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%I #13 Sep 19 2023 15:32:04
%S 1,1,3,10,37,144,579,2379,9927,41905,178498,765876,3305926,14342635,
%T 62496000,273347910,1199555865,5279624910,23298471120,103057254030,
%U 456833819925,2028998199360,9027655451400,40232167135245
%N Eigenvector of triangle A119245, so that a(n) = Sum_{k=0..floor(n/2)} A119245(n,k)*a(k).
%C Self-convolution equals A119243, which is the eigenvector of triangle A118919.
%H Andrew Howroyd, <a href="/A119244/b119244.txt">Table of n, a(n) for n = 0..1000</a>
%F G.f. A(x) satisfies: A(x) = A(-x/(1-4*x))/sqrt(1-4*x).
%F 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.
%F From _Peter Bala_, Sep 16 2023: (Start)
%F It appears that the g.f. A(x) satisfies A(x^2) = 1/(1 + x)*A(x/(1 + x)^2).
%F 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)
%o (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)))}
%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]*(4*k+1)*binomial(2*n+1,n-2*k))/(2*n+1)); a} \\ _Andrew Howroyd_, Sep 19 2023
%Y Cf. A119245, A119243 (self-convolution).
%K nonn
%O 0,3
%A _Paul D. Hanna_, May 10 2006