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%I #40 May 05 2022 08:08:21
%S 1,1,2,1,3,6,1,4,19,20,1,5,34,141,70,1,6,51,328,1107,252,1,7,70,587,
%T 3334,8953,924,1,8,91,924,7123,34904,73789,3432,1,9,114,1345,12870,
%U 89055,372436,616227,12870,1,10,139,1856,20995,184756,1135005,4027216,5196627,48620
%N Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt((1+(k-4)*x+sqrt(1-2*(k+4)*x+((k-4)*x)^2)) / (2 * (1-2*(k+4)*x+((k-4)*x)^2))).
%H Seiichi Manyama, <a href="/A337389/b337389.txt">Antidiagonals n = 0..139, flattened</a>
%F T(n,k) = Sum_{j=0..n} k^(n-j) * binomial(2*j,j) * binomial(2*n,2*j).
%F T(0,k) = 1, T(1,k) = k+2 and n * (2*n-1) * (4*n-5) * T(n,k) = (4*n-3) * (4*(k+4)*n^2-6*(k+4)*n+k+6) * T(n-1,k) - (k-4)^2 * (n-1) * (2*n-3) * (4*n-1) * T(n-2,k) for n > 1. - _Seiichi Manyama_, Aug 28 2020
%F For fixed k > 0, T(n,k) ~ (2 + sqrt(k))^(2*n + 1/2) / sqrt(8*Pi*n). - _Vaclav Kotesovec_, Aug 31 2020
%F Conjecture: the k-th column entries, k >= 0, are given by [x^n] ( (1 + (k-2)*x + x^2)*(1 + x)^2/(1 - x)^2 )^n. This is true for k = 0 and k = 4. - _Peter Bala_, May 03 2022
%e Square array begins:
%e 1, 1, 1, 1, 1, 1, ...
%e 2, 3, 4, 5, 6, 7, ...
%e 6, 19, 34, 51, 70, 91, ...
%e 20, 141, 328, 587, 924, 1345, ...
%e 70, 1107, 3334, 7123, 12870, 20995, ...
%e 252, 8953, 34904, 89055, 184756, 337877, ...
%t T[n_, k_] := Sum[If[k == 0, Boole[n == j], k^(n - j)] * Binomial[2*j, j] * Binomial[2*n, 2*j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* _Amiram Eldar_, Aug 25 2020 *)
%o (PARI) {T(n, k) = sum(j=0, n, k^(n-j)*binomial(2*j, j)*binomial(2*n, 2*j))}
%Y Columns k=0..5 give A000984, A082758, A337390, A245926, A001448, A243946.
%Y Main diagonal gives A337388.
%Y Cf. A337369, A337419.
%K nonn,tabl
%O 0,3
%A _Seiichi Manyama_, Aug 25 2020
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