Side Proof 9

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(11)=f(17)=f(29)=1, f(7)=f(13)=f(23)=-1.

Proof

s(36) = 3+f(31), so f(31)=-1. f[185,190] = 5-f(37), so f(37)=1.

s(44) = 4+f(41)+f(43), so f(41)=f(43)=-1.

We have two equations:

1) f[423,438] = 7-f(61)-f(71)-f(73)+f(107)+f(109)+f(431)+f(433) <= 4

2) s(72) = 5+f(59)+f(61)+f(67)+f(71)+f(73) <= 2

(1)+(2)-12: f(59)+f(67)+f(107)+f(109)+f(431)+f(433) <= -6

Therefore, f(59)=f(67)=f(107)=f(109)=f(431)=f(433)=-1. However, now f[107,112] = -6, which forces the discrepancy above 3. Therefore, f(37)=-1.