A119243 Eigenvector of triangle A118919, so that a(n) = Sum_{k=0..floor(n/2)} A118919(n,k)*a(k).

1, 2, 7, 26, 103, 422, 1768, 7520, 32335, 140174, 611530, 2681516, 11807683, 52177166, 231262945, 1027703054, 4577477065, 20429990450, 91348096963, 409110897122, 1834954888618, 8241277167236, 37059369415102

Offset: 0

Author

Paul D. Hanna, May 10 2006

Keywords

• nonn

Comments

Equals the self-convolution of A119244, which is the eigenvector of triangle A119245.

Links

• Andrew Howroyd, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A118919, A119244 (A(x)^(1/2)), A119245.

Programs

PARI

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

PARI

seq(n) = {my(a=vector(n+1)); a[1]=1; for(n=1, n, a[1+n] = sum(k=0, n\2, a[1+k]*(2*k+1)*binomial(2*n+2,n-2*k))/(n+1)); a} \\ Andrew Howroyd, Sep 19 2023

Formula

G.f. A(x) satisfies: A(x) = A(-x/(1-4*x))/(1-4*x).

Eigenvector: a(n) = Sum_{k=0..floor(n/2)} a(k)*(2*k+1)*binomial(2*n+2,n-2*k)/(n+1) for n>=0, with a(0)=1.

It appears that the g.f. A(x) satisfies A(x^2) = 1/(1 + x)^2*A(x/(1 + x)^2). - Peter Bala, Sep 16 2023