A111884 E.g.f.: exp(x/(1+x)).

1, 1, -1, 1, 1, -19, 151, -1091, 7841, -56519, 396271, -2442439, 7701409, 145269541, -4833158329, 104056218421, -2002667085119, 37109187217649, -679877731030049, 12440309297451121, -227773259993414719, 4155839606711748061, -74724654677947488521, 1293162252850914402221

Offset: 0

Author

Wolfdieter Lang, Aug 23 2005

Keywords

• sign
• easy

Comments

Row sums of triangle A111596.

With different signs see A066668.

From Peter Bala, Aug 15 2022: (Start)

The congruence a(n+k) == a(n) (mod k) holds for all n and k.

It follows that the sequence obtained by taking a(n) modulo a fixed positive integer k is periodic with period dividing k. For example, taken modulo 10 the sequence becomes [1, 1, 9, 1, 1, 1, 1, 9, 1, 1, ...], a purely periodic sequence with period 5. More generally, the same property holds for any sequence with an e.g.f. of the form F(x)*exp(x*G(x)), where F(x) and G(x) are power series with integer coefficients and G(0) = 1. (End)

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..450
• Paul Barry, The Restricted Toda Chain, Exponential Riordan Arrays, and Hankel Transforms, J. Int. Seq. 13 (2010) # 10.8.4, example 4.
• Paul Barry, Exponential Riordan Arrays and Permutation Enumeration, J. Int. Seq. 13 (2010) # 10.9.1, example 6.
• Paul Barry, Riordan Arrays, Orthogonal Polynomials as Moments, and Hankel Transforms, J. Int. Seq. 14 (2011) # 11.2.2, example 20.
• Paul Barry, Combinatorial Polynomials as Moments, Hankel Transforms, and Exponential Riordan Arrays, J. Int. Seq. 14 (2011) 11.6.7, example 10.
• Bérénice Delcroix-Oger and Clément Dupont, Lie-operads and operadic modules from poset cohomology, arXiv:2505.06094 [math.CO], 2025. See p. 32.
• Aoife Hennessy and Paul Barry, Generalized Stirling Numbers, Exponential Riordan Arrays, and Orthogonal Polynomials, J. Int. Seq. 14 (2011) # 11.8.2.

Crossrefs

Unsigned row sums of A111596: A000262.

Programs

Mathematica

nn=30; CoefficientList[Series[Exp[x/(1+x)],{x,0,nn}], x] Range[0,nn]! (* Harvey P. Dale, Jul 21 2011 *)

Sage

A111884 = lambda n: hypergeometric([-n+1,-n], [], -1)
[Integer(A111884(n).n(100)) for n in (0..23)] # Peter Luschny, Sep 23 2014

Formula

E.g.f.: exp(x/(1+x)).

From Sergei N. Gladkovskii, Jul 21 2012: (Start)

Let E(x) be the e.g.f., then

E(x) = 1/G(0) where G(k)= 1 - x/((1+x)*(2*k+1) - x*(1+x)*(2*k+1)/(x - (1+x)*(2*k+2)/G(k+1))); (continued fraction, 3rd kind, 3-step).

E(x) = 1 + x/(G(0)-x) where G(k)= 1 + 2*x + (1+x)*k - x*(1+x)*(k+1)/G(k+1); (continued fraction, Euler's 1st kind, 1-step).

E(x) = G(0) where G(k)= 1 + x/((1+x)*(2*k+1) - x*(1+x)*(2*k+1)/(x + 2*(1+x)*(k+1)/G(k+1))); (continued fraction, 3rd kind, 3-step).

(End)

E.g.f.: 1 + x*(E(0)-1)/(x+1) where E(k) = 1 + 1/(k+1)/(1+x)/(1-x/(x+1/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 27 2013

E.g.f.: E(0)/2, where E(k)= 1 + 1/(1 - x/(x + (k+1)*(1+x)/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 31 2013

a(n) = sum(k=0..n, (-1)^(n-k)*L(n,k)); L(n,k) the unsigned Lah numbers. - Peter Luschny, Oct 18 2014

a(n) = hypergeom([-n+1,-n],[],-1). - Peter Luschny, Apr 08 2015

D-finite with recurrence a(n) +(2*n-3)*a(n-1) +(n-1)*(n-2)*a(n-2)=0. - R. J. Mathar, Jul 20 2017