

A233579


Numbers n such that the denominator/6 of Bernoulli(n) is congruent to {11, 17, 23, 25 or 29} modulo 30.


3



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

1,1


COMMENTS

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 StaudtClausen 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.


LINKS

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


EXAMPLE

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.


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=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)$


CROSSREFS

Cf. A233578, subsequence of A005843.
Sequence in context: A228491 A295667 A241763 * A291639 A004260 A129848
Adjacent sequences: A233576 A233577 A233578 * A233580 A233581 A233582


KEYWORD

nonn


AUTHOR

Michael G. Kaarhus, Dec 13 2013


STATUS

approved



