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%I #35 Jan 06 2018 07:13:28
%S 1,3,50,2100,166824,21538440,4115105280,1091804313600,384202115256960,
%T 173201547619900800,97349279409046828800,66747386996603337024000,
%U 54838533307770850530816000,53185913922332495626882560000
%N G.f. A(x) = sum(n>0, a(n)*x^n/(2*n-1)!) is the inverse function to x*Bernoulli(x).
%C r(n)=sum(A191578(n,k)*k!/(n!*(n-k)!)*a(k)/(2*k-1)!,k,1,n)=0, n>1. r(1)=1.
%C The central column of the Worpitzky triangle, a(n) = A028246(2n, n). _Peter Luschny_, Jul 17 2012
%H G. C. Greubel, <a href="/A185157/b185157.txt">Table of n, a(n) for n = 1..215</a>
%H Vladimir Kruchinin, <a href="http://arxiv.org/abs/1211.3244">The method for obtaining expressions for coefficients of reverse generating functions</a>, arXiv:1211.3244 [math.CO], 2012.
%F a(n) = (n-1)!*stirling2(2*n-1,n).
%F a(n) = (1/n)*sum{i=0..n}(-1)^(n-i)*binomial(n,i)*i^(2*n-1) - _Peter Luschny_, Jul 17 2012
%F O.g.f.: Sum_{n>=1} n^(2*n-2)*x^n/(1 + n^2*x)^n = Sum_{n>=1} a(n)*x^n. - _Paul D. Hanna_, Jan 06 2018
%t a[n_] := (n-1)!*StirlingS2[2*n-1, n]; Table[a[n], {n, 1, 14}] (* _Jean-François Alcover_, Feb 21 2013, from 1st formula *)
%o (Maxima) a(n)=(n-1)!*stirling2(2*n-1,n);
%o (Sage)
%o def A185157(n) :
%o return (1/n)*add((-1)^(n-i)*binomial(n,i)*i^(2*n-1) for i in (0..n))
%o [A185157(n) for n in (1..14)] # _Peter Luschny_, Jul 17 2012
%Y Cf. A028246, A191578.
%K nonn
%O 1,2
%A _Vladimir Kruchinin_, Jan 23 2012
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