A339768 Square array read by descending antidiagonals. T(n,k) is the number of acyclic k-multidigraphs on n labeled vertices, n>=0,k>=0.
1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 25, 1, 1, 1, 7, 109, 543, 1, 1, 1, 9, 289, 9449, 29281, 1, 1, 1, 11, 601, 63487, 3068281, 3781503, 1, 1, 1, 13, 1081, 267249, 69711361, 3586048685, 1138779265, 1
Offset: 0
Examples
1, 1, 1, 1, 1, 1, ... 1, 1, 1, 1, 1, 1, ... 1, 3, 5, 7, 9, 11, ... 1, 25, 109, 289, 601, 1081, ... 1, 543, 9449, 63487, 267249, 849311, ... 1, 29281, 3068281, 69711361, 742650001, 5004309601, ...
Links
- Seiichi Manyama, Antidiagonals n = 0..50, flattened
- R. P. Stanley, Acyclic orientation of graphs, Discrete Math. 5 (1973), 171-178.
Crossrefs
Programs
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Mathematica
nn = 5; Table[g[n_] := q^Binomial[n, 2] n!; e[z_] := Sum[z^k/g[k], {k, 0, nn}]; Table[g[n], {n, 0, nn}] CoefficientList[Series[1/e[-z], {z, 0, nn}], z], {q, 1, nn + 1}] //Transpose // Grid
Formula
Let E(x) = Sum_{n>=0} x^n/(n!*(k+1)^binomial(n,2)). Then 1/E(-x) = Sum_{n>=0} T(n,k)x^n/(n!*(k+1)^binomial(n,2)).
T(0,k) = 1 and T(n,k) = Sum_{j=1..n} (-1)^(j+1) * (k+1)^(j*(n-j)) * binomial(n,j) * T(n-j,k) for n > 0. - Seiichi Manyama, Jun 13 2022
Comments