Difference between revisions of "Human proof that completely multiplicative sequences have discrepancy greater than 2"

Revision as of 03:09, 30 January 2010

It is 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 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
-