

A176898


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


2



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

1,1


COMMENTS

During April 2628, 2010, ZhiWei 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)
ZhiWei 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..p1} 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, T. Hessami Pilehrood, Jacobi polynomials and congruences involving some higherorder Catalan numbers and binomial coefficients, preprint, arXiv:1504.07944 [math.NT], 2015.
K. H. Pilehrood, T. H. Pilehrood, Jacobi Polynomials and Congruences Involving Some HigherOrder 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.
ZhiWei Sun, Products and sums divisible by central binomial coefficients, preprint, arXiv:1004.4623 [math.NT], 2010.
ZhiWei Sun, Open conjectures on congruences, preprint, arXiv:0911.5665 [math.NT], 20092011.
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.: (16*s)/((12*s1)*(8*s2))  1/2, where x+(3*s1)*(12*s1)^2*s*(4*s1)^2 = 0.  Mark van Hoeij, May 06 2013


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((16*s)/((12*s1)*(8*s2))  1/2, s=RootOf(x+(3*s1)*(12*s1)^2*s*(4*s1)^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(nr)
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, A173774, A176285, A176477.
Sequence in context: A157776 A147540 A187366 * A274996 A142668 A330308
Adjacent sequences: A176895 A176896 A176897 * A176899 A176900 A176901


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Apr 28 2010


STATUS

approved



