A306245 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where A(0,k) = 1 and A(n,k) = Sum_{j=0..n-1} k^j * binomial(n-1,j) * A(j,k) for n > 0.

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 5, 1, 1, 1, 4, 17, 15, 1, 1, 1, 5, 43, 179, 52, 1, 1, 1, 6, 89, 1279, 3489, 203, 1, 1, 1, 7, 161, 5949, 108472, 127459, 877, 1, 1, 1, 8, 265, 20591, 1546225, 26888677, 8873137, 4140, 1

Offset

0,9

Links

Seiichi Manyama, Antidiagonals n = 0..55, flattened

Formula

G.f. A_k(x) of column k satisfies A_k(x) = 1 + x * A_k(k * x / (1 - x)) / (1 - x). - Seiichi Manyama, Jun 18 2022

Example

Square array begins:
   1,  1,    1,      1,       1,        1, ...
   1,  1,    1,      1,       1,        1, ...
   1,  2,    3,      4,       5,        6, ...
   1,  5,   17,     43,      89,      161, ...
   1, 15,  179,   1279,    5949,    20591, ...
   1, 52, 3489, 108472, 1546225, 12950796, ...

Maple

A:= proc(n, k) option remember; `if`(n=0, 1,
      add(k^j*binomial(n-1, j)*A(j, k), j=0..n-1))
    end:
seq(seq(A(n, d-n), n=0..d), d=0..12);  # Alois P. Heinz, Jul 28 2019

Mathematica

A[0, _] = 1;
A[n_, k_] := A[n, k] = Sum[k^j Binomial[n-1, j] A[j, k], {j, 0, n-1}];
Table[A[n-k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, May 29 2020 *)

Crossrefs

Columns k=0..4 give A000012, A000110, A126443, A355081, A355082.

Rows n=0+1, 2 give A000012, A000027(n+1).

Main diagonal gives A309401.

Cf. A309386.

Sequence in context: A124560 A368025 A290759 * A275043 A227061 A201949

Adjacent sequences: A306242 A306243 A306244 * A306246 A306247 A306248

Keyword

nonn,tabl

Author

Seiichi Manyama, Jul 28 2019

Status

approved