A009235 E.g.f. exp( sinh(x) / exp(x) ) = exp( (1-exp(-2*x))/2 ).
1, 1, -1, -1, 9, -23, -25, 583, -3087, 4401, 79087, -902097, 4783801, 2361049, -348382697, 4102879415, -24288551071, -47413121055, 3214104039007, -44472852461857, 326386562502889, 417716032223049, -55104307651136313, 962111031220099495
Offset: 0
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..529
- I. M. Gessel, Applications of the classical umbral calculus, arXiv:math/0108121 [math.CO], 2001.
Programs
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Maple
a := n -> (-2)^n*add(Stirling2(n,k)*(-1/2)^k, k=0..n): seq(a(n), n=0..23); # Peter Luschny, Jan 06 2020
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Mathematica
With[{nn=30},CoefficientList[Series[Exp[Sinh[x]/Exp[x]],{x,0,nn}],x]Range[0,nn]!] (* Harvey P. Dale, Jan 07 2013 *) Table[(-2)^n BellB[n, -1/2], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 20 2015 *) -
PARI
x='x+O('x^66); /* that many terms */ v=Vec(serlaplace(exp(sinh(x)/exp(x)))) /* Joerg Arndt, May 19 2012 */
Formula
a(n) = Sum_{k=0..n} (-2)^(n-k)*Stirling2(n, k). - Vladeta Jovovic, Apr 04 2003
From Peter Bala, May 16 2012: (Start)
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.
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)
From Sergei N. Gladkovskii, Sep 21 2012 - Oct 24 2013: (Start)
Continued fractions:
G.f.: (1/E(0)-1)/x where E(k)= 1 - x/(1 - 2*x + 2*x*(k+1)/E(k+1));
G.f.: 1 +x/G(0) where G(k)= 1 + 2*x/(1 + 1/(1 + 4*x*(k+1)/G(k+1)));
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1+x*2*k)/(1-x/(x-1/G(k+1)));
G.f.: 1/Q(0) where Q(k)= 1 - x/(1 + 2*x*(k+1)/Q(k+1) );
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)
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
a(n) = (-2)^n*B_n(-1/2), where B_n(x) is n-th Bell polynomial. - Vladimir Reshetnikov, Oct 20 2015
G.f. A(x) satisfies: A(x) = 1 + x*A(x/(1 + 2*x))/(1 + 2*x). - Ilya Gutkovskiy, May 02 2019
Extensions
Extended with signs by Olivier Gérard, Mar 15 1997
Comments