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Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt(2 / ( (1-2*(k+4)*x+((k-4)*x)^2) * (1+(k-4)*x+sqrt(1-2*(k+4)*x+((k-4)*x)^2)) )).
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%I #30 Aug 31 2020 04:24:16

%S 1,1,6,1,7,30,1,8,51,140,1,9,74,393,630,1,10,99,736,3139,2772,1,11,

%T 126,1175,7606,25653,12012,1,12,155,1716,14499,80464,212941,51480,1,

%U 13,186,2365,24310,183195,864772,1787607,218790,1,14,219,3128,37555,352716,2351805,9400192,15134931,923780

%N Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt(2 / ( (1-2*(k+4)*x+((k-4)*x)^2) * (1+(k-4)*x+sqrt(1-2*(k+4)*x+((k-4)*x)^2)) )).

%H Seiichi Manyama, <a href="/A337369/b337369.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+1,2*j).

%F T(0,k) = 1, T(1,k) = k+6 and n * (2*n+1) * (4*n-3) * T(n,k) = (4*n-1) * (4*(k+4)*n^2-2*(k+4)*n-k-2) * T(n-1,k) - (k-4)^2 * (n-1) * (2*n-1) * (4*n+1) * T(n-2,k) for n > 1. - _Seiichi Manyama_, Aug 29 2020

%F For fixed k > 0, T(n,k) ~ (2 + sqrt(k))^(2*n + 3/2) / sqrt(8*k*Pi*n). - _Vaclav Kotesovec_, Aug 31 2020

%e Square array begins:

%e 1, 1, 1, 1, 1, 1, ...

%e 6, 7, 8, 9, 10, 11, ...

%e 30, 51, 74, 99, 126, 155, ...

%e 140, 393, 736, 1175, 1716, 2365, ...

%e 630, 3139, 7606, 14499, 24310, 37555, ...

%e 2772, 25653, 80464, 183195, 352716, 610897, ...

%t T[n_, k_] := Sum[If[k == 0, Boole[n == j], k^(n - j)] * Binomial[2*j, j] * Binomial[2*n + 1, 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+1, 2*j))}

%Y Columns k=0..5 give A002457, A273055, A337370, A245927, A002458, A243947.

%Y Main diagonal gives A337387.

%Y Cf. A337389, A337464.

%K nonn,tabl

%O 0,3

%A _Seiichi Manyama_, Aug 25 2020