Side Proof 8

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(23)=1, f(7)=f(13)=-1.

Proof

s(36) = 4+f(29)+f(31), so f(29)=f(31)=-1. This seems to be as far as we can get just with the original assumptions.

Case 1: f(2)=f(11)=f(17)=f(23)=f(37)=1, f(7)=f(13)=-1

s(50)=5+f(41)+f(43)+f(47), so f(41)=f(43)=f(47)=-1. f[339,344] = -5-f(113), so f(113)=-1. f[107,116] = -6+f(107)+f(109), so f(107)=f(109)=1. However, now f[319,328] = -7+f(163), which forces the discrepancy above 3.

Case 2: f(2)=f(11)=f(17)=f(23)=f(41)=1, f(7)=f(13)=f(37)=-1

s(50)=4+f(43)+f(47), so f(43)=f(47)=-1. f[339,344] = -5-f(113), so f(113)=-1. f[81,100] = 7+f(83)+f(89)+f(97), so f(83)=f(89)=f(97)=-1. f[101,126] = -10+f(59)+f(61)+f(101)+f(103)+f(107)+f(109), so f(59)=f(61)=f(101)=f(103)=f(107)=f(109)=1. f[575,586] = 8+f(73)-f(193)+f(293)+f(577), so f(73)=f(293)=f(577)=-1, f(193)=1. f[293,300] = -6-f(59)+f(149), so f(59)=-1, f(149)=1. s(76) = -5+f(61)+f(67)+f(71), so f(61)=f(67)=f(71). However, now f[115,126] = -6, which forces the discrepancy above 3.

Case 3: f(2)=f(11)=f(17)=f(23)=f(43)=1, f(7)=f(13)=f(37)=f(41)=-1

s(50) = 3+f(47), so f(47)=-1. f[83,100] = 7+f(83)+f(89)+f(97), so f(83)=f(89)=f(97)=-1. f[287,292] = 5+f(73), so f(73)=-1. f[73,80] = -5+f(79), so f(79)=1. f[573,586] = 8-f(191)-f(193)+f(293)+f(577), so f(191)=f(193)=1, f(293)=f(577)=-1. f[293,304] = -8-f(59)-f(101)+f(149)+f(151), so f(59)=f(101)=-1, f(149)=f(151)=1. s(76) = -5+f(61)+f(67)+f(71), so f(61)=f(67)=f(71)=1. s(118) = -6+f(103)+f(107)+f(109)+f(113), so f(103)=f(107)=f(109)=f(113)=1. But now, f[319,328] = -7+f(163), which forces the discrepancy above 3.

Case 4: f(2)=f(11)=f(17)=f(23)=1, f(7)=f(13)=f(37)=f(41)=f(43)=-1

s(56) = -3+f(47), so f(47)=1. However, now f[91,100] = 7+f(97), which forces the discrepancy above 3.