Side Proof 3
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)=f(43)=1, f(29)=f(31)=-1.
Proof
If we also assume that f(47) is also positive, then the discrepancy at 66 is equal to 5+f(53)+f(59)+f(61), so we have that f(53)=f(59)=f(61)=-1. Given this, we have that f[527,534]=5-f(89), so f(89)=1. But then f[169,178]=7+f(173), forcing the discrepancy to be greater than 3. Therefore f(47)=-1. Similarly, if f(61)=1, then f(53)=f(59)=-1, and the same problem occurs, so f(61)=-1. If f(59)=1, then f(53)=-1, and f[527,534]=7-f(89), which is impossible. Therefore f(59)=-1. The discrepancy up to 62 is -3+f(53), so f(53)=1. Now, since f[113,118]=-5+f(113), 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 drift f[169,178] = 6+f(89)+f(173), so f(89)=f(173)=-1. Also, f[243,250]=-5-f(83), so f(83)=-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 - + + ? - - - ? - + 160-169 + + + - + + - + - ? 170-179
It seems that once again we hit a point where nothing can be derived.
Case 1: f(2)=f(7)=f(19)=f(23)=f(37)=f(43)=f(67)=f(71)=1, f(29)=f(31)=-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 - + + ? - - - ? - + 160-169 + + + - + + - + - ? 170-179
The discrepancy up to 74 is 3+f(73), so f(73)=-1. Now f[425,436]=-8+f(107)+f(109)+f(431)+f(433), so f(107)=f(109)=f(431)=f(433)=1. Also, because of the cut at 90, we must have that f(79)=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 - + + ? - - - ? - + 160-169 + + + - + + - + - ? 170-179
Now, since f[319,336]=-7+f(163)+f(167)+f(331), we must have that f(163)=f(167)=f(331)=1. We now have that f[141,172]=7+f(149)+f(151)+f(157), so f(149)=f(151)=f(157)=-1. However, we now have that f[437,454]=11+f(223)+f(227)+f(439)+f(443)+f(449)>=6, which is a contradiction. Therefore f(71)=-1
Case 2: f(2)=f(7)=f(19)=f(23)=f(37)=f(43)=f(67)=1, f(29)=f(31)=f(71)=-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 - + + ? - - - ? - + 160-169 + + + - + + - + - ? 170-179
Because of the cut at 84, f(73)=f(79)=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 - + + ? - - - ? - + 160-169 + + + - + + - + - ? 170-179
Now, f[285,298]=7-f(97)+f(149)+f(293). Therefore f(97)=1, f(149)=f(293)=-1. We also have that f[341,358]=8+f(179)+f(347)+f(349)+f(353). Therefore, f(179)=f(347)=f(349)=f(353)=-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 - + + ? - - - ? - + 160-169 + + + - + + - + - - 170-179
We also have that f(223)=f(449)=-1, because f[445,452]=6+f(223)+f(449). Now, we have six blocks:
(1) f[715,726] = -4+f(103)+f(181)-f(239)-f(241)+f(359)+f(719) >= -4 (2) f[341,362] = 4+f(181)+f(359) <= 4 (3) f[215,248] = -8+f(109)+f(227)+f(229)+f(233)+f(239)+f(241) >= -4 (4) f[325,336] = -6-f(109)+f(163)+f(167)+f(331) >= -4 (5) f[143,172] = 4+f(151)+f(157)+f(163)+f(167) <= 4 (6) f[447,454] = 4-f(151)+f(227) <= 4
Moving the constants to the RHS, we have:
(1) f(103)+f(181)-f(239)-f(241)+f(359)+f(719) >= 0 (2) f(181)+f(359) <= 0 (3) f(109)+f(227)+f(229)+f(233)+f(239)+f(241) >= 4 (4) -f(109)+f(163)+f(167)+f(331) >= 2 (5) f(151)+f(157)+f(163)+f(167) <= 0 (6) -f(151)+f(227) <= 0
adding and subtracting the inequalities, like (1)-(2)+(3)+(4)-(5)-(6), we get:
f(103)-f(157)+f(229)+f(233)+f(331)+f(719) >= 6. Therefore f(103)=f(229)=f(233)=f(331)=f(719)=1, f(157)=-1. Now, the discrepancy up to 106 is 3+f(101), so f(101)=-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 - + + ? - - - ? - + 160-169 + + + - + + - + - - 170-179
Since f[285,304]=5+f(151)<=4, we have that f(151)=-1. Now, from f[447,454] = 5+f(227), we get f(227)=-1. Also, from f[215,248], we have: -7+f(109)+f(239)+f(241)>=-4, so f(109)=f(239)=f(241)=1. By f[325,336] = -6+f(163)+f(167), f(163)=f(167)=1. The discrepancy up to 110 is 3+f(107), so f(107)=-1. However, now f[527,536]=6, which is a contradiction.
Case 3: f(2)=f(7)=f(19)=f(23)=f(37)=f(43)=1, f(29)=f(31)=f(67)=-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 - + + ? - - - ? - + 160-169 +|+ + - + + - + - ? 170-179
The discrepancy up to 96 is -5+f(71)+f(73)+f(79). Therefore, f(71)=f(73)=f(79)=1 (remember f(83)=-1 because of f[243,250]). However, we now have that f[141,146] = 6, which is a contradiction. We can therefore conclude that the assumption f(43)=1 fails at 726.
