A337389 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))).
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, 3334, 8953, 924, 1, 8, 91, 924, 7123, 34904, 73789, 3432, 1, 9, 114, 1345, 12870, 89055, 372436, 616227, 12870, 1, 10, 139, 1856, 20995, 184756, 1135005, 4027216, 5196627, 48620
Offset: 0
Examples
Square array begins:
1, 1, 1, 1, 1, 1, ...
2, 3, 4, 5, 6, 7, ...
6, 19, 34, 51, 70, 91, ...
20, 141, 328, 587, 924, 1345, ...
70, 1107, 3334, 7123, 12870, 20995, ...
252, 8953, 34904, 89055, 184756, 337877, ...
Links
- Seiichi Manyama, Antidiagonals n = 0..139, flattened
Crossrefs
Programs
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Mathematica
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 *) -
PARI
{T(n, k) = sum(j=0, n, k^(n-j)*binomial(2*j, j)*binomial(2*n, 2*j))}
Formula
T(n,k) = Sum_{j=0..n} k^(n-j) * binomial(2*j,j) * binomial(2*n,2*j).
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
For fixed k > 0, T(n,k) ~ (2 + sqrt(k))^(2*n + 1/2) / sqrt(8*Pi*n). - Vaclav Kotesovec, Aug 31 2020
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