%I #69 Jan 06 2020 14:13:19
%S 1,1,-1,-1,9,-23,-25,583,-3087,4401,79087,-902097,4783801,2361049,
%T -348382697,4102879415,-24288551071,-47413121055,3214104039007,
%U -44472852461857,326386562502889,417716032223049,-55104307651136313,962111031220099495
%N E.g.f. exp( sinh(x) / exp(x) ) = exp( (1-exp(-2*x))/2 ).
%C Hankel transform is (-1)^binomial(n+1,2)*A108400. - _Paul Barry_, Apr 15 2010
%H Seiichi Manyama, <a href="/A009235/b009235.txt">Table of n, a(n) for n = 0..529</a>
%H I. M. Gessel, <a href="http://arxiv.org/abs/math.CO/0108121">Applications of the classical umbral calculus</a>, arXiv:math/0108121 [math.CO], 2001.
%F a(n) = Sum_{k=0..n} (-2)^(n-k)*Stirling2(n, k). - _Vladeta Jovovic_, Apr 04 2003
%F From _Peter Bala_, May 16 2012: (Start)
%F Recurrence equation: a(n+1) = Sum_{k = 0..n} (-2)^(n-k)*C(n,k)*a(k). Written umbrally this is a(n+1) = (a-2)^n (expand the binomial and replace a^k with a(k)). More generally, a*f(a) = f(a-2) holds umbrally for any polynomial f(x). An inductive argument then establishes the umbral recurrence a*(a+2)*(a+4)*...*(a+2*(n-1)) = 1 with a(0) = 1. Cf. A004211.
%F Touchard's congruence holds for odd prime p: a(p+k) = (a(k) + a(k+1)) (mod p) for k = 0,1,2, ... (adapt the proof of Theorem 10.1 in Gessel). In particular, a(p) = 2 (mod p) for odd prime p. (End)
%F From _Sergei N. Gladkovskii_, Sep 21 2012 - Oct 24 2013: (Start)
%F Continued fractions:
%F G.f.: (1/E(0)-1)/x where E(k)= 1 - x/(1 - 2*x + 2*x*(k+1)/E(k+1));
%F G.f.: 1 +x/G(0) where G(k)= 1 + 2*x/(1 + 1/(1 + 4*x*(k+1)/G(k+1)));
%F G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1+x*2*k)/(1-x/(x-1/G(k+1)));
%F G.f.: 1/Q(0) where Q(k)= 1 - x/(1 + 2*x*(k+1)/Q(k+1) );
%F G.f.: Q(0)/(1-x), where Q(k) = 1 - 2*x^2*(k+1)/( 2*x^2*(k+1) + (1-x+2*x*k)*(1+x+2*x*k)/Q(k+1)). (End)
%F Lim sup n->infinity (abs(a(n))/n!)^(1/n) / (2*abs(exp(1/LambertW(-2*n)) / LambertW(-2*n))) = 1. - _Vaclav Kotesovec_, Aug 04 2014
%F a(n) = (-2)^n*B_n(-1/2), where B_n(x) is n-th Bell polynomial. - _Vladimir Reshetnikov_, Oct 20 2015
%F G.f. A(x) satisfies: A(x) = 1 + x*A(x/(1 + 2*x))/(1 + 2*x). - _Ilya Gutkovskiy_, May 02 2019
%p a := n -> (-2)^n*add(Stirling2(n,k)*(-1/2)^k, k=0..n):
%p seq(a(n), n=0..23); # _Peter Luschny_, Jan 06 2020
%t With[{nn=30},CoefficientList[Series[Exp[Sinh[x]/Exp[x]],{x,0,nn}],x]Range[0,nn]!] (* _Harvey P. Dale_, Jan 07 2013 *)
%t Table[(-2)^n BellB[n, -1/2], {n, 0, 20}] (* _Vladimir Reshetnikov_, Oct 20 2015 *)
%o (PARI) x='x+O('x^66); /* that many terms */
%o v=Vec(serlaplace(exp(sinh(x)/exp(x)))) /* _Joerg Arndt_, May 19 2012 */
%Y Cf. A004211, A317996, A318179, A318180, A318181.
%K sign,easy
%O 0,5
%A _R. H. Hardin_
%E Extended with signs by _Olivier GĂ©rard_, Mar 15 1997