A380865 Triangle read by rows: T(n, k) = 2^(2*n)*JacobiP(n - k, k, -1/2 - n, -1).

1, 2, 4, 6, 24, 16, 20, 120, 160, 64, 70, 560, 1120, 896, 256, 252, 2520, 6720, 8064, 4608, 1024, 924, 11088, 36960, 59136, 50688, 22528, 4096, 3432, 48048, 192192, 384384, 439296, 292864, 106496, 16384, 12870, 205920, 960960, 2306304, 3294720, 2928640, 1597440, 491520, 65536

Offset: 0

Author

Peter Luschny, Feb 07 2025

Keywords

• nonn
• tabl

Examples

			Triangle begins:
  [0]    1;
  [1]    2,     4;
  [2]    6,    24,     16;
  [3]   20,   120,    160,     64;
  [4]   70,   560,   1120,    896,    256;
  [5]  252,  2520,   6720,   8064,   4608,   1024;
  [6]  924, 11088,  36960,  59136,  50688,  22528,   4096;
  [7] 3432, 48048, 192192, 384384, 439296, 292864, 106496, 16384;
		

Crossrefs

Cf. A038234, A380851, A097807, A128908, A380864 (row sums).

Programs

Maple

T := (n, k) -> 2^(2*n)*JacobiP(n - k, k, -1/2 - n, -1):
seq(print(seq(simplify(T(n, k)), k=0..n)), n=0..9);

Mathematica

T[n_, k_] := 4^n Binomial[n, k] Hypergeometric2F1[1/2, k - n, k + 1, 1];
Table[T[n, k], {n, 0, 7}, {k, 0, n}] // Flatten

Formula

Consider a family of Jacobi polynomials defined with a rational number r as

J(n, k, r, x) = denominator(r)^(2*n)*JacobiP(n - k, k, r - n, x).

For r = -1/2 and x = -1 is J(n, k, r, x) = T(n, k).

For r = 1/2 and x = -1 is J(n, k, r, x) = A380851(n, k).

For r = 1/2 or r = -1/2 and x = 1 is J(n, k, r, x) = A038234(n, k).

The choice r = n and x = -1 gives Riordan array A097807, (1/(1 + x), 1).

The choice r = k and x = -1 gives Riordan array A128908, (1, x/(1 - x)^2).

The choice r = n and x = 1 gives the Pascal triangle.

T(n, k) = 4^n*binomial(n, k)*hypergeom([1/2, k - n], [k + 1], 1).