OFFSET
0,2
COMMENTS
Let h(m) denote the sequence whose n-th term is Sum_{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(2).
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..825
Pedro J. Miana, Natalia Romero, Moments of combinatorial and Catalan numbers, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Remark 3 p. 1882. Omega2(n) = a(n-1).
Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv:1305.2017 [math.CO], 2013.
Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33.
FORMULA
Conjecture: n*(2*n+1)*a(n) + 2*(-26*n^2+25*n-11)*a(n-1) + 20*(4*n-5)*(4*n-7)*a(n-2) = 0. - R. J. Mathar, Sep 08 2013
a(n) = ((4n)!*(3n+1))/((2n)!^2*(2n+1)) = binomial(4n,2n)*(3n+1)/(2n+1). - Philippe Deléham, Nov 25 2013
Therefore a(n) = A051960(2*n) / 2. - F. Chapoton, Jun 14 2024
From Peter Luschny, Nov 26 2013: (Start)
a(n) = 16^n*(3*n+1)*gamma(2*n+1/2)/(sqrt(Pi)*gamma(2*n+2)).
a(n) = a(n-1)*(6*n+2)*(4*n-3)*(4*n-1)/(n*(2*n+1)*(3*n-2)) if n > 0 else 1.
a(n) = [x^n] I*HeunG(8/5,0,-1/4,1/4,3/2,1/2,16*x)/sqrt(16*x-1) where [x^n] f(x) is the coefficient of x^n in f(x) and HeunG is the Heun general function. (End)
MAPLE
B:=(n, k)->binomial(2*n, n-k) - binomial(2*n, n-k-2); #A039598
Omega:=(m, n)->add((k+1)^m*B(n, k)^2, k=0..n);
h:=m->[seq(Omega(m, n), n=0..20)];
h(2);
# Second solution:
h := n -> I*HeunG(8/5, 0, -1/4, 1/4, 3/2, 1/2, 16*x)/sqrt(16*x-1);
seq(coeff(series(h(x), x, n+2), x, n), n=0..19); # Peter Luschny, Nov 26 2013
MATHEMATICA
a[n_] := Binomial[4n, 2n] (3n+1)/(2n+1);
Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Jul 30 2018, after Philippe Deléham *)
PROG
(Sage)
@CachedFunction
def A228329(n):
return A228329(n-1)*(6*n+2)*(4*n-3)*(4*n-1)/(n*(2*n+1)*(3*n-2)) if n>0 else 1
[A228329(n) for n in (0..19)] # Peter Luschny, Nov 26 2013
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Aug 26 2013
STATUS
approved