OFFSET
0,3
COMMENTS
Also the number of maximum independent vertex sets in the 3(n-1)-triangular honeycomb acute knight graph. - Eric W. Weisstein, Dec 31 2017
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..500
Eric Weisstein's World of Mathematics, Catalan Number.
Eric Weisstein's World of Mathematics, Maximum Independent Vertex Set.
FORMULA
From Ilya Gutkovskiy, Jan 23 2017: (Start)
O.g.f.: (1 - 3F2(-1/2,-1/2,-1/2; 1,1; 64*x))/(8*x).
E.g.f.: 3F3(1/2,1/2,1/2; 2,2,2; 64*x).
a(n) ~ 64^n/(Pi^(3/2)*n^(9/2)). (End)
From Amiram Eldar, Mar 27 2022: (Start)
a(n) = A000108(n)^3.
Sum_{n>=0} a(n)/64^n = 8 - 16*Gamma(3/4)*Gamma(7/4)/(Pi*Gamma(5/4)^2). (End)
MAPLE
seq((binomial(2*n, n)/(n+1))^3, n = 0..20); # G. C. Greubel, Oct 14 2019
MATHEMATICA
Table[CatalanNumber@n^3, {n, 0, 20}] (* Vincenzo Librandi, Nov 13 2012 *)
CatalanNumber[Range[0, 20]]^3 (* Eric W. Weisstein, Dec 31 2017 *)
PROG
(MuPAD) combinat::dyckWords::count(n)^3 $ n = 0..16; // Zerinvary Lajos, Feb 15 2007
(Sage) [catalan_number(i)^3 for i in range(0, 17)] # Zerinvary Lajos, May 17 2009
(Magma) [Catalan(n)^3: n in [0..20]]; // Vincenzo Librandi, Nov 13 2012
(PARI) a(n) = (binomial(2*n, n)/(n+1))^3; \\ Altug Alkan, Dec 31 2017
(Sage) [catalan_number(n)^3 for n in (0..20)] # G. C. Greubel, Oct 14 2019
(GAP) List([0..20], n-> (Binomial(2*n, n)/(n+1))^3); # G. C. Greubel, Oct 14 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Dec 10 1999
STATUS
approved