A336574 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where T(n,k) = Sum_{j=0..n} 2^j * binomial(n,j) * binomial(k*n+j+1,n)/(k*n+j+1).
1, 1, 3, 1, 3, 6, 1, 3, 15, 12, 1, 3, 24, 93, 24, 1, 3, 33, 255, 645, 48, 1, 3, 42, 498, 3102, 4791, 96, 1, 3, 51, 822, 8691, 40854, 37275, 192, 1, 3, 60, 1227, 18708, 164937, 566934, 299865, 384, 1, 3, 69, 1713, 34449, 464115, 3305868, 8164263, 2474025, 768
Offset: 0
Examples
Square array begins: 1, 1, 1, 1, 1, 1, ... 3, 3, 3, 3, 3, 3, ... 6, 15, 24, 33, 42, 51, ... 12, 93, 255, 498, 822, 1227, ... 24, 645, 3102, 8691, 18708, 34449, ... 48, 4791, 40854, 164937, 464115, 1055838, ...
Links
- Seiichi Manyama, Antidiagonals n = 0..139, flattened
Crossrefs
Programs
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Mathematica
T[n_, k_] := Sum[2^j * Binomial[n, j] * Binomial[k*n + j + 1, n]/(k*n + j + 1), {j, 0, n}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, Jul 27 2020 *) -
PARI
T(n, k) = sum(j=0, n, 2^j*binomial(n, j)*binomial(k*n+j+1, n)/(k*n+j+1));
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PARI
T(n, k) = my(A=1+x*O(x^n)); for(i=0, n, A=1+x*A^k*(1+2*A)); polcoeff(A, n);
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PARI
T(n, k) = sum(j=0, n, 2^(n-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 * (1 + 2 * A_k(x)).
T(n,k) = (1/(k*n+1)) * Sum_{j=0..n} 2^(n-j) * binomial(k*n+1,j) * binomial((k+1)*n-j,n-j).
From Seiichi Manyama, Aug 10 2023: (Start)
T(n,k) = (1/n) * Sum_{j=0..n-1} (-1)^j * 3^(n-j) * binomial(n,j) * binomial((k+1)*n-j,n-1-j) for n > 0.
T(n,k) = (1/n) * Sum_{j=1..n} 3^j * 2^(n-j) * binomial(n,j) * binomial(k*n,j-1) for n > 0. (End)
T(n,k) = binomial(1+k*n, n)*hypergeom([-n, 1+k*n], [2+(k-1)*n], -2)/(1 + k*n) for k > 0. - Stefano Spezia, Aug 09 2025