A119244 Eigenvector of triangle A119245, so that a(n) = Sum_{k=0..floor(n/2)} A119245(n,k)*a(k).

1, 1, 3, 10, 37, 144, 579, 2379, 9927, 41905, 178498, 765876, 3305926, 14342635, 62496000, 273347910, 1199555865, 5279624910, 23298471120, 103057254030, 456833819925, 2028998199360, 9027655451400, 40232167135245

Offset: 0

Author

Paul D. Hanna, May 10 2006

Keywords

• nonn

Comments

Self-convolution equals A119243, which is the eigenvector of triangle A118919.

Links

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

Crossrefs

Cf. A119245, A119243 (self-convolution).

Programs

PARI

{a(n)=if(n==0,1,sum(k=0,n\2,a(k)*(4*k+1)*binomial(2*n+1,n-2*k)/(2*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]*(4*k+1)*binomial(2*n+1,n-2*k))/(2*n+1)); a} \\ Andrew Howroyd, Sep 19 2023

Formula

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

Eigenvector: a(n) = Sum_{k=0..floor(n/2)} a(k)*(4*k+1)*binomial(2*n+1,n-2*k)/(2*n+1) for n>=0, with a(0)=1.

From Peter Bala, Sep 16 2023: (Start)

It appears that the g.f. A(x) satisfies A(x^2) = 1/(1 + x)*A(x/(1 + x)^2).

If true, then F(x) := A(x^2) satisfies the functional equation F(x^2) = 1/(1 + x^2)*F(x/(1 + x^2)). Compare with Landen's transformation K(x^2) = 1/(1 + x^2)*K(2*x/(1 + x^2)) (for 0 <= x < 1) for the complete elliptic integral of the first kind K(x) = Integral_{t = 0..Pi/2} dt/sqrt(1 - x^2*sin^2(t)). (End)