A337419 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt((1-(k+4)*x+sqrt(1+2*(k-4)*x+((k+4)*x)^2)) / (2 * (1+2*(k-4)*x+((k+4)*x)^2))).

1, 1, 2, 1, 1, 6, 1, 0, -5, 20, 1, -1, -14, -41, 70, 1, -2, -21, -48, -125, 252, 1, -3, -26, -7, 198, 131, 924, 1, -4, -29, 76, 739, 2080, 3301, 3432, 1, -5, -30, 195, 1222, 1629, 1780, 15625, 12870, 1, -6, -29, 344, 1395, -3772, -26859, -57120, 16115, 48620

Offset: 0

Author

Seiichi Manyama, Aug 27 2020

Keywords

• sign
• tabl

Examples

			Square array begins:
    1,    1,    1,    1,     1,      1, ...
    2,    1,    0,   -1,    -2,     -3, ...
    6,   -5,  -14,  -21,   -26,    -29, ...
   20,  -41,  -48,   -7,    76,    195, ...
   70, -125,  198,  739,  1222,   1395, ...
  252,  131, 2080, 1629, -3772, -14873, ...
		

Links

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

Crossrefs

Columns k=0..4 give A000984, A337393, A337421, A337422, A337396.

Main diagonal gives A337420.

Cf. A337389, A337464.

Programs

Mathematica

T[n_, k_] := Sum[If[k == 0, Boole[n == j], (-k)^(n - j)] * Binomial[2*j, j] * Binomial[2*n, 2*j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, Aug 27 2020 *)

PARI

{T(n, k) = sum(j=0, n, (-k)^(n-j)*binomial(2*j, j)*binomial(2*n, 2*j))}

Formula

T(n,k) = Sum_{j=0..n} (-k)^(n-j) * binomial(2*j,j) * binomial(2*n,2*j).

T(0,k) = 1, T(1,k) = 2-k and n * (2*n-1) * (4*n-5) * T(n,k) = (4*n-3) * (-4*(k-4)*n^2+6*(k-4)*n-k+6) * T(n-1,k) - (k+4)^2 * (n-1) * (2*n-3) * (4*n-1) * T(n-2,k) for n > 1. - Seiichi Manyama, Aug 28 2020