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%I #23 May 06 2023 04:16:48
%S 1,1,4,15,70,336,1716,9009,48620,267410,1494844,8465184,48466796,
%T 280073300,1631408400,9568812015,56466198990,335002137360,
%U 1997007404700,11955535480350,71850862117320,433322055191220,2621615826231480,15906988165723200,96775058652983100
%N Product of Fibonacci and Catalan numbers: a(n) = A000045(n+1)*A000108(n).
%C Radius of convergence: r = (sqrt(5)-1)/8; A(r) = sqrt(2+2/sqrt(5)). More generally, given {S} such that: S(n) = b*S(n-1) + c*S(n-2), |b|>0, |c|>0, then Sum_{n>=0} S(n)*Catalan(n)*x^n = sqrt( (1-2*b*x - sqrt(1-4*b*x-16*c*x^2))/(2*b^2+8*c) )/x.
%H Paul D. Hanna, <a href="/A098614/b098614.txt">Table of n, a(n) for n = 0..1000</a>
%H Paul Barry, <a href="https://arxiv.org/abs/2101.10218">On the duals of the Fibonacci and Catalan-Fibonacci polynomials and Motzkin paths</a>, arXiv:2101.10218 [math.CO], 2021.
%H Paul Barry and Arnauld Mesinga Mwafise, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL21/Barry/barry362.html">Classical and Semi-Classical Orthogonal Polynomials Defined by Riordan Arrays, and Their Moment Sequences</a>, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.5.
%F G.f.: A(x) = sqrt( (1-2*x - sqrt(1-4*x-16*x^2))/10 )/x.
%F G.f. satisfies: A(x) = sqrt( 1 + 2*x*A(x)^2 + 5*x^2*A(x)^4 ).
%F a(n) == 1 (mod 2) iff n = 2^k - 1 for k>=0.
%F n*(n+1)*a(n) -2*n*(2*n-1)*a(n-1) -4*(2*n-1)*(2*n-3)*a(n-2)=0. - _R. J. Mathar_, Nov 17 2018
%F Sum_{n>=0} a(n)/8^n = 2*sqrt(2/5). - _Amiram Eldar_, May 06 2023
%e Begins: {1*1, 1*1, 2*2, 3*5, 5*14, 8*42, 13*132, 21*429,...}.
%t With[{nn=30},Times@@@Thread[{Fibonacci[Range[nn]],CatalanNumber[ Range[ 0,nn-1]]}]] (* _Harvey P. Dale_, Nov 14 2011 *)
%o (PARI) {a(n)=local(X=x+O(x^(n+3)), A); A = sqrt( (1-2*x - sqrt(1-4*X-16*x^2)) / (10*x^2)); polcoeff(A, n)}
%o for(n=0, 30, print1(a(n), ", "))
%o (PARI) {a(n)=binomial(2*n,n)/(n+1)*round(((1+sqrt(5))^(n+1)-(1-sqrt(5))^(n+1))/(2^(n+1)*sqrt(5)))}
%Y Cf. A000045, A000108, A098615, A098616, A098618, A249925.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Oct 09 2004
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