A144304 Square array A(n,m), n>=0, m>=0, read by antidiagonals: A(n,m) = n-th number of the m-th iteration of the hyperbinomial transform on sequence A001858.
1, 1, 1, 1, 2, 2, 1, 3, 7, 7, 1, 4, 14, 38, 38, 1, 5, 23, 93, 291, 291, 1, 6, 34, 178, 822, 2932, 2932, 1, 7, 47, 299, 1763, 9193, 36961, 36961, 1, 8, 62, 462, 3270, 21504, 125292, 561948, 561948, 1, 9, 79, 673, 5523, 43135, 313585, 2022555, 10026505, 10026505, 1
Offset: 0
Examples
Square array begins: 1, 1, 1, 1, 1, ... 1, 2, 3, 4, 5, ... 2, 7, 14, 23, 34, ... 7, 38, 93, 178, 299, ... 38, 291, 822, 1763, 3270, ...
Links
- Alois P. Heinz, Antidiagonals n = 0..140, flattened
- N. J. A. Sloane, Transforms
Crossrefs
Programs
-
Maple
hymtr:= proc(p) proc(n,m) `if`(m=0, p(n), m*add(p(k) *binomial(n, k) *(n-k+m)^(n-k-1), k=0..n)) end end: f:= proc(n) option remember; add(add(binomial(m, j) *binomial(n-1, n-m-j) *n^(n-m-j) *(m+j)!/ (-2)^j/ m!, j=0..m), m=0..n) end: A:= hymtr(f): seq(seq(A(n, d-n), n=0..d), d=0..12);
-
Mathematica
hymtr[p_] := Function[{n, m}, If[m == 0, p[n], m*Sum[p[k]*Binomial[n, k]*(n-k+m)^(n-k-1), {k, 0, n}]]]; f[0] = 1; f[n_] := f[n] = Sum[Sum[Binomial[m, j]*Binomial[n-1, n-m-j]*n^(n-m-j)*(m+j)!/(-2)^j/m!, {j, 0, m}], {m, 0, n}]; A[0, ] = 1; A[1, k] := k+1; A[n_, m_] := hymtr[f][n, m]; Table[Table[A[n, d-n], {n, 0, d}], {d, 0, 12}] // Flatten (* Jean-François Alcover, Dec 27 2013, translated from Maple *)