A100615 Let N(n)(x) be the Nørlund polynomials as defined in A001898, with N(n)(1) equal to the usual Bernoulli numbers A027641/A027642. Sequence gives numerators of N(n)(2).

1, -1, 5, -1, 1, 1, -5, -1, 7, 3, -15, -5, 7601, 691, -91, -35, 3617, 3617, -745739, -43867, 3317609, 1222277, -5981591, -854513, 5436374093, 1181820455, -213827575, -76977927, 213745149261, 23749461029, -249859397004145, -8615841276005, 238988952277727, 84802531453387

Offset: 0

Author

N. J. A. Sloane, Dec 03 2004

Keywords

• sign
• frac

Comments

With the signs of A359738, the rational sequence reflects the identity B(z)^2 = (z + 1)*B(z) - z*B'(z), that goes back to Euler, where B(z) = z/(1 - e^(-z)) is the e.g.f. of the Bernoulli numbers with B(1) = 1/2. - Peter Luschny, Jan 23 2023

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..575
• Madeline Beals-Reid, A Quadratic Relation in the Bernoulli Numbers, The Pump Journal of Undergraduate Research, 6 (2023), 29-39.

Crossrefs

Cf. A001898, A027641, A027642, A100616, A359738.

Programs

Maple

S:= series((x/(exp(x)-1))^2, x, 41):
seq(numer(coeff(S,x,j)*j!), j=0..40); # Robert Israel, Jun 02 2015
# Second program:
a := n -> if n = 0 then 1 else numer(-n*bernoulli(n-1) - (n-1)*bernoulli(n)) fi:
seq(a(n), n = 0..33);  # Peter Luschny, May 18 2023

Mathematica

Table[Numerator@NorlundB[n, 2], {n, 0, 32}] (* Arkadiusz Wesolowski, Oct 22 2012 *)
Table[If[n == 0, 1, -Numerator[n*BernoulliB[n - 1] + (n - 1)*BernoulliB[n]]], {n, 0,  33}] (* Peter Luschny, May 18 2023 *)

Maxima

a(n):=sum((-1)^k*k!/(k+1)*sum(binomial(n,j)*stirling2(n-j,k)*bern(j),j,0,n-k),k,0,n); /* Vladimir Kruchinin, Jun 02 2015 */

PARI

a(n) = numerator(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) = numerator(Sum_{k=0..n}(-1)^k*k!/(k+1)*Sum_{j=0..n-k} C(n,j)*Stirling2(n-j,k)*B(j)), where B(n) is Bernoulli numbers. - Vladimir Kruchinin, Jun 02 2015

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

a(n) = -numerator(n*B(n-1) + (n-1)*B(n)) for n >= 1, where B(n) = Bernoulli(n, 0). - Peter Luschny, May 18 2023