A233578 n >= 2 such that the denominator/6 of Bernoulli(n) is congruent to {1, 5, 7, 13 or 19} modulo 30.

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

Offset: 1

Author

Michael G. Kaarhus, Dec 13 2013

Keywords

• nonn

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 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.

Examples

			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.
		

Links

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

Crossrefs

Cf. A233579, subsequence of A005843.

Programs

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)$