A176898 a(n) = binomial(6*n, 3*n)*binomial(3*n, n)/(2*(2*n+1)*binomial(2*n, n)).

5, 231, 14586, 1062347, 84021990, 7012604550, 607892634420, 54200780036595, 4938927219474990, 457909109348466930, 43057935618181929900, 4096531994713828810686, 393617202432246696493436, 38142088615983865845923052, 3723160004902167033863327592

Offset

1,1

Comments

During April 26-28, 2010, Zhi-Wei Sun introduced this new sequence and proved that a(n) = binomial(6n,3n)*binomial(3n,n)/(2*(2n+1)*binomial(2n,n)) is a positive integer for every n=1,2,3,... He also observed that a(n) is odd if and only if n is a power of two, and that 3a(n)=0 (mod 2n+3). By Stirling's formula, we have lim_n (8n*sqrt(n*Pi)a(n)/108^n) = 1. It is interesting to find a combinatorial interpretation or recursion for the sequence.

From Tatiana Hessami Pilehrood, Dec 01 2015: (Start)

Zhi-Wei Sun formulated two conjectures concerning a(n) (see Conjectures 1.1 and 1.2 in Z.-W. Sun, "Products and sums divisible by central binomial coefficients" and Conjecture A89 in "Open conjectures on congruences"). The first conjecture states that Sum_{n=1..p-1} a(n)/(108^n) is congruent to 0 or -1 modulo a prime p > 3 depending on whether p is congruent to +-1 or +-5 modulo 12, respectively.

The second conjecture asks about an exact formula for a companion sequence of a(n). Both conjectures as well as many numerical congruences involving a(n) and (2n+1)a(n) were solved by Kh. Hessami Pilehrood and T. Hessami Pilehrood, see the link below. (End)

Links

Indranil Ghosh, Table of n, a(n) for n = 1..450

Kh. Hessami Pilehrood and T. Hessami Pilehrood, Jacobi polynomials and congruences involving some higher-order Catalan numbers and binomial coefficients, preprint, arXiv:1504.07944 [math.NT], 2015.

K. H. Pilehrood and T. H. Pilehrood, Jacobi Polynomials and Congruences Involving Some Higher-Order Catalan Numbers and Binomial Coefficients, J. Int. Seq. 18 (2015) 15.11.7

M. R. Sepanski, On Divisibility of Convolutions of Central Binomial Coefficients, Electronic Journal of Combinatorics, 21 (1) 2014, #P1.32.

Zhi-Wei Sun, Products and sums divisible by central binomial coefficients, preprint, arXiv:1004.4623 [math.NT], 2010.

Zhi-Wei Sun, Open conjectures on congruences, preprint, arXiv:0911.5665 [math.NT], 2009-2011.

Brian Y. Sun, J. X. Meng, Proof of a Conjecture of Z.-W. Sun on Trigonometric Series, arXiv preprint arXiv:1606.08153 [math.CO], 2016.

Formula

G.f.: (1-6*s)/((12*s-1)*(8*s-2)) - 1/2, where x+(3*s-1)*(12*s-1)^2*s*(4*s-1)^2 = 0. - Mark van Hoeij, May 06 2013

a(n) ~ 2^(2*n-3) * 3^(3*n) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Jan 09 2023

From Peter Bala, Feb 21 2023: (Start)

a(n+1) = 6*(6*n + 1)*(6*n + 5)/((n + 1)*(2*n + 3))*a(n).

a(n) = (2^(2*n-1)) * Product_{1 <= i <= j <= 2*n-1} (2*i + j + 2)/(2*i + j - 1). Cf. A006013. (End)

Example

For n=2 we have a(2) = binomial(12,6)*binomial(6,2)/(2*(2*2+1)*binomial(4,2)) = 231.

Maple

ogf := eval((1-6*s)/((12*s-1)*(8*s-2)) - 1/2, s=RootOf(x+(3*s-1)*(12*s-1)^2*s*(4*s-1)^2, s));
series(ogf, x=0, 30); # Mark van Hoeij, May 06 2013

Mathematica

S[n_]:=Binomial[6n, 3n]Binomial[3n, n]/(2(2n+1)Binomial[2n, n]) Table[S[n], {n, 1, 50}]

Prog

(Magma) [Binomial(6*n, 3*n)*Binomial(3*n, n)/(2*(2*n+1)*Binomial(2*n, n)): n in [1..15]]; // Vincenzo Librandi, Dec 02 2015
(PARI) a(n) = binomial(6*n, 3*n) * binomial(3*n, n) / (2*(2*n+1) * binomial(2*n, n)); \\ Indranil Ghosh, Mar 05 2017
(Python)
import math
f=math.factorial
def C(n, r): return f(n)/f(r)/f(n-r)
def A176898(n): return C(6*n, 3*n) * C(3*n, n) / (2*(2*n+1) * C(2*n, n)) # Indranil Ghosh, Mar 05 2017

Crossrefs

Cf. A000984, A006013, A173774, A176285, A176477.

Sequence in context: A157776 A147540 A187366 * A274996 A142668 A330308

Adjacent sequences: A176895 A176896 A176897 * A176899 A176900 A176901

Keyword

nonn,easy

Author

Zhi-Wei Sun, Apr 28 2010

Status

approved