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

%I A380865 #11 Feb 08 2025 04:48:20
%S A380865 1,2,4,6,24,16,20,120,160,64,70,560,1120,896,256,252,2520,6720,8064,
%T A380865 4608,1024,924,11088,36960,59136,50688,22528,4096,3432,48048,192192,
%U A380865 384384,439296,292864,106496,16384,12870,205920,960960,2306304,3294720,2928640,1597440,491520,65536
%N A380865 Triangle read by rows: T(n, k) = 2^(2*n)*JacobiP(n - k, k, -1/2 - n, -1).
%F A380865 Consider a family of Jacobi polynomials defined with a rational number r as
%F A380865 J(n, k, r, x) = denominator(r)^(2*n)*JacobiP(n - k, k, r - n, x).
%F A380865 For r = -1/2 and x = -1 is J(n, k, r, x) = T(n, k).
%F A380865 For r = 1/2 and x = -1 is J(n, k, r, x) = A380851(n, k).
%F A380865 For r = 1/2 or r = -1/2 and x = 1 is J(n, k, r, x) = A038234(n, k).
%F A380865 The choice r = n and x = -1 gives Riordan array A097807, (1/(1 + x), 1).
%F A380865 The choice r = k and x = -1 gives Riordan array A128908, (1, x/(1 - x)^2).
%F A380865 The choice r = n and x = 1 gives the Pascal triangle.
%F A380865 T(n, k) = 4^n*binomial(n, k)*hypergeom([1/2, k - n], [k + 1], 1).
%e A380865 Triangle begins:
%e A380865   [0]    1;
%e A380865   [1]    2,     4;
%e A380865   [2]    6,    24,     16;
%e A380865   [3]   20,   120,    160,     64;
%e A380865   [4]   70,   560,   1120,    896,    256;
%e A380865   [5]  252,  2520,   6720,   8064,   4608,   1024;
%e A380865   [6]  924, 11088,  36960,  59136,  50688,  22528,   4096;
%e A380865   [7] 3432, 48048, 192192, 384384, 439296, 292864, 106496, 16384;
%p A380865 T := (n, k) -> 2^(2*n)*JacobiP(n - k, k, -1/2 - n, -1):
%p A380865 seq(print(seq(simplify(T(n, k)), k=0..n)), n=0..9);
%t A380865 T[n_, k_] := 4^n Binomial[n, k] Hypergeometric2F1[1/2, k - n, k + 1, 1];
%t A380865 Table[T[n, k], {n, 0, 7}, {k, 0, n}] // Flatten
%Y A380865 Cf. A038234, A380851, A097807, A128908, A380864 (row sums).
%K A380865 nonn,tabl
%O A380865 0,2
%A A380865 _Peter Luschny_, Feb 07 2025