%I #79 Jun 18 2022 23:03:43
%S 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,
%T 89,1279,3489,203,1,1,1,7,161,5949,108472,127459,877,1,1,1,8,265,
%U 20591,1546225,26888677,8873137,4140,1
%N 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.
%H Seiichi Manyama, <a href="/A306245/b306245.txt">Antidiagonals n = 0..55, flattened</a>
%F 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
%e Square array begins:
%e 1, 1, 1, 1, 1, 1, ...
%e 1, 1, 1, 1, 1, 1, ...
%e 1, 2, 3, 4, 5, 6, ...
%e 1, 5, 17, 43, 89, 161, ...
%e 1, 15, 179, 1279, 5949, 20591, ...
%e 1, 52, 3489, 108472, 1546225, 12950796, ...
%p A:= proc(n, k) option remember; `if`(n=0, 1,
%p add(k^j*binomial(n-1, j)*A(j, k), j=0..n-1))
%p end:
%p seq(seq(A(n, d-n), n=0..d), d=0..12); # _Alois P. Heinz_, Jul 28 2019
%t A[0, _] = 1;
%t A[n_, k_] := A[n, k] = Sum[k^j Binomial[n-1, j] A[j, k], {j, 0, n-1}];
%t Table[A[n-k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* _Jean-François Alcover_, May 29 2020 *)
%Y Columns k=0..4 give A000012, A000110, A126443, A355081, A355082.
%Y Rows n=0+1, 2 give A000012, A000027(n+1).
%Y Main diagonal gives A309401.
%Y Cf. A309386.
%K nonn,tabl
%O 0,9
%A _Seiichi Manyama_, Jul 28 2019