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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).

%I A100616 #22 Mar 26 2020 03:21:47
%S A100616 1,1,6,2,10,6,42,6,30,10,22,6,2730,210,6,2,34,30,798,42,330,110,46,6,
%T A100616 2730,546,6,2,290,30,14322,462,510,170,2,6,54834,51870,6,2,4510,330,
%U A100616 1806,42,690,46,94,6,46410,6630,66,22,530,30,798,798,174,290,118,6,56786730
%N 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).
%D A100616 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)}.]
%H A100616 Robert Israel, <a href="/A100616/b100616.txt">Table of n, a(n) for n = 0..1000</a>
%F A100616 E.g.f.: (x/(exp(x)-1))^2. - _Vladeta Jovovic_, Feb 27 2006
%F A100616 a(n) = denominator(Sum_{j=0..n} binomial(n,j)*Bernoulli(n-j)*Bernoulli(j)). - _Fabián Pereyra_, Mar 02 2020
%e A100616 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.
%p A100616 S:= series((x/(exp(x)-1))^2, x, 101):
%p A100616 seq(denom(coeff(S,x,n)*n!), n=0..100); # _Robert Israel_, Jun 02 2015
%t A100616 Table[Denominator@NorlundB[n, 2], {n, 0, 59}] (* _Arkadiusz Wesolowski_, Oct 22 2012 *)
%o A100616 (PARI) a(n) = denominator(sum(j=0, n, binomial(n,j)*bernfrac(n-j)*bernfrac(j))); \\ _Michel Marcus_, Mar 03 2020
%Y A100616 Cf. A001898, A027641, A027642, A100615.
%K A100616 nonn,frac
%O A100616 0,3
%A A100616 _N. J. A. Sloane_, Dec 03 2004