A060311 Expansion of e.g.f. exp((exp(x)-1)^2/2).

1, 0, 1, 3, 10, 45, 241, 1428, 9325, 67035, 524926, 4429953, 40010785, 384853560, 3925008361, 42270555603, 478998800290, 5693742545445, 70804642315921, 918928774274028, 12419848913448565, 174467677050577515, 2542777209440690806, 38388037137038323353

Offset: 0

Author

Vladeta Jovovic, Mar 27 2001

Keywords

• nonn

Comments

After the first term, this is the Stirling transform of the sequence of moments of the standard normal (or "Gaussian") probability distribution. It is not itself a moment sequence of any probability distribution. - Michael Hardy (hardy(AT)math.umn.edu), May 29 2005

a(n) is the number of simple labeled graphs on n nodes in which each component is a complete bipartite graph. - Geoffrey Critzer, Dec 03 2011

References

• I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, John Wiley and Sons, N.Y., 1983, Ex. 3.3.5b.

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..518 (first 101 terms from Harry J. Smith)
• Paul Barry, Constructing Exponential Riordan Arrays from Their A and Z Sequences, Journal of Integer Sequences, 17 (2014), #14.2.6.
• Vaclav Kotesovec, Asymptotic solution of the equations using the Lambert W-function

Crossrefs

Column k=2 of A324162.

Cf. A052859, A330047.

Programs

Maple

a:= proc(n) option remember; `if`(n=0, 1, add(a(n-j)
      *binomial(n-1, j-1)*Stirling2(j, 2), j=2..n))
    end:
seq(a(n), n=0..25);  # Alois P. Heinz, Sep 02 2019

Mathematica

a = Exp[x] - 1; Range[0, 20]! CoefficientList[Series[Exp[a^2/2], {x, 0, 20}], x] (* Geoffrey Critzer, Dec 03 2011 *)

PARI

a(n)=if(n<0, 0, n!*polcoeff( exp((exp(x+x*O(x^n))-1)^2/2), n)) /* Michael Somos, Jun 01 2005 */

PARI

{ for (n=0, 100, write("b060311.txt", n, " ", n!*polcoeff(exp((exp(x + x*O(x^n)) - 1)^2/2), n)); ) } \\ Harry J. Smith, Jul 03 2009

PARI

a(n) = sum(k=0, n\2, (2*k)!*stirling(n, 2*k, 2)/(2^k*k!)); \\ Seiichi Manyama, May 07 2022

Formula

E.g.f. A(x) = B(exp(x)-1) where B(x)=exp(x^2/2) is e.g.f. of A001147(2n), hence a(n) is the Stirling transform of A001147(2n). - Michael Somos, Jun 01 2005

From Vaclav Kotesovec, Aug 06 2014: (Start)

a(n) ~ exp(1/2*(exp(r)-1)^2 - n) * n^(n+1/2) / (r^n * sqrt(exp(r)*r*(-1-r+exp(r)*(1+2*r)))), where r is the root of the equation exp(r)*(exp(r) - 1)*r = n.

(a(n)/n!)^(1/n) ~ 2*exp(1/LambertW(2*n)) / LambertW(2*n).

(End)

a(n) = Sum_{k=0..floor(n/2)} (2*k)! * Stirling2(n,2*k)/(2^k * k!). - Seiichi Manyama, May 07 2022