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%I #23 Sep 11 2019 08:24:36
%S 1,1,4,1,27,36,1,172,864,576,1,1125,17500,36000,14400,1,7591,351000,
%T 1746000,1944000,518400,1,52479,7197169,80262000,191394000,133358400,
%U 25401600,1,369580,151633440,3691514176,17188416000,23866214400,11379916800,1625702400
%N Triangle of generalized Stirling numbers.
%H Alois P. Heinz, <a href="/A061692/b061692.txt">Rows n = 1..100, flattened</a>
%H J.-M. Sixdeniers, K. A. Penson and A. I. Solomon, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL4/SIXDENIERS/bell.html">Extended Bell and Stirling Numbers From Hypergeometric Exponentiation</a>, J. Integer Seqs. Vol. 4 (2001), #01.1.4.
%F T(n, k) = 1/k!*Sum multinomial(n, n_1, n_2, ..n_k)^3, where the sum extends over all compositions (n_1, n_2, .., n_k) of n into exactly k nonnegative parts. - _Vladeta Jovovic_, Apr 23 2003
%F The row polynomials R(n,x) satisfy the recurrence equation R(n,x) = x*( sum {k = 0..n-1} binomial(n,k)^2*binomial(n-1,k)*R(k,x) ) with R(0,x) = 1. Also R(n,x + y) = sum {k = 0..n} binomial(n,k)^3*R(k,x)*R(n-k,y). - _Peter Bala_, Sep 17 2013
%e 1; 1,4; 1,27,36; 1,172,864,576; ...
%p b:= proc(n) option remember; expand(
%p `if`(n=0, 1, add(x*b(n-i)/i!^3, i=1..n)))
%p end:
%p T:= n-> (p-> seq(coeff(p, x, i)/i!, i=1..n))(b(n)*n!^3):
%p seq(T(n), n=1..10); # _Alois P. Heinz_, Sep 10 2019
%t R[0, _] = 1; R[n_, x_] := R[n, x] = x*Sum[Binomial[n, k]^2*Binomial[n-1, k]*R[k, x], {k, 0, n-1}]; Table[CoefficientList[R[n, x], x] // Rest, {n, 1, 8}] // Flatten (* _Jean-François Alcover_, Sep 01 2015, after _Peter Bala_ *)
%Y Diagonals give A001044, A061695, A061693, A061694. Cf. A061691.
%Y Row sums give A061684.
%K nonn,tabl
%O 1,3
%A _N. J. A. Sloane_, Jun 19 2001
%E More terms from _Vladeta Jovovic_, Apr 23 2003
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