A100616 Let B(n)(x) be the Bernoulli polynomials as defined in A001898, with B(n)(1) equal to the usual Bernoulli numbers A027641/A027642. Sequence gives denominators of B(n)(2).

1, 1, 6, 2, 10, 6, 42, 6, 30, 10, 22, 6, 2730, 210, 6, 2, 34, 30, 798, 42, 330, 110, 46, 6, 2730, 546, 6, 2, 290, 30, 14322, 462, 510, 170, 2, 6, 54834, 51870, 6, 2, 4510, 330, 1806, 42, 690, 46, 94, 6, 46410, 6630, 66, 22, 530, 30, 798, 798, 174, 290, 118, 6, 56786730

Offset

0,3

References

F. N. David, Probability Theory for Statistical Methods, Cambridge, 1949; see pp. 103-104. [There is an error in the recurrence for B_s^{(r)}.]

Links

Robert Israel, Table of n, a(n) for n = 0..1000

Formula

E.g.f.: (x/(exp(x)-1))^2. - Vladeta Jovovic, Feb 27 2006

a(n) = denominator(Sum_{j=0..n} binomial(n,j)*Bernoulli(n-j)*Bernoulli(j)). - Fabián Pereyra, Mar 02 2020

Example

1, -1, 5/6, -1/2, 1/10, 1/6, -5/42, -1/6, 7/30, 3/10, -15/22, -5/6, 7601/2730, 691/210, -91/6, -35/2, 3617/34, 3617/30, -745739/798, -43867/42, ... = A100615/A100616.

Maple

S:= series((x/(exp(x)-1))^2, x, 101):
seq(denom(coeff(S, x, n)*n!), n=0..100); # Robert Israel, Jun 02 2015

Mathematica

Table[Denominator@NorlundB[n, 2], {n, 0, 59}] (* Arkadiusz Wesolowski, Oct 22 2012 *)

Prog

(PARI) a(n) = denominator(sum(j=0, n, binomial(n, j)*bernfrac(n-j)*bernfrac(j))); \\ Michel Marcus, Mar 03 2020

Crossrefs

Cf. A001898, A027641, A027642, A100615.

Sequence in context: A260329 A141379 A357396 * A097474 A194351 A040035

Adjacent sequences: A100613 A100614 A100615 * A100617 A100618 A100619

Keyword

nonn,frac

Author

N. J. A. Sloane, Dec 03 2004

Status

approved