# Side Proof 7

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

## Proof

We have a few equations:

1) f[243,262] = -8+f(31)-f(37)-f(41)-f(43)+f(61)-f(83)+f(127)+f(131)+f(251)+f(257) >= -4

2) f[117,146] = 7+f(31)-f(41)-f(43)-f(47)+f(59)+f(61)+f(67)+f(71)+f(73)+f(127)+f(131)+f(137)+f(139) <= 4

3) f[65,78] = -4+f(37)+f(67)+f(71)+f(73) >= -4

(1)-(2)+(3)+19: f(47)-f(59)-f(83)-f(137)-f(139)+f(251)+f(257) >= 7

Therefore, f(47)=f(251)=f(257)=1, f(59)=f(83)=f(137)=f(139)=-1.

It seems this is as far as we can get with these assumptions, so we will have to split this into cases:

## Case 1: f(31)=1

Now, f[147,156] = -7+f(37)+f(149)+f(151), so f(37)=f(149)=f(151)=1. We then have s(52)=4+f(41)+f(43), so f(41)=f(43)=-1. f[119,124] = 5+f(61), so f(61)=-1. We then have f[55,68] = -5+f(67), so f(67)=1.

We now have three inequalities:

1) f[243,262] = -4+f(127)+f(131) >= -4

2) f[119,146] = 6+f(71)+f(73)+f(127)+f(131) <= 4

3) f[55,78] = -5+f(67)+f(71)+f(73) >= -4

(1)-(2)+(3)+15: f(67) >= 3

Which is impossible, therefore, f(31)=-1.

## Case 2: f(31)=-1

s(40)=-3+f(37), so f(37)=1. f[55,68] = -5+f(61)+f(67), so f(61)=f(67)=1. s(46)=-2+f(41)+f(43), and s(52)=2+f(41)+f(43), so f(41)+f(43)=0. f[243,262] = -6+f(127)+f(131), so f(127)=f(131)=1. f[119,146] = 6+f(71)+f(73), so f(71)=f(73)=-1. However, now f[55,78] = -6, which forces he discrepancy above 3.

Therefore, the assumption f(29)=-1 fails at 262.