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A119243
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Eigenvector of triangle A118919, so that a(n) = Sum_{k=0..floor(n/2)} A118919(n,k)*a(k).
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3
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1, 2, 7, 26, 103, 422, 1768, 7520, 32335, 140174, 611530, 2681516, 11807683, 52177166, 231262945, 1027703054, 4577477065, 20429990450, 91348096963, 409110897122, 1834954888618, 8241277167236, 37059369415102
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OFFSET
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0,2
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COMMENTS
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Equals the self-convolution of A119244, which is the eigenvector of triangle A119245.
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LINKS
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FORMULA
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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
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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