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%I #11 Jan 22 2021 21:24:52
%S 1,1,0,1,1,0,1,2,3,0,1,3,10,25,0,1,4,21,122,543,0,1,5,36,339,3550,
%T 29281,0,1,6,55,724,12477,241442,3781503,0,1,7,78,1325,32316,1035843,
%U 37717630,1138779265,0
%N Square array read by descending antidiagonals. Let G be a simple labeled graph on n nodes. T(n,k) is the number of ways to give G an acyclic orientation and a coloring function C:V(G) -> {1,2,...,k} so that u->v implies C(u) >= C(v) for all u,v in V(G), n >= 0, k >= 0.
%H R. P. Stanley, <a href="https://doi.org/10.1016/j.disc.2006.03.010">Acyclic orientation of graphs</a>, Discrete Math. 5 (1973), 171-178.
%F Let E(x) = Sum_{n>=0} x^n/(n!*2^binomial(n,2)). Then Sum_{n>=0} T(n,k)*x^n/(n!*2^binomial(n,2)) = 1/E(-x)^k.
%F T(n,k) = (-1)^n p_n(-k) where p_n is the n-th polynomial described in A219765.
%e Array begins
%e 1, 1, 1, 1, 1, 1, ...
%e 0, 1, 2, 3, 4, 5, ...
%e 0, 3, 10, 21, 36, 55, ...
%e 0, 25, 122, 339, 724, 1325, ...
%e 0, 543, 3550, 12477, 32316, 69595, ...
%e 0, 29281, 241442, 1035843, 3180484, 7934885, ...
%e ...
%t nn = 6; e[x_] := Sum[x^n/(n! 2^Binomial[n, 2]), {n, 0, nn}];
%t Prepend[Table[Table[n! 2^Binomial[n, 2], {n, 0, nn}] CoefficientList[
%t Series[1/e[-x]^k, {x, 0, nn}], x], {k, 1, nn}],PadRight[{1}, nn + 1]] // Transpose // Grid
%Y Cf. A003024 (column k=1), A339934 (column k=2), A322280, A219765.
%K nonn,tabl
%O 0,8
%A _Geoffrey Critzer_, Jan 21 2021