# Side Proof 1

This page will handle one of the long cases in the Human proof that completely multiplicative sequences have discrepancy greater than 3, so that the page can be shorter and not have so many long sections. Specifically, this page will take care of the case where we assume: f(2)=f(7)=f(19)=f(23)=f(29)=1.

Looking at the table:

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


We can first see that the discrepancy up to 32 is 3+f(31), so f(31)=-1. Also, there is a cut after 64, because of f[63,66].

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


Now the discrepancy up to 38 is 3+f(37), therefore f(37)=-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


Here is where things start to get interesting. The discrepancy up to 64, which is 4+f(41)+f(43)+f(47)+f(53)+f(59)+f(61), must be zero because of the cut. Therefore exactly one of f(41), f(43), f(47), f(53), f(59), f(61) must be 1, and the others -1. However, the discrepancy f[243,248] = -4-f(41)+f(61), so if f(41) = 1, f(61) must be -1, but then the discrepancy f[243,248] = -6, forcing the discrepancy to be greater than 3. Therefore f(41)=-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



Now, if f(43) = 1, then f(61)=-1, and considering f[243,250] = -5-f(83), we have also that f(83)=-1. Also, f(47)=f(53)=f(59)=f(61)=-1, and the table would look like this:

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 discrepancy up to 96 is now -7+f(67)+f(71)+f(73)+f(79)+f(89). Therefore f(67)=f(71)=f(73)=f(79)=f(89)=1. If we now say that f(107)=a, f(109)=b, f(163)=c, f(167)=d, f(331)=e. We have three choke points to consider here. First, f[155,172] = 5+f(157)+c+d, which implies that c+d<=0. Second, f[213,218] = -4+a+b implies that a+b>=0. Third, f[319,334]=-7-a-b+c+d+e. Plugging the results from the first two equations in, we get that f[319,334]<=-7+e, which is impossible, since that would force the discrepancy to be greater than 3.

Therefore f(43) = -1. The table is now:

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 discrepancy up to 48 is now -3-f(47), so f(47)=1. Therefore, f(53)=f(59)=f(61)=-1. But now, f[183,188]=6, which is a contradiction.

Therefore f(29)=-1.