%I #13 Sep 19 2023 15:32:00

%S 1,2,7,26,103,422,1768,7520,32335,140174,611530,2681516,11807683,

%T 52177166,231262945,1027703054,4577477065,20429990450,91348096963,

%U 409110897122,1834954888618,8241277167236,37059369415102

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

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

%H Andrew Howroyd, <a href="/A119243/b119243.txt">Table of n, a(n) for n = 0..1000</a>

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

%F 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.

%F 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

%o (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)))}

%o (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

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

%K nonn

%O 0,2

%A _Paul D. Hanna_, May 10 2006