%I #43 Apr 14 2024 17:09:40
%S 1,1,1,1,1,1,1,1,2,1,1,1,5,4,1,1,1,16,21,7,1,1,1,65,142,63,11,1,1,1,
%T 326,1201,709,151,16,1,1,1,1957,12336,9709,2521,311,22,1,1,1,13700,
%U 149989,157971,50045,7186,575,29,1,1,1,109601,2113546,2993467,1158871,193765,17536,981,37,1
%N Array described in comments to A053482, here read by increasing antidiagonals. See comments below.
%C We denote by a(n,k) the number in row number n >= 0 and column number k >= 0. The recurrence which defines the array is a(n,k) = n*(k-1)*a(n-1,k) + a(n,k-1). The initial values are given by a(n,0) = 1 = a(0,k) for all n >= 0 and k >= 0.
%F If we consider the e.g.f. Psi(k) of column number k we have: Psi(k)(z) = Psi(k-1)(z)/(1-(k-1)*z) with Psi(1)(z) = exp(z). Then Psi(k)(z) = exp(z)/Product_{j=0..k-1} (1 - j*z). We conclude that a(n,k) = n!*Sum_{m=0..n} Sum_{j=1..k-1} (-1)^(k-1-j)*j^(m+k-2)/((n-m)!*(j-1)!*(k-1-j)!). It seems after the recurrence (and its proof) in A053482 that:
%F A(n,k) = -Sum_{j=1..k-1} s1(k,k-j)*n*(n-1)*...*(n-k+1)*a(n-j,k) + 1 where s1(m,n) are the classical Stirling numbers of the first kind.
%F A(n,1) = 1 for every n.
%F A(1,k) = 1 + k*(k-1)/2 for every k.
%F A(n, k+1) = A371898(n+k, k) * n! / ((n+k)! * k!). - _Werner Schulte_, Apr 14 2024
%e Array read row after row:
%e 1, 1, 1, 1, 1, 1, 1, ...
%e 1, 1, 2, 4, 7, 11, 16, ...
%e 1, 1, 5, 21, 63, 151, 311, ...
%e 1, 1, 16, 142, 709, 2521, 7186, ...
%e 1, 1, 65, 1201, 9709, 50045, 193765, ...
%e 1, 1, 326, 12336, 157971, 1158871, 6002996, ...
%e 1, 1, 1957, 149989, 2993467, 30806371, 210896251, ...
%e ...
%e A(4,3) = 1201.
%p A181783 := proc(n,k)
%p option remember;
%p if n =0 or k = 0 then
%p 1;
%p else
%p n*(k-1)*procname(n-1,k)+procname(n,k-1) ;
%p end if;
%p end proc:
%p seq(seq(A181783(d-k,k),k=0..d),d=0..12) ; # _R. J. Mathar_, Mar 02 2016
%t T[n_, k_] := T[n, k] = If[n == 0 || k == 0, 1, n (k - 1) T[n - 1, k] + T[n, k - 1]];
%t Table[T[n - k, k], {n, 0, 12}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Mar 10 2023 *)
%Y Cf. A000522, A053482, A185106, A371898.
%K nonn,tabl,easy
%O 0,9
%A _Richard Choulet_, Dec 23 2012
%E Edited by _N. J. A. Sloane_, Dec 24 2012