Side Proof 10

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(31)=1, f(7)=f(13)=f(23)=f(29)=-1.

Proof

It seems that we can't derive anything just from these assumptions.

Case 1: f(37)=1

Now, s(44) = 4+f(41)+f(43), so f(41)=f(43)=-1. We have two inequalities:

1) f[423,430] = 5-f(61)-f(71)+f(107) <= 4

2) s(74) = 5+f(59)+f(61)+f(67)+f(71)+f(73) <= 2

3) f[207,222] = -7-f(71)-f(73)+f(107)+f(109)+f(211) >= -4

(1)+(2)-(3)-17: f(59)+f(67)+f(71)+2f(73)-f(109)-f(211) <= -7, so f(59)=f(67)=f(71)=f(73)=-1, f(109)=f(211)=1.

We have another two inequalities:

1) f[141,160] = -6+f(79)+f(149)+f(151)+f(157) >= -4

2) f[471,476] = -4-f(79)-f(157) >= -4

(1)+(2)+10: f(149)+f(151) >= 2

Therefore, f(149)=f(151)=1

f[287,302] = 6-f(97)+f(293), so f(97)=1, f(293)=-1.

We have another two inequalities:

1) f[187,206] = 8+f(101)+f(103)+f(191)+f(193)+f(197)+f(199) <= 4

2) f[101,112] = -5+f(101)+f(103)+f(107) >= -4

(2)-(1)+13: f(107)-f(191)-f(193)-f(197)-f(199) >= 5, so f(107)=1, f(191)=f(193)=f(197)=f(199)=-1. However, now f[423,430] = 7-f(61), which forces the discrepancy above 3. Therefore, f(37)=-1.

Case 2: f(41)=1

Now, s(44) = 4+f(41)+f(43), so f(41)=f(43)=-1. We have two inequalities:

1) f[423,430] = 5-f(61)-f(71)+f(107) <= 4

2) s(72) = 4+f(59)+f(61)+f(67)+f(71) <= 2

(1)+(2)-9: f(59)+f(67)+f(107) <= -3. Therefore, f(59)=f(67)=f(107)=-1.

f[373,378] = 5+f(373), so f(373)=-1.