A228329 a(n) = Sum_{k=0..n} (k+1)^2*T(n,k)^2 where T(n,k) is the Catalan triangle A039598.

1, 8, 98, 1320, 18590, 268736, 3952228, 58837680, 883941750, 13373883600, 203487733020, 3110407163760, 47726453450988, 734694122886080, 11341161925265480, 175489379096245984, 2721169178975361702, 42273090191785999728, 657788911222324942060, 10250564041646388681200

Offset: 0

Author

N. J. A. Sloane, Aug 26 2013

Keywords

• nonn

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 and 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.

Crossrefs

Cf. A039598, A000108, A024492 (h(0)), A000894 (h(1)), A000515 (h(3)), A228330 (h(4)), A228331 (h(5)), A228332 (h(6)), A228333 (h(7)).

Cf. A000142, A007318, A051960.

Programs

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 *)

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

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)