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.

Case 1.1.1: f(2)=f(7)=f(19)=f(23)=f(37)=f(53)=f(67)=f(71)=f(79)=1, f(29)=f(31)=f(43)=-1

Now, f[283,290] = 5+f(283), so f(283)=-1. Also, f[235,248] = -6+f(239)+f(241), so f(239)=f(241)=1. Another easy deduction is: f[227,238] = -7+f(227)+f(229)+f(233), so f(227)=f(229)=f(233)=1.

There are three block inequalities that need to be resolved:

1) f[169,188] = 6+f(89)+f(173)+f(179)+f(181) <= 4

2) f[715,726] = -5+f(103)+f(179)+f(181)+f(359)+f(719) >= -4

3) s(106) = 3+f(89)+f(101)+f(103) <= 2

The most useful thing I can get out of these equations is:

(2)-(1)-(3)+14: -2f(89)-f(101)-f(173)+f(359)+f(719) >= 4. Therefore, f(89)=-1.

Now, f[437,454] = 10-f(149)-f(151)+f(223)+f(439)+f(443)+f(449) <= 4, so f(149)=f(151)=1, and f(223)=f(439)=f(443)=f(449)=-1. Then we have that: f[129,152] = 7+f(131)+f(137)+f(139), so f(131)=f(137)=f(139)=-1. Also, f[127,134]=5+f(127), so f(127)=-1, and f[143,158]=5+f(157), so f(157)=-1. f[867,872] = -6+f(109), so f(109)=1, and f[527,536] = 5-f(107), so f(107)=1. Therefore, since s(110)=4+f(101)+(103), so f(101)=f(103)=-1. Finally, f[543,552] = -6-f(181)+f(547), so f(181)=-1, and f(547)=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
- - + + + + - - + +   140-149
- + + - - + + + + -   150-159

Now, f[271,282] = -7+f(271)+f(277)+f(281), so f(271)=f(277)=f(281)=1. Also, f[723,728] = -5+f(727), so f(727)=1.

We have two inequalities that need to be satisfied:

1) f[1073,1078] = -3-f(179)+f(269)-f(359) >= -4 2) f[715,726] = -7+f(179)+f(359)+f(719) >= -4

(1)+(2): -10+f(269)+f(719) >= -8

Therefore, f(269)=f(719)=1. Now, f[263,270] = 5+f(263), so f(263) = -1. Also, f[527,542] = 5+f(541), so f(541)=-1. f[783,796] = 7+f(197)+f(199)+f(787), so f(197)=f(199)=f(787)=-1. Also, f[189,202] = -6+f(191)+f(193), so f(191)=f(193)=1.