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A233578
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n >= 2 such that the denominator/6 of Bernoulli(n) is congruent to {1, 5, 7, 13 or 19} modulo 30.
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3
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2, 4, 6, 8, 12, 14, 18, 24, 26, 34, 36, 38, 40, 42, 54, 62, 68, 70, 72, 74, 76, 78, 86, 88, 94, 98, 100, 102, 108, 110, 114, 118, 120, 122, 124, 126, 130, 134, 142, 146, 152, 158, 162, 182, 186, 188, 190, 194, 196, 202, 204, 206, 208, 210, 214, 216, 218, 220, 222, 228, 230, 232, 234
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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 1 modulo 6. If my conjecture is true, then you can obtain the residue modulo 6 of the abs. value of Bernoulli numerators by calculating their denominators/6 modulo 30. Program uses the von Staudt-Clausen Theorem. None of these n are in the complementary sequence, A233579 (n such that the denominator/6 of Bernoulli(n) is congruent to {11, 17, 23, 25 or 29} 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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100 is in this sequence, because the denominator of Bernoulli(100) = 33330, and 33330/6 = 5555, and 5555 is congruent to 5 modulo 30. As for the conjecture, the abs. val. of the numerator of Bernoulli(100) is congruent to 1 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=1 or x=5 or x=7 or x=13 or x=19) 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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