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)
This is the even bisection of A078531 divided by 2. The odd bisection divided by 2 is A281733. - Akiva Weinberger, Dec 09 2024
LINKS
Indranil Ghosh, Table of n, a(n) for n = 1..450
Khodabakhsh Hessami Pilehrood and Tatiana Hessami Pilehrood, Jacobi polynomials and congruences involving some higher-order Catalan numbers and binomial coefficients, preprint, arXiv:1504.07944 [math.NT], 2015.
Khodabakhsh Hessami Pilehrood and Tatiana Hessami Pilehrood, Jacobi Polynomials and Congruences Involving Some Higher-Order Catalan Numbers and Binomial Coefficients, J. Int. Seq. 18 (2015) 15.11.7.
Guo-Shuai Mao, Congruences involving Fermat quotients and Euler polynomials, ResearchGate 2025. See p. 2.
Mark Roger 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 and J. X. Meng, Proof of a Conjecture of Z.-W. Sun on Trigonometric Series, arXiv preprint arXiv:1606.08153 [math.CO], 2016.
Chen Wang and Hui-Li Han, Supercongruences involving binomial coefficients and Euler polynomials, arXiv:2407.19882 [math.NT], 2024. See p. 2.
Xiran Zhang and Guo-Shuai Mao, Congruences involving binomial coefficients and Legendre symbol, ResearchGate (2024). See p. 10.
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)
D-finite with recurrence n*(2*n+1)*a(n) -6*(6*n-1)*(6*n-5)*a(n-1)=0. - R. J. Mathar, Nov 22 2024
a(n) = 2^(4n-1) * binomial(3n-1/2, 2n)/(2n+1). - Akiva Weinberger, Dec 09 2024
a(n) = A078531(2*n)/2. - Akiva Weinberger, Dec 09 2024
E.g.f.: (hypergeom([1/6, 5/6], [1, 3/2], 108*x) - 1)/2. - Stefano Spezia, Nov 24 2025
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
KEYWORD
nonn,easy
AUTHOR
Zhi-Wei Sun, Apr 28 2010
STATUS
approved