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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).
5
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
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
KEYWORD
nonn,frac
AUTHOR
N. J. A. Sloane, Dec 03 2004
STATUS
approved