A025168 Expansion of e.g.f.: exp(x/(1-2*x)).

1, 1, 5, 37, 361, 4361, 62701, 1044205, 19748177, 417787921, 9770678101, 250194150581, 6959638411705, 208919770666777, 6729933476435261, 231512615111396221, 8469125401589550241, 328241040596380393505, 13434223364220816489637, 578931271898150002093381

Offset: 0

Author

Wouter Meeussen

Keywords

• nonn
• easy

Comments

From Peter Bala, Nov 21 2017: (Start)

The sequence terms have the form 4*m + 1 (follows from the recurrence).

For k = 2,3,4,... the difference a(n+k) - a(n) is divisible by k (proof by induction on n making use of the recurrence - the starting case a(k) == a(0) (mod k) for all k follows from the sum formula for a(k)). Hence for each k, the sequence b(n) == a(n) (mod k) is periodic with the exact period dividing k. (End)

Compound Poisson distribution with parameter 1 and distribution Geometric(1/2) has a probability mass function p_n = a(n)*e^(-1/2)/(4^n*n!). More specifically, let S = Sum_{i=0..N} X_i where X_i's are i.i.d. random variables with Geometric(1/2) distribution (i.e., Pr{X_i = k} = 1/2^(k+1) for k=0,1,2...) and N is a random variable with Poisson(1) distribution independent of all X_i's. Then Pr{S=n} = a(n)*e^(-1/2)/(4^n*n!) = a(n)*e^(-1/2)/A047053(n) for nonnegative integers n. - Xiaohan Zhang, Nov 16 2022

Links

• Robert Israel, Table of n, a(n) for n = 0..400
• Norihiro Nakashima and Shuhei Tsujie, Enumeration of Flats of the Extended Catalan and Shi Arrangements with Species, arXiv:1904.09748 [math.CO], 2019.
• K. A. Penson, P. Blasiak, G. Duchamp, A. Horzela and A. I. Solomon, Hierarchical Dobinski-type relations via substitution and the moment problem, arXiv:quant-ph/0312202, 2003; J. Phys. A 37 (2004), 3475-3487.
• N. J. A. Sloane, Transforms
• N. J. A. Sloane and Thomas Wieder, The Number of Hierarchical Orderings, arXiv:math/0307064 [math.CO], 2003; Order 21 (2004), 83-89.
• Thomas Wieder, Expanded definitions of A103446 and A025168

Crossrefs

Cf. A000012, A000262, A103446.

Programs

Maple

with(combstruct); SetSeqSeqL := [T, {T=Set(S), S=Sequence(U,card >= 1), U=Sequence(Z,card >=1)},labeled];
f:= gfun:-rectoproc({a(n) = (4*n-3)*a(n-1) - 4*(n-2)*(n-1)*a(n-2),a(0)=1,a(1)=1},a(n),remember):
map(f, [$0..30]); # Robert Israel, Nov 21 2017

Mathematica

Table[ n! 2^n LaguerreL[ n, 1, -1/2 ], {n, 0, 12} ]
With[{nn=20},CoefficientList[Series[Exp[x/(1-2x)],{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Aug 12 2012 *)

Sage

A025168 = lambda n: hypergeometric([-n,-n+1], [], 2)
[Integer(A025168(n).n(100)) for n in range(20)] # Peter Luschny, Sep 22 2014

Formula

Second LAH transform of A000012. LAH transform of A000262. a(n) = Sum_{k=0..n} 2^(n-k)*n!/k!*binomial(n-1, k-1). - Vladeta Jovovic, Oct 17 2003

Define f_1(x), f_2(x), ... such that f_1(x) = e^x, f_{n+1}(x) = (d/dx)(x^2*f_n(x)), for n=2,3,.... Then a(n) = e^(-1/2)*4*(n-1)*f_n(1/2). - Milan Janjic, May 30 2008

From Vaclav Kotesovec, Jun 22 2013: (Start)

D-finite with recurrence: a(n) = (4*n-3)*a(n-1) - 4*(n-2)*(n-1)*a(n-2).

a(n) ~ 2^(n-3/4)*n^(n-1/4)*exp(sqrt(2*n)-n-1/4) * (1-1/(3*sqrt(2*n))).

(End)

E.g.f.: E(0)/2, where E(k) = 1 + 1/(1 - x/(x + (k+1)*(1-2*x)/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 09 2013

a(n) = hypergeometric([-n,-n+1],[],2). - Peter Luschny, Sep 22 2014

Sum_{n>=0} a(n)/(4^n*n!) = sqrt(e) = A019774. -Xiaohan Zhang, Nov 16 2022

Extensions

Corrected and extended by Vladeta Jovovic, Sep 08 2002