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A185289
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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.
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1
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1, 1, 6, 75, 1612, 52805, 2442666, 151382959, 12093970008, 1209295535049, 147859385866390, 21692929137930611, 3759744512444581860, 759740612270504941453, 177000400360669503651138, 47085371754008630756331255, 14182051733113750632290151856
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OFFSET
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0,3
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COMMENTS
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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
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REFERENCES
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Paolo Hägler, Il problema dei pasti, Bollettino dei docenti di matematica, 63 (2011), 101-108
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LINKS
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FORMULA
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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!.
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EXAMPLE
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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.
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MAPLE
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f:= n-> add(add((-1)^(i-j)*j^n*n!*i!/(j!*(i-j)!*(n-i)!), j=0..i), i=0..n);
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MATHEMATICA
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Table[Sum[Binomial[n, k] StirlingS2[n, k] k!^2, {k, 0, n}], {n, 0, 20}] (* Geoffrey Critzer, Dec 23 2011 *)
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PROG
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(PARI) a(n) = sum(k=0, n, binomial(n, k)*stirling(n, k, 2)*k!^2); \\ Michel Marcus, Mar 23 2016
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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