OFFSET
0,2
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..200
Jocelyn Quaintance and Harris Kwong, A combinatorial interpretation of the Catalan and Bell number difference tables, Integers, 13 (2013), #A29.
FORMULA
a(n) = ( 27*n^3 + 81*n^2 + 108*n + 24 )*binomial(2*n, n)/( (n+1)*(n+2)*(n+3)*(n+4) ). - Benoit Cloitre, Jun 11 2004
a(n) = -binomial(2*n,n)/(n+1)*hypergeom([-3,n+1/2],[n+2],4). - Peter Luschny, Aug 15 2012
G.f.: (1 + x + x^2*C(x)^3)*C(x)^3 where C(x) is the g.f. of A000108. - Philippe Deléham, Feb 04 2014
From G. C. Greubel, May 03 2021: (Start)
G.f.: (x + (1-x)*C(x))*C(x)^3, where C(x) is the g.f. of A000108.
E.g.f.: exp(2*x)*(BesselI(0, 2*x) +2*BesselI(1, 2*x) -BesselI(2, 2*x) -BesselI(3, 2*x) - BesselI(4, 2*x)).
a(n) = Sum_{k=0..3} (-1)^k*binomial(3,k)*C(n-k+3), where C(n) = A000108(n). (End)
Sum_{n>=0} a(n)/4^n = 14. - Amiram Eldar, Jul 10 2023
a(n) ~ 27 * 4^n / (n^(3/2) * sqrt(Pi)). - Amiram Eldar, Sep 30 2025
MAPLE
C:= n-> binomial(2*n, n)/(n+1);
a:= n-> add((-1)^j*binomial(3, j)*C(n-j+3), j=0..3);
seq(a(n), n=0..30); # G. C. Greubel, May 03 2021
MATHEMATICA
Table[(27n^3 +81n^2 +108n +24)*n!*Binomial[2n, n]/(n+4)!, {n, 0, 40}] (* Vincenzo Librandi, Feb 05 2014 *)
Differences[CatalanNumber[Range[0, 40]], 3] (* Harvey P. Dale, Jul 05 2020 *)
PROG
(PARI) a(n)=( 27*n^3 + 81*n^2 + 108*n + 24)*n!*binomial(2*n, n)/(n+4)!;
(Magma) [(27*n^3+81*n^2+108*n+24)*Binomial(2*n, n)/((n+1)*(n+2)*(n+3)*(n+4)): n in [0..30]]; // Vincenzo Librandi, Feb 05 2014
(SageMath) [sum((-1)^j*binomial(3, j)*catalan_number(n-j+3) for j in (0..3)) for n in (0..40)] # G. C. Greubel, May 03 2021
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved