A185289 In a bipartite graph with 2n vertices (|V_1|=|V_2|=n), this sequence gives the number of ways to create n edges, one for each vertex of V_1, and to rank the vertices of V_2 which have incident edges.

1, 1, 6, 75, 1612, 52805, 2442666, 151382959, 12093970008, 1209295535049, 147859385866390, 21692929137930611, 3759744512444581860, 759740612270504941453, 177000400360669503651138, 47085371754008630756331255, 14182051733113750632290151856

Offset: 0

Author

Paolo Hägler, Feb 20 2011

Keywords

• nonn

Comments

a(n) is the number of ways to choose a function f:{1,2,...,n}->{1,2,...,n} and then linearly order the blocks of the coimage of f. - Geoffrey Critzer, Dec 23 2011

Examples

			For n=2 the a(2)=6 solutions are Aab, Bab, AaBb, AbBa, BbAa, BaAb. The capital letters are the vertices of V_2, in order, and the lower-case letters are the vertices of V_1 joined to the vertex of V_2 represented by the capital letter.
		

References

• Paolo Hägler, Il problema dei pasti, Bollettino dei docenti di matematica, 63 (2011), 101-108

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..200

Programs

Maple

f:= n-> add(add((-1)^(i-j)*j^n*n!*i!/(j!*(i-j)!*(n-i)!),j=0..i),i=0..n);
[seq(f(n),n=0..20)]; # N. J. A. Sloane, Mar 08 2011

Mathematica

Table[Sum[Binomial[n, k] StirlingS2[n, k] k!^2, {k, 0, n}], {n, 0,20}]  (* Geoffrey Critzer, Dec 23 2011 *)

PARI

a(n) = sum(k=0, n, binomial(n,k)*stirling(n,k,2)*k!^2); \\ Michel Marcus, Mar 23 2016

Formula

a(n) = Sum_{k=0..n} Sum_{j=0..k} (-1)^(k-j)*j^n*n!/(n-k)!*k!/(j!*(k-j)!). Paolo Hägler, Feb 20 2011

a(n) = Sum_{k=0..n} C(n,k)*Stirling2(n,k)*k!^2.

a(n) = Sum_{k=0..n} A090657(n,k)*k!.