Side Proof 9: Difference between revisions
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(1)+(2)-12: f(59)+f(67)+f(107)+f(109)+f(431)+f(433) <= -6 | (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(59)=f(67)=f(107)=f(109)=f(431)=f(433)=-1. However, now f[107,112] = -6, which forces the discrepancy above 3. | ||
Latest revision as of 19:41, 21 June 2015
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.
