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

Author

N. J. A. Sloane, Dec 03 2004

Keywords

• nonn
• frac

Examples

			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.
		

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

Crossrefs

Cf. A001898, A027641, A027642, A100615.

Programs

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 *)

PARI

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

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