Side Proof 4
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(7)=f(19)=f(23)=f(37)=1, f(29)=f(31)=f(43)=-1.
Proof
Looking at the table:
0 1 2 3 4 5 6 7 8 9
0|+ + - + - - + + + 0-9 - - - - + + + - + + 10-19 - - - + - + - - + - 20-29 + - + + - - + + + + 30-39 - - -|- - - + ? - + 40-49 + + - ? - + + - - ? 50-59 + ? - + +|+ + ? - - 60-69 - ? + ? + - + - + ? 70-79 - + - ? - + - + - ? 80-89 - - + + ? - - ? + - 90-99 + ? + ? - + ? ? - ? 100-109 + - + ? - - - - ? - 110-119 + + ? + - - + ? + ? 120-129 + ? + + ? + - ? - + 130-139
The discrepancy up to 48 is -3+f(47), so f(47)=1. The discrepancy up to 66 is 3+f(53)+f(59)+f(61), so only one of those is positive, the others are negative. Updating the table:
0 1 2 3 4 5 6 7 8 9
0|+ + - + - - + + + 0-9 - - - - + + + - + + 10-19 - - - + - + - - + - 20-29 + - + + - - + + + + 30-39 - - -|- - - + + - + 40-49 + + - ? - + + - - ? 50-59 + ? - + +|+ + ? - - 60-69 - ? + ? + - + - + ? 70-79 - + - ? - + - + - ? 80-89 - - + + + - - ? + - 90-99 + ? + ? - + ? ? - ? 100-109 + - + ? - - - - ? - 110-119 + + ? + - - + ? + ? 120-129 + ? + + ? + - ? - + 130-139
It seems like we can't get very far with these assumptions, so we will now assume f(53)=1.
Case 1: f(2)=f(7)=f(19)=f(23)=f(37)=f(53)=1, f(29)=f(31)=f(43)=-1
If f(53)=1, then f(59)=f(61)=-1, so updating the table:
0 1 2 3 4 5 6 7 8 9
0|+ + - + - - + + + 0-9 - - - - + + + - + + 10-19 - - - + - + - - + - 20-29 + - + + - - + + + + 30-39 - - -|- - - + + - + 40-49 + + - + - + + - - - 50-59 + - - + +|+ + ? - - 60-69 - ? + ? + - + - + ? 70-79 - + - ? - + - + - ? 80-89 - - + + + - - ? + - 90-99 + ? + ? - + ? ? - ? 100-109 + - + ? - - -|- - - 110-119 + + - + - - + ? + ? 120-129 + ? + + ? + - ? - + 130-139
f[243,250] = -5-f(83), so f(83)=-1. Also, f[113,118] = -5+f(113), so f(113)=1. Updating the table:
0 1 2 3 4 5 6 7 8 9
0|+ + - + - - + + + 0-9 - - - - + + + - + + 10-19 - - - + - + - - + - 20-29 + - + + - - + + + + 30-39 - - -|- - - + + - + 40-49 + + - + - + + - - - 50-59 + - - + +|+ + ? - - 60-69 - ? + ? + - + - + ? 70-79 - + - - - + - + - ? 80-89 - - + + + - - ? + - 90-99 + ? + ? - + + ? - ? 100-109 + - + + - - - - - - 110-119 + + - + - - + ? + ? 120-129 + ? + + ? + - ? - + 130-139
It again seems like no more deductions can be made, so we will make more assumptions.
Case 1.1: f(2)=f(7)=f(19)=f(23)=f(37)=f(53)=f(67)=f(71)=1, f(29)=f(31)=f(43)=-1
Assume f(67)=f(71)=1. The discrepancy up to 74 is 3+f(73), so f(73)=-1. Updating the table:
0 1 2 3 4 5 6 7 8 9
0|+ + - + - - + + + 0-9 - - - - + + + - + + 10-19 - - - + - + - - + - 20-29 + - + + - - + + + + 30-39 - - - - - - + + - + 40-49 + + - + - + + - - - 50-59 + - - + + + + + - - 60-69 -|+ + - + - + - + ? 70-79 - + - - - + - + - ? 80-89 - - + + + - - ? + - 90-99 + ? + ? - + + ? - ? 100-109 + - + + - - - - - - 110-119 + + - + - - + ? + ? 120-129 + ? + + + + - ? - + 130-139
Now, f[775,782] = -6+f(97)+f(389), so f(97)=f(389)=1. Updating the table:
0 1 2 3 4 5 6 7 8 9
0|+ + - + - - + + + 0-9 - - - - + + + - + + 10-19 - - - + - + - - + - 20-29 + - + + - - + + + + 30-39 - - - - - - + + - + 40-49 + + - + - + + - - - 50-59 + - - + + + + + - - 60-69 -|+ + - + - + - + ? 70-79 - + - - - + - + - ? 80-89 - - + + + - - + + - 90-99 + ? + ? - + + ? - ? 100-109 + - + + - - - - - - 110-119 + + - + - - + ? + ? 120-129 + ? + + + + - ? - + 130-139
It seems we can't get much further with this assumption. However, when we assume f(79)=1, suddenly things fall into place.
