Experimental results: Difference between revisions
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* Find long/longest quasi-multiplicative sequences with some fixed group G, function <math>G\to \{-1,1\}</math> and maximal discrepancy C | * 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) | ** <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 | * 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 done this for k=1 and k=3. | *. 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 done this [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4563 for k=1] and 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. | *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. | ||
* ... you are welcome to add more. | * ... you are welcome to add more. | ||
Revision as of 00:14, 10 January 2010
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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]\displaystyle{ \mathbb{T} }[/math]:
Wish list
- Find long/longest quasi-multiplicative sequences with some fixed group G, function [math]\displaystyle{ G\to \{-1,1\} }[/math] and maximal discrepancy C
- [math]\displaystyle{ 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. [1] Also 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]\displaystyle{ C_{2k} }[/math]. Then set [math]\displaystyle{ 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 done this for k=1 and k=3.
- Search for the longest sequence of discrepancy 2 with the property that [math]\displaystyle{ 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]\displaystyle{ x_{32n} }[/math] then that will have some influence on several other choices, such as [math]\displaystyle{ x_{4n},x_{8n} }[/math] and [math]\displaystyle{ x_{16n} }[/math], so maybe it will lead to something interesting.
- ... you are welcome to add more.
