A128079 a(n) = Sum_{k=0..n} A000984(k)*A001263(n+1,k+1), where A000984 is the central binomial coefficients and A001263 is the Narayana triangle.

1, 3, 13, 69, 411, 2633, 17739, 124029, 892327, 6567285, 49235715, 374841195, 2890994445, 22545855855, 177524073021, 1409591810133, 11275693221519, 90792020672429, 735367765159347, 5987665336600683, 48987680485918149

Offset: 0

Author

Paul D. Hanna, Feb 23 2007

Keywords

• nonn

Examples

			Illustrate a(n) = Sum_{k=0..n} A000984(k)*A001263(n+1,k+1) by:
a(2) = 1*(1) + 2*(3) + 6*(1) = 13;
a(3) = 1*(1) + 2*(6) + 6*(6) + 20*(1) = 69;
a(4) = 1*(1) + 2*(10)+ 6*(20)+ 20*(10)+ 70*(1) = 411.
The Narayana triangle A001263(n+1,k+1) = C(n,k)*C(n+1,k)/(k+1) begins:
1;
1, 1;
1, 3, 1;
1, 6, 6, 1;
1, 10, 20, 10, 1;
1, 15, 50, 50, 15, 1; ...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200

Crossrefs

Cf. A000984, A001263.

Programs

Mathematica

Table[Sum[Binomial[2*k,k]*Binomial[n,k]*Binomial[n+1,k]/(k+1),{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 20 2012 *)

PARI

{a(n)=sum(k=0,n,binomial(2*k,k)*binomial(n,k)*binomial(n+1,k)/(k+1))}

Formula

a(n) = Sum_{k=0..n} C(2k,k)*C(n,k)*C(n+1,k)/(k+1).

Recurrence: (n+1)*(n+2)*a(n) = (7*n^2+11*n+6)*a(n-1) + 3*(7*n^2-19*n+6)*a(n-2) - 27*(n-2)*(n-1)*a(n-3) . - Vaclav Kotesovec, Oct 20 2012

a(n) ~ 3^(2*n+7/2)/(8*Pi*n^2) . - Vaclav Kotesovec, Oct 20 2012

a(n) = ((n+3)^2*A005802(n+1)-(n-3)*(n+1)*A005802(n))/12. - Mark van Hoeij, Nov 12 2023