A240936 Number of ways to partition the (vertex) set {1,2,...,n} into any number of classes and then select some unordered pairs (edges) such that a and b are in distinct classes of the partition.

1, 1, 3, 21, 337, 11985, 930241, 155643329, 55638770689, 42200814258433, 67536939792143361, 227017234854393949185, 1596674435594864988020737, 23421099407847007850007154689, 714530983411175509576743561314305, 45227689798343820164634911814524846081

Offset: 0

Author

Geoffrey Critzer, Aug 03 2014

Keywords

• nonn

Comments

The elements of a class are allowed to be used multiple times to form the unordered pairs.

Equivalently, a(n) is the sum of the number of k-colored graphs on n labeled nodes taken over k colors, 1<=k<=n, where labeled graphs using k colors that differ only by a permutation of the k colors are considered to be the same.

Also the number of ways to choose a stable partition of a simple graph on n vertices. A stable partition of a graph is a set partition of the vertices where no edge has both ends in the same block. - Gus Wiseman, Nov 24 2018

Examples

			a(2)=3 because the empty graph with 2 nodes is counted twice (once for each partition of 2) and the complete graph is counted once. 2+1=3.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..75

Crossrefs

Cf. A000569, A001187, A006125, A058843, A277203, A321979.

Programs

Maple

b:= proc(n, k) b(n, k):= `if`(k=1, 1, add(binomial(n, i)*
      2^(i*(n-i))*b(i, k-1)/k, i=1..n-1))
    end:
a:= n-> `if`(n=0, 1, add(b(n, k), k=1..n)):
seq(a(n), n=0..20);  # Alois P. Heinz, Aug 04 2014

Mathematica

nn=15;e[x_]:=Sum[x^n/(n!*2^Binomial[n,2]),{n,0,nn}];Table[n!2^Binomial[n,2],{n,0,nn}]CoefficientList[Series[Exp[(e[x]-1)],{x,0,nn}],x]

PARI

seq(n)={Vec(serconvol(sum(j=0, n, x^j*j!*2^binomial(j,2)) + O(x*x^n), exp(sum(j=1, n, x^j/(j!*2^binomial(j, 2))) + O(x*x^n))))} \\ Andrew Howroyd, Dec 01 2018

Formula

a(n) = n! * 2^C(n,2) * [x^n] exp(E(x)-1) where E(x) is Sum_{n>=0} x^n/(n!*2^C(n,2)).

a(n) = Sum_{k=1..n} A058843(n,k) for n>0.