A061692 Triangle of generalized Stirling numbers.
1, 1, 4, 1, 27, 36, 1, 172, 864, 576, 1, 1125, 17500, 36000, 14400, 1, 7591, 351000, 1746000, 1944000, 518400, 1, 52479, 7197169, 80262000, 191394000, 133358400, 25401600, 1, 369580, 151633440, 3691514176, 17188416000, 23866214400, 11379916800, 1625702400
Offset: 1
Examples
1; 1,4; 1,27,36; 1,172,864,576; ...
Links
- Alois P. Heinz, Rows n = 1..100, flattened
- J.-M. Sixdeniers, K. A. Penson and A. I. Solomon, Extended Bell and Stirling Numbers From Hypergeometric Exponentiation, J. Integer Seqs. Vol. 4 (2001), #01.1.4.
Programs
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Maple
b:= proc(n) option remember; expand( `if`(n=0, 1, add(x*b(n-i)/i!^3, i=1..n))) end: T:= n-> (p-> seq(coeff(p, x, i)/i!, i=1..n))(b(n)*n!^3): seq(T(n), n=1..10); # Alois P. Heinz, Sep 10 2019 -
Mathematica
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 *)
Formula
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
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
Extensions
More terms from Vladeta Jovovic, Apr 23 2003