - + 40-49
+ +|+ - - - - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - + 70-79
- + +|? + + - + + ? 80-89
- - + + ? + - ? + + 90-99
-

But this creates a contradiction at the block [45,51]. Thus this case is impossible.

== Case 1.a f(2)=f(37)=1 ==

We have thus concluded in this case that f(13)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + +|e - + -|- - f 20-29
+|- + - - + +|+ - + 30-39
-|? + ? + - e ? - + 40-49
+|+ - ? -|- - + f ? 50-59
+ ? -|- + + -|? - E 60-69
+ ? + ? + - -|- + ? 70-79
- + ? ? +|+ ? F + ? 80-89
- + +|+ ? + - ? + + 90-99
-

From the [23,26] and [27,30] blocks we have e=+1 and f=+1; from the block [51,54] we have f(53)=+1. Looking at [93,95] we see that f(94)=-1 hence f(47)=+1, and we have a cut at 92-93, hence also at 96-97:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? +|- + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + -|? - - 60-69
+ ? + ? + - -|- +|? 70-79
- + ? ? +|+ ? - + ? 80-89
- + +|+ - + -|? + + 90-99
-

from f[97,99] we have f(97)=-1. From the [41,44] block we have f(41)=f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + -|? -|- 60-69
+ ? + ? + - -|- +|? 70-79
- + - ? +|+ - - +|? 80-89
- + +|+ - + - - + + 90-99
-|

Using the cuts we can fill in f(67)=+1 and f(89)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+|+ - ? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + +|+ - + - - + + 90-99
-|

The 41-50 block then forces f(41)=-1, hence f(82)=-1.

If 79 and 83 are both positive the sum at 85 is 3, and if 79 and 83 are both negative the sum at 82 is -3, so 79 and 83 must be alternate signs, so there is a cut before 85:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + -|? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

Looking at 53-56 we see that f(53)=+1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + + ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

Suppose f(59)=+1. Then f(118)=f(120)=f(121)=1 and f(119)=-f(17), so f(17)=+1. Also, f(57)=f(59)=f(60)=1 and f(58)=f(29), hence f(29)=-1.
Next, f(15)+…+f(22) = f(11)+2, hence f(11)=-1; and then f(1)+…+f(13) = f(13)-2, so f(13)=+1. But then f(1)+…+f(23)=f(23)+2, so f(23)=-1. But then f(1)+…+f(29)=-3, contradiction.

Thus we must also have f(59)=-1, so f(61)=+1.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

The remaining three question marks are unconstrained other than requiring every block to sum to 0. Now consider the sequence up to 119, with the multiplicative values filled in:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? - - + ? - ? 100-109
- - - ? + - + - - + 110-119

109 is clearly + (otherwise there’s a sequence of 5 -s).

Consider 101 and 103.

Both can’t be + because the sum at 103 would reach 3.

Both can’t be – because then at 105 the sum would reach -3.

Therefore one has to be + and the other -, so the sum is 0 immediately before 105 and 0 immediately before 107.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|? - + 100-109
- - - ? + - + - - + 110-119

Since the sum is 0 before 107, 107 must be + (otherwise the sum is -3 at 111).

Also, the sum is -2 before 103, so 103 is +.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119

Continuing with known multiplicative values and values set by the sum being 2 or -2 up to 149:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+ + ? - + - ? - + ? 140-149

107 and 109 can't both be + (sum reaches 3 at 141) and can't both be - (5 -s in a row) so before 141 the sum must be 0:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+|+ ? - + - ? - + ? 140-149

Extending through multiplicative knowledge ...

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+|+ ? - + - ? - + ? 140-149
- ? - - - + + ? - - 150-159

We know 71 and 73 must be alternate signs, so by multiplicative properties 142 (2 * 71) and 146 2 * 73) must have alternate signs, so the sum is zero immediately before 147 and 149. Now extending the list to 159:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+|+ ? - + - ?|- +|? 140-149
- ? - - - + + ? - - 150-159

151 must be + because otherwise we have 5 -s in a row, and 149 must be + because otherwise the sum will be -3 at 153, so immediately before 157 the sum is 0:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + - ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+|+ ? - + - ?|- +|+ 140-149
- + - - - + +|? - - 150-159

Human proof that completely multiplicative sequences have discrepancy greater than 2

2010-01-29T15:44:41Z

Jbd:

It is [[multiplicative sequences|known from computer search]] that the longest completely multiplicative sequence <math>f: [N] \to \{-1,+1\}</math> of discrepancy 2 has length 246. Here we record the attempts to prove this by hand. Notation: we display our function in rows of 10, starting with 0. We use the notation that + and - denote values which are known to equal +1 or -1 respectively, and ? for unknown values. I'll also put a bar | after the last place where it is known that f(1)+...f(n) sums to 0; these can only occur after an even number, and will be automatic in the middle of any ++++ or ---- string. Thus, initially our array looks like this:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

since f(n^2) = 1 for all square numbers.

Now, let us '''assume''' f(2)=1 and that the discrepancy is at most 2 and see what happens. From multiplicativity we obtain f(2n^2)=1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + ? + ? ? ? + + 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

and then by using discrepancy we get f(3)=f(5)=-1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + - + - ? ? + + 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

which then by multiplicativity extends to

0 1 2 3 4 5 6 7 8 9

0 + + - + - -|? + + 0-9
- ? - ? ? + + ? + ? 10-19
- ? ? ? - + ? - ? ? 20-29
+ ? + ? ? ? + ? ? ? 30-39

Looking at f(1)+...+f(9) we see that f(7) is forced to be -1; from this and multiplicativity, and showing rows up to 100, we arrive at

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + ? 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? ? ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - ? ? ? 50-59
+ ? ? ? + ? ? ? ? ? 60-69
+ ? + ? ? - ? ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? ? - ? + ? 90-99
-

This seems to be about as far as one can get just from the assumption f(2)=+1.

Let's write f(11)=a, f(13)=b, f(17)=c, f(19)=d, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, D=-d, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + d 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? d B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - D f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - d A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? D - ? + a 90-99
-

The most profitable '''assumption''' here seems to be f(37)=+1. This forces f(1)+...+f(37) to equal +1, which then forces f(34)=-1 and then f(33)=-1 also. Filling in these values gives

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + d 10-19
- + + e - + b - - f 20-29
+ ? + - - + + + d B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? - - - D f ? 50-59
+ ? ? - + B - ? - E 60-69
+ ? + ? + - d - B ? 70-79
- + ? ? + + ? F + ? 80-89
- B + ? ? D - ? + + 90-99
-

looking at f(30)+...+f(37) we thus have f(31)=-1,

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + d 10-19
- + + e - + b - - f 20-29
+ - + - - + + + d B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? - - - D f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - d - B ? 70-79
- + ? ? + + ? F + ? 80-89
- B + + ? D - ? + + 90-99
-

from f(1)+...+f(19) we have b+d <= 0, while from f(35)+...+f(39) we have d-b <= 0. Thus d must be -1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+ - + - - + + + - B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? - - - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - - - B ? 70-79
- + ? ? + + ? F + ? 80-89
- B + + ? + - ? + + 90-99
-

Now suppose we '''assume''' b=+1. Then by looking at f(1)+...+f(23) we have e=-1, and by looking at f(1)+...+f(37) we then have f=-1, leaving us with

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + -|- 30-39
- ? + ? + - - ? - + 40-49
+ + + ? - - - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - ? 70-79
- + ? ? + + ? + + ? 80-89
- - + + ? + - ? + + 90-99
-

from f(1)+...+f(41) we have f(41)=+1, and from f(75)+...+f(79) we have f(79)=+1, and from f(84)+...+f(88) we have f(86)=-1, hence f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + - - 30-39
- + + - +|- - ? - + 40-49
+ + + ? - - - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - + 70-79
- + +|? + + - + + ? 80-89
- - + + ? + - ? + + 90-99
-

from f(81)+...+f(85) we must have f(83)=-1, and from f(81)+...+f(89) we must have f(89)=-1, and from f(1)+...+f(47) we have f(47)=+1, hence f(94)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + - - 30-39
- + + - + - - + - + 40-49
+|+ + ? - - - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - + 70-79
- + + - + + - + + - 80-89
- - +|+ + + - ? + + 90-99
-

But this creates a contradiction at 95, we cannot have three +++ after a |. So we conclude (under the assumptions f(2)=+1 and f(37)=+1) that b=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + +|e - + - - - f 20-29
+ - + - - + + + - + 30-39
- ? + ? + - e ? - + 40-49
+ + - ? - - - + f ? 50-59
+ ? - - + + - ? - E 60-69
+ ? + ? + - - - + ? 70-79
- + ? ? + + ? F + ? 80-89
- + + + ? + - ? + + 90-99
-

From f(1)+...+f(29) we have e=+1 and f=+1. Looking at f(91)+...+f(95) we see that f(94)=-1 hence f(47)=+1, and we have a cut at 92-93, hence also at 96-97:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? + - + - - + 40-49
+ + - ? - - - + + ? 50-59
+ ? - - + + - ? - - 60-69
+ ? + ? + - - - + ? 70-79
- + ? ? + + ? - + ? 80-89
- + +|+ - + -|? + + 90-99
-

from f(91)+...+f(99) we have f(97)=-1. From the f(75)+...+f(77) chain we must have a cut at 76-77, hence also at 78-79:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? + - + - - + 40-49
+ + - ? - - - + + ? 50-59
+ ? - - + + - ? - - 60-69
+ ? + ? + - -|- +|? 70-79
- + ? ? + + ? - + ? 80-89
- + +|+ - + - - + + 90-99
-|

Any time we have two repeated signs at an even number followed by an odd number, there must be a cut in between, so we have a few more cuts to put in here:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? + - + - - + 40-49
+|+ - ? - - - + + ? 50-59
+ ? -|- +|+ - ? -|- 60-69
+|? + ? + - -|- +|? 70-79
- + ? ? +|+ ? - + ? 80-89
- + +|+ - + - - + + 90-99
-|

Using the cuts we can fill in f(67)=+1 and f(86)=f(89)=-1, hence f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + - + - + - - + 40-49
+|+ - ? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + ? ? +|+ - - + - 80-89
- + +|+ - + - - + + 90-99
-|

The 41-50 block then forces f(41)=-1, hence f(82)=+1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + -|? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + + ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

The 79-84 block then forces f(79)=f(83)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + -|? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|

Looking at 53-56 we see that f(53)=+1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|

Suppose f(59)=+1. Then f(118)=f(120)=f(121)=1 and f(119)=-f(17), so f(17)=+1. Also, f(57)=f(59)=f(60)=1 and f(58)=f(29), hence f(29)=-1.
Next, f(15)+…+f(22) = f(11)+2, hence f(11)=-1; and then f(1)+…+f(13) = f(13)-2, so f(13)=+1. But then f(1)+…+f(23)=f(23)+2, so f(23)=-1. But then f(1)+…+f(29)=-3, contradiction.

Thus we must also have f(59)=-1, so f(61)=+1.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|

The remaining three question marks are unconstrained other than requiring every block to sum to 0. Now consider the sequence up to 119, with the multiplicative values filled in:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? - - + ? - ? 100-109
- - - ? + - + - - + 110-119

109 is clearly + (otherwise there’s a sequence of 5 -s).

Consider 101 and 103.

Both can’t be + because the sum at 103 would reach 3.

Both can’t be – because then at 105 the sum would reach -3.

Therefore one has to be + and the other -, so the sum is 0 immediately before 105 and 0 immediately before 107.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|? - + 100-109
- - - ? + - + - - + 110-119

Since the sum is 0 before 107, 107 must be + (otherwise the sum is -3 at 111).

Also, the sum is -2 before 103, so 103 is +.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119

Continuing with known multiplicative values and values set by the sum being 2 or -2 up to 149:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+ + ? - + - ? - + ? 140-149

107 and 109 can't both be + (sum reaches 3 at 141) and can't both be - (5 -s in a row) so before 141 the sum must be 0:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+|+ ? - + - ? - + ? 140-149

Extending through multiplicative knowledge ...

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+|+ ? - + - ? - + ? 140-149
- ? - - - + + ? - - 150-159

We know 71 and 73 must be alternate signs, so by multiplicative properties 142 (2 * 71) and 146 2 * 73) must have alternate signs, so the sum is zero immediately before 147 and 149. Now extending the list to 159:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+|+ ? - + - ?|- +|? 140-149
- ? - - - + + ? - - 150-159

151 must be + because otherwise we have 5 -s in a row, and 149 must be + because otherwise the sum will be -3 at 153, so immediately before 157 the sum is 0:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+|+ ? - + - ?|- +|+ 140-149
- + - - - + +|? - - 150-159

Human proof that completely multiplicative sequences have discrepancy greater than 2

2010-01-29T15:43:38Z

Jbd:

It is [[multiplicative sequences|known from computer search]] that the longest completely multiplicative sequence <math>f: [N] \to \{-1,+1\}</math> of discrepancy 2 has length 246. Here we record the attempts to prove this by hand. Notation: we display our function in rows of 10, starting with 0. We use the notation that + and - denote values which are known to equal +1 or -1 respectively, and ? for unknown values. I'll also put a bar | after the last place where it is known that f(1)+...f(n) sums to 0; these can only occur after an even number, and will be automatic in the middle of any ++++ or ---- string. Thus, initially our array looks like this:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

since f(n^2) = 1 for all square numbers.

Now, let us '''assume''' f(2)=1 and that the discrepancy is at most 2 and see what happens. From multiplicativity we obtain f(2n^2)=1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + ? + ? ? ? + + 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

and then by using discrepancy we get f(3)=f(5)=-1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + - + - ? ? + + 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

which then by multiplicativity extends to

0 1 2 3 4 5 6 7 8 9

0 + + - + - -|? + + 0-9
- ? - ? ? + + ? + ? 10-19
- ? ? ? - + ? - ? ? 20-29
+ ? + ? ? ? + ? ? ? 30-39

Looking at f(1)+...+f(9) we see that f(7) is forced to be -1; from this and multiplicativity, and showing rows up to 100, we arrive at

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + ? 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? ? ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - ? ? ? 50-59
+ ? ? ? + ? ? ? ? ? 60-69
+ ? + ? ? - ? ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? ? - ? + ? 90-99
-

This seems to be about as far as one can get just from the assumption f(2)=+1.

Let's write f(11)=a, f(13)=b, f(17)=c, f(19)=d, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, D=-d, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + d 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? d B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - D f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - d A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? D - ? + a 90-99
-

The most profitable '''assumption''' here seems to be f(37)=+1. This forces f(1)+...+f(37) to equal +1, which then forces f(34)=-1 and then f(33)=-1 also. Filling in these values gives

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + d 10-19
- + + e - + b - - f 20-29
+ ? + - - + + + d B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? - - - D f ? 50-59
+ ? ? - + B - ? - E 60-69
+ ? + ? + - d - B ? 70-79
- + ? ? + + ? F + ? 80-89
- B + ? ? D - ? + + 90-99
-

looking at f(30)+...+f(37) we thus have f(31)=-1,

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + d 10-19
- + + e - + b - - f 20-29
+ - + - - + + + d B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? - - - D f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - d - B ? 70-79
- + ? ? + + ? F + ? 80-89
- B + + ? D - ? + + 90-99
-

from f(1)+...+f(19) we have b+d <= 0, while from f(35)+...+f(39) we have d-b <= 0. Thus d must be -1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+ - + - - + + + - B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? - - - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - - - B ? 70-79
- + ? ? + + ? F + ? 80-89
- B + + ? + - ? + + 90-99
-

Now suppose we '''assume''' b=+1. Then by looking at f(1)+...+f(23) we have e=-1, and by looking at f(1)+...+f(37) we then have f=-1, leaving us with

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + -|- 30-39
- ? + ? + - - ? - + 40-49
+ + + ? - - - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - ? 70-79
- + ? ? + + ? + + ? 80-89
- - + + ? + - ? + + 90-99
-

from f(1)+...+f(41) we have f(41)=+1, and from f(75)+...+f(79) we have f(79)=+1, and from f(84)+...+f(88) we have f(86)=-1, hence f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + - - 30-39
- + + - +|- - ? - + 40-49
+ + + ? - - - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - + 70-79
- + +|? + + - + + ? 80-89
- - + + ? + - ? + + 90-99
-

from f(81)+...+f(85) we must have f(83)=-1, and from f(81)+...+f(89) we must have f(89)=-1, and from f(1)+...+f(47) we have f(47)=+1, hence f(94)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + - - 30-39
- + + - + - - + - + 40-49
+|+ + ? - - - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - + 70-79
- + + - + + - + + - 80-89
- - +|+ + + - ? + + 90-99
-

But this creates a contradiction at 95, we cannot have three +++ after a |. So we conclude (under the assumptions f(2)=+1 and f(37)=+1) that b=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + +|e - + - - - f 20-29
+ - + - - + + + - + 30-39
- ? + ? + - e ? - + 40-49
+ + - ? - - - + f ? 50-59
+ ? - - + + - ? - E 60-69
+ ? + ? + - - - + ? 70-79
- + ? ? + + ? F + ? 80-89
- + + + ? + - ? + + 90-99
-

From f(1)+...+f(29) we have e=+1 and f=+1. Looking at f(91)+...+f(95) we see that f(94)=-1 hence f(47)=+1, and we have a cut at 92-93, hence also at 96-97:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? + - + - - + 40-49
+ + - ? - - - + + ? 50-59
+ ? - - + + - ? - - 60-69
+ ? + ? + - - - + ? 70-79
- + ? ? + + ? - + ? 80-89
- + +|+ - + -|? + + 90-99
-

from f(91)+...+f(99) we have f(97)=-1. From the f(75)+...+f(77) chain we must have a cut at 76-77, hence also at 78-79:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? + - + - - + 40-49
+ + - ? - - - + + ? 50-59
+ ? - - + + - ? - - 60-69
+ ? + ? + - -|- +|? 70-79
- + ? ? + + ? - + ? 80-89
- + +|+ - + - - + + 90-99
-|

Any time we have two repeated signs at an even number followed by an odd number, there must be a cut in between, so we have a few more cuts to put in here:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? + - + - - + 40-49
+|+ - ? - - - + + ? 50-59
+ ? -|- +|+ - ? -|- 60-69
+|? + ? + - -|- +|? 70-79
- + ? ? +|+ ? - + ? 80-89
- + +|+ - + - - + + 90-99
-|

Using the cuts we can fill in f(67)=+1 and f(86)=f(89)=-1, hence f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + - + - + - - + 40-49
+|+ - ? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + ? ? +|+ - - + - 80-89
- + +|+ - + - - + + 90-99
-|

The 41-50 block then forces f(41)=-1, hence f(82)=+1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + -|? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + + ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

The 79-84 block then forces f(79)=f(83)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + -|? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|

Looking at 53-56 we see that f(53)=+1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|

Suppose f(59)=+1. Then f(118)=f(120)=f(121)=1 and f(119)=-f(17), so f(17)=+1. Also, f(57)=f(59)=f(60)=1 and f(58)=f(29), hence f(29)=-1.
Next, f(15)+…+f(22) = f(11)+2, hence f(11)=-1; and then f(1)+…+f(13) = f(13)-2, so f(13)=+1. But then f(1)+…+f(23)=f(23)+2, so f(23)=-1. But then f(1)+…+f(29)=-3, contradiction.

Thus we must also have f(59)=-1, so f(61)=+1.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|

The remaining three question marks are unconstrained other than requiring every block to sum to 0. Now consider the sequence up to 119, with the multiplicative values filled in:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? - - + ? - ? 100-109
- - - ? + - + - - + 110-119

109 is clearly + (otherwise there’s a sequence of 5 -s).

Consider 101 and 103.

Both can’t be + because the sum at 103 would reach 3.

Both can’t be – because then at 105 the sum would reach -3.

Therefore one has to be + and the other -, so the sum is 0 immediately before 105 and 0 immediately before 107.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|? - + 100-109
- - - ? + - + - - + 110-119

Since the sum is 0 before 107, 107 must be + (otherwise the sum is -3 at 111).

Also, the sum is -2 before 103, so 103 is +.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119

Continuing with known multiplicative values and values set by the sum being 2 or -2 up to 149:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+ + ? - + - ? - + ? 140-149

107 and 109 can't both be + (sum reaches 3 at 141) and can't both be - (5 -s in a row) so before 141 the sum must be 0:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+|+ ? - + - ? - + ? 140-149

Extending through multiplicative knowledge ...

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+|+ ? - + - ? - + ? 140-149
- ? - - - + + ? - - 150-159

We know 71 and 73 must be alternate signs, so by multiplicative properties 142 (2 * 71) and 146 2 * 73) must have alternate signs, so the sum is zero immediately before 147 and 149. Now extendind the list to 159:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+|+ ? - + - ?|- +|? 140-149
- ? - - - + + ? - - 150-159

151 must be + because otherwise we have 5 -s in a row, and 149 must be + because otherwise the sum will be -3 at 153, so immediately before 157 the sum is 0:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+|+ ? - + - ?|- +|+ 140-149
- + - - - + +|? - - 150-159

Human proof that completely multiplicative sequences have discrepancy greater than 2

2010-01-29T15:03:45Z

Jbd:

It is [[multiplicative sequences|known from computer search]] that the longest completely multiplicative sequence <math>f: [N] \to \{-1,+1\}</math> of discrepancy 2 has length 246. Here we record the attempts to prove this by hand. Notation: we display our function in rows of 10, starting with 0. We use the notation that + and - denote values which are known to equal +1 or -1 respectively, and ? for unknown values. I'll also put a bar | after the last place where it is known that f(1)+...f(n) sums to 0; these can only occur after an even number, and will be automatic in the middle of any ++++ or ---- string. Thus, initially our array looks like this:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

since f(n^2) = 1 for all square numbers.

Now, let us '''assume''' f(2)=1 and that the discrepancy is at most 2 and see what happens. From multiplicativity we obtain f(2n^2)=1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + ? + ? ? ? + + 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

and then by using discrepancy we get f(3)=f(5)=-1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + - + - ? ? + + 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

which then by multiplicativity extends to

0 1 2 3 4 5 6 7 8 9

0 + + - + - -|? + + 0-9
- ? - ? ? + + ? + ? 10-19
- ? ? ? - + ? - ? ? 20-29
+ ? + ? ? ? + ? ? ? 30-39

Looking at f(1)+...+f(9) we see that f(7) is forced to be -1; from this and multiplicativity, and showing rows up to 100, we arrive at

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + ? 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? ? ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - ? ? ? 50-59
+ ? ? ? + ? ? ? ? ? 60-69
+ ? + ? ? - ? ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? ? - ? + ? 90-99
-

This seems to be about as far as one can get just from the assumption f(2)=+1.

Let's write f(11)=a, f(13)=b, f(17)=c, f(19)=d, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, D=-d, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + d 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? d B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - D f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - d A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? D - ? + a 90-99
-

The most profitable '''assumption''' here seems to be f(37)=+1. This forces f(1)+...+f(37) to equal +1, which then forces f(34)=-1 and then f(33)=-1 also. Filling in these values gives

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + d 10-19
- + + e - + b - - f 20-29
+ ? + - - + + + d B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? - - - D f ? 50-59
+ ? ? - + B - ? - E 60-69
+ ? + ? + - d - B ? 70-79
- + ? ? + + ? F + ? 80-89
- B + ? ? D - ? + + 90-99
-

looking at f(30)+...+f(37) we thus have f(31)=-1,

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + d 10-19
- + + e - + b - - f 20-29
+ - + - - + + + d B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? - - - D f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - d - B ? 70-79
- + ? ? + + ? F + ? 80-89
- B + + ? D - ? + + 90-99
-

from f(1)+...+f(19) we have b+d <= 0, while from f(35)+...+f(39) we have d-b <= 0. Thus d must be -1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+ - + - - + + + - B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? - - - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - - - B ? 70-79
- + ? ? + + ? F + ? 80-89
- B + + ? + - ? + + 90-99
-

Now suppose we '''assume''' b=+1. Then by looking at f(1)+...+f(23) we have e=-1, and by looking at f(1)+...+f(37) we then have f=-1, leaving us with

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + -|- 30-39
- ? + ? + - - ? - + 40-49
+ + + ? - - - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - ? 70-79
- + ? ? + + ? + + ? 80-89
- - + + ? + - ? + + 90-99
-

from f(1)+...+f(41) we have f(41)=+1, and from f(75)+...+f(79) we have f(79)=+1, and from f(84)+...+f(88) we have f(86)=-1, hence f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + - - 30-39
- + + - +|- - ? - + 40-49
+ + + ? - - - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - + 70-79
- + +|? + + - + + ? 80-89
- - + + ? + - ? + + 90-99
-

from f(81)+...+f(85) we must have f(83)=-1, and from f(81)+...+f(89) we must have f(89)=-1, and from f(1)+...+f(47) we have f(47)=+1, hence f(94)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + - - 30-39
- + + - + - - + - + 40-49
+|+ + ? - - - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - + 70-79
- + + - + + - + + - 80-89
- - +|+ + + - ? + + 90-99
-

But this creates a contradiction at 95, we cannot have three +++ after a |. So we conclude (under the assumptions f(2)=+1 and f(37)=+1) that b=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + +|e - + - - - f 20-29
+ - + - - + + + - + 30-39
- ? + ? + - e ? - + 40-49
+ + - ? - - - + f ? 50-59
+ ? - - + + - ? - E 60-69
+ ? + ? + - - - + ? 70-79
- + ? ? + + ? F + ? 80-89
- + + + ? + - ? + + 90-99
-

From f(1)+...+f(29) we have e=+1 and f=+1. Looking at f(91)+...+f(95) we see that f(94)=-1 hence f(47)=+1, and we have a cut at 92-93, hence also at 96-97:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? + - + - - + 40-49
+ + - ? - - - + + ? 50-59
+ ? - - + + - ? - - 60-69
+ ? + ? + - - - + ? 70-79
- + ? ? + + ? - + ? 80-89
- + +|+ - + -|? + + 90-99
-

from f(91)+...+f(99) we have f(97)=-1. From the f(75)+...+f(77) chain we must have a cut at 76-77, hence also at 78-79:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? + - + - - + 40-49
+ + - ? - - - + + ? 50-59
+ ? - - + + - ? - - 60-69
+ ? + ? + - -|- +|? 70-79
- + ? ? + + ? - + ? 80-89
- + +|+ - + - - + + 90-99
-|

Any time we have two repeated signs at an even number followed by an odd number, there must be a cut in between, so we have a few more cuts to put in here:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? + - + - - + 40-49
+|+ - ? - - - + + ? 50-59
+ ? -|- +|+ - ? -|- 60-69
+|? + ? + - -|- +|? 70-79
- + ? ? +|+ ? - + ? 80-89
- + +|+ - + - - + + 90-99
-|

Using the cuts we can fill in f(67)=+1 and f(86)=f(89)=-1, hence f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + - + - + - - + 40-49
+|+ - ? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + ? ? +|+ - - + - 80-89
- + +|+ - + - - + + 90-99
-|

The 41-50 block then forces f(41)=-1, hence f(82)=+1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + -|? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + + ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

The 79-84 block then forces f(79)=f(83)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + -|? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|

Looking at 53-56 we see that f(53)=+1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|

Suppose f(59)=+1. Then f(118)=f(120)=f(121)=1 and f(119)=-f(17), so f(17)=+1. Also, f(57)=f(59)=f(60)=1 and f(58)=f(29), hence f(29)=-1.
Next, f(15)+…+f(22) = f(11)+2, hence f(11)=-1; and then f(1)+…+f(13) = f(13)-2, so f(13)=+1. But then f(1)+…+f(23)=f(23)+2, so f(23)=-1. But then f(1)+…+f(29)=-3, contradiction.

Thus we must also have f(59)=-1, so f(61)=+1.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|

The remaining three question marks are unconstrained other than requiring every block to sum to 0. Now consider the sequence up to 119, with the multiplicative values filled in:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? - - + ? - ? 100-109
- - - ? + - + - - + 110-119

109 is clearly + (otherwise there’s a sequence of 5 -s).

Consider 101 and 103.

Both can’t be + because the sum at 103 would reach 3.

Both can’t be – because then at 105 the sum would reach -3.

Therefore one has to be + and the other -, so the sum is 0 immediately before 105 and 0 immediately before 107.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|? - + 100-109
- - - ? + - + - - + 110-119

Since the sum is 0 before 107, 107 must be + (otherwise the sum is -3 at 111).

Also, the sum is -2 before 103, so 103 is +.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119

Continuing with known multiplicative values and values set by the sum being 2 or -2 up to 149:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+ + ? - + - ? - + ? 140-149

107 and 109 can't both be + (sum reaches 3 at 141) and can't both be - (5 -s in a row) so before 141 the sum must be 0:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+|+ ? - + - ? - + ? 140-149

Human proof that completely multiplicative sequences have discrepancy greater than 2

2010-01-29T14:57:59Z

Jbd:

It is [[multiplicative sequences|known from computer search]] that the longest completely multiplicative sequence <math>f: [N] \to \{-1,+1\}</math> of discrepancy 2 has length 246. Here we record the attempts to prove this by hand. Notation: we display our function in rows of 10, starting with 0. We use the notation that + and - denote values which are known to equal +1 or -1 respectively, and ? for unknown values. I'll also put a bar | after the last place where it is known that f(1)+...f(n) sums to 0; these can only occur after an even number, and will be automatic in the middle of any ++++ or ---- string. Thus, initially our array looks like this:

0 1 2 3 4 5 6 7 8 9

0|+ ? ? + ? ? ? ? + 0-9
? ? ? ? ? ? + ? ? ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? ? ? ? ? + ? ? ? 30-39

since f(n^2) = 1 for all square numbers.

Now, let us '''assume''' f(2)=1 and that the discrepancy is at most 2 and see what happens. From multiplicativity we obtain f(2n^2)=1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + ? + ? ? ? + + 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

and then by using discrepancy we get f(3)=f(5)=-1, thus

0 1 2 3 4 5 6 7 8 9

0|+ + - + - ? ? + + 0-9
? ? ? ? ? ? + ? + ? 10-19
? ? ? ? ? + ? ? ? ? 20-29
? ? + ? ? ? + ? ? ? 30-39

which then by multiplicativity extends to

0 1 2 3 4 5 6 7 8 9

0 + + - + - -|? + + 0-9
- ? - ? ? + + ? + ? 10-19
- ? ? ? - + ? - ? ? 20-29
+ ? + ? ? ? + ? ? ? 30-39

Looking at f(1)+...+f(9) we see that f(7) is forced to be -1; from this and multiplicativity, and showing rows up to 100, we arrive at

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|? - ? - + + ? + ? 10-19
- + ? ? - + ? - - ? 20-29
+ ? + ? ? + + ? ? ? 30-39
- ? + ? ? - ? ? - + 40-49
+ ? ? ? - ? - ? ? ? 50-59
+ ? ? ? + ? ? ? ? ? 60-69
+ ? + ? ? - ? ? ? ? 70-79
- + ? ? ? ? ? ? ? ? 80-89
- ? ? ? ? ? - ? + ? 90-99
-

This seems to be about as far as one can get just from the assumption f(2)=+1.

Let's write f(11)=a, f(13)=b, f(17)=c, f(19)=d, f(23)=e, f(29)=f and A = -a, B = -b, C = -c, D=-d, E=-e, F=-f. Then we can fill in more:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
-|a - b - + + c + d 10-19
- + a e - + b - - f 20-29
+ ? + A c + + ? d B 30-39
- ? + ? a - e ? - + 40-49
+ C b ? - A - D f ? 50-59
+ ? ? - + B A ? c E 60-69
+ ? + ? ? - d A B ? 70-79
- + ? ? + C ? F a ? 80-89
- B C ? ? D - ? + a 90-99
-

The most profitable '''assumption''' here seems to be f(37)=+1. This forces f(1)+...+f(37) to equal +1, which then forces f(34)=-1 and then f(33)=-1 also. Filling in these values gives

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + d 10-19
- + + e - + b - - f 20-29
+ ? + - - + + + d B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? - - - D f ? 50-59
+ ? ? - + B - ? - E 60-69
+ ? + ? + - d - B ? 70-79
- + ? ? + + ? F + ? 80-89
- B + ? ? D - ? + + 90-99
-

looking at f(30)+...+f(37) we thus have f(31)=-1,

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + d 10-19
- + + e - + b - - f 20-29
+ - + - - + + + d B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? - - - D f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - d - B ? 70-79
- + ? ? + + ? F + ? 80-89
- B + + ? D - ? + + 90-99
-

from f(1)+...+f(19) we have b+d <= 0, while from f(35)+...+f(39) we have d-b <= 0. Thus d must be -1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + -|b - + + - + - 10-19
- + + e - + b - - f 20-29
+ - + - - + + + - B 30-39
- ? + ? + - e ? - + 40-49
+ + b ? - - - + f ? 50-59
+ ? - - + B - ? - E 60-69
+ ? + ? + - - - B ? 70-79
- + ? ? + + ? F + ? 80-89
- B + + ? + - ? + + 90-99
-

Now suppose we '''assume''' b=+1. Then by looking at f(1)+...+f(23) we have e=-1, and by looking at f(1)+...+f(37) we then have f=-1, leaving us with

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + -|- 30-39
- ? + ? + - - ? - + 40-49
+ + + ? - - - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - ? 70-79
- + ? ? + + ? + + ? 80-89
- - + + ? + - ? + + 90-99
-

from f(1)+...+f(41) we have f(41)=+1, and from f(75)+...+f(79) we have f(79)=+1, and from f(84)+...+f(88) we have f(86)=-1, hence f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + - - 30-39
- + + - +|- - ? - + 40-49
+ + + ? - - - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - + 70-79
- + +|? + + - + + ? 80-89
- - + + ? + - ? + + 90-99
-

from f(81)+...+f(85) we must have f(83)=-1, and from f(81)+...+f(89) we must have f(89)=-1, and from f(1)+...+f(47) we have f(47)=+1, hence f(94)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - + - + + - + - 10-19
- + + - - + + - - - 20-29
+ - + - - + + + - - 30-39
- + + - + - - + - + 40-49
+|+ + ? - - - + - ? 50-59
+ ? - - + - - ? - + 60-69
+ ? + ? + - -|- - + 70-79
- + + - + + - + + - 80-89
- - +|+ + + - ? + + 90-99
-

But this creates a contradiction at 95, we cannot have three +++ after a |. So we conclude (under the assumptions f(2)=+1 and f(37)=+1) that b=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + +|e - + - - - f 20-29
+ - + - - + + + - + 30-39
- ? + ? + - e ? - + 40-49
+ + - ? - - - + f ? 50-59
+ ? - - + + - ? - E 60-69
+ ? + ? + - - - + ? 70-79
- + ? ? + + ? F + ? 80-89
- + + + ? + - ? + + 90-99
-

From f(1)+...+f(29) we have e=+1 and f=+1. Looking at f(91)+...+f(95) we see that f(94)=-1 hence f(47)=+1, and we have a cut at 92-93, hence also at 96-97:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? + - + - - + 40-49
+ + - ? - - - + + ? 50-59
+ ? - - + + - ? - - 60-69
+ ? + ? + - - - + ? 70-79
- + ? ? + + ? - + ? 80-89
- + +|+ - + -|? + + 90-99
-

from f(91)+...+f(99) we have f(97)=-1. From the f(75)+...+f(77) chain we must have a cut at 76-77, hence also at 78-79:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? + - + - - + 40-49
+ + - ? - - - + + ? 50-59
+ ? - - + + - ? - - 60-69
+ ? + ? + - -|- +|? 70-79
- + ? ? + + ? - + ? 80-89
- + +|+ - + - - + + 90-99
-|

Any time we have two repeated signs at an even number followed by an odd number, there must be a cut in between, so we have a few more cuts to put in here:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + ? + - + - - + 40-49
+|+ - ? - - - + + ? 50-59
+ ? -|- +|+ - ? -|- 60-69
+|? + ? + - -|- +|? 70-79
- + ? ? +|+ ? - + ? 80-89
- + +|+ - + - - + + 90-99
-|

Using the cuts we can fill in f(67)=+1 and f(86)=f(89)=-1, hence f(43)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
-|? + - + - + - - + 40-49
+|+ - ? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + ? ? +|+ - - + - 80-89
- + +|+ - + - - + + 90-99
-|

The 41-50 block then forces f(41)=-1, hence f(82)=+1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + -|? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- +|? 70-79
- + + ? +|+ - - + - 80-89
- + + + - + - - + + 90-99
-|

The 79-84 block then forces f(79)=f(83)=-1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + -|? - - - + + ? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|

Looking at 53-56 we see that f(53)=+1:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|? 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|

Suppose f(59)=+1. Then f(118)=f(120)=f(121)=1 and f(119)=-f(17), so f(17)=+1. Also, f(57)=f(59)=f(60)=1 and f(58)=f(29), hence f(29)=-1.
Next, f(15)+…+f(22) = f(11)+2, hence f(11)=-1; and then f(1)+…+f(13) = f(13)-2, so f(13)=+1. But then f(1)+…+f(23)=f(23)+2, so f(23)=-1. But then f(1)+…+f(29)=-3, contradiction.

Thus we must also have f(59)=-1, so f(61)=+1.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|

The remaining three question marks are unconstrained other than requiring every block to sum to 0. Now consider the sequence up to 119, with the multiplicative values filled in:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? - - + ? - ? 100-109
- - - ? + - + - - + 110-119

109 is clearly + (otherwise there’s a sequence of 5 -s).

Consider 101 and 103.

Both can’t be + because the sum at 103 would reach 3.

Both can’t be – because then at 105 the sum would reach -3.

Therefore one has to be + and the other -, so the sum is 0 immediately before 105 and 0 immediately before 107.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|? - + 100-109
- - - ? + - + - - + 110-119

Since the sum is 0 before 107, 107 must be + (otherwise the sum is -3 at 111).

Also, the sum is -2 before 103, so 103 is +.

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ + -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? -|- +|+ - + 100-109
- - - + +|- +|- - + 110-119

Continuing with known multiplicative values and values set by the sum being 2 or -2 up to 149:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? - - + ? - ? 100-109
- - - + + - + - - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+ + ? - + - ? - + ? 140-149

107 and 109 can't both be + (sum reaches 3 at 141) and can't both be - (5 -s in a row) so before 141 the sum must be 0:

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? - - + ? - ? 100-109
- - - + + - + - - + 110-119
+ + + - - - - + + + 120-129
+ - - + + - -|? - ? 130-139
+|+ ? - + - ? - + ? 140-149

Human proof that completely multiplicative sequences have discrepancy greater than 2

2010-01-29T14:40:51Z

Jbd:

Human proof that completely multiplicative sequences have discrepancy greater than 2

2010-01-29T14:19:26Z

Jbd:

Talk:Human proof that completely multiplicative sequences have discrepancy greater than 2

2010-01-29T04:23:10Z

Jbd: New page: 0 1 2 3 4 5 6 7 8 9 0 + + - + - - - + + 0-9 - + - - - + + - + - 10-19 - + + + - + - - - + 20-29 + - + - - + + + - + 30-39 - - + - + - + - - + 40-49 + + - + - - - + +|- ...

0 1 2 3 4 5 6 7 8 9

0 + + - + - - - + + 0-9
- + - - - + + - + - 10-19
- + + + - + - - - + 20-29
+ - + - - + + + - + 30-39
- - + - + - + - - + 40-49
+ + - + - - - + +|- 50-59
+ ? -|- + + - + - - 60-69
+|? + ? + - -|- + - 70-79
- + + - + + - - + - 80-89
- + + + - + - - + + 90-99
-|? + ? - - + ? - ? 100-109
- - - ? + - + - - + 110-119
+ ? ? ? - ? - ? + ? 120-129
+ ? - ? + ? - ? - ? 130-139
+ ? ? ? + ? ? ? + ? 140-149
- ? - ? - ? + ? - ? 150-159
- ? + ? + ? - ? + ? 160-169
+ ? - ? - ? + ? - ? 170-179
- ? + ? + ? + ? - ? 180-189
+ ? - ? - ? + ? + ? 190-199
- ? ? ? + ? ? ? - ? 200-209
? ? + ? ? ? - ? ? ? 210-219

Human proof that completely multiplicative sequences have discrepancy greater than 2

2010-01-29T03:15:06Z

Jbd:

Convert raw input string into CSV table

2010-01-11T03:24:26Z

Jbd:

<code>
/* This code (in C) reads a file input.txt, numbers it, and arranges it in a table. */
/* It prints the code in a CSV format which can be piped into a file. */

#include <stdio.h>

#define TABLE_WRAP 24

int main(void)
{
FILE *stream;
char ch;
int i, j;

stream = fopen("input.txt", "r");

/* seek to the beginning of the file */
fseek(stream, 0, SEEK_SET);
i = 0; j = 0;

printf("0");

do
{
/* read a char from the file */
ch = fgetc(stream);

/* display the character */
if((ch == '+') || (ch == '-')) {

i++; j++;
if (j == TABLE_WRAP) {
j = 0;
printf("\n");
}
else printf(", ");
printf("%i%c",i,ch);

}

} while (ch != EOF);

fclose(stream);

return 0;
}
</code>

Convert raw input string into CSV table

2010-01-11T03:19:22Z

Jbd:

<nowiki>
/* This code (in C) reads a file input.txt, numbers it, and arranges it in a table. */
/* It prints the code in a CSV format which can be piped into a file. */

#include <stdio.h>

#define TABLE_WRAP 24

int main(void)
{
FILE *stream;
char ch;
int i, j;

stream = fopen("input.txt", "r");

/* seek to the beginning of the file */
fseek(stream, 0, SEEK_SET);
i = 0; j = 0;

printf("0");

do
{
/* read a char from the file */
ch = fgetc(stream);

/* display the character */
if((ch == '+') || (ch == '-')) {

i++; j++;
if (j == TABLE_WRAP) {
j = 0;
printf("\n");
}
else printf(", ");
printf("%i%c",i,ch);

}

} while (ch != EOF);

fclose(stream);

return 0;
}
</nowiki>

Convert raw input string into CSV table

2010-01-11T03:18:56Z

Jbd: New page: /* This code (in C) reads a file input.txt, numbers it, and arranges it in a table. */ /* It prints the code in a CSV format which can be piped into a file. */ #include <stdio.h> #define...

/* This code (in C) reads a file input.txt, numbers it, and arranges it in a table. */
/* It prints the code in a CSV format which can be piped into a file. */

#include <stdio.h>

#define TABLE_WRAP 24

int main(void)
{
FILE *stream;
char ch;
int i, j;

stream = fopen("input.txt", "r");

/* seek to the beginning of the file */
fseek(stream, 0, SEEK_SET);
i = 0; j = 0;

printf("0");

do
{
/* read a char from the file */
ch = fgetc(stream);

/* display the character */
if((ch == '+') || (ch == '-')) {

i++; j++;
if (j == TABLE_WRAP) {
j = 0;
printf("\n");
}
else printf(", ");
printf("%i%c",i,ch);

}

} while (ch != EOF);

fclose(stream);

return 0;
}

Experimental results

2010-01-11T03:16:20Z

Jbd:

To return to the main Polymath5 page, click [[The Erdős discrepancy problem|here]].

Perhaps we should have two kinds of subpages to this page: Pages about finding examples, and pages about analyzing them?

== Experimental data==
* [[The first 1124-sequence]] with discrepancy 2. ''Some more description''
* Other [[length 1124 sequences]] with discrepancy 2. ''Some more description''
* Some data about the problem with [[different upper and lower bound]]. ''Some more description''
* Sequences taking values in <math>\mathbb{T}</math>:
** [[4th roots of unity]]
** [[6th roots of unity]]
* [http://thomas1111.wordpress.com/2010/01/10/tables-for-a-c10-candidate/ A sequence of length 407] with discrepancy 2 such that <math>x_n=x_{32 n}</math> for every n. [[The HAP-subsequence structure of that sequence]].
* More [[T32-invariant sequences]].
* Long [[multiplicative sequences]].
* Long sequences satisfying [[T2(x) = -x]].
* Long sequences satisfying [[T2(x) = T5(x) = -x]]
* Long sequences satisfying [[T2(x) = -T3(x)]].

==Source code==

* [[Convert raw input string into CSV table]]

==Wish list==

* Find long/longest quasi-multiplicative sequences with some fixed group G, function <math>G\to \{-1,1\}</math> and maximal discrepancy C
** <math>G=C_6</math> and the function that sends 0,1 and 2 to 1 (because this seems to be a good choice)
* Do a "Mark-Bennet-style analysis" of one of the new 1124-sequences. [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4827] Also [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4842 done] (by Mark Bennet).
*. Take a moderately large k and search for the longest sequence of discrepancy 2 that's constructed as follows. First, pick a completely multiplicative function f to the group <math>C_{2k}</math>. Then set <math>x_n</math> to be 1 if f(n) lies between 0 and k-1, and -1 if f(n) lies between k and 2k-1. Alec has already [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4563 done this for k=1] and [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4734 partially done it for k=3].
*Search for the longest sequence of discrepancy 2 with the property that <math>x_n=x_{32n}</math> for every n. The motivation for this is to produce a fundamentally different class of examples (different because their group structure would include an element of order 5). It's not clear that it will work, since 32 is a fairly large number. However, if you've chosen <math>x_{32n}</math> then that will have some influence on several other choices, such as <math>x_{4n},x_{8n}</math> and <math>x_{16n}</math>, so maybe it will lead to something interesting. Alec [http://gowers.wordpress.com/2010/01/09/erds-discrepancy-problem-continued/#comment-4873 has made a start on this] and an [http://gowers.wordpress.com/2010/01/09/erds-discrepancy-problem-continued/#comment-4874 initial investigation] suggests that the sequence he has found does indeed have some <math>C_{10}</math>-related structure.
*Here's another experiment that should be pretty easy to program and might yield something interesting. It's to look at the how the discrepancy appears to grow when you define a sequence using a greedy algorithm. I say "a" greedy algorithm because there are various algorithms that could reasonably be described as greedy. Here are a few.

<blockquote>
1. For each n let <math>x_n</math> be chosen so as to minimize the discrepancy so far, given the choices already made for <math>x_1,\dots,x_{n-1}</math>. (If this does not uniquely determine <math>x_n</math> then choose it arbitrarily, or randomly, or according to some simple rule like always equalling 1.)
</blockquote>

<blockquote>
2. Same as 1 but with additional constraints, in the hope that these make the sequence more likely to be good. For instance, one might insist that <math>x_{2k}=x_{3k}</math> for every k. Here, when choosing <math>x_n</math> one would probably want to minimize the discrepancy up to <math>x_{n+k}</math> if <math>x_{n+1},\dots,x_{n+k}</math> had already been chosen. Another obvious constraint to try is complete multiplicativity.
</blockquote>

<blockquote>
3. A greedy algorithm of sorts, but this time trying to minimize a different parameter. The first algorithm will do this: when you pick <math>x_n</math> you look, for each factor d of n, at the partial sum along the multiples of d up to but not including n. This will give you a set A of numbers (the possible partial sums). If max(A) is greater than max(-A) then you set <math>x_n=-1</math>, if max(-A) is greater than max(A) then you let <math>x_n=1</math>, and if they are equal then you make the decision according to some rule that seems sensible. But it might be that you would end up with a slower-growing discrepancy if you regarded A as a multiset and made the decision on some other basis. For instance, you could take the sum of <math>2^k</math> over all positive elements <math>k\in A</math> (with multiplicity) and the sum of <math>2^{-k}</math> over all negative elements and choose <math>x_n</math> according to which was bigger. Although that wouldn't minimize the discrepancy at each stage, it might make the sequence better for future development because it wouldn't sacrifice the needs of an overwhelming majority to those of a few rogue elements.

The motivation for these experiments is to see whether they, or some variants, appear to lead to sublogarithmic growth. If they do, then we could start trying to prove rigorously that sublogarithmic growth is possible. I still think that a function that arises in nature and satisfies f(1124)=2 ought to be sublogarithmic.
</blockquote>

*What happens if one applies a backtracking algorithm to try to extend the following discrepancy-2 sequence, which satisfies <math>x_{2n}=-x_n</math> for every n, to a much longer discrepancy-2 sequence: + - - + - + + - - + + - + - + + - + - - + - - + + - - + + - + - - + - - + + + + - - - + + + - - + - + + - + - - + - ? This question has been answered [http://gowers.wordpress.com/2010/01/09/erds-discrepancy-problem-continued/#comment-4893 in the comments following the asking of the question on the blog].

* Investigate what happens if our HAPs are restricted to allow differences divisible only by 2 or 3 [and then other sets of primes including 2] - {2,3,5,7} would be interesting - is there an infinite sequence of discrepancy 2 in these simple cases - is it easy to find an infinite sequence with finite discrepancy in these cases? [for sets of odd primes, take a sequence which is 1 on odd numbers, -1 on even numbers. Including 2 is the non-trivial case]. It is possible that completely multiplicative sequences could be found for some of these cases.

* ... you are welcome to add more.

Length 1124 sequences

2010-01-09T21:38:31Z

Jbd: /* The data */

''This page is about a large family of length 1124 sequences. The family does not include [[the first 1124-sequence]], since we used another method for that one (correct me if I'm wrong).''

''If you can find better name for the pages I create, you are more than welcome.''

==Method==

Here should be a short description of the way the sequence was found. (The code(s) used should be further down this page.)

== Status ==

Is the data still relevant (e.g. longest known)? Is the method still relevant, or have we found a better method? Is the program still running on a computer somewhere?

These sequences are among the longest known sequences with discrepancy 2.

== The data==

There are at least 428 995 120 sequences in this family. See [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4831 this comment].

0 + + - - + + - + - - + - - - + + - + - - + + + - + - - + -
+ + - - + - + - - + + + - + - - + - + + - - + + - + - + + -
- + - + + - + - - - + + - - + + + - - + - + - - + + - - + -
+ + - + - - - + + - + + + - - - + - + + - + - - - + + + - +
+ - + - - - - + + + + - - + - - - + + - + + - + + - + - + -
- + - - - + + + + - - - - + + + - - + - + + - + - - + + - +
- - + - + - + - - + + - - + + - - + + - + + - + + - - - - +
- + + - + + - - + + - + - - + + + - + - - + - - - + + - - +
+ - + + - + - - + + - + - - + - - + + - + + + - - - + + - -
+ + + - - + - + - - - + + - + - - + + - + - - + + + + - - +
- - + + - - - + + + - - + - + + - + - - + - - + + - + - - +
+ - + + - + - + + - - + - - + + - - + - + + - + - - + + + -
- + - - - + + - + + - + - - + + - + + - + - - - + - + - - +
- + + - - + + + - - - + + - + - - + - + + + - - + - + + - +
+ - + - - + - - + + - + - + + - + - + - + - - + - - + + - +
+ - + - - + + - - + - - - + + + - - - + + - - + + - + + + -
- - + + - + - + + - - - + - + - - + - + + + - - + - + - - +
+ - + + - + - - + + - + - + - - + - + - + + - + - - + - - +
+ + - + - - + + + - - + - - + + - + + - + - - + + - + - - +
- - + + - + - - + - - + + - + + + - - - + + - + - - + + - +
+ - + - - + - - + + - + + - + - - + + - + - - + - - + + - +
- - + - - + + - + + - - - + + + - + + - + - - - - + + - - +
- + + + - + + - + - - + + - + - - + - - + + - + - - + + - -
+ - + + - + - - + + - - + - + - - + + + + - - + - - + - - +
+ - + - - + + - + + - + - - + + - + - + + - - - + - + + - +
- - + + - + - + + - - - + + + - - + - - + + - - + - + - - +
+ - + + - + + - + - - + - + + - - - + - + + - + - - + + - +
- - + + - + + - + - - - - + + + - + + - + - - + + - + - - +
- - + + - + - - + - - + + - + + - + - - + + - + + - + - - +
+ - - - - + - + + - - + + + + - - - + - + + - + - - + + - +
- - + - - + + - + + - + - - + + + - - + + - - + + - + - - +
- - - + + + + - + - - - + - + + - + - - + + - + - + + - - -
+ + - + + - - - + + - + - + + - - - + - + + + - - - + + - +
+ - + - - + + - - - - + - + + + - + - - + - - + + - + + - -
- + + + - + + - + - - - + - + - - + - + + + - + + - - - - +
+ + + - - - + - + - - + - + + + + - - - + + - + - - + + - +
- + + - - + + - + - - - - + + + + - + - + - - + - - + + - +
+ - - + - + + - + - - + - + +

[http://thomas1111.wordpress.com/2010/01/09/tables-for-the-second-1124-sequence/ The sequence divided into groups of 24, and also with multiples of 8 only].

[http://spreadsheets.google.com/ccc?key=0AkbsKAn5VTtvdFhYQUlaQ3J5OV83czdDWTJDVmJwRmc&hl=en The sequence compared with the first 1124 sequence we obtained.]

== Relevant Code ==
The code(s) (or a link to the code(s)) used to find this sequence should be posted here.

Experimental results

2010-01-09T17:07:59Z

Jbd: /* Another sequence of length 1124 */

To return to the main Polymath5 page, click [[The Erdős discrepancy problem|here]].

==Different displays of a long low-discrepancy sequence==

It would be good to have some long sequences with low discrepancy displayed nicely here (or on subsidiary pages) in tabular form, together with discussions about the structures that can be found in these sequences. For now, here is the current record not displayed in a particularly helpful way. It is a sequence of discrepancy 2 and length 1124.

===The raw sequence===

+ - + + - - - - + + - + + + - - + - + + - - - + - + + - + -
- + + - + - - + - - + + - + + - - - + + + - - + - + - - + +
+ + - - + - - + + - - + + + - - + + - + - - + - - + + - + -
- + - + + + - - + - + - - + + - - - + + - + + + - - - + - -
+ - + + + + - - - - + + - + + + - - + - - + - - + - + - + +
- + + + - - - - + + + + - - - + + - + - - + - + + - - + - +
+ - + - + - + + - - - + + - + + - - + - - + - - + + + + - +
- - + - - + + - - + - + + - - - + - + + - + + + - - + + - -
+ - - + - + + - - + - + + - + + - - + - - + - + + - - + + -
+ - + - - + - + + + - - + - + + - - - - + + - - + - + + - +
+ - - + - + + - - + + - + - - + - + - - + + - - + - + + + -
+ - - + - + - - + + + + - - - + + - + - - + - + + - - + - +
+ - + + - - + - - + - + + - - + + - - - + + - - + - + + - +
+ - - + - - + + + + - - - - + + - + + - + + - - + - - + + -
- - + + - + + - - + - + + - + - - - + - + + - + + - - + - -
+ - + + - - + + - + + + - - - + - + + - + + - - + - - - + +
+ - - + - + - - + + + - + - + + - + - - - + + - + - + + - -
+ - - + - + + - - + - + - + + - + - + - - + - + + - - + - -
+ - + + + - - - + + - + + + - + - - + - + + - - + - + + - +
- - - + - + + - + + - - + - - - + + + - - + - + + - - + + -
+ - + + - + - - - + - - + - + + - - + - + + - + + - - + - +
+ - + + - - + - - + - + + - - + - - + - + + - + - - + + - +
+ - - + - - + - + + - - + - + + - + + - - + - + + - - + + -
+ - - + - + + - - + - - + - + + - - + - + + - + + - + + - -
+ - + + - - + - - + - + + - - + - + - - + + + - + - - + - +
+ - - + - + - - + + + - - - + + - + + - - + + - + - + + - -
+ - - + - - + - + + - + - - + + + - + - - + - + + - - + - +
- - + + - - + - + + - + + - - + + - + - - + - - + - + + - +
+ - - + - + + - + + - - + - - + + + - - - + - - + - + + - -
+ + + + - + - - + + - - + - + + - - + - - + - + + - - + - +
+ - + + - - + - - + - + + - - - + + - - + + - - + - + + - +
+ - - + - - + - + + + - + - - + - + + - - + - + + - + - + -
- + - + - + + - - + - + - - + + + - + - - - + + + - - + - -
+ - + + - - + + + + - + - - - + - + + - + + - - + - - + + +
- - - + - - + - + + + - + - + + - + - - - + - - + + + + - -
+ - + + - - + - + + - + - - + - + + - - - + - - + + - + - +
+ - + - - - + - + + + + - + - - - - + - + + - + + - - + + -
+ + - + - - + - + + - + - -

===The sequence, together with the corresponding integers===

1+ 2- 3+ 4+ 5- 6- 7- 8- 9+ 10+ 11- 12+ 13+ 14+ 15- 16- 17+ 18- 19+ 20+ 21- 22- 23- 24+ 25- 26+ 27+ 28- 29+ 30- 31- 32+ 33+ 34- 35+ 36- 37- 38+ 39- 40- 41+ 42+ 43- 44+ 45+ 46- 47- 48- 49+ 50+ 51+ 52- 53- 54+ 55- 56+ 57- 58- 59+ 60+ 61+ 62+ 63- 64- 65+ 66- 67- 68+ 69+ 70- 71- 72+ 73+ 74+ 75- 76- 77+ 78+ 79- 80+ 81- 82- 83+ 84- 85- 86+ 87+ 88- 89+ 90- 91- 92+ 93- 94+ 95+ 96+ 97- 98- 99+ 100- 101+ 102- 103- 104+ 105+ 106- 107- 108- 109+ 110+ 111- 112+ 113+ 114+ 115- 116- 117- 118+ 119- 120- 121+ 122- 123+ 124+ 125+ 126+ 127- 128- 129- 130- 131+ 132+ 133- 134+ 135+ 136+ 137- 138- 139+ 140- 141- 142+ 143- 144- 145+ 146- 147+ 148- 149+ 150+ 151- 152+ 153+ 154+ 155- 156- 157- 158- 159+ 160+ 161+ 162+ 163- 164- 165- 166+ 167+ 168- 169+ 170- 171- 172+ 173- 174+ 175+ 176- 177- 178+ 179- 180+ 181+ 182- 183+ 184- 185+ 186- 187+ 188+ 189- 190- 191- 192+ 193+ 194- 195+ 196+ 197- 198- 199+ 200- 201- 202+ 203- 204- 205+ 206+ 207+ 208+ 209- 210+ 211- 212- 213+ 214- 215- 216+ 217+ 218- 219- 220+ 221- 222+ 223+ 224- 225- 226- 227+ 228- 229+ 230+ 231- 232+ 233+ 234+ 235- 236- 237+ 238+ 239- 240- 241+ 242- 243- 244+ 245- 246+ 247+ 248- 249- 250+ 251- 252+ 253+ 254- 255+ 256+ 257- 258- 259+ 260- 261- 262+ 263- 264+ 265+ 266- 267- 268+ 269+ 270- 271+ 272- 273+ 274- 275- 276+ 277- 278+ 279+ 280+ 281- 282- 283+ 284- 285+ 286+ 287- 288- 289- 290- 291+ 292+ 293- 294- 295+ 296- 297+ 298+ 299- 300+ 301+ 302- 303- 304+ 305- 306+ 307+ 308- 309- 310+ 311+ 312- 313+ 314- 315- 316+ 317- 318+ 319- 320- 321+ 322+ 323- 324- 325+ 326- 327+ 328+ 329+ 330- 331+ 332- 333- 334+ 335- 336+ 337- 338- 339+ 340+ 341+ 342+ 343- 344- 345- 346+ 347+ 348- 349+ 350- 351- 352+ 353- 354+ 355+ 356- 357- 358+ 359- 360+ 361+ 362- 363+ 364+ 365- 366- 367+ 368- 369- 370+ 371- 372+ 373+ 374- 375- 376+ 377+ 378- 379- 380- 381+ 382+ 383- 384- 385+ 386- 387+ 388+ 389- 390+ 391+ 392- 393- 394+ 395- 396- 397+ 398+ 399+ 400+ 401- 402- 403- 404- 405+ 406+ 407- 408+ 409+ 410- 411+ 412+ 413- 414- 415+ 416- 417- 418+ 419+ 420- 421- 422- 423+ 424+ 425- 426+ 427+ 428- 429- 430+ 431- 432+ 433+ 434- 435+ 436- 437- 438- 439+ 440- 441+ 442+ 443- 444+ 445+ 446- 447- 448+ 449- 450- 451+ 452- 453+ 454+ 455- 456- 457+ 458+ 459- 460+ 461+ 462+ 463- 464- 465- 466+ 467- 468+ 469+ 470- 471+ 472+ 473- 474- 475+ 476- 477- 478- 479+ 480+ 481+ 482- 483- 484+ 485- 486+ 487- 488- 489+ 490+ 491+ 492- 493+ 494- 495+ 496+ 497- 498+ 499- 500- 501- 502+ 503+ 504- 505+ 506- 507+ 508+ 509- 510- 511+ 512- 513- 514+ 515- 516+ 517+ 518- 519- 520+ 521- 522+ 523- 524+ 525+ 526- 527+ 528- 529+ 530- 531- 532+ 533- 534+ 535+ 536- 537- 538+ 539- 540- 541+ 542- 543+ 544+ 545+ 546- 547- 548- 549+ 550+ 551- 552+ 553+ 554+ 555- 556+ 557- 558- 559+ 560- 561+ 562+ 563- 564- 565+ 566- 567+ 568+ 569- 570+ 571- 572- 573- 574+ 575- 576+ 577+ 578- 579+ 580+ 581- 582- 583+ 584- 585- 586- 587+ 588+ 589+ 590- 591- 592+ 593- 594+ 595+ 596- 597- 598+ 599+ 600- 601+ 602- 603+ 604+ 605- 606+ 607- 608- 609- 610+ 611- 612- 613+ 614- 615+ 616+ 617- 618- 619+ 620- 621+ 622+ 623- 624+ 625+ 626- 627- 628+ 629- 630+ 631+ 632- 633+ 634+ 635- 636- 637+ 638- 639- 640+ 641- 642+ 643+ 644- 645- 646+ 647- 648- 649+ 650- 651+ 652+ 653- 654+ 655- 656- 657+ 658+ 659- 660+ 661+ 662- 663- 664+ 665- 666- 667+ 668- 669+ 670+ 671- 672- 673+ 674- 675+ 676+ 677- 678+ 679+ 680- 681- 682+ 683- 684+ 685+ 686- 687- 688+ 689+ 690- 691+ 692- 693- 694+ 695- 696+ 697+ 698- 699- 700+ 701- 702- 703+ 704- 705+ 706+ 707- 708- 709+ 710- 711+ 712+ 713- 714+ 715+ 716- 717+ 718+ 719- 720- 721+ 722- 723+ 724+ 725- 726- 727+ 728- 729- 730+ 731- 732+ 733+ 734- 735- 736+ 737- 738+ 739- 740- 741+ 742+ 743+ 744- 745+ 746- 747- 748+ 749- 750+ 751+ 752- 753- 754+ 755- 756+ 757- 758- 759+ 760+ 761+ 762- 763- 764- 765+ 766+ 767- 768+ 769+ 770- 771- 772+ 773+ 774- 775+ 776- 777+ 778+ 779- 780- 781+ 782- 783- 784+ 785- 786- 787+ 788- 789+ 790+ 791- 792+ 793- 794- 795+ 796+ 797+ 798- 799+ 800- 801- 802+ 803- 804+ 805+ 806- 807- 808+ 809- 810+ 811- 812- 813+ 814+ 815- 816- 817+ 818- 819+ 820+ 821- 822+ 823+ 824- 825- 826+ 827+ 828- 829+ 830- 831- 832+ 833- 834- 835+ 836- 837+ 838+ 839- 840+ 841+ 842- 843- 844+ 845- 846+ 847+ 848- 849+ 850+ 851- 852- 853+ 854- 855- 856+ 857+ 858+ 859- 860- 861- 862+ 863- 864- 865+ 866- 867+ 868+ 869- 870- 871+ 872+ 873+ 874+ 875- 876+ 877- 878- 879+ 880+ 881- 882- 883+ 884- 885+ 886+ 887- 888- 889+ 890- 891- 892+ 893- 894+ 895+ 896- 897- 898+ 899- 900+ 901+ 902- 903+ 904+ 905- 906- 907+ 908- 909- 910+ 911- 912+ 913+ 914- 915- 916- 917+ 918+ 919- 920- 921+ 922+ 923- 924- 925+ 926- 927+ 928+ 929- 930+ 931+ 932- 933- 934+ 935- 936- 937+ 938- 939+ 940+ 941+ 942- 943+ 944- 945- 946+ 947- 948+ 949+ 950- 951- 952+ 953- 954+ 955+ 956- 957+ 958- 959+ 960- 961- 962+ 963- 964+ 965- 966+ 967+ 968- 969- 970+ 971- 972+ 973- 974- 975+ 976+ 977+ 978- 979+ 980- 981- 982- 983+ 984+ 985+ 986- 987- 988+ 989- 990- 991+ 992- 993+ 994+ 995- 996- 997+ 998+ 999+ 1000+ 1001- 1002+ 1003- 1004- 1005- 1006+ 1007- 1008+ 1009+ 1010- 1011+ 1012+ 1013- 1014- 1015+ 1016- 1017- 1018+ 1019+ 1020+ 1021- 1022- 1023- 1024+ 1025- 1026- 1027+ 1028- 1029+ 1030+ 1031+ 1032- 1033+ 1034- 1035+ 1036+ 1037- 1038+ 1039- 1040- 1041- 1042+ 1043- 1044- 1045+ 1046+ 1047+ 1048+ 1049- 1050- 1051+ 1052- 1053+ 1054+ 1055- 1056- 1057+ 1058- 1059+ 1060+ 1061- 1062+ 1063- 1064- 1065+ 1066- 1067+ 1068+ 1069- 1070- 1071- 1072+ 1073- 1074- 1075+ 1076+ 1077- 1078+ 1079- 1080+ 1081+ 1082- 1083+ 1084- 1085- 1086- 1087+ 1088- 1089+ 1090+ 1091+ 1092+ 1093- 1094+ 1095- 1096- 1097- 1098- 1099+ 1100- 1101+ 1102+ 1103- 1104+ 1105+ 1106- 1107- 1108+ 1109+ 1110- 1111+ 1112+ 1113- 1114+ 1115- 1116- 1117+ 1118- 1119+ 1120+ 1121- 1122+ 1123- 1124-

===Links to other displays and visually displayed information about the sequence===

Here is an alternate formatting. [http://numberwarrior.wordpress.com/files/2009/12/longsequence.png Link].

[http://spreadsheets.google.com/ccc?key=0AkbsKAn5VTtvdGpoOG9xYWlUYTNpa1I0UktEMUsxZmc&hl=en The sequence in groups of 24] (Google Docs format).
[http://go2.wordpress.com/?id=725X1342&site=gowers.wordpress.com&url=http%3A%2F%2Fspreadsheets.google.com%2Fccc%3Fkey%3D0AkbsKAn5VTtvdDdrTDd1YmM3bGZESEFwZWhnSVBZMEE%26hl%3Den Multiples of 2 only].

[http://thomas1111.wordpress.com/2010/01/09/erdos-discrepancy-a-program-to-get-html-tables/ The sequence in groups of 24, also multiples of 8 only] (HTML format).

Here is a Rauzy tree that conveys information about subsequences of the sequence: [http://obryant.wordpress.com/2010/01/09/a-rauzy-tree/ Link].

==Multiplicativity properties of the 1124 sequence==

The 1124 sequence is not weakly multiplicative, but it does appear to be close to a weakly multiplicative sequence. That is, if you are prepared to disregard a few "errors", then the number of distinct HAP-subsequences is 6. One can use this information to identify a quasi-multiplicative sequence that is in some sense close to the 1124 sequence. Some details can be found in the sequences of comments that begin [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4682 here] and [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4709 here].

== Another sequence of length 1124 ==

Here is another sequence of length <math>1124</math>:

0 + + - - + + - + - - + - - - + + - + - - + + + - + - - + -
+ + - - + - + - - + + + - + - - + - + + - - + + - + - + + -
- + - + + - + - - - + + - - + + + - - + - + - - + + - - + -
+ + - + - - - + + - + + + - - - + - + + - + - - - + + + - +
+ - + - - - - + + + + - - + - - - + + - + + - + + - + - + -
- + - - - + + + + - - - - + + + - - + - + + - + - - + + - +
- - + - + - + - - + + - - + + - - + + - + + - + + - - - - +
- + + - + + - - + + - + - - + + + - + - - + - - - + + - - +
+ - + + - + - - + + - + - - + - - + + - + + + - - - + + - -
+ + + - - + - + - - - + + - + - - + + - + - - + + + + - - +
- - + + - - - + + + - - + - + + - + - - + - - + + - + - - +
+ - + + - + - + + - - + - - + + - - + - + + - + - - + + + -
- + - - - + + - + + - + - - + + - + + - + - - - + - + - - +
- + + - - + + + - - - + + - + - - + - + + + - - + - + + - +
+ - + - - + - - + + - + - + + - + - + - + - - + - - + + - +
+ - + - - + + - - + - - - + + + - - - + + - - + + - + + + -
- - + + - + - + + - - - + - + - - + - + + + - - + - + - - +
+ - + + - + - - + + - + - + - - + - + - + + - + - - + - - +
+ + - + - - + + + - - + - - + + - + + - + - - + + - + - - +
- - + + - + - - + - - + + - + + + - - - + + - + - - + + - +
+ - + - - + - - + + - + + - + - - + + - + - - + - - + + - +
- - + - - + + - + + - - - + + + - + + - + - - - - + + - - +
- + + + - + + - + - - + + - + - - + - - + + - + - - + + - -
+ - + + - + - - + + - - + - + - - + + + + - - + - - + - - +
+ - + - - + + - + + - + - - + + - + - + + - - - + - + + - +
- - + + - + - + + - - - + + + - - + - - + + - - + - + - - +
+ - + + - + + - + - - + - + + - - - + - + + - + - - + + - +
- - + + - + + - + - - - - + + + - + + - + - - + + - + - - +
- - + + - + - - + - - + + - + + - + - - + + - + + - + - - +
+ - - - - + - + + - - + + + + - - - + - + + - + - - + + - +
- - + - - + + - + + - + - - + + + - - + + - - + + - + - - +
- - - + + + + - + - - - + - + + - + - - + + - + - + + - - -
+ + - + + - - - + + - + - + + - - - + - + + + - - - + + - +
+ - + - - + + - - - - + - + + + - + - - + - - + + - + + - -
- + + + - + + - + - - - + - + - - + - + + + - + + - - - - +
+ + + - - - + - + - - + - + + + + - - - + + - + - - + + - +
- + + - - + + - + - - - - + + + + - + - + - - + - - + + - +
+ - - + - + + - + - - + - + +

[http://thomas1111.wordpress.com/2010/01/09/tables-for-the-second-1124-sequence/ The sequence divided into groups of 24, and also with multiples of 8 only].

== Varying the upper and lower bounds ==

If <math>N(a,b)</math> is the maximum length of a <math>\pm 1</math> sequence with the partial sums along its HAPs bounded below by <math>-a</math> and above by <math>b</math>, then:

<math>N(a, b) = N(b, a)</math>

<math>N(0, b) = b</math> (everything must be <math>+1</math>)

<math>N(1, 1) = 11</math> (there are <math>4</math> such sequences: choose <math>x_1</math>, and use the constraints <math>x_n +x_{2n} = 0</math> and <math>x_1 + \ldots + x_{2n} = 0</math> to determine the entries up to <math>10</math>; then choose <math>x_{11}</math>)

<math>N(1, 2) = 41</math> (there are <math>4</math> such sequences -- example below)

0 + - - + - + + - + + - - +
- + + - + - - - + - + + - -
+ + - + - + + - - - + - + +

<math>N(1, 3) = 83</math> (there are <math>216</math> such sequences -- example below)

0 - + - + - + + - + + - - +
- + + - - + - - + - + + - +
+ + - - - + + - + + - - + -
+ - - + + - - + - + + + - +
- - - + + - + - + - - + - +
+ - + + - - + - + - - - + +

<math>N(1, 4) = 131</math> (there are <math>87144</math> such sequences -- example below)

0 + - - + - + + - - + -
+ + - + + + + - - - + -
- + - + + - - + + + - -
- + + - + - + - - + + -
+ + - - + - - + - + + +
+ - - + - + - - + + + -
+ - - - - + + - - - + +
- - + + + + - - - - + +
- + - + + + + - - + + -
+ - - - + + - - - + - -
+ + + - + + - - + - - +

<math>N(2, 2) \geq 1124</math>

--[[User:Alec|Alec]] 13:46, 9 January 2010 (UTC)

== Geometric variations ==

It has been pointed out that the problem can be generalized to higher dimensions, for example by considering sequences with <math>\Vert x_n \Vert_2 = 1</math> having partial sums lying within a sphere. It is difficult to do much in the way of computation when the <math>x_n</math> can vary continuously, but if they are restricted to some finite set the problem becomes purely combinatorial and one can do more.

=== The seven-point hexagon ===

If the <math>x_n</math> are allowed to be any of the six points of a regular hexagon, and one requires all sums along HAPs to be zero or one of those same points, the maximum length of a sequence is <math>116</math>. The following sequence achieves this, where the numbers index the points in order around the hexagon:

-, 0, 3, 3, 0, 3, 0, 2, 4, 0, 1, 4, 3,
1, 5, 0, 2, 4, 3, 1, 5, 5, 1, 4, 1, 2,
5, 3, 2, 5, 4, 1, 4, 1, 2, 5, 0, 2, 4,
4, 1, 5, 2, 1, 4, 3, 1, 5, 5, 2, 4, 1,
2, 4, 0, 1, 5, 4, 1, 4, 2, 2, 4, 0, 1,
5, 4, 2, 5, 1, 2, 4, 3, 1, 5, 5, 1, 4,
1, 3, 4, 2, 1, 0, 4, 1, 4, 3, 1, 5, 0,
2, 4, 5, 2, 4, 1, 1, 5, 3, 2, 0, 3, 4,
5, 1, 1, 3, 4, 1, 5, 0, 2, 4, 1, 3, 5

Here are some of its HAP subseqences, which seem to show the presence of some kind of multiplicative structure, though the structure seems to degenerate as the sequences progress.

====The 2-sequence====

3, 0, 0, 4, 1, 3, 5, 2, 3, 5, 1, 1, 5, 2, 4, 4, 2, 0, 4, 1, 2, 4, 1, 5, 4, 2, 0, 5, 1, 2, 4, 1, 4, 5, 2, 3, 5, 1, 1, 4, 1, 4, 4, 1, 0, 4, 2, 1, 5, 2, 3, 5, 1, 4, 5, 2, 1, 5

====The 3-sequence====

3, 0, 0, 3, 0, 3, 5, 1, 3, 4, 1, 0, 4, 2, 3, 5, 1, 0, 4, 2, 0, 4, 1, 3, 5, 1, 2, 4, 3, 0, 5, 1, 3, 3, 1, 4, 0, 1

====The 4-sequence====

0, 4, 3, 2, 5, 1, 2, 4, 0, 1, 4, 5, 2, 5, 2, 1, 5, 3, 1, 4, 4, 1, 4, 1, 2, 5, 4, 2, 5

====The 5-sequence====

3, 1, 0, 5, 2, 4, 5, 1, 3, 4, 1, 2, 5, 2, 5, 4, 1, 0, 4, 2, 1, 5, 3

====The 6-sequence====

0, 3, 3, 1, 4, 0, 2, 5, 0, 2, 4, 3, 1, 4, 0, 1, 3, 4, 1

====The 8-sequence====

4, 2, 1, 4, 1, 5, 5, 1, 3, 4, 1, 1, 5, 2

====The 9-sequence====

0, 3, 3, 0, 3, 0, 0, 3, 2, 0, 3, 4

====The 10-sequence====

1, 5, 4, 1, 4, 2, 2, 4, 0, 2, 5

====The 12-sequence====

3, 1, 0, 5, 2, 3, 4, 1, 4

=== The nine-point square ===

If the <math>x_n</math> are allowed to be any of the four points <math>(\pm 1, 0)</math> and <math>(0, \pm 1)</math>, and one requires all sums along HAPs to belong to one of the nine points at unit spacing centred on the origin, the maximum length of a sequence is ''at least'' <math>314</math>. The following sequence achieves this:

(1, 0), (0, 1), (-1, 0), (1, 0), (-1, 0), (0, -1), (0, 1), (-1, 0), (1, 0), (1, 0), (0, -1), (-1, 0), (-1, 0), (0, -1), (1, 0), (1, 0), (-1, 0), (0, 1), (1, 0), (-1, 0), (0, -1), (-1, 0), (0, 1), (1, 0), (0, -1), (1, 0), (-1, 0), (0, 1), (1, 0), (-1, 0), (0, -1), (-1, 0), (0, 1), (0, -1), (1, 0), (1, 0), (-1, 0), (-1, 0), (1, 0), (1, 0), (0, 1), (0, 1), (-1, 0), (0, -1), (-1, 0), (1, 0), (1, 0), (-1, 0), (0, -1), (0, 1), (1, 0), (-1, 0), (1, 0), (0, -1), (0, 1), (0, -1), (-1, 0), (0, 1), (-1, 0), (1, 0), (0, -1), (-1, 0), (0, 1), (1, 0), (0, -1), (1, 0), (0, 1), (0, 1), (0, -1), (-1, 0), (1, 0), (-1, 0), (-1, 0), (1, 0), (0, 1), (1, 0), (-1, 0), (-1, 0), (1, 0), (-1, 0), (1, 0), (0, -1), (0, 1), (0, -1), (0, -1), (0, 1), (-1, 0), (0, 1), (0, -1), (1, 0), (1, 0), (-1, 0), (0, 1), (0, -1), (0, 1), (1, 0), (0, -1), (0, 1), (0, -1), (0, -1), (0, 1), (0, 1), (-1, 0), (1, 0), (-1, 0), (0, -1), (1, 0), (-1, 0), (-1, 0), (0, -1), (1, 0), (0, 1), (-1, 0), (1, 0), (1, 0), (0, -1), (-1, 0), (0, 1), (1, 0), (-1, 0), (1, 0), (-1, 0), (0, -1), (1, 0), (0, 1), (0, -1), (-1, 0), (-1, 0), (1, 0), (0, 1), (0, -1), (0, 1), (-1, 0), (1, 0), (1, 0), (0, -1), (0, 1), (-1, 0), (0, -1), (1, 0), (-1, 0), (0, 1), (0, 1), (1, 0), (-1, 0), (0, -1), (0, 1), (-1, 0), (1, 0), (0, -1), (1, 0), (-1, 0), (-1, 0), (1, 0), (0, 1), (1, 0), (0, -1), (-1, 0), (-1, 0), (1, 0), (0, -1), (0, 1), (0, 1), (-1, 0), (0, -1), (0, -1), (0, 1), (0, 1), (0, -1), (0, 1), (1, 0), (1, 0), (-1, 0), (0, -1), (-1, 0), (0, -1), (1, 0), (0, 1), (1, 0), (-1, 0), (0, -1), (-1, 0), (0, 1), (1, 0), (0, -1), (1, 0), (0, 1), (0, 1), (0, -1), (0, -1), (0, 1), (-1, 0), (-1, 0), (1, 0), (0, 1), (0, -1), (1, 0), (-1, 0), (0, -1), (0, 1), (0, -1), (-1, 0), (0, 1), (0, -1), (1, 0), (1, 0), (0, 1), (-1, 0), (0, -1), (1, 0), (0, 1), (0, 1), (-1, 0), (-1, 0), (0, -1), (1, 0), (-1, 0), (1, 0), (1, 0), (0, 1), (0, -1), (-1, 0), (0, 1), (0, -1), (0, -1), (1, 0), (0, 1), (-1, 0), (1, 0), (-1, 0), (1, 0), (0, 1), (-1, 0), (1, 0), (-1, 0), (1, 0), (-1, 0), (-1, 0), (1, 0), (1, 0), (0, -1), (-1, 0), (-1, 0), (0, -1), (1, 0), (0, 1), (0, 1), (-1, 0), (0, -1), (0, -1), (1, 0), (0, 1), (-1, 0), (1, 0), (0, 1), (1, 0), (-1, 0), (-1, 0), (1, 0), (-1, 0), (1, 0), (1, 0), (-1, 0), (0, -1), (0, -1), (0, 1), (0, 1), (0, -1), (1, 0), (-1, 0), (-1, 0), (0, 1), (0, -1), (1, 0), (0, 1), (1, 0), (-1, 0), (-1, 0), (0, -1), (0, -1), (1, 0), (0, 1), (0, -1), (-1, 0), (1, 0), (1, 0), (0, 1), (-1, 0), (0, -1), (1, 0), (-1, 0), (1, 0), (0, 1), (0, -1), (-1, 0), (0, 1), (0, 1), (0, -1), (1, 0), (0, 1), (-1, 0), (-1, 0), (1, 0), (0, -1), (0, 1), (1, 0), (-1, 0), (-1, 0), (0, -1), (1, 0), (1, 0), (0, -1), (0, 1), (0, 1)

--[[User:Alec|Alec]] 14:42, 9 January 2010 (UTC)

Experimental results

2010-01-09T16:50:17Z

Jbd: /* Links to other displays and visually displayed information about the sequence */

To return to the main Polymath5 page, click [[The Erdős discrepancy problem|here]].

==Different displays of a long low-discrepancy sequence==

It would be good to have some long sequences with low discrepancy displayed nicely here (or on subsidiary pages) in tabular form, together with discussions about the structures that can be found in these sequences. For now, here is the current record not displayed in a particularly helpful way. It is a sequence of discrepancy 2 and length 1124.

===The raw sequence===

+ - + + - - - - + + - + + + - - + - + + - - - + - + + - + -
- + + - + - - + - - + + - + + - - - + + + - - + - + - - + +
+ + - - + - - + + - - + + + - - + + - + - - + - - + + - + -
- + - + + + - - + - + - - + + - - - + + - + + + - - - + - -
+ - + + + + - - - - + + - + + + - - + - - + - - + - + - + +
- + + + - - - - + + + + - - - + + - + - - + - + + - - + - +
+ - + - + - + + - - - + + - + + - - + - - + - - + + + + - +
- - + - - + + - - + - + + - - - + - + + - + + + - - + + - -
+ - - + - + + - - + - + + - + + - - + - - + - + + - - + + -
+ - + - - + - + + + - - + - + + - - - - + + - - + - + + - +
+ - - + - + + - - + + - + - - + - + - - + + - - + - + + + -
+ - - + - + - - + + + + - - - + + - + - - + - + + - - + - +
+ - + + - - + - - + - + + - - + + - - - + + - - + - + + - +
+ - - + - - + + + + - - - - + + - + + - + + - - + - - + + -
- - + + - + + - - + - + + - + - - - + - + + - + + - - + - -
+ - + + - - + + - + + + - - - + - + + - + + - - + - - - + +
+ - - + - + - - + + + - + - + + - + - - - + + - + - + + - -
+ - - + - + + - - + - + - + + - + - + - - + - + + - - + - -
+ - + + + - - - + + - + + + - + - - + - + + - - + - + + - +
- - - + - + + - + + - - + - - - + + + - - + - + + - - + + -
+ - + + - + - - - + - - + - + + - - + - + + - + + - - + - +
+ - + + - - + - - + - + + - - + - - + - + + - + - - + + - +
+ - - + - - + - + + - - + - + + - + + - - + - + + - - + + -
+ - - + - + + - - + - - + - + + - - + - + + - + + - + + - -
+ - + + - - + - - + - + + - - + - + - - + + + - + - - + - +
+ - - + - + - - + + + - - - + + - + + - - + + - + - + + - -
+ - - + - - + - + + - + - - + + + - + - - + - + + - - + - +
- - + + - - + - + + - + + - - + + - + - - + - - + - + + - +
+ - - + - + + - + + - - + - - + + + - - - + - - + - + + - -
+ + + + - + - - + + - - + - + + - - + - - + - + + - - + - +
+ - + + - - + - - + - + + - - - + + - - + + - - + - + + - +
+ - - + - - + - + + + - + - - + - + + - - + - + + - + - + -
- + - + - + + - - + - + - - + + + - + - - - + + + - - + - -
+ - + + - - + + + + - + - - - + - + + - + + - - + - - + + +
- - - + - - + - + + + - + - + + - + - - - + - - + + + + - -
+ - + + - - + - + + - + - - + - + + - - - + - - + + - + - +
+ - + - - - + - + + + + - + - - - - + - + + - + + - - + + -
+ + - + - - + - + + - + - -

===The sequence, together with the corresponding integers===

1+ 2- 3+ 4+ 5- 6- 7- 8- 9+ 10+ 11- 12+ 13+ 14+ 15- 16- 17+ 18- 19+ 20+ 21- 22- 23- 24+ 25- 26+ 27+ 28- 29+ 30- 31- 32+ 33+ 34- 35+ 36- 37- 38+ 39- 40- 41+ 42+ 43- 44+ 45+ 46- 47- 48- 49+ 50+ 51+ 52- 53- 54+ 55- 56+ 57- 58- 59+ 60+ 61+ 62+ 63- 64- 65+ 66- 67- 68+ 69+ 70- 71- 72+ 73+ 74+ 75- 76- 77+ 78+ 79- 80+ 81- 82- 83+ 84- 85- 86+ 87+ 88- 89+ 90- 91- 92+ 93- 94+ 95+ 96+ 97- 98- 99+ 100- 101+ 102- 103- 104+ 105+ 106- 107- 108- 109+ 110+ 111- 112+ 113+ 114+ 115- 116- 117- 118+ 119- 120- 121+ 122- 123+ 124+ 125+ 126+ 127- 128- 129- 130- 131+ 132+ 133- 134+ 135+ 136+ 137- 138- 139+ 140- 141- 142+ 143- 144- 145+ 146- 147+ 148- 149+ 150+ 151- 152+ 153+ 154+ 155- 156- 157- 158- 159+ 160+ 161+ 162+ 163- 164- 165- 166+ 167+ 168- 169+ 170- 171- 172+ 173- 174+ 175+ 176- 177- 178+ 179- 180+ 181+ 182- 183+ 184- 185+ 186- 187+ 188+ 189- 190- 191- 192+ 193+ 194- 195+ 196+ 197- 198- 199+ 200- 201- 202+ 203- 204- 205+ 206+ 207+ 208+ 209- 210+ 211- 212- 213+ 214- 215- 216+ 217+ 218- 219- 220+ 221- 222+ 223+ 224- 225- 226- 227+ 228- 229+ 230+ 231- 232+ 233+ 234+ 235- 236- 237+ 238+ 239- 240- 241+ 242- 243- 244+ 245- 246+ 247+ 248- 249- 250+ 251- 252+ 253+ 254- 255+ 256+ 257- 258- 259+ 260- 261- 262+ 263- 264+ 265+ 266- 267- 268+ 269+ 270- 271+ 272- 273+ 274- 275- 276+ 277- 278+ 279+ 280+ 281- 282- 283+ 284- 285+ 286+ 287- 288- 289- 290- 291+ 292+ 293- 294- 295+ 296- 297+ 298+ 299- 300+ 301+ 302- 303- 304+ 305- 306+ 307+ 308- 309- 310+ 311+ 312- 313+ 314- 315- 316+ 317- 318+ 319- 320- 321+ 322+ 323- 324- 325+ 326- 327+ 328+ 329+ 330- 331+ 332- 333- 334+ 335- 336+ 337- 338- 339+ 340+ 341+ 342+ 343- 344- 345- 346+ 347+ 348- 349+ 350- 351- 352+ 353- 354+ 355+ 356- 357- 358+ 359- 360+ 361+ 362- 363+ 364+ 365- 366- 367+ 368- 369- 370+ 371- 372+ 373+ 374- 375- 376+ 377+ 378- 379- 380- 381+ 382+ 383- 384- 385+ 386- 387+ 388+ 389- 390+ 391+ 392- 393- 394+ 395- 396- 397+ 398+ 399+ 400+ 401- 402- 403- 404- 405+ 406+ 407- 408+ 409+ 410- 411+ 412+ 413- 414- 415+ 416- 417- 418+ 419+ 420- 421- 422- 423+ 424+ 425- 426+ 427+ 428- 429- 430+ 431- 432+ 433+ 434- 435+ 436- 437- 438- 439+ 440- 441+ 442+ 443- 444+ 445+ 446- 447- 448+ 449- 450- 451+ 452- 453+ 454+ 455- 456- 457+ 458+ 459- 460+ 461+ 462+ 463- 464- 465- 466+ 467- 468+ 469+ 470- 471+ 472+ 473- 474- 475+ 476- 477- 478- 479+ 480+ 481+ 482- 483- 484+ 485- 486+ 487- 488- 489+ 490+ 491+ 492- 493+ 494- 495+ 496+ 497- 498+ 499- 500- 501- 502+ 503+ 504- 505+ 506- 507+ 508+ 509- 510- 511+ 512- 513- 514+ 515- 516+ 517+ 518- 519- 520+ 521- 522+ 523- 524+ 525+ 526- 527+ 528- 529+ 530- 531- 532+ 533- 534+ 535+ 536- 537- 538+ 539- 540- 541+ 542- 543+ 544+ 545+ 546- 547- 548- 549+ 550+ 551- 552+ 553+ 554+ 555- 556+ 557- 558- 559+ 560- 561+ 562+ 563- 564- 565+ 566- 567+ 568+ 569- 570+ 571- 572- 573- 574+ 575- 576+ 577+ 578- 579+ 580+ 581- 582- 583+ 584- 585- 586- 587+ 588+ 589+ 590- 591- 592+ 593- 594+ 595+ 596- 597- 598+ 599+ 600- 601+ 602- 603+ 604+ 605- 606+ 607- 608- 609- 610+ 611- 612- 613+ 614- 615+ 616+ 617- 618- 619+ 620- 621+ 622+ 623- 624+ 625+ 626- 627- 628+ 629- 630+ 631+ 632- 633+ 634+ 635- 636- 637+ 638- 639- 640+ 641- 642+ 643+ 644- 645- 646+ 647- 648- 649+ 650- 651+ 652+ 653- 654+ 655- 656- 657+ 658+ 659- 660+ 661+ 662- 663- 664+ 665- 666- 667+ 668- 669+ 670+ 671- 672- 673+ 674- 675+ 676+ 677- 678+ 679+ 680- 681- 682+ 683- 684+ 685+ 686- 687- 688+ 689+ 690- 691+ 692- 693- 694+ 695- 696+ 697+ 698- 699- 700+ 701- 702- 703+ 704- 705+ 706+ 707- 708- 709+ 710- 711+ 712+ 713- 714+ 715+ 716- 717+ 718+ 719- 720- 721+ 722- 723+ 724+ 725- 726- 727+ 728- 729- 730+ 731- 732+ 733+ 734- 735- 736+ 737- 738+ 739- 740- 741+ 742+ 743+ 744- 745+ 746- 747- 748+ 749- 750+ 751+ 752- 753- 754+ 755- 756+ 757- 758- 759+ 760+ 761+ 762- 763- 764- 765+ 766+ 767- 768+ 769+ 770- 771- 772+ 773+ 774- 775+ 776- 777+ 778+ 779- 780- 781+ 782- 783- 784+ 785- 786- 787+ 788- 789+ 790+ 791- 792+ 793- 794- 795+ 796+ 797+ 798- 799+ 800- 801- 802+ 803- 804+ 805+ 806- 807- 808+ 809- 810+ 811- 812- 813+ 814+ 815- 816- 817+ 818- 819+ 820+ 821- 822+ 823+ 824- 825- 826+ 827+ 828- 829+ 830- 831- 832+ 833- 834- 835+ 836- 837+ 838+ 839- 840+ 841+ 842- 843- 844+ 845- 846+ 847+ 848- 849+ 850+ 851- 852- 853+ 854- 855- 856+ 857+ 858+ 859- 860- 861- 862+ 863- 864- 865+ 866- 867+ 868+ 869- 870- 871+ 872+ 873+ 874+ 875- 876+ 877- 878- 879+ 880+ 881- 882- 883+ 884- 885+ 886+ 887- 888- 889+ 890- 891- 892+ 893- 894+ 895+ 896- 897- 898+ 899- 900+ 901+ 902- 903+ 904+ 905- 906- 907+ 908- 909- 910+ 911- 912+ 913+ 914- 915- 916- 917+ 918+ 919- 920- 921+ 922+ 923- 924- 925+ 926- 927+ 928+ 929- 930+ 931+ 932- 933- 934+ 935- 936- 937+ 938- 939+ 940+ 941+ 942- 943+ 944- 945- 946+ 947- 948+ 949+ 950- 951- 952+ 953- 954+ 955+ 956- 957+ 958- 959+ 960- 961- 962+ 963- 964+ 965- 966+ 967+ 968- 969- 970+ 971- 972+ 973- 974- 975+ 976+ 977+ 978- 979+ 980- 981- 982- 983+ 984+ 985+ 986- 987- 988+ 989- 990- 991+ 992- 993+ 994+ 995- 996- 997+ 998+ 999+ 1000+ 1001- 1002+ 1003- 1004- 1005- 1006+ 1007- 1008+ 1009+ 1010- 1011+ 1012+ 1013- 1014- 1015+ 1016- 1017- 1018+ 1019+ 1020+ 1021- 1022- 1023- 1024+ 1025- 1026- 1027+ 1028- 1029+ 1030+ 1031+ 1032- 1033+ 1034- 1035+ 1036+ 1037- 1038+ 1039- 1040- 1041- 1042+ 1043- 1044- 1045+ 1046+ 1047+ 1048+ 1049- 1050- 1051+ 1052- 1053+ 1054+ 1055- 1056- 1057+ 1058- 1059+ 1060+ 1061- 1062+ 1063- 1064- 1065+ 1066- 1067+ 1068+ 1069- 1070- 1071- 1072+ 1073- 1074- 1075+ 1076+ 1077- 1078+ 1079- 1080+ 1081+ 1082- 1083+ 1084- 1085- 1086- 1087+ 1088- 1089+ 1090+ 1091+ 1092+ 1093- 1094+ 1095- 1096- 1097- 1098- 1099+ 1100- 1101+ 1102+ 1103- 1104+ 1105+ 1106- 1107- 1108+ 1109+ 1110- 1111+ 1112+ 1113- 1114+ 1115- 1116- 1117+ 1118- 1119+ 1120+ 1121- 1122+ 1123- 1124-

===Links to other displays and visually displayed information about the sequence===

Here is an alternate formatting. [http://numberwarrior.wordpress.com/files/2009/12/longsequence.png Link].

[http://spreadsheets.google.com/ccc?key=0AkbsKAn5VTtvdGpoOG9xYWlUYTNpa1I0UktEMUsxZmc&hl=en The sequence in groups of 24] (Google Docs format).
[http://go2.wordpress.com/?id=725X1342&site=gowers.wordpress.com&url=http%3A%2F%2Fspreadsheets.google.com%2Fccc%3Fkey%3D0AkbsKAn5VTtvdDdrTDd1YmM3bGZESEFwZWhnSVBZMEE%26hl%3Den Multiples of 2 only].

[http://thomas1111.wordpress.com/2010/01/09/erdos-discrepancy-a-program-to-get-html-tables/ The sequence in groups of 24, also multiples of 8 only] (HTML format).

Here is a Rauzy tree that conveys information about subsequences of the sequence: [http://obryant.wordpress.com/2010/01/09/a-rauzy-tree/ Link].

==Multiplicativity properties of the 1124 sequence==

The 1124 sequence is not weakly multiplicative, but it does appear to be close to a weakly multiplicative sequence. That is, if you are prepared to disregard a few "errors", then the number of distinct HAP-subsequences is 6. One can use this information to identify a quasi-multiplicative sequence that is in some sense close to the 1124 sequence. Some details can be found in the sequences of comments that begin [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4682 here] and [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4709 here].

== Another sequence of length 1124 ==

Here is another sequence of length <math>1124</math>:

0 + + - - + + - + - - + - - - + + - + - - + + + - + - - + -
+ + - - + - + - - + + + - + - - + - + + - - + + - + - + + -
- + - + + - + - - - + + - - + + + - - + - + - - + + - - + -
+ + - + - - - + + - + + + - - - + - + + - + - - - + + + - +
+ - + - - - - + + + + - - + - - - + + - + + - + + - + - + -
- + - - - + + + + - - - - + + + - - + - + + - + - - + + - +
- - + - + - + - - + + - - + + - - + + - + + - + + - - - - +
- + + - + + - - + + - + - - + + + - + - - + - - - + + - - +
+ - + + - + - - + + - + - - + - - + + - + + + - - - + + - -
+ + + - - + - + - - - + + - + - - + + - + - - + + + + - - +
- - + + - - - + + + - - + - + + - + - - + - - + + - + - - +
+ - + + - + - + + - - + - - + + - - + - + + - + - - + + + -
- + - - - + + - + + - + - - + + - + + - + - - - + - + - - +
- + + - - + + + - - - + + - + - - + - + + + - - + - + + - +
+ - + - - + - - + + - + - + + - + - + - + - - + - - + + - +
+ - + - - + + - - + - - - + + + - - - + + - - + + - + + + -
- - + + - + - + + - - - + - + - - + - + + + - - + - + - - +
+ - + + - + - - + + - + - + - - + - + - + + - + - - + - - +
+ + - + - - + + + - - + - - + + - + + - + - - + + - + - - +
- - + + - + - - + - - + + - + + + - - - + + - + - - + + - +
+ - + - - + - - + + - + + - + - - + + - + - - + - - + + - +
- - + - - + + - + + - - - + + + - + + - + - - - - + + - - +
- + + + - + + - + - - + + - + - - + - - + + - + - - + + - -
+ - + + - + - - + + - - + - + - - + + + + - - + - - + - - +
+ - + - - + + - + + - + - - + + - + - + + - - - + - + + - +
- - + + - + - + + - - - + + + - - + - - + + - - + - + - - +
+ - + + - + + - + - - + - + + - - - + - + + - + - - + + - +
- - + + - + + - + - - - - + + + - + + - + - - + + - + - - +
- - + + - + - - + - - + + - + + - + - - + + - + + - + - - +
+ - - - - + - + + - - + + + + - - - + - + + - + - - + + - +
- - + - - + + - + + - + - - + + + - - + + - - + + - + - - +
- - - + + + + - + - - - + - + + - + - - + + - + - + + - - -
+ + - + + - - - + + - + - + + - - - + - + + + - - - + + - +
+ - + - - + + - - - - + - + + + - + - - + - - + + - + + - -
- + + + - + + - + - - - + - + - - + - + + + - + + - - - - +
+ + + - - - + - + - - + - + + + + - - - + + - + - - + + - +
- + + - - + + - + - - - - + + + + - + - + - - + - - + + - +
+ - - + - + + - + - - + - + +

== Varying the upper and lower bounds ==

If <math>N(a,b)</math> is the maximum length of a <math>\pm 1</math> sequence with the partial sums along its HAPs bounded below by <math>-a</math> and above by <math>b</math>, then:

<math>N(a, b) = N(b, a)</math>

<math>N(0, b) = b</math> (everything must be <math>+1</math>)

<math>N(1, 1) = 11</math> (there are <math>4</math> such sequences: choose <math>x_1</math>, and use the constraints <math>x_n +x_{2n} = 0</math> and <math>x_1 + \ldots + x_{2n} = 0</math> to determine the entries up to <math>10</math>; then choose <math>x_{11}</math>)

<math>N(1, 2) = 41</math> (there are <math>4</math> such sequences -- example below)

0 + - - + - + + - + + - - +
- + + - + - - - + - + + - -
+ + - + - + + - - - + - + +

<math>N(1, 3) = 83</math> (there are <math>216</math> such sequences -- example below)

0 - + - + - + + - + + - - +
- + + - - + - - + - + + - +
+ + - - - + + - + + - - + -
+ - - + + - - + - + + + - +
- - - + + - + - + - - + - +
+ - + + - - + - + - - - + +

<math>N(1, 4) = 131</math> (there are <math>87144</math> such sequences -- example below)

0 + - - + - + + - - + -
+ + - + + + + - - - + -
- + - + + - - + + + - -
- + + - + - + - - + + -
+ + - - + - - + - + + +
+ - - + - + - - + + + -
+ - - - - + + - - - + +
- - + + + + - - - - + +
- + - + + + + - - + + -
+ - - - + + - - - + - -
+ + + - + + - - + - - +

<math>N(2, 2) \geq 1124</math>

--[[User:Alec|Alec]] 13:46, 9 January 2010 (UTC)

== Geometric variations ==

It has been pointed out that the problem can be generalized to higher dimensions, for example by considering sequences with <math>\Vert x_n \Vert_2 = 1</math> having partial sums lying within a sphere. It is difficult to do much in the way of computation when the <math>x_n</math> can vary continuously, but if they are restricted to some finite set the problem becomes purely combinatorial and one can do more.

=== The seven-point hexagon ===

If the <math>x_n</math> are allowed to be any of the six points of a regular hexagon, and one requires all sums along HAPs to be zero or one of those same points, the maximum length of a sequence is <math>116</math>. The following sequence achieves this, where the numbers index the points in order around the hexagon:

-, 0, 3, 3, 0, 3, 0, 2, 4, 0, 1, 4, 3,
1, 5, 0, 2, 4, 3, 1, 5, 5, 1, 4, 1, 2,
5, 3, 2, 5, 4, 1, 4, 1, 2, 5, 0, 2, 4,
4, 1, 5, 2, 1, 4, 3, 1, 5, 5, 2, 4, 1,
2, 4, 0, 1, 5, 4, 1, 4, 2, 2, 4, 0, 1,
5, 4, 2, 5, 1, 2, 4, 3, 1, 5, 5, 1, 4,
1, 3, 4, 2, 1, 0, 4, 1, 4, 3, 1, 5, 0,
2, 4, 5, 2, 4, 1, 1, 5, 3, 2, 0, 3, 4,
5, 1, 1, 3, 4, 1, 5, 0, 2, 4, 1, 3, 5

Here are some of its HAP subseqences, which seem to show the presence of some kind of multiplicative structure, though the structure seems to degenerate as the sequences progress.

====The 2-sequence====

3, 0, 0, 4, 1, 3, 5, 2, 3, 5, 1, 1, 5, 2, 4, 4, 2, 0, 4, 1, 2, 4, 1, 5, 4, 2, 0, 5, 1, 2, 4, 1, 4, 5, 2, 3, 5, 1, 1, 4, 1, 4, 4, 1, 0, 4, 2, 1, 5, 2, 3, 5, 1, 4, 5, 2, 1, 5

====The 3-sequence====

3, 0, 0, 3, 0, 3, 5, 1, 3, 4, 1, 0, 4, 2, 3, 5, 1, 0, 4, 2, 0, 4, 1, 3, 5, 1, 2, 4, 3, 0, 5, 1, 3, 3, 1, 4, 0, 1

====The 4-sequence====

0, 4, 3, 2, 5, 1, 2, 4, 0, 1, 4, 5, 2, 5, 2, 1, 5, 3, 1, 4, 4, 1, 4, 1, 2, 5, 4, 2, 5

====The 5-sequence====

3, 1, 0, 5, 2, 4, 5, 1, 3, 4, 1, 2, 5, 2, 5, 4, 1, 0, 4, 2, 1, 5, 3

====The 6-sequence====

0, 3, 3, 1, 4, 0, 2, 5, 0, 2, 4, 3, 1, 4, 0, 1, 3, 4, 1

====The 8-sequence====

4, 2, 1, 4, 1, 5, 5, 1, 3, 4, 1, 1, 5, 2

====The 9-sequence====

0, 3, 3, 0, 3, 0, 0, 3, 2, 0, 3, 4

====The 10-sequence====

1, 5, 4, 1, 4, 2, 2, 4, 0, 2, 5

====The 12-sequence====

3, 1, 0, 5, 2, 3, 4, 1, 4

=== The nine-point square ===

If the <math>x_n</math> are allowed to be any of the four points <math>(\pm 1, 0)</math> and <math>(0, \pm 1)</math>, and one requires all sums along HAPs to belong to one of the nine points at unit spacing centred on the origin, the maximum length of a sequence is ''at least'' <math>314</math>. The following sequence achieves this:

(1, 0), (0, 1), (-1, 0), (1, 0), (-1, 0), (0, -1), (0, 1), (-1, 0), (1, 0), (1, 0), (0, -1), (-1, 0), (-1, 0), (0, -1), (1, 0), (1, 0), (-1, 0), (0, 1), (1, 0), (-1, 0), (0, -1), (-1, 0), (0, 1), (1, 0), (0, -1), (1, 0), (-1, 0), (0, 1), (1, 0), (-1, 0), (0, -1), (-1, 0), (0, 1), (0, -1), (1, 0), (1, 0), (-1, 0), (-1, 0), (1, 0), (1, 0), (0, 1), (0, 1), (-1, 0), (0, -1), (-1, 0), (1, 0), (1, 0), (-1, 0), (0, -1), (0, 1), (1, 0), (-1, 0), (1, 0), (0, -1), (0, 1), (0, -1), (-1, 0), (0, 1), (-1, 0), (1, 0), (0, -1), (-1, 0), (0, 1), (1, 0), (0, -1), (1, 0), (0, 1), (0, 1), (0, -1), (-1, 0), (1, 0), (-1, 0), (-1, 0), (1, 0), (0, 1), (1, 0), (-1, 0), (-1, 0), (1, 0), (-1, 0), (1, 0), (0, -1), (0, 1), (0, -1), (0, -1), (0, 1), (-1, 0), (0, 1), (0, -1), (1, 0), (1, 0), (-1, 0), (0, 1), (0, -1), (0, 1), (1, 0), (0, -1), (0, 1), (0, -1), (0, -1), (0, 1), (0, 1), (-1, 0), (1, 0), (-1, 0), (0, -1), (1, 0), (-1, 0), (-1, 0), (0, -1), (1, 0), (0, 1), (-1, 0), (1, 0), (1, 0), (0, -1), (-1, 0), (0, 1), (1, 0), (-1, 0), (1, 0), (-1, 0), (0, -1), (1, 0), (0, 1), (0, -1), (-1, 0), (-1, 0), (1, 0), (0, 1), (0, -1), (0, 1), (-1, 0), (1, 0), (1, 0), (0, -1), (0, 1), (-1, 0), (0, -1), (1, 0), (-1, 0), (0, 1), (0, 1), (1, 0), (-1, 0), (0, -1), (0, 1), (-1, 0), (1, 0), (0, -1), (1, 0), (-1, 0), (-1, 0), (1, 0), (0, 1), (1, 0), (0, -1), (-1, 0), (-1, 0), (1, 0), (0, -1), (0, 1), (0, 1), (-1, 0), (0, -1), (0, -1), (0, 1), (0, 1), (0, -1), (0, 1), (1, 0), (1, 0), (-1, 0), (0, -1), (-1, 0), (0, -1), (1, 0), (0, 1), (1, 0), (-1, 0), (0, -1), (-1, 0), (0, 1), (1, 0), (0, -1), (1, 0), (0, 1), (0, 1), (0, -1), (0, -1), (0, 1), (-1, 0), (-1, 0), (1, 0), (0, 1), (0, -1), (1, 0), (-1, 0), (0, -1), (0, 1), (0, -1), (-1, 0), (0, 1), (0, -1), (1, 0), (1, 0), (0, 1), (-1, 0), (0, -1), (1, 0), (0, 1), (0, 1), (-1, 0), (-1, 0), (0, -1), (1, 0), (-1, 0), (1, 0), (1, 0), (0, 1), (0, -1), (-1, 0), (0, 1), (0, -1), (0, -1), (1, 0), (0, 1), (-1, 0), (1, 0), (-1, 0), (1, 0), (0, 1), (-1, 0), (1, 0), (-1, 0), (1, 0), (-1, 0), (-1, 0), (1, 0), (1, 0), (0, -1), (-1, 0), (-1, 0), (0, -1), (1, 0), (0, 1), (0, 1), (-1, 0), (0, -1), (0, -1), (1, 0), (0, 1), (-1, 0), (1, 0), (0, 1), (1, 0), (-1, 0), (-1, 0), (1, 0), (-1, 0), (1, 0), (1, 0), (-1, 0), (0, -1), (0, -1), (0, 1), (0, 1), (0, -1), (1, 0), (-1, 0), (-1, 0), (0, 1), (0, -1), (1, 0), (0, 1), (1, 0), (-1, 0), (-1, 0), (0, -1), (0, -1), (1, 0), (0, 1), (0, -1), (-1, 0), (1, 0), (1, 0), (0, 1), (-1, 0), (0, -1), (1, 0), (-1, 0), (1, 0), (0, 1), (0, -1), (-1, 0), (0, 1), (0, 1), (0, -1), (1, 0), (0, 1), (-1, 0), (-1, 0), (1, 0), (0, -1), (0, 1), (1, 0), (-1, 0), (-1, 0), (0, -1), (1, 0), (1, 0), (0, -1), (0, 1), (0, 1)

--[[User:Alec|Alec]] 14:42, 9 January 2010 (UTC)

Experimental results

2010-01-09T12:02:15Z

Jbd: /* Links to other displays and visually displayed information about the sequence */

Experimental results

2010-01-09T11:12:01Z

Jbd: /* The sequence, together with the corresponding integers */

Experimental results

2010-01-09T11:01:31Z

Jbd: /* The sequence, together with the corresponding integers */

Experimental results

2010-01-09T03:04:40Z

Jbd:

To return to the main Polymath5 page, click [[The Erdős discrepancy problem|here]].

It would be good to have some long sequences with low discrepancy displayed nicely here (or on subsidiary pages) in tabular form, together with discussions about the structures that can be found in these sequences. For now, here is the current record not displayed in a particularly helpful way. It is a sequence of discrepancy 2 and length 1124.

+ - + + - - - - + + - + + + - - + - + + - - - + - + + - + -
- + + - + - - + - - + + - + + - - - + + + - - + - + - - + +
+ + - - + - - + + - - + + + - - + + - + - - + - - + + - + -
- + - + + + - - + - + - - + + - - - + + - + + + - - - + - -
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- + + + - - - - + + + + - - - + + - + - - + - + + - - + - +
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- - + - - + + - - + - + + - - - + - + + - + + + - - + + - -
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- - - + - + + - + + - - + - - - + + + - - + - + + - - + + -
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Here's a slightly more helpful (to start with -- it would be great if we could all chip in a bit and finish it) display:

1+ 2- 3+ 4+ 5- 6- 7- 8- 9+ 10+ 11- 12+ 13+ 14+ 15- 16- 17+ 18- 19+ 20+ 21- 22- 23- 24+ 25- 26+ 27+ 28- 29+ 30-
31- 32+ 33+ 34- 35+ 36- 37- 38+ 39- 40- 41+ 42+ 43- 44+ 45+ 46- 47- 48- 49+ 50+ 51+ 52- 53- 54+ 55- 56+ 57- 58- 59+ 60+
61+ 62+ 63- 64- 65+ 66- 67- 68+ 69+ 70- 71- 72+ 73+ 74+ 75- 76- 77+ 78+ 79- 80+ 81- 82- 83+ 84- 85- 86+ 87+ 88- 89+ 90-
91- 92+ 93- 94+ 95+ 96+ 97- 98- 99+ 100- 101+ 102- 103- 104+ 105+ 106- 107- 108- 109+ 110+ 111- 112+ 113+ 114+ 115- 116- 117- 118+ 119- 120-
121+ 122- 123+ 124+ 125+ 126+ 127- 128- 129- 130- 131+ 132+ 133- 134+ 135+ 136+ 137- 138- 139+ 140- 141- 142+ 143- 144- 145+ 146- 147+ 148- 149+ 150+
151- 152+ 153+ 154+ 155- 156- 157- 158- 159+ 160+ 161+ 162+ 163- 164- 165- 166+ 167+ 168- 169+ 170- 171- 172+ 173- 174+ 175+ 176- 177- 178+ 179- 180+
181+ 182- 183+ 184- 185+ 186- 187+ 188+ 189- 190- 191- 192+ 193+ 194- 195+ 196+ 197- 198- 199+ 200- 201- 202+ 203- 204- 205+ 206+ 207+ 208+ 209- 210+
211- 212- 213+ 214- 215- 216+ 217+ 218- 219- 220+ 221- 222+ 223+ 224- 225- 226- 227+ 228- 229+ 230+ 231- 232+ 233+ 234+ 235- 236- 237+ 238+ 239- 240-
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+ + - + - - + - + + - + - -

Here is an alternate formatting. [http://numberwarrior.wordpress.com/files/2009/12/longsequence.png Link].

The Erdős discrepancy problem

2010-01-07T14:41:03Z

Jbd: Stub for annotated bibliography

==Introduction and statement of problem==

The Erdős discrepancy problem is a problem that has been around since the 1930s and has resisted attack ever since. It is extremely easy to state, and a very appealing problem, so perhaps Polymath can succeed where individual mathematicians have failed. If so, then, given the notoriety of the problem, it would be a significant coup for the multiply collaborative approach to mathematics. But even if a full solution is not reached, preliminary investigations have already thrown up some interesting ideas, which could perhaps be developed into publishable results.

To state the problem, we first define a homogeneous arithmetic progression to be a set of positive integers of the form {d,2d,3d,...,nd}. (It is also reasonable to define it to be a set of non-negative integers of the form {0,d,2d,3d,...,nd}: at the time of writing it is not clear what the "official" definition should be.) Suppose that we have a sequence <math>x_1,x_2,x_3,\dots</math> of elements of the set <math>\{-1,1\}</math> (henceforth to be referred to as a <math>\pm 1</math> sequence). We say that the discrepancy of the sequence <math>(x_n)</math> with respect to a set A of positive integers is <math>|\sum_{n\in A}x_n|</math>. The reason for this name is that we can think of the sequence as a red/blue-colouring of the positive integers, and the discrepancy with respect to A is then the difference between the number of red elements of A and the number of blue elements of A. The Erdős discrepancy problem is the following question.

Problem. Let <math>(x_n)</math> be a <math>\pm 1</math> sequence. Must it be the case that for every constant C there exists a homogeneous arithmetic progression P such that the discrepancy of the sequence <math>(x_n)</math> with respect to P is at least C?

==Some definitions and notational conventions==

==Simple observations==

It might be good to have links from some of these observations to more detailed discussions of them.

*To answer the problem negatively, it is sufficient to find a completely multiplicative sequence (that is, a sequence <math>(x_n)</math> such that <math>x_{mn}=x_mx_n</math> for every m,n) such that its partial sums <math>x_1+\dots+x_n</math> are bounded.

*The function that takes n to 1 if the last non-zero digit of n in its ternary representation is 1 and -1 if the last non-zero digit is 2 is completely multiplicative and the partial sum up to n is easily shown to be at most <math>\log_3n +1</math>. Therefore, the rate at which the worst discrepancy grows, as a function of the length of the homogeneous progression, can be as slow as logarithmic. It turns out that the base can be made significantly higher than 3, so this example is not best possible.

*To answer the problem negatively, it is also sufficient to find a completely multiplicative function f from <math>\mathbb{N}</math> to a finite Abelian group G (meaning that f(mn)=f(m)+f(n) for every m,n) and a function h from G to {-1,1} such that for each g in G there exists d with f(d)=g and with the partial sums hf(d)+hf(2d)+...+hf(nd) bounded. In that case, one can set <math>x_n=hf(n)</math>. (Though this observation is simple with the benefit of hindsight, it might not have been made had it not been for the fact that this sort of structure was identified in a long sequence that had been produced experimentally. [A link to the appropriate comments on the blog would be useful here.])

*It can be shown that if a <math>\pm 1</math> sequence <math>(x_n)</math> starts with 1 and has only finitely many distinct subsequences of the form <math>(x_d,x_{2d},x_{3d},\dots)</math>, then it must arise in the above way.

*The problem for the positive integers is equivalent to the problem for the positive rationals. (Sketch proof: if you have a function f from the positive integers to {-1,1} that works, then define <math>f_q(x)</math> to be f(q!x) whenever q!x is an integer. Then take a pointwise limit of a subsequence of the functions <math>f_q</math>.)

==Experimental evidence==

A nice feature of the problem is that it lends itself well to investigation on a computer. Although the number of <math>\pm 1</math> sequences grows exponentially, the constraints on these sequences are such that depth-first searches can lead quite quickly to interesting examples. For instance, at the time of writing they have yielded a sequence of length 1124 with discrepancy 2. See [[Experimental results|this page]] for more details about this and for further links.

==Annoted Bibliography==

Mathias, A. R. D. [http://www.dpmms.cam.ac.uk/~ardm/erdoschu.pdf On a conjecture of Erdős and Čudakov]. Combinatorics, geometry and probability (Cambridge, 1993).

This is a short paper establishing the maximal length of sequence for the case where C=1 is 11, and is the starting point for our experimental studies.

Fujimura's problem

2009-03-30T21:45:58Z

Jbd:

Let <math>\overline{c}^\mu_n</math> the largest subset of the triangular grid

:<math>\Delta_n := \{ (a,b,c) \in {\Bbb Z}_+^3: a+b+c=n \}</math>

which contains no equilateral triangles <math>(a+r,b,c), (a,b+r,c), (a,b,c+r)</math> with <math>r > 0</math>; call such sets ''triangle-free''. (It is an interesting variant to also allow negative r, thus allowing "upside-down" triangles, but this does not seem to be as closely connected to DHJ(3).) Fujimura's problem is to compute <math>\overline{c}^\mu_n</math> ([http://www.research.att.com/~njas/sequences/A157795 OEIS A157795]). This quantity is relevant to a certain [[hyper-optimistic conjecture]].

We are also exploring issues raised by [[higher-dimensional Fujimura]].

{|
| n || 0 || 1 || 2 || 3 || 4 || 5 || 6 || 7 || 8 || 9 || 10 || 11 || 12 || 13
|-
| <math>\overline{c}^\mu_n</math> || 1 || 2 || 4 || 6 || 9 || 12 || 15 || 18 || 22 || 26 || 31 || 35 || 40 || 46
|}

== n=0 ==

<math>\overline{c}^\mu_0 = 1</math>:

This is clear.

== n=1 ==

<math>\overline{c}^\mu_1 = 2</math>:

This is clear.

== n=2 ==

<math>\overline{c}^\mu_2 = 4</math>:

This is clear (e.g. remove (0,2,0) and (1,0,1) from <math>\Delta_2</math>).

== n=3 ==

:<math>\overline{c}^\mu_3 = 6</math>:

For the lower bound, delete (0,3,0), (0,2,1), (2,1,0), (1,0,2) from <math>\Delta_3</math>.

For the upper bound: observe that with only three removals each of these (non-overlapping) triangles must have one removal:

* set A: (0,3,0) (0,2,1) (1,2,0)
* set B: (0,1,2) (0,0,3) (1,0,2)
* set C: (2,1,0) (2,0,1) (3,0,0)

Consider choices from set A:

* (0,3,0) leaves triangle (0,2,1) (1,2,0) (1,1,1)
* (0,2,1) forces a second removal at (2,1,0) [otherwise there is triangle at (1,2,0) (1,1,1) (2,1,0)] but then none of the choices for third removal work
* (1,2,0) is symmetrical with (0,2,1)

== n=4 ==

:<math>\overline{c}^\mu_4=9</math>:

The set of all <math>(a,b,c)</math> in <math>\Delta_4</math> with exactly one of a,b,c =0, has 9 elements and is triangle-free.
(Note that it does contain the equilateral triangle (2,2,0),(2,0,2),(0,2,2), so would not qualify for the generalised version of Fujimura's problem in which <math>r</math> is allowed to be negative.)

Let <math>S\subset \Delta_4</math> be a set without equilateral triangles. If <math>(0,0,4)\in S</math>, there can only be one of <math>(0,x,4-x)</math> and <math>(x,0,4-x)</math> in S for <math>x=1,2,3,4</math>. Thus there can only be 5 elements in S with <math>a=0</math> or <math>b=0</math>. The set of elements with <math>a,b>0</math> is isomorphic to <math>\Delta_2</math>, so S can at most have 4 elements in this set. So <math>|S|\leq 4+5=9</math>. Similar if S contain (0,4,0) or (4,0,0). So if <math>|S|>9</math> S doesn’t contain any of these. Also, S can’t contain all of <math>(0,1,3), (0,3,1), (2,1,1)</math>. Similar for <math>(3,0,1), (1,0,3),(1,2,1)</math> and <math>(1,3,0), (3,1,0), (1,1,2)</math>. So now we have found 6 elements not in S, but <math>|\Delta_4|=15</math>, so <math>S\leq 15-6=9</math>.

''Remark'': curiously, the best constructions for <math>c_4</math> uses only 7 points instead of 9.

== n=5 ==

:<math>\overline{c}^\mu_5=12</math>:

The set of all (a,b,c) in <math>\Delta_5</math> with exactly one of a,b,c=0 has 12 elements and doesn’t contain any equilateral triangles.

Let <math>S\subset \Delta_5</math> be a set without equilateral triangles. If <math>(0,0,5)\in S</math>, there can only be one of (0,x,5-x) and (x,0,5-x) in S for x=1,2,3,4,5. Thus there can only be 6 elements in S with a=0 or b=0. The set of element with a,b>0 is isomorphic to <math>\Delta_3</math>, so S can at most have 6 elements in this set. So <math>|S|\leq 6+6=12</math>. Similar if S contain (0,5,0) or (5,0,0). So if |S| >12 S doesn’t contain any of these. S can only contain 2 point in each of the following equilateral triangles:

(3,1,1),(0,4,1),(0,1,4)

(4,1,0),(1,4,0),(1,1,3)

(4,0,1),(1,3,1),(1,0,4)

(1,2,2),(0,3,2),(0,2,3)

(3,2,0),(2,3,0),(2,2,1)

(3,0,2),(2,1,2),(2,0,3)

So now we have found 9 elements not in S, but <math>|\Delta_5|=21</math>, so <math>S\leq 21-9=12</math>.

== n=6 ==

:<math>\overline{c}^\mu_6 = 15</math>:

<math>\overline{c}^\mu_6 \geq 15</math> from the bound for general n.

Note that there are ten extremal solutions to <math>\overline{c}^\mu_3</math>:

Solution I: remove 300, 020, 111, 003 
Solution II (and 2 rotations): remove 030, 111, 201, 102 
Solution III (and 2 rotations): remove 030, 021, 210, 102 
Solution III' (and 2 rotations): remove 030, 120, 012, 201 

Also consider the same triangular lattice with the point 020 removed, making a trapezoid. Solutions based on I-III are:

Solution IV: remove 300, 111, 003 
Solution V: remove 201, 111, 102 
Solution VI: remove 210, 021, 102 
Solution VI': remove 120, 012, 201 

The on the 7x7x7 triangular lattice triangle 141-411-114 must have at least one point removed. Remove 141, noting by symmetry any logic that follows will also work for either of the other two points.

Suppose we can remove all equilateral triangles on our 7×7x7 triangular lattice with only 12 removals.

Here, "top triangle" means the top four rows of the lattice (with 060 at top) and "bottom trapezoid" means the bottom three rows.

At least 4 of those removals must come from the top triangle (the solutions of <math> \overline{c}^\mu_3</math> mentioned above).

The bottom trapezoid includes the overlapping trapezoids 600-420-321-303 and 303-123-024-006. If the solutions of these trapezoids come from V, VI, or VI', then 6 points have been removed. Suppose the trapezoid 600-420-321-303 uses the solution IV (by symmetry the same logic will work with the other trapezoid). Then there are 3 disjoint triangles 402-222-204, 213-123-114, and 105-015-006. Then 6 points have been removed. Therefore at least six removals must come from the bottom trapezoid.

To make a total of 12 removals there must be either: 
Case A: 4 removals from the top triangle and 8 from the bottom trapezoid. 
Case B: 5 removals from the top triangle and 7 from the bottom trapezoid. 
Case C: 6 removals from the top triangle and 6 from the bottom trapezoid. 

* Suppose case A is true.

Because 141 is already removed, the solution to the top triangle must remove either solution I (remove 060, 330, 033), solution II (remove 060, 231, 132), solution IIb (remove 033, 150, 240) or solution IIc (remove 330, 051, 042)

Suppose I is the solution for the top triangle.

Suppose 222 is open. Then 420, 321, 123, and 024 must all be removed. This leaves five disjoint triangles which require removals in the bottom trapezoid (150-600-105, 051-501-006, 222-402-204, 231-411-213, 132-312-114); therefore the bottom trapezoid needs at least 9 removals, but we can only make 8, therefore 222 is closed.

Suppose 411 is open. Then 213 and 015 must be removed. This leaves five disjoint triangles such that each triangle must have exactly one removal (420-150-123, 321-051,024, 600-510-501, 402-312-303, 204-114-105), so the remaining point (006) must be open, forcing 501 to be removed. This makes 600 and 510 open, and based on the triangles 600-240-204 and 510-150-114 both 204 and 114 must both be removed, but 204 and 114 are on the same disjoint triangle, contradicting the statement that each triangle must have exactly one removal. So 411 is closed.

This leaves six disjoint triangles each which must have at least one removal (420-123-150, 321-024-051, 510-213-240, 312-015-042, 501-204-231, 402-105-132). This forces 600 and 006 to be open. Based on the triangles 006-501-051 and 600-204-240, this forces 501 and 204 to be open. But then there are no removals from the triangle 501-204-231, which is a contradiction. Therefore the solution of the top triangle cannot be I.

Suppose II is the solution for the top triangle.

There are seven disjoint triangles (150-600-105, 051-501-006, 222-402-204, 240-510-213, 042-312-015, 330-420-321, 033-123-024), therefore of the three points remaining in the bottom trapezoid (411, 303, 114) exactly one must be removed.

Suppose 411 is removed. Then 114 and 303 are open; 114 open forces 510 to be removed, forcing 213 to be open. 114 and 213 open force 123 to be closed, forcing 024 to be open. 024 open forces 222 to be closed, which forces 204 to be open, which leaves the triangle 213-303-204 open so we have a contradiction.

By symmetry 114 also can't be removed. Therefore we must remove 303. This leaves 411 and 114 open, forcing 510 and 015 closed. 510 and 015 closed forces 312 and 213 open, forcing 222 closed. 222 closed forces 402 and 204 open. 402 and 204 open force 501 and 105 closed. 510 and 105 closed force 600 and 006 open, leaving equilateral triangles 600-204-240 and 402-006-042 open. Therefore 303 can't be removed, and so the solution of the top triangle can't be II.

Suppose IIb is the solution for the top triangle.

Suppose 024 is open. This forces 420, 321, and 222 closed. Five disjoint triangles remain (510-600-501, 231-411-213, 132-312-114, 123-303-105, 024-204-006) so each must have exactly one point removed, and the remaining points in the bottom trapezoid (402, 015) must be open. This forces 312, 411, 510, and 006 closed, which then force the the other points in their disjoint triangles open (600, 501, 231, 213, 132, 114, 204). The triangle 501-231-204 is therefore open, so we have a contradiction, therefore 024 is closed.

Suppose 006 is open. This forces 600, 501 and 402 to be closed, and leaves five disjoint triangles (510-420-411, 321-231-222, 312-132-114, 303-213-204, 105-015-006) and so 123 must be open. This forces 222 closed, which forces 321 open, which forces 420 closed, which forced 510 and 411 open, which forces 213 closed, which forces 303 and 204 open, which forces 105 closed, which forces 015 and 006 to be open, leaving an open triangle at 411-015-051. Therefore we have a contradiction, so 006 is closed.

Given 024 and 006 closed, now note there are six disjoint triangles (600-510-501, 402-312-303, 204-114-105, 420-330-321, 222-132-123, 411-231-213). Therefore the remaining point in the bottom trapezoid 015 must be open, forcing 510, 411, and 312 to be closed. Using the disjoint triangles this forces 600, 501, 402, 303 and 213 to be open, which then forces 420 and 321 to be closed. Both 420 and 321 are on the same disjoint triangle, therefore we have a contradiction, so IIb can't be solution.

Note by symmetry, the same logic for IIb will apply for IIc. Therefore case A isn't true.

* Suppose case B is true.

The row 330-231-132-033 has ten possible solutions (excluding reflections, which by symmetry will be handled by the same logic):

Case Q: open-open-open-open 
Case R: closed-closed-closed-closed 
Case S: closed-open-open-open 
Case T: closed-open-closed-closed 
Case U: open-closed-closed-closed 
Case V: closed-open-closed-closed 
Case W: closed-closed-open-open 
Case X: closed-open-closed-open 
Case Y: open-closed-closed-open 
Case Z: closed-open-open-closed 

Consider all 10 cases. Note in all these cases we are still assuming 141 is removed.

(Case Q) 330, 231, 132, 033 open forces 060, 150, 051, 240, 141, 042 closed, but the top triangle only allows 5 removals, so case Q forms a contradiction.

(Case R) 060-240-042 is left open, so case R forms a contradiction.

(Case S) 231, 132, and 033 open force 031 and 042 removed, which means from 240, 150, 061 there must be exactly one removal.

Suppose 213 is open. Then 411 and 015 are closed, and five disjoint triangles are left where each triangle must have exactly one removal (600-420-402, 501-321-303, 312-222-213, 204-105-114, 123-024-033). So the two remaining points in the bottom trapezoid (510 and 006) must be open. 510 and 006 open force 303 closed, which forces 321 and 501 open, which forces 204 closed, which forces 114 and 103 open, which forces 402 and 312 closed, forcing 222 open, leaving 321-231-222 as an open triangle, so we have a contradiction.

Therefore 213 is closed. This leaves six disjoint triangles (600-420-402, 501-231-204, 303-033-006, 411-321-312, 222-132-123, 114-024-015) which each have exactly one removal. Therefore 510 from the bottom trapezoid is open. Note since one of 150 and 060 must be open, then one of 114 or 015 must be closed; therefore 024 is open. This forces 123 closed, forcing 222 to be open, forcing 321 to be closed, forcing 411 and 312 open, forcing 421 closed, forcing 600 and 402 open, forcing 303 closed, forcing 006 open, forcing 204 closed, forcing 501 open, leaving the triangle 600-510-501 open and a contradiction.

(Case T) 231 is closed, and 330, 132, and 033 open force 060, 150, and 042 closed. Since 141 is already closed we have five removals from the top triangle, so 240 and 051 are open.

Suppose 024 is open. This forces 321 and 123 closed, leaving five disjoint triangles (600-510-501, 402-312-303, 204-114-105, 420-240-222, 213-033-015). Therefore 006 and 411 remaining in the bottom trapezoid are open, forcing 501, 303 and 015 closed, forcing 600, 510, 312, 402, and 213 open, forcing 222 closed, forcing 420 open, leaving the open triangle 420-510-411, so we have a contradiction.

Therefore 024 is closed. There are now six disjoint triangles (510-600-501, 330-420-321, 312-402-303, 132-222-123, 033-213-015, 114-204-105). This leaves 411 and 006 open, which forces 015 and 501 closed, which forces 510 and 213 open, leaving the triangle 240-510-213 open, so we have a contradiction.

(Case U) Given the removals 231, 132, 033, and 141, the only possible removal to not leave any equilateral triangles in the top triangle is 060. So 330, 240, 150, 051, and 042 are open.

Suppose 420 is open. This forces 321, 222, and 123 closed. This leaves five disjoint triangles (600-330-303, 510-420-411, 204-105-114, 501-051-006, and 312-042-015), so we have a contradiction.

Therefore 420 is closed. This leaves six disjoint triangles (600-330-303, 501-051-006, 204-114-105, 510-240-213, 411-321-312, 222-042-024). 015 in the bottom trapezoid is therefore open, forcing 312 and 411 closed, but 312 and 411 are on the same disjoint triangle, so we have a contradiction.

(Case V) Given the removals 330, 132, and 033, the only possible removal to not leave any equilateral triangles in the top triangle is 060. So 150, 051, 240, 042, and 231 are open.

Suppose 420 is open. Then 222 and 123 are closed, and we are left with 6 disjoint triangles (150-600-105, 240-510-213, 231-501-204, 024-114-015, 042-402-006, 411-321-312). This exceeds our limit of 7 removals in the bottom trapezoid, so we have a contradiction.

Therefore 420 is closed. Again we are left with the same 6 disjoint triangles (150-600-105, 240-510-213, 231-501-204, 024-114-015, 042-402-006, 411-321-312). Therefore 222, 123, and 303 are open. This forces 321, 024, and 105 to be closed, which then forces 411 and 015 to be open, forming an open triangle at 051-411-015, so we have a contradiction.

(Case W) Given the removals 330 and 231 with 132 and 033 open, 132 and 033 open force 042 closed. Given 141 is already closed, that leaves one more removal on the top triangle, which must be one of 060-150-051, so 240 must be open.

Suppose 024 is open. This forces 123 to be closed, and leaves 6 disjoint triangles (600-510-501, 420-240-222, 411-321-312, 402-132-105, 213-033-015, 204-024-006). So 303 and 114 in the bottom trapezoid are left open, forcing 312 and 005 to be closed, forcing 204 and 321 to be open, forcing 600 and 501 to be closed. 600 and 501 are on the same disjoint triangle so we have a contradiction.

Therefore 024 is closed. Suppose 312 is open. This forces 114 to be closed, and leaevs 5 disjoint triangles (600-510-501, 411-321-312, 303-213-204, 222-132-123, 105-015-006). Therefore 402 in the bottom trapezoid is open, which forces 105 to be closed, which forces 015 and 006 to be open, which forces 213 and 303 to be closed, which are on the same disjoint triangle, so we have a contradiction.

Therefore 312 is closed. This leaves five disjoint triangles (510-240-213, 501-411-402, 222-132-123, 204-114-105, 303-033-006), and leaves 600, 420, 321, and 015 in the bottom trapezoid open. This forces 402 and 213 to be closed, which forces 510 and 411 to be open, leaving an open triangle at 510-420-411, so we have a contradiction.

(Case X) 330 and 132 closed and 231 and 033 open imply 051 is closed, and since 141 is closed and we have only one more removal in the top triangle it must be one of 240, 042, or 060. This leaves 150 open.

Suppose 321 is open. This forces 222 to be closed and leaves six disjoint triangles (600-420-402, 510-150-114, 411-321-312, 303-213-204, 105-015-006, 123-033-024), leaving 501 in the bottom trapezoid open. This forces 204 to be closed, which forces 303 to be open, leaving an open triangle at 501-321-303.

So 321 is closed. This leaves six disjoint triangles (600-420-402, 510-150-114, 501-231-204, 312-222-213, 123-033-024, 105-015-006) and leaves 303 and 411 in the bottom trapezoid open. This forces 213 and 006 to be closed, and forces 312, 222, 105, and 015 to be open. This forces 600 to be closed and 402 to be open, leaving an open triangle at 402-312-303. So we have a contradiction.

(Case Y) 330 and 033 are open, 330 and 033 open force 060 to be closed. With 141 closed there can be only one more removal, so suppose 150 is open (and note that if 150 was closed, we can use symmetry considering 051 to be open).

Suppose 600 is open. This forces 303 to be closed, and leaves six disjoint triangles (510-150-114, 420-330-321, 411-501-402, 222-312-213, 033-123-024, 015-105-006). So 204 in the bottom trapezoid is left open. 600 and 150 open implies 105 is closed, forcing 015 and 006 to be open, forcing 024 and 213 to be closed, forcing 312, 222, and 123 to be open, forcing 042 in the top triangle to be closed (so 051 and 240 are open). 051 and 015 open force 411 to be closed, which forces 501 to be open and leaves the open triangle 051-501-006.

So 600 is closed. This leaves six disjoint triangles (510-150-114, 420-330-321, 411-501-402, 222-312-213, 033-123-024, 015-105-006). 015 is the bottom trapezoid is then left open, forcing 312 and 411 to be closed. However, 312 and 411 are on the same disjoint triangle, so we have a contradiction.

(Case Z) 330, 141, and 033 are closed, and 231 and 132 are open. Suppose 051 is open (note that if this forms a contradiction, by symmetrical argument we can say both 150 and 051 are closed).

Suppose 114 is closed. This leaves six disjoint triangles (600-510-501, 402-312-303, 105-015-006, 411-231-213, 222-132-123, 321-051-024) and so 420 and 204 are open. This forces 501 to be closed, which forces 510 and 600 to be open, which forces 411 and 402 to be closed, which forces 213 and 303 to be open, leaving an open triangle at 213-303-204.

So 114 is open. Suppose 213 is closed. This leaves five disjoint triangles (600-420-402, 501-321-303, 411-051-015, 222-132-123, 204-024-006) and in the bottom trapezoid 510 and 105 are open. This forces 204 to be closed, forcing 024 and 006 to be open, forcing 015 to be closed, forcing 411 to be open, forcing 420 to be closed, forcing 402 to be open, leaving the open triangle 402-132-105.

So both 114 and 213 are open. Therefore 411 and 312 are closed. This leaves five disjoint triangles (600-510-501, 303-213-204, 105-015-006, 222-132-123, 321-051-024). The remaining points in the bottom trapezoid (420, 402) are then open, forcing 105 to be closed, forcing 015 and 006 to be open, forcing 024 to be closed. Also note 114 and 213 open force 123 to be closed, which forces 222 to be open, which forces 321 to be closed. 321 and 024 are on the same disjoint triangle, so we have a contradiction.

So 150 and 051 are both closed. However, this leaves the triangle 240-042-060 open, so case Z is impossible.

Since all ten solutions have been eliminated, case B is impossible.

* Suppose case C is true.

Suppose the trapezoid 600-420-321-303 used solution IV. There are three disjoint triangles 402-222-204, 213-123-114, and 105-015-006. The remainder of the points in the bottom trapezoid (420, 321, 510, 501, 402, 312, 024) must be left open. 024 being open forces either 114 or 015 to be removed.

Suppose 114 is removed. Then 213 is open, and with 312 open that forces 222 to be removed. Then 204 is open, and with 024 that forces 006 to be removed. So the bottom trapezoid is a removal configuration of 600-411-303-222-114-006, and the rest of the points in the bottom trapezoid are open. All 10 points in the top triangle form equilateral triangles with bottom trapezoid points, hence 10 removals in the top triangle would be needed (more than the 6 allowed), so 114 being removed doesn't work.

Suppose 015 is removed. Then 006-024 forces 204 to be removed. Regardless of where the removal in 123-213-114, the points 420, 321, 222, 024, 510, 312, 501, 402, 105, and 006 must be open. This forces top triangle removals at 330, 231, 042, 060, 051, 132, our remaining 6 removals on the top triangle. However, we have already removed 141, forcing one removal too many, so the trapezoid 600-420-321-303 doesn't use solution IV.

Suppose the trapezoid 600-420-321-303 uses solution VI. The trapezoid 303-123-024-006 can't be IV (already eliminated by symmetry) or VI' (leaves the triangle 402-222-204). Suppose the trapezoid 303-123-024-006 is solution VI. The removals from the bottom trapezoid are then 420, 501, 312, 123, 204, and 015, leaving the remaining points in the bottom trapezoid open. The remaining open points is forces 10 top triangle removals, so the trapezoid 600-420-321-303 doesn't use solution VI. Therefore the trapezoid 303-123-024-006 is solution V. The removals from the bottom trapezoid are then 420, 510, 312, 204, 114, and 105. The remaining points in the bottom trapezoid are open, and force 9 top triangle removals, hence the trapezoid 303-123-024-006 can't be V, and the solution for 600-420-321-303 can't be VI.

The solution VI' for the trapezoid 600-420-321-303 can be eliminated by the same logic by symmetry.

Therefore it is impossible for the bottom trapezoid to use only 6 removals.

We have determined cases A, B, and C to be impossible, therefore it impossible to form a triangle free configuration on the 7x7x7 lattice with only 12 removals. Therefore <math>\overline{c}^\mu_6 = 15</math>.

== n = 7 ==

<math>\overline{c}^\mu_{7} \leq 22</math>:

Using the same ten extremal solutions to <math> \overline{c}^\mu_3 </math> as previous proofs:

Solution I: remove 300, 020, 111, 003 
Solution II (and 2 rotations): remove 030, 111, 201, 102 
Solution III (and 2 rotations): remove 030, 021, 210, 102 
Solution III' (and 2 rotations): remove 030, 120, 012, 201

Suppose the 8x8x8 lattice can be triangle-free with only 13 removals.

Slice the lattice into region A (070-340-043) region B (430-700-403) and region C (034-304-007). Each region must have at least 4 points removed. Note there is an additional disjoint triangle 232-322-223 that also must have a point removed. Therefore the points 331, 133, and 313 are open. 331-313 open means 511 must be removed, 331-133 open means 151 must be removed, and 133-313 open means 115 must be removed. Based on the three removals, the solutions for regions A, B, and C must be either I or II. All possible combinations for the solutions leave several triangles open (for example 160-520-124). So we have a contradiction, and <math> \overline{c}^\mu_7 \leq 22 </math>.

== n = 8 ==

== n = 9 ==

== n = 10 ==

== Computer data ==

From integer programming, we have

* n=3, maximum 6 points, [http://abel.math.umu.se/~klasm/solutions-n=3-k=3-FUJ 10 solutions]
* n=4, maximum 9 points, [http://abel.math.umu.se/~klasm/solutions-n=4-k=3-FUJ 1 solution]
* n=5, maximum 12 points, [http://abel.math.umu.se/~klasm/solutions-n=5-k=3-FUJ 1 solution]
* n=6, maximum 15 points, [http://abel.math.umu.se/~klasm/solutions-n=6-k=3-FUJ 4 solutions]
* n=7, maximum 18 points, [http://abel.math.umu.se/~klasm/solutions-n=7-k=3-FUJ 85 solutions]
* n=8, maximum 22 points, [http://abel.math.umu.se/~klasm/solutions-n=8-k=3-FUJ 72 solutions]
* n=9, maximum 26 points, [http://abel.math.umu.se/~klasm/solutions-n=9-k=3-FUJ 183 solutions]
* n=10, maximum 31 points, [http://abel.math.umu.se/~klasm/solutions-n=10-k=3-FUJ 6 solutions]
* n=11, maximum 35 points, [http://abel.math.umu.se/~klasm/solutions-n=11-k=3-FUJ 576 solutions]
* n=12, maximum 40 points, [http://abel.math.umu.se/~klasm/solutions-n=12-k=3-FUJ 876 solutions]

== General n ==

A lower bound for <math>\overline{c}^\mu_n</math> is 2n for <math>n \geq 1</math>, by removing (n,0,0), the triangle (n-2,1,1) (0,n-1,1) (0,1,n-1), and all points on the edges of and inside the same triangle. In a similar spirit, we have the lower bound

:<math>\overline{c}^\mu_{n+1} \geq \overline{c}^\mu_n + 2</math>

for <math>n \geq 1</math>, because we can take an example for <math>\overline{c}^\mu_n</math> (which cannot be all of <math>\Delta_n</math>) and add two points on the bottom row, chosen so that the triangle they form has third vertex outside of the original example.

An asymptotically superior lower bound for <math>\overline{c}^\mu_n</math> is 3(n-1), made of all points in <math>\Delta_n</math> with exactly one coordinate equal to zero.

A trivial upper bound is

:<math>\overline{c}^\mu_{n+1} \leq \overline{c}^\mu_n + n+2</math>

since deleting the bottom row of a equilateral-triangle-free-set gives another equilateral-triangle-free-set. We also have the asymptotically superior bound

:<math>\overline{c}^\mu_{n+2} \leq \overline{c}^\mu_n + \frac{3n+2}{2}</math>

which comes from deleting two bottom rows of a triangle-free set and counting how many vertices are possible in those rows.

Another upper bound comes from counting the triangles. There are <math>\binom{n+2}{3}</math> triangles, and each point belongs to n of them. So you must remove at least (n+2)(n+1)/6 points to remove all triangles, leaving (n+2)(n+1)/3 points as an upper bound for <math>\overline{c}^\mu_n</math>.

== Asymptotics ==

The [[corners theorem]] tells us that <math>\overline{c}^\mu_n = o(n^2)</math> as <math>n \to \infty</math>.

By looking at those triples (a,b,c) with a+2b inside a Behrend set, one can obtain the lower bound <math>\overline{c}^\mu_n \geq n^2 \exp(-O(\sqrt{\log n}))</math>.

Higher-dimensional Fujimura

2009-03-30T13:37:43Z

Jbd:

Let <math>\overline{c}^\mu_{n,4}</math> be the largest subset of the tetrahedral grid:

:<math> \{ (a,b,c,d) \in {\Bbb Z}_+^4: a+b+c+d=n \}</math>

which contains no tetrahedrons <math>(a+r,b,c,d), (a,b+r,c,d), (a,b,c+r,d), (a,b,c,d+r)</math> with <math>r > 0</math>; call such sets ''tetrahedron-free''.

These are the currently known values of the sequence:

{|
| n || 0 || 1 || 2 || 3 || 4 || 5 || 6
|-
| <math>\overline{c}^\mu_{n,4}</math> || 1 || 3 || 7 || 14 || 24 || 37 || 55
|}

== n=0 ==

<math>\overline{c}^\mu_{0,4} = 1</math>:

There are no tetrahedrons, so no removals are needed.

== n=1 ==

<math>\overline{c}^\mu_{1,4} = 3</math>:

Removing any one point on the grid will leave the set tetrahedron-free.

== n=2 ==

<math>\overline{c}^\mu_{2,4} = 7</math>:

Suppose the set can be tetrahedron-free in two removals. One of (2,0,0,0), (0,2,0,0), (0,0,2,0), and (0,0,0,2) must be removed. Removing any one of the four leaves three tetrahedrons to remove. However, no point coincides with all three tetrahedrons, therefore there must be more than two removals.

Three removals (for example (0,0,0,2), (1,1,0,0) and (0,0,2,0)) leaves the set tetrahedron-free with a set size of 7.

== General n ==

A lower bound of 2(n-1)(n-2) can be obtained by keeping all points with exactly one coordinate equal to zero.

You get a non-constructive quadratic lower bound for the quadruple problem by taking a random subset of size <math>cn^2</math>. If c is not too large the linearity of expectation shows that the expected number of tetrahedrons in such a set is less than one, and so there must be a set of that size with no tetrahedrons. I think <math> c = \frac{24^{1/4}}{6} + o(\frac{1}{n})</math>.

With coordinates (a,b,c,d), take the value a+2b+3c. This forms an arithmetic progression of length 4 for any of the tetrahedrons we are looking for. So we can take subsets of the form a+2b+3c=k, where k comes from a set with no such arithmetic progressions. [[http://arxiv.org/PS_cache/arxiv/pdf/0811/0811.3057v2.pdf This paper]] gives a complicated formula for the possible number of subsets.

One upper bound can be found by counting tetrahedrons. For a given n the tetrahedral grid has <math>\frac{1}{24}n(n+1)(n+2)(n+3)</math> tetrahedrons. Each point on the grid is part of n tetrahedrons, so <math>\frac{1}{24}(n+1)(n+2)(n+3)</math> points must be removed to remove all tetrahedrons. This gives an upper bound of <math>\frac{1}{8}(n+1)(n+2)(n+3)</math>.

Main Page

2009-03-30T02:10:14Z

Jbd: /* Threads and further problems */

== The Problem ==

Initially, the basic problem to be considered by the Polymath1 project was to explore a particular [http://gowers.wordpress.com/2009/02/01/a-combinatorial-approach-to-density-hales-jewett/ combinatorial approach] to the [[density Hales-Jewett theorem]] for k=3 (DHJ(3)), suggested by Tim Gowers. The [[Furstenberg-Katznelson argument|original proof of DHJ(3) used arguments from ergodic theory]]. Fairly soon, the scope of the project expanded and the main aim became that of discovering any combinatorial argument for the theorem. This aim appears to have been achieved but the proof has not yet been fully written up.

==Basic definitions==

*[[line|Algebraic line]]

*[[line|Combinatorial line]]

*[[Combinatorial subspace]]

*[[corners|Corner]]

*[[Density]]

*[[line|Geometric line]]

*[[Slice]]

== Useful background materials ==

Here is [http://gowers.wordpress.com/2009/01/30/background-to-a-polymath-project/ some background to the project.] There is also a [http://gowers.wordpress.com/2009/01/27/is-massively-collaborative-mathematics-possible/ general discussion on massively collaborative "polymath" projects.] This is [http://meta.wikimedia.org/wiki/File:MediaWikiRefCard.png a cheatsheet for editing the wiki.] Here is a [http://michaelnielsen.org/blog/?p=582 python script] which can help convert sizeable chunks of LaTeX into wiki-tex. Finally, here is the general [http://meta.wikimedia.org/wiki/Help:Contents Wiki user's guide].

== Threads and further problems==

* [http://gowers.wordpress.com/2009/01/27/is-massively-collaborative-mathematics-possible/ Is massively collaborative mathematics possible?] (inactive)
* (1-199) [http://gowers.wordpress.com/2009/02/01/a-combinatorial-approach-to-density-hales-jewett/ A combinatorial approach to density Hales-Jewett] (inactive)
* (200-299) [http://terrytao.wordpress.com/2009/02/05/upper-and-lower-bounds-for-the-density-hales-jewett-problem/ Upper and lower bounds for the density Hales-Jewett problem] (inactive)
* (300-399) [http://gowers.wordpress.com/2009/02/06/dhj-the-triangle-removal-approach/ The triangle-removal approach] (inactive)
* (400-499) [http://gowers.wordpress.com/2009/02/08/dhj-quasirandomness-and-obstructions-to-uniformity Quasirandomness and obstructions to uniformity] (inactive)
* (500-599) [http://gowers.wordpress.com/2009/02/13/dhj-possible-proof-strategies/#more-441/ Possible proof strategies] (inactive)
* (600-699) [http://terrytao.wordpress.com/2009/02/11/a-reading-seminar-on-density-hales-jewett/ A reading seminar on density Hales-Jewett] (inactive)
* (700-799) [http://terrytao.wordpress.com/2009/02/13/bounds-for-the-first-few-density-hales-jewett-numbers-and-related-quantities/ Bounds for the first few density Hales-Jewett numbers, and related quantities] (inactive)
* (800-849) [http://gowers.wordpress.com/2009/02/23/brief-review-of-polymath1/ Brief review of polymath1] (inactive)
* (850-899) [http://gowers.wordpress.com/2009/03/02/dhj3-851-899/ DHJ(3): 851-899] (inactive)
* (900-999) [http://terrytao.wordpress.com/2009/03/04/dhj3-900-999-density-hales-jewett-type-numbers/ DHJ(3): 900-999 (Density Hales-Jewett type numbers)] (inactive)
* (1000-1049) [http://gowers.wordpress.com/2009/03/10/problem-solved-probably/ Problem solved (probably)] (inactive)
* [http://gowers.wordpress.com/2009/03/10/polymath1-and-open-collaborative-mathematics/ Polymath1 and open collaborative mathematics] (active)
* (1050-1099) [http://gowers.wordpress.com/2009/03/16/dhj3-and-related-results-1050-1099/ DHJ(3) and related results: 1050-1099] (active)
* (1100-1199) [http://terrytao.wordpress.com/2009/03/14/dhj3-1100-1199-density-hales-jewett-type-numbers/ DHJ(3): 1100-1199 (Density Hales-Jewett type numbers)] (active)

[http://blogsearch.google.com/blogsearch?hl=en&ie=UTF-8&q=polymath1&btnG=Search+Blogs Here is a further list of blog posts related to the Polymath1 project]. [http://en.wordpress.com/tag/polymath1/ Here is wordpress's list]. Here is a [[timeline]] of progress so far.

A spreadsheet containing the latest upper and lower bounds for <math>c_n</math> can be found [http://spreadsheets.google.com/ccc?key=p5T0SktZY9DsU-uZ1tK7VEg here]. Here are the proofs of our [[upper and lower bounds]] for these constants.

We are also collecting bounds for [[Fujimura's problem]], motivated by a [[hyper-optimistic conjecture]]. We are additionally investigating [[higher-dimensional Fujimura]].

There is also a chance that we will be able to improve the known bounds on [[Moser's cube problem]] or the [[Kakeya problem]].

Here are some [[unsolved problems]] arising from the above threads.

Here is a [[tidy problem page]].

== Proof strategies ==

It is natural to look for strategies based on one of the following:

* [[Szemerédi's original proof of Szemerédi's theorem]].
* [[Szemerédi's combinatorial proof of Roth's theorem]].
* [[Ajtai-Szemerédi's proof of the corners theorem]].
* The [[density increment method]].
* The [[triangle removal lemma]].
* [[Ergodic-inspired methods]].
* The [[Furstenberg-Katznelson argument]].
* Use of [[equal-slices measure]].

== Related theorems ==

* [[Carlson's theorem]].
* The [[Carlson-Simpson theorem]].
* [[Folkman's theorem]].
* The [[Graham-Rothschild theorem]].
* The colouring [[Hales-Jewett theorem]].
* The [[Kruskal-Katona theorem]].
* [[Roth's theorem]].
* The [[IP-Szemerédi theorem]].
* [[Sperner's theorem]].
* [[Szemerédi's regularity lemma]].
* [[Szemerédi's theorem]].
* The [[triangle removal lemma]].

All these theorems are worth knowing. The most immediately relevant are Roth's theorem, Sperner's theorem, Szemerédi's regularity lemma and the triangle removal lemma, but some of the others could well come into play as well.

==Important concepts related to possible proofs==

* [[Complexity of a set]]
* [[Concentration of measure]]
* [[Influence of variables]]
* [[Obstructions to uniformity]]
* [[Quasirandomness]]

==Complete proofs or detailed sketches of potentially useful results==

*[[Sperner's theorem|The multidimensional Sperner theorem]]
*[[Line-free sets correlate locally with complexity-1 sets]]
*[[Correlation with a 1-set implies correlation with a subspace]] (Superseded)
*[[Fourier-analytic_proof_of_Sperner|A Fourier-analytic proof of Sperner's theorem]]
*[[A second Fourier decomposition related to Sperner's theorem]]
*[[A Hilbert space lemma]]
*A [[Modification of the Ajtai-Szemerédi argument]]
*An [[abstract regularity lemma]]
*[[A general result about density increments]]

==Attempts at proofs of DHJ(3)==

*[[An outline of a density-increment argument]] (ultimately didn't work)
*[[A second outline of a density-increment argument]] (seems to be OK but more checking needed)
*[[Furstenberg-Katznelson argument]] (very sketchy)
*[[Austin's proof]] (very sketchy)
*[[Austin's proof II]] (mostly complete, though more explanation and motivation needed)

==Generalizing to DHJ(k)==

*[[DHJ(k) implies multidimensional DHJ(k)]]
*[[Line free sets correlate locally with dense sets of complexity k-2]]
*[[A general partitioning principle]]

== Bibliography ==

Here is a [[Bibliography]] of relevant papers in the field.

== How to help out ==

There are a number of ways that even casual participants can help contribute to the Polymath1 project:

* Expand the [[bibliography]]
* Join the [http://gowers.wordpress.com/2009/03/10/polymath1-and-open-collaborative-mathematics metadiscussion thread]
* Add some more Ramsey theorems to this wiki; one could hope to flesh out this wiki into a Ramsey theory resource at some point.
* Suggest a logo for this wiki!
* Suggest a way to speed up our [[genetic algorithm]]
* Point out places where the exposition could be improved
* Add to this list