

A233578


n >= 2 such that the denominator/6 of Bernoulli(n) is congruent to {1, 5, 7, 13 or 19} modulo 30.


3



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

1,1


COMMENTS

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 StaudtClausen 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.


LINKS

Michael G. Kaarhus, Table of n, a(n) for n = 1..10000
M. G. Kaarhus, Splitting the Bernoulli Numbers


EXAMPLE

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.


PROG

(Maxima) float(true)$ load(basic)$ i:[1]$ n:2$ for r:1 thru 10000 step 0 do (for p:3 while p1<=n step 0 do (p:next_prime(p), if mod(n, p1)=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)$


CROSSREFS

Cf. A233579, subsequence of A005843.
Sequence in context: A089681 A227308 A214294 * A057220 A294847 A082742
Adjacent sequences: A233575 A233576 A233577 * A233579 A233580 A233581


KEYWORD

nonn


AUTHOR

Michael G. Kaarhus, Dec 13 2013


STATUS

approved



