A307910 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 - 2*k*x + k*(k-4)*x^2).
1, 1, 0, 1, 1, 0, 1, 2, 3, 0, 1, 3, 8, 7, 0, 1, 4, 15, 32, 19, 0, 1, 5, 24, 81, 136, 51, 0, 1, 6, 35, 160, 459, 592, 141, 0, 1, 7, 48, 275, 1120, 2673, 2624, 393, 0, 1, 8, 63, 432, 2275, 8064, 15849, 11776, 1107, 0, 1, 9, 80, 637, 4104, 19375, 59136, 95175, 53344, 3139, 0
Offset: 0
Examples
Square array begins: 1, 1, 1, 1, 1, 1, 1, ... 0, 1, 2, 3, 4, 5, 6, ... 0, 3, 8, 15, 24, 35, 48, ... 0, 7, 32, 81, 160, 275, 432, ... 0, 19, 136, 459, 1120, 2275, 4104, ... 0, 51, 592, 2673, 8064, 19375, 40176, ... 0, 141, 2624, 15849, 59136, 168125, 400896, ... 0, 393, 11776, 95175, 439296, 1478125, 4053888, ...
Links
- Seiichi Manyama, Antidiagonals n = 0..139, flattened
Crossrefs
Programs
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Mathematica
A[n_, k_] := k^n Hypergeometric2F1[(1-n)/2, -n/2, 1, 4/k]; A[0, ] = 1; A[, 0] = 0; Table[A[n-k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, May 07 2019 *)
Formula
A(n,k) is the coefficient of x^n in the expansion of (1 + k*x + k*x^2)^n.
A(n,k) = Sum_{j=0..floor(n/2)} k^(n-j) * binomial(n,j) * binomial(n-j,j) = Sum_{j=0..floor(n/2)} k^(n-j) * binomial(n,2*j) * binomial(2*j,j).
n * A(n,k) = k * (2*n-1) * A(n-1,k) - k * (k-4) * (n-1) * A(n-2,k).