A336575 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where T(0,k) = 1 and T(n,k) = (1/n) * Sum_{j=1..n} 3^j * binomial(n,j) * binomial(k*n,j-1) for n > 0.
1, 1, 3, 1, 3, 3, 1, 3, 12, 3, 1, 3, 21, 57, 3, 1, 3, 30, 192, 300, 3, 1, 3, 39, 408, 2001, 1686, 3, 1, 3, 48, 705, 6402, 22539, 9912, 3, 1, 3, 57, 1083, 14799, 109137, 267276, 60213, 3, 1, 3, 66, 1542, 28488, 338430, 1964010, 3287496, 374988, 3, 1, 3, 75, 2082, 48765, 817743, 8181597, 36718680, 41556585, 2381322, 3
Offset: 0
Examples
Square array begins: 1, 1, 1, 1, 1, 1, ... 3, 3, 3, 3, 3, 3, ... 3, 12, 21, 30, 39, 48, ... 3, 57, 192, 408, 705, 1083, ... 3, 300, 2001, 6402, 14799, 28488, ... 3, 1686, 22539, 109137, 338430, 817743, ...
Links
- Seiichi Manyama, Antidiagonals n = 0..139, flattened
Crossrefs
Programs
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Mathematica
T[0, k_] := 1; T[n_, k_] := Sum[3^j * Binomial[n, j] * Binomial[k*n, j - 1], {j, 1, n}]/n; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Jul 27 2020 *) -
PARI
T(n, k) = if(n==0, 1, sum(j=1, n, 3^j*binomial(n, j)*binomial(k*n, j-1))/n);
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PARI
T(n, k) = my(A=1+x*O(x^n)); for(i=0, n, A=1+x*A^k*(2+A)); polcoeff(A, n);
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PARI
T(n, k) = sum(j=0, n, 2^(n-j)*binomial(n, j)*binomial(k*n+j+1, n)/(k*n+j+1));
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PARI
T(n, k) = sum(j=0, n, 2^j*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 * (2 + A_k(x)).
T(n,k) = Sum_{j=0..n} 2^(n-j) * binomial(n,j) * binomial(k*n+j+1,n)/(k*n+j+1).
T(n,k) = (1/(k*n+1)) * Sum_{j=0..n} 2^j * binomial(k*n+1,j) * binomial((k+1)*n-j,n-j).
T(n,k) = (1/n) * Sum_{j=0..n-1} (-2)^j * 3^(n-j) * binomial(n,j) * binomial((k+1)*n-j,n-1-j) for n > 0. - Seiichi Manyama, Aug 10 2023
T(n,k) = 3*hypergeom([1-n, -k*n], [2], 3) for n > 0. - Stefano Spezia, Aug 09 2025