Andrew Howroyd, <a href="/A119243/b119243_1.txt">Table of n, a(n) for n = 0..1000</a>
Andrew Howroyd, <a href="/A119243/b119243_1.txt">Table of n, a(n) for n = 0..1000</a>
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Andrew Howroyd, <a href="/A119243/b119243_1.txt">Table of n, a(n) for n = 0..1000</a>
G.f. A(x) satisfies: A(x) = A(-x/(1-4*x))/(1-4*x).
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)*Cbinomial(2*n+2,n-2*k)/(n+1) for n>=0, with a(0)=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
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Apparently, 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
Apparently, the g.f. A(x) satisfies A(x^2) = 1/(1 + x)^2*A(x/(1 + x)^2). - Peter Bala, Sep 16 2023
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