OFFSET
0,5
COMMENTS
Included for completeness, normally alternating zeros like this are omitted. A001896 is the official version of this sequence.
The sequence {a(n)/A141459(n)} gives the generalized Bernoulli numbers B[2,1] obtained from the generalized Stirling2 triangle S3[2,1] = A154537. See the formula section. - Wolfdieter Lang, Apr 27 2017
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..250
Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli Numbers, arXiv:math/1707.04451 [math.NT], July 2017.
FORMULA
Let P(x) = Sum_{n>=0} x^(2*n+1)/(2*n+1)!; then a(n) = numerator( n! [x^n] x/P(x) ). - Peter Luschny, Jul 05 2016
a(n) = numerator(r(n)) with the rationals r(n) = Sum_{k=0..n} ((-1)^k / (k+1))*A154537(n, k)*k! = Sum_{k=0..n} ((-1)^k/(k+1))*A145901(n, k). The denominators are in A141459. r(n) = B[2,1](n) = 2^n*B(n, 1/2) with the Bernoulli polynomials A196838/A196839 or A053382/A053383. - Wolfdieter Lang, Apr 27 2017
a(n) = numerator(-(1-2^(1-n))*Bernoulli(n)). - Fabián Pereyra, Dec 31 2022
MATHEMATICA
Numerator[BernoulliB[Range[0, 40], 1/2]] (* Harvey P. Dale, May 04 2013 *)
PROG
(Sage)
def A157779_list(size):
f = x / sum(x^(n*2+1)/factorial(n*2+1) for n in (0..2*size))
t = taylor(f, x, 0, size)
return [(factorial(n)*s).numerator() for n, s in enumerate(t.list())]
print(A157779_list(33)) # Peter Luschny, Jul 05 2016
(PARI) a(n) = numerator(subst(bernpol(n, x), x, 1/2)); \\ Altug Alkan, Jul 05 2016
CROSSREFS
KEYWORD
sign,frac
AUTHOR
N. J. A. Sloane, Nov 08 2009
STATUS
approved