A181783 Array described in comments to A053482, here read by increasing antidiagonals. See comments below.
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, 326, 1201, 709, 151, 16, 1, 1, 1, 1957, 12336, 9709, 2521, 311, 22, 1, 1, 1, 13700, 149989, 157971, 50045, 7186, 575, 29, 1, 1, 1, 109601, 2113546, 2993467, 1158871, 193765, 17536, 981, 37, 1
Offset: 0
Examples
Array read row after row: 1, 1, 1, 1, 1, 1, 1, ... 1, 1, 2, 4, 7, 11, 16, ... 1, 1, 5, 21, 63, 151, 311, ... 1, 1, 16, 142, 709, 2521, 7186, ... 1, 1, 65, 1201, 9709, 50045, 193765, ... 1, 1, 326, 12336, 157971, 1158871, 6002996, ... 1, 1, 1957, 149989, 2993467, 30806371, 210896251, ... ... A(4,3) = 1201.
Programs
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Maple
A181783 := proc(n,k) option remember; if n =0 or k = 0 then 1; else n*(k-1)*procname(n-1,k)+procname(n,k-1) ; end if; end proc: seq(seq(A181783(d-k,k),k=0..d),d=0..12) ; # R. J. Mathar, Mar 02 2016
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Mathematica
T[n_, k_] := T[n, k] = If[n == 0 || k == 0, 1, n (k - 1) T[n - 1, k] + T[n, k - 1]]; Table[T[n - k, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 10 2023 *)
Formula
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:
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.
A(n,1) = 1 for every n.
A(1,k) = 1 + k*(k-1)/2 for every k.
A(n, k+1) = A371898(n+k, k) * n! / ((n+k)! * k!). - Werner Schulte, Apr 14 2024
Extensions
Edited by N. J. A. Sloane, Dec 24 2012
Comments