A336534 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, 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

0,3

Links

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Formula

G.f. of column k A_k(x) 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).

Example

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, ...

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 *)

Prog

(PARI) {T(n, k) = sum(j=0, n, binomial(k*n+1, j)*binomial((k+1)*n-j, n-j))/(k*n+1)}

Crossrefs

Column k=0-3 give A040000, A006318, A027307, A144097.

If Michael D. Weiner's conjecture on A260332 is correct, column 4 is A260332 for n > 0.

Main diagonal gives A336537.

Cf. A336521, A336574. A336575.

Sequence in context: A271362 A354555 A263643 * A271852 A224362 A350560

Adjacent sequences: A336531 A336532 A336533 * A336535 A336536 A336537

Keyword

nonn,tabl

Author

Seiichi Manyama, Jul 25 2020

Status

approved