%I #23 May 30 2026 17:02:47
%S 1,1,0,1,1,0,1,2,7,0,1,3,20,100,0,1,4,39,386,2197,0,1,5,64,948,11232,
%T 65536,0,1,6,95,1876,34293,440162,2476099,0,1,7,132,3260,81088,
%U 1659648,21779392,113379904,0,1,8,175,5190,163845,4665124,101033859,1305082130,6103515625,0
%N Square array A(n,k), n>=0, k>=0, read by antidiagonals downwards, where A(n,k) = n! * [x^n] (F_k(x)/x)^(1/2) and F_k(x) is the k-th iteration of x*G(x)^2 with G(x) = exp(x*G(x)^3).
%F E.g.f. of column k: ((1/x) * Series_Reversion( H_k(x) ))^(1/2), where H_k(x) is the k-th iterate of U(x)*exp(-3*U(x)) and U(x) = -LambertW(-x).
%F A(n,k) = Sum_{0 = x_0 <= x_1 <= ... <= x_{k-1} <= x_k = n} Product_{j=0..k-1} (2*x_j + 1) * (3*x_{j+1} - x_j + 1)^(x_{j+1} - x_j - 1) * binomial(x_{j+1},x_j).
%F A(n,0) = 0^n; A(n,k) = Sum_{j=0..n} (2*j+1) * (3*n-j+1)^(n-j-1) * binomial(n,j) * A(j,k-1) for k > 0.
%e Square array begins:
%e 1, 1, 1, 1, 1, 1, ...
%e 0, 1, 2, 3, 4, 5, ...
%e 0, 7, 20, 39, 64, 95, ...
%e 0, 100, 386, 948, 1876, 3260, ...
%e 0, 2197, 11232, 34293, 81088, 163845, ...
%e 0, 65536, 440162, 1659648, 4665124, 10916480, ...
%e ...
%o (PARI)
%o a(n, k, p=3, s=2, r=1) = {
%o my(T=matrix(n+1, n+1, row, col, my(xr=row-1, xc=col-1); if(xc<xr, 0, (s*xr+r)*(p*xc-(p-s)*xr+r)^(xc-xr-1)*binomial(xc, xr))));
%o my(TK=T^k);
%o TK[1, n+1];
%o };
%Y Columns k=0..1 give A000007, A052752.
%Y Cf. A396446, A396503.
%K nonn,tabl
%O 0,8
%A _Seiichi Manyama_, May 28 2026