A233579 -id:A233579 - cp's OEIS frontend

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

Michael G. Kaarhus, Dec 13 2013

nonn

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.

			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.

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

Cf. A233579, subsequence of A005843.

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)$

A234788 Solutions to numerator(Bernoulli(k)) == denominator/6 (Bernoulli(k)) (mod 30).

Original entry on oeis.org

1, 23, 1, 1, 23, 1, 1, 1, 19, 1, 1, 1, 23, 23, 1, 1, 17, 13, 1, 23, 7, 1, 23, 23, 1, 23, 11, 1, 23, 7, 1, 1, 1, 1, 19, 7, 1, 1, 7, 17, 1, 1, 1, 23, 19, 19, 1, 7, 1, 23, 7, 1, 1, 19, 1, 7, 23, 1, 1, 7, 1, 23, 29, 1, 23, 13, 1, 23, 7, 1, 1, 19, 1, 1, 19, 1, 23

Offset: 1

Michael G. Kaarhus, Dec 30 2013

nonn

Conjecture: the residues mod 30 of the numerator and the denominator/6 of Bernoulli(20(n-1) + 2) are equal. Conjecture: the only solutions to the above equation are {1, 7, 11, 13, 17, 19, 23 or 29}. Observation: the differences between these solutions are (6, 4, 2, 4, 2, 4, 6), a sequence with bilateral symmetry. Program checks all nonzero Bernoulli numbers except B(1), but if the above conjecture is true, then it needs check only every 20th Bernoulli Number starting with B(2).

			13 is in this sequence because both the numerator and the denominator/6 of a Bernoulli Number are congruent to 13 mod 30. Using my conjectural formula, you can find which Bernoulli Number: 13 is the 18th number in this sequence. k = 20(18-1) + 2. k = 342. So, both the numerator and the denominator/6 of Bernoulli(342) are congruent to 13 mod 30.

Michael G. Kaarhus, Table of n, a(n) for n = 1..300

Similar to A233578 and A233579.

k:-2$ for n:1 thru 300 step 0 do (k:k+2, b:bern(k), f:mod(num(b), 30), a:mod(denom(b)/6, 30), if f=a then (print(n, ", ", a), if 20*(n-1)+2#k then (print("Exception at k=", k, " n=", n), n:4000), n:n+1))$

This formula is conjectural, but the program verified it for each of the first 300 numbers in this sequence: to obtain k from the n of this sequence, k = 20(n-1) + 2.

cp's OEIS Frontend

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

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