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%I #21 Dec 08 2023 12:05:19
%S 1,1,5,10,21,51,122,295,725,1792,4455,11133,27930,70305,177483,449160,
%T 1139157,2894625,7367720,18781387,47941271,122524216,313484385,
%U 802877055,2058184346,5280670051,13559216117,34841384560,89587774395
%N Row squared sums of triangle of Lucas polynomials (A034807) for n>0: Sum_{k=0..floor(n/2)} A034807(n,k)^2, with a(0)=1.
%C Also equals row squared sums of triangle A132460 and so equals the sum of the initial floor(n/2)+1 squared terms of 1/C(x)^n where C(x) is the g.f. of the Catalan numbers (A000108).
%H G. C. Greubel, <a href="/A132461/b132461.txt">Table of n, a(n) for n = 0..1000</a>
%H Marcos Mariño and Claudia Rella, <a href="https://arxiv.org/abs/2312.00624">On the structure of wave functions in complex Chern-Simons theory</a>, arXiv:2312.00624 [hep-th], 2023. See p. 23.
%F a(n) = Sum_{k=0..floor(n/2)} ( C(n-k, k) + C(n-k-1, k-1) )^2.
%F Ignoring initial term, equals the logarithmic derivative of A093128, which gives the number of dissections of a polygon using strictly disjoint diagonals. - _Paul D. Hanna_, Nov 09 2013
%F From _Vaclav Kotesovec_, Feb 28 2014: (Start)
%F Recurrence (for n>=5): (n-3)*n*a(n) = (2*n^2 - 7*n + 4)*a(n-1) + (n-4)*n*a(n-2) + (2*n^2 - 9*n + 8)*a(n-3) - (n-4)*(n-1)*a(n-4).
%F G.f.: (2-x+2*x^2)/sqrt((x^2+x+1)*(x^2-3*x+1))-1.
%F a(n) ~ 5^(3/4) * ((3+sqrt(5))/2)^n / (2*sqrt(Pi*n)).
%F (End)
%t Flatten[{1,Table[Sum[(Binomial[n-k,k] + Binomial[n-k-1,k-1])^2,{k,0,Floor[n/2]}],{n,1,20}]}] (* _Vaclav Kotesovec_, Feb 28 2014 *)
%o (PARI) {a(n)=sum(k=0,n\2,(binomial(n-k, k)+binomial(n-k-1, k-1))^2)}
%o for(n=0, 30, print1(a(n), ", "))
%o (PARI) /* squared sums of negative powers of Catalan series: */
%o {a(n)=local(Catalan=2/(1+sqrt(1-4*x +x*O(x^n)))); sum(k=0,n\2,polcoeff(Catalan^-n,k)^2)}
%o for(n=0,30,print1(a(n),", "))
%Y Cf. A034807 (Lucas polynomials); A093128, A132460, A132459; A000108 (Catalan).
%K nonn
%O 0,3
%A _Paul D. Hanna_, Aug 21 2007
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