%I #28 May 30 2026 17:02:25
%S 1,1,0,1,2,0,1,4,20,0,1,6,56,392,0,1,8,108,1432,11664,0,1,10,176,3408,
%T 54144,468512,0,1,12,260,6608,156240,2730784,23762752,0,1,14,360,
%U 11320,355968,9468096,172930816,1458000000,0,1,16,476,17832,700560,25170368,716663232,13214614144,105046700288,0
%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 and F_k(x) is the k-th iteration of x*G(x)^2 with G(x) = exp(x*G(x)^4).
%F E.g.f. of column k: (1/x) * Series_Reversion( H_k(x) ), where H_k(x) is the k-th iterate of U(x)*exp(-4*U(x)) and U(x) = -LambertW(-2*x)/2.
%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) * (4*x_{j+1} - 2*x_j + 2)^(x_{j+1} - x_j - 1) * binomial(x_{j+1},x_j).
%F A(n,0) = 0^n; A(n,k) = 2 * Sum_{j=0..n} (j+1) * (4*n-2*j+2)^(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, 2, 4, 6, 8, 10, ...
%e 0, 20, 56, 108, 176, 260, ...
%e 0, 392, 1432, 3408, 6608, 11320, ...
%e 0, 11664, 54144, 156240, 355968, 700560, ...
%e 0, 468512, 2730784, 9468096, 25170368, 56596960, ...
%e ...
%o (PARI)
%o a(n, k, p=4, s=2, r=2) = {
%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, A396509.
%Y Cf. A396430, A396502.
%K nonn,tabl
%O 0,5
%A _Seiichi Manyama_, May 28 2026