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%I #53 May 31 2023 16:23:34
%S 1,0,1,0,1,1,0,3,2,1,0,16,8,3,1,0,125,50,15,4,1,0,1296,432,108,24,5,1,
%T 0,16807,4802,1029,196,35,6,1,0,262144,65536,12288,2048,320,48,7,1,0,
%U 4782969,1062882,177147,26244,3645,486,63,8,1,0,100000000,20000000,3000000,400000,50000,6000,700,80,9,1
%N Triangular array read by rows: T(n,k) is the number of simple labeled graphs on vertex set {1,2,...,n} with exactly k components (all of which are trees) such that the labels {1,2,...,k} are all in distinct components (trees), n >= 0, 0 <= k <= n.
%C Row sums = (n^n-n)/(n-1)^2 = A058128(n).
%C Column k without leading zeros is the k-th exponential (also called binomial) convolution of the sequence {A000272(n+1)} = {A232006(n+1, 1)}, for n >= 0, with e.g.f. LamberW(-x)/(-x), where LambertW is the principal branch of the Lambert W-function. This is also the row polynomial P(n, x) of the unsigned triangle A137452, evaluated at x = k. - _Wolfdieter Lang_, Apr 24 2023
%D R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Proposition 5.3.2.
%H G. C. Greubel, <a href="/A232006/b232006.txt">Rows n=0..75 of triangle, flattened</a>
%H Chad Casarotto, <a href="http://www.math.uchicago.edu/~may/VIGRE/VIGRE2006/PAPERS/Casarotto.pdf">Graph Theory and Cayley's Formula</a>, 2006
%H Alan D. Sokal, <a href="https://arxiv.org/abs/1910.14519">A remark on the enumeration of rooted labeled trees</a>, arXiv:1910.14519 [math.CO], 2019.
%H Marc van Leeuwen, <a href="http://math.stackexchange.com/questions/2280076/">I am stuck with a combinatoric problem...</a> Math Stackexchange, Answer May 14 2017.
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/LambertW-Function.html">Lambert W-function</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Lambert_W_function">Lambert W function</a>
%F T(n, k) = k*n^(n-k-1).
%F T(n, k) = Sum_{i=0..n-k} T(n-1, k-1+i)*C(n-k,i), T(0, 0) = 1, T(n, 0) = 0 when n >= 1.
%F From _Wolfdieter Lang_, Apr 24 2023: (Start)
%F E.g.f. for {T(n+k, k)}_{n>=0} is (LambertW(-x)/(-x))^k, for k >= 0.
%F T(n, k) = Sum_{m=0..n-k} |A137452(n-k, m)|*k^m, for n >= 0 and k = 0..n. That is, T(n, n) = 1, for n >= 0, and T(n, k) = Sum_{m=1..n-k} binomial(n-k-1, m-1)*(n-k)^(n-k-m)*k^m, for k = 0..n-1 and n >= k+1. (End)
%e The triangle begins:
%e n\k 0 1 2 3 4 5 6 7 8 9 10 ...
%e 0: 1
%e 1: 0 1
%e 2: 0 1 1
%e 3: 0 3 2 1
%e 4: 0 16 8 3 1
%e 5: 0 125 50 15 4 1
%e 6: 0 1296 432 108 24 5 1
%e 7: 0 16807 4802 1029 196 35 6 1
%e 8: 0 262144 65536 12288 2048 320 48 7 1
%e 9: 0 4782969 1062882 177147 26244 3645 486 63 8 1
%e 10: 0 100000000 20000000 3000000 400000 50000 6000 700 80 9 1
%e ... Reformatted by _Wolfdieter Lang_, Apr 24 2023
%t Prepend[Table[Table[k n^(n-k-1),{k,0,n}],{n,1,8}],{1}]//Grid
%o (PARI) {T(n, k) = if( k<0 || k>n, 0, n^(n-k-1))}; /* _Michael Somos_, May 15 2017 */
%Y Cf. A058128, |A137452|.
%Y Columns give A000007, A000272, A007334, A362354, A362355, A362356, ...
%K nonn,tabl,easy
%O 0,8
%A _Geoffrey Critzer_, Nov 16 2013