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A233579
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Numbers n such that the denominator/6 of Bernoulli(n) is congruent to {11, 17, 23, 25 or 29} modulo 30.
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3
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10, 16, 20, 22, 28, 30, 32, 44, 46, 48, 50, 52, 56, 58, 60, 64, 66, 80, 82, 84, 90, 92, 96, 104, 106, 112, 116, 128, 132, 136, 138, 140, 144, 148, 150, 154, 156, 160, 164, 166, 168, 170, 172, 174, 176, 178, 180, 184, 192, 198, 200, 212, 224, 226, 238, 240, 242, 246, 252, 260, 262, 268
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OFFSET
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1,1
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COMMENTS
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Conjecture: for these and only these n, the absolute value of the numerator of Bernoulli(n) is congruent 5 modulo 6. If this is true, then you can obtain the residue modulo 6 of the absolute value of Bernoulli numerators by calculating their denominators/6 modulo 30. The program uses the von Staudt-Clausen Theorem. None of these n are in the complementary sequence, A233578 (n >= 2 such that the denominator/6 of Bernoulli_n is congruent to {1, 5, 7, 13 or 19} modulo 30). I have checked and verified that, up to n = 50446, the union of A233578 and A233579 is all even numbers >= 2.
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LINKS
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EXAMPLE
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112 is in this sequence, because the denominator of Bernoulli(112) = 1671270, and 1671270/6 = 278545, and 278545 is congruent to 25 modulo 30. As for the conjecture, the absolute value of the numerator of Bernoulli(112) is congruent to 5 modulo 6.
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PROG
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(Maxima) float(true)$ load(basic)$ i:[1]$ n:2$ for r:1 thru 10000 step 0 do (for p:3 while p-1<=n step 0 do (p:next_prime(p), if mod(n, p-1)=0 then push(p, i)), d:(product(i[k], k, 1, length(i))), x:mod(d, 30), if (x=11 or x=17 or x=23 or x=25 or x=29) then (print(r, ", ", n), r:r+1), i:[1], n:n+2)$
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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