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%I #35 Jan 01 2023 09:03:41
%S 1,0,-1,0,7,0,-31,0,127,0,-2555,0,1414477,0,-57337,0,118518239,0,
%T -5749691557,0,91546277357,0,-1792042792463,0,1982765468311237,0,
%U -286994504449393,0,3187598676787461083,0,-4625594554880206790555,0,16555640865486520478399,0
%N Numerator of Bernoulli(n, 1/2).
%C Included for completeness, normally alternating zeros like this are omitted. A001896 is the official version of this sequence.
%C 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
%H Vincenzo Librandi, <a href="/A157779/b157779.txt">Table of n, a(n) for n = 0..250</a>
%H Wolfdieter Lang, <a href="http://arXiv.org/abs/1707.04451">On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli Numbers</a>, arXiv:math/1707.04451 [math.NT], July 2017.
%F 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
%F 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
%F a(n) = numerator(-(1-2^(1-n))*Bernoulli(n)). - _Fabián Pereyra_, Dec 31 2022
%t Numerator[BernoulliB[Range[0,40],1/2]] (* _Harvey P. Dale_, May 04 2013 *)
%o (Sage)
%o def A157779_list(size):
%o f = x / sum(x^(n*2+1)/factorial(n*2+1) for n in (0..2*size))
%o t = taylor(f, x, 0, size)
%o return [(factorial(n)*s).numerator() for n,s in enumerate(t.list())]
%o print(A157779_list(33)) # _Peter Luschny_, Jul 05 2016
%o (PARI) a(n) = numerator(subst(bernpol(n, x), x, 1/2)); \\ _Altug Alkan_, Jul 05 2016
%Y For denominators see A157780 and A141459.
%K sign,frac
%O 0,5
%A _N. J. A. Sloane_, Nov 08 2009
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