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

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 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.

Examples

			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.
		

Links

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

Crossrefs

Cf. A233578, 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=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)$