A233579 Numbers n such that the denominator/6 of Bernoulli(n) is congruent to {11, 17, 23, 25 or 29} modulo 30.

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

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 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.

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 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)$

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