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A132461
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.
5
1, 1, 5, 10, 21, 51, 122, 295, 725, 1792, 4455, 11133, 27930, 70305, 177483, 449160, 1139157, 2894625, 7367720, 18781387, 47941271, 122524216, 313484385, 802877055, 2058184346, 5280670051, 13559216117, 34841384560, 89587774395
OFFSET
0,3
COMMENTS
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).
LINKS
Marcos MariƱo and Claudia Rella, On the structure of wave functions in complex Chern-Simons theory, arXiv:2312.00624 [hep-th], 2023. See p. 23.
FORMULA
a(n) = Sum_{k=0..floor(n/2)} ( C(n-k, k) + C(n-k-1, k-1) )^2.
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
From Vaclav Kotesovec, Feb 28 2014: (Start)
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).
G.f.: (2-x+2*x^2)/sqrt((x^2+x+1)*(x^2-3*x+1))-1.
a(n) ~ 5^(3/4) * ((3+sqrt(5))/2)^n / (2*sqrt(Pi*n)).
(End)
MATHEMATICA
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 *)
PROG
(PARI) {a(n)=sum(k=0, n\2, (binomial(n-k, k)+binomial(n-k-1, k-1))^2)}
for(n=0, 30, print1(a(n), ", "))
(PARI) /* squared sums of negative powers of Catalan series: */
{a(n)=local(Catalan=2/(1+sqrt(1-4*x +x*O(x^n)))); sum(k=0, n\2, polcoeff(Catalan^-n, k)^2)}
for(n=0, 30, print1(a(n), ", "))
CROSSREFS
Cf. A034807 (Lucas polynomials); A093128, A132460, A132459; A000108 (Catalan).
Sequence in context: A295952 A132174 A297301 * A255150 A266580 A271257
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Aug 21 2007
STATUS
approved