A112519 Riordan array (1, x*c(x)*c(-x*c(x))), c(x) the g.f. of A000108.

1, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 1, 4, 0, 1, 0, 12, 2, 6, 0, 1, 0, 14, 28, 3, 8, 0, 1, 0, 100, 32, 48, 4, 10, 0, 1, 0, 180, 249, 54, 72, 5, 12, 0, 1, 0, 990, 440, 455, 80, 100, 6, 14, 0, 1, 0, 2310, 2552, 792, 726, 110, 132, 7, 16, 0, 1, 0, 10920, 5876, 4836, 1248, 1070, 144, 168, 8, 18, 0, 1

Offset: 0

Author

Paul Barry, Sep 09 2005

Keywords

• easy
• nonn
• tabl

Comments

Row sums are A112520. Second column is essentially A055392. Inverse is A112517. Riordan array product (1, x*c(x))*(1, x*c(-x)).

Examples

			Triangle begins
  1;
  0,   1;
  0,   0,   1;
  0,   2,   0,   1;
  0,   1,   4,   0,  1;
  0,  12,   2,   6,  0,   1;
  0,  14,  28,   3,  8,   0,  1;
  0, 100,  32,  48,  4,  10,  0,  1;
  0, 180, 249,  54, 72,   5, 12,  0, 1;
  0, 990, 440, 455, 80, 100,  6, 14, 0, 1;
		

Links

• G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Crossrefs

Cf. A000108, A055392, A112517, A112520.

Programs

Magma

A112519:= func< n,k | n eq 0 and k eq 0 select 1 else (k/n)*(&+[(-1)^j*Binomial(2*n-k-j-1, n-k-j)*Binomial(2*j+k-1, j): j in [0..n-k]]) >;
[A112519(n,k): k in [0..n], n in [0..10]]; // G. C. Greubel, Jan 12 2022

Mathematica

(* First program *)
c[x_]:= (1 - Sqrt[1-4x])/(2x);
(* The function RiordanArray is defined in A256893. *)
RiordanArray[1&, # c[#] c[-# c[#]]&, 12] // Flatten (* Jean-François Alcover, Jul 16 2019 *)
(* Second program *)
T[n_, k_]:= If[k==n, 1, (k/n)*Binomial[2*n-k-1, n-1]*HypergeometricPFQ[{k-n, k/2, (1+k)/2}, {k-2*n+1, k}, -4]];
Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Jan 12 2022 *)

Sage

@CachedFunction
def A112519(n,k):
    if (k==n): return 1
    else: return (k/n)*sum( (-1)^j*binomial(2*n-k-j-1, n-k-j)*binomial(2*j+k-1, j) for j in (0..n-k) )
flatten([[A112519(n,k) for k in (0..n)] for n in (0..10)]) # G. C. Greubel, Jan 12 2022

Formula

Riordan array (1, (sqrt(3-2*sqrt(1-4*x)) - 1)/2).

T(n, k) = (k/n)*Sum_{j=0..n} (-1)^(j-k)*C(2*n-j-1, n-j)*C(2*j-k-1, j-k), with T(0, 0) = 1.

T(n, k) = (k/n)*binomial(2*n-k-1, n-1)*Hypergoemetric3F2([k-n, k/2, (1+k)/2], [k-2*n+1, k], -4), with T(0, 0) = 1. - G. C. Greubel, Jan 12 2022