A108459 Number of labeled partitions of (n,n) into pairs (i,j).

1, 1, 5, 52, 855, 19921, 614866, 24040451, 1152972925, 66200911138, 4465023867757, 348383154017581, 31052765897026352, 3128792250765898965, 353179564583216567917, 44320731930172534543092, 6141797839043095806714667, 934330605640859569909566925

Offset: 0

Author

Christian G. Bower, Jun 03 2005

Keywords

• nonn

Comments

Partitions of n black objects labeled 1..n and n white objects labeled 1..n. Each partition must have at least one white object.

a(n) is also the number of elements of the partition monoid P_n with domain {1,...,n}. Elements of P_n are set partitions of {1,1',...,n,n'}, and the domain of such a partition is the set of all points in {1,...,n} that belong to a block containing a dashed element. - James East, Apr 10 2018

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..200

Crossrefs

Main diagonal of A108458. Cf. A108461.

Cf. A048993 (Stirling2), A068424 (falling factorial).

Cf. A326600, A326270, A326271, A326288.

Bisection of A124421 (even part).

Programs

Maple

b:= proc(n) option remember; expand(`if`(n=0, 1,
      x*add(b(n-j)*binomial(n-1, j-1), j=1..n)))
    end:
a:= n-> add(coeff(b(n), x, j)*j^n, j=0..n):
seq(a(n), n=0..21);  # Alois P. Heinz, Dec 02 2023

Mathematica

a[n_] := If[n == 0, 1, Sum[k^n*StirlingS2[n, k], {k, 0, n}]];
Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Dec 10 2024 *)

PARI

{a(n)=polcoeff(sum(m=0, n, m^m*x^m/prod(k=1, m, 1-m*k*x +x*O(x^n))), n)} \\ Paul D. Hanna, Sep 17 2013

PARI

{a(n)=n!*polcoeff(sum(m=0, n, (exp(m*x+x*O(x^n))-1)^m/m!), n)} \\ Paul D. Hanna, Sep 17 2013

Formula

a(n) = Sum_{k=0..n} k^n*Stirling2(n,k). - Vladeta Jovovic, Aug 31 2006

E.g.f.: Sum_{n>=0} (exp(n*x)-1)^n / n!. - Vladeta Jovovic, Jul 12 2007

E.g.f.: Sum_{n>=0} exp(n^2*x) * exp( -exp(n*x) ) / n!. - Paul D. Hanna, Jun 28 2019

O.g.f.: Sum_{n>=0} n^n * x^n / Product_{k=1..n} (1 - n*k*x). - Paul D. Hanna, Sep 17 2013

a(n) = Sum_{k=0..n} Stirling2(n,k) * Sum_{l=k..n} Stirling2(n,l)*T(l,k). Here T(l,k) are the falling factorials. - James East, Apr 10 2018