Side Proof 2

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(31)=1, f(29)=-1

Assuming that f(31)=1, the table is now:

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

The discrepancy up to 38 is 3+f(37), therefore f(37) = -1. Also, f[285,290]=5+f(41), therefore f(41)=-1 (actually, f(41) is forced to be -1 before this, but the reasoning is more complex). f[61,66] = 5+f(61), so f(61)=-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

Now, the discrepancy up to 64 is 2+f(43)+f(47)+f(53)+f(59), which must be 0, because of the cut. Therefore, exactly one of f(43), f(47), f(53), f(59) is one, and the others are -1. If f(43)=1, then f(47)=f(59)=-1, and f[89,96]=-5+f(89), and so f(89)=1. However, then f[169,178]=7+f(173), which is a contradiction, therefore f(43)=-1. Updating the diagram:

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

From f[123,134]=7+f(67)+f(127)+f(131), we can deduce that f(67)=f(127)=f(131)=-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

Also, we have that the discrepancy up to 48 is -5+f(47), so f(47)=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

We can now deduce (using the cut at 64), that f(53)=f(59)=-1. Also, we can now see that, from f[113,118]=-5+f(113), that 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

The discrepancy up to 110 (the cut) is -6+f(71)+f(73)+f(79)+f(83)+f(89)+f(97)+f(101)+f(103)+f(107)+f(109)=0. Therefore, exactly two of those primes are negative, and the other eight are positive. We also have that:

f[183,226] = 9-f(71)-f(73)+f(97)+f(101)+f(103)+f(107)+f(109)+f(191)+f(193)+f(197)+f(199)+f(211)+f(223)<=4. The last condition that we have is:

f[567,584]=7+f(71)+f(73)+f(83)-f(97)-f(191)-f(193)+f(569)+f(571)+f(577)<=4.

Setting f(197)=f(199)=f(211)=f(223)=f(569)=f(571)=f(577)=-1 to make the equations as easy as possible to satisfy, and to simplify them (because these terms only occur in one equation each), we now have that:

1-f(71)-f(73)+f(97)+f(101)+f(103)+f(107)+f(109)+f(191)+f(193)<=0,

f(71)+f(73)+f(83)-f(97)-f(191)-f(193)<=0

Adding the equations together, we have that:

1+f(83)+f(101)+f(103)+f(107)+f(109)<=0

Amazingly, almost all of the terms have cancelled. In order for this equation to hold, we must have that at least 3 of f(83), f(101), f(103), f(107), f(109) are negative. However, this contradicts the first condition (the cut at 110). Therefore, these conditions are impossible to satisfy. Therefore, f(31) must be -1 if we are to progress past 584.