A336534 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) = Sum_{j=0..n} binomial(n,j) * binomial(k*n+j+1,n)/(k*n+j+1).
1, 1, 2, 1, 2, 2, 1, 2, 6, 2, 1, 2, 10, 22, 2, 1, 2, 14, 66, 90, 2, 1, 2, 18, 134, 498, 394, 2, 1, 2, 22, 226, 1482, 4066, 1806, 2, 1, 2, 26, 342, 3298, 17818, 34970, 8558, 2, 1, 2, 30, 482, 6202, 52450, 226214, 312066, 41586, 2, 1, 2, 34, 646, 10450, 122762, 881970, 2984206, 2862562, 206098, 2
Offset: 0
Examples
Square array begins: 1, 1, 1, 1, 1, 1, ... 2, 2, 2, 2, 2, 2, ... 2, 6, 10, 14, 18, 22, ... 2, 22, 66, 134, 226, 342, ... 2, 90, 498, 1482, 3298, 6202, ... 2, 394, 4066, 17818, 52450, 122762, ...
Links
- Seiichi Manyama, Antidiagonals n = 0..139, flattened
Crossrefs
Programs
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Mathematica
T[n_, k_] := Sum[Binomial[n, j] * Binomial[k*n+j+1, n]/(k*n+j+1), {j, 0, n}]; Table[T[k, n-k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, May 01 2021 *)
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PARI
T(n, k) = sum(j=0, n, binomial(k*n+1, j)*binomial((k+1)*n-j, n-j))/(k*n+1);
Formula
G.f. A_k(x) of column k satisfies A_k(x) = 1 + x * A_k(x)^k * (1 + A_k(x)).
T(n,k) = (1/n) * Sum_{j=1..n} 2^j * binomial(n,j) * binomial(k*n,j-1) for n > 0.
T(n,k) = (1/(k*n+1)) * Sum_{j=0..n} binomial(k*n+1,j) * binomial((k+1)*n-j,n-j).
T(n,k) = binomial(1+k*n, n)*hypergeom([-n, 1+k*n], [2+(k-1)*n], -1)/(1 + k*n) for k > 0. - Stefano Spezia, Aug 09 2025