A184423 - OEIS

A184423

a(n) = (2*n)!*(3*n)!/n!^5.

1, 12, 540, 33600, 2425500, 190702512, 15849497664, 1369618398720, 121821136479900, 11079206239530000, 1025579963180813040, 96310511463483233280, 9152842704012278107200, 878622906816654279840000, 85070165409274310884800000, 8298008596757445137115033600, 814671580329898664032798508700

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

COMMENTS

Denoted by h_3[n] by T. Piezas III. He also gives formulas for 1/Pi such as 1/Pi = 2 * Sum_{n>=0} a(n) * (-1)^n * (51*n + 7) / (12^3)^(n + 1/2). - Michael Somos, May 31 2012

Diagonal of the rational function R(x,y,z,w) = 1/(1-(w*y+w*z+x+y+z)). - Gheorghe Coserea, Jul 15 2016

From Peter Bala, Jun 28 2023: (Start)

The supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for primes p >= 5 and positive integers n and r. This follows from Meštrović equation 39, since a(n) = binomial(3*n,n) * binomial(2*n,n)^2.

Inductively define a family of sequences {a(i,n) : n >= 0}, i >= 1, by setting a(1,n) = a(n) and, for i >= 2, a(i,n) = [x^n] ( exp(Sum_{k >= 1} a(i-1,k)*x^k/k) )^n. We conjecture that the sequences {a(i,n) : n >= 0}, i >= 2, also satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for primes p >= 5 and positive integers n and r. Cf. A362730 and A362732. (End)

LINKS

Gheorghe Coserea, Table of n, a(n) for n = 0..200

Timothy Huber, Daniel Schultz, and Dongxi Ye, Ramanujan-Sato series for 1/pi, Acta Arith. (2023) Vol. 207, 121-160. See p. 11.

Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012), arXiv:1111.3057 [math.NT], 2011.

Titus Piezas III, 0013: Article 3 (Pi Formulas and the Monster Group).

FORMULA

Self-convolution of A184424:

a(n) = Sum_{k=0..n} A184424(k)*A184424(n-k) where A184424(n) = (3^n/n!^2)*Product_{k=1..n} (6*k-4)*(6*k-5).

a(n) = 6 * (2*n - 1) * (3*n - 1) * (3*n - 2) / n^3 * a(n-1) if n>0. - Michael Somos, May 31 2012

0 = (x^2-108*x^3)*y''' + (3*x-486*x^2)*y''+ (1-348*x)*y' - 12*y, where y is g.f. - Gheorghe Coserea, Jul 15 2016

a(n) ~ 3^(1/2)/(2*Pi^(3/2)) * n^(-3/2) * 108^n. - Ilya Gutkovskiy, Jul 15 2016

a(n) = C(2*n,n)^2 * C(3*n,n) = ( [x^n](1 + x)^(2*n) )^2 * ( [x^n](1 + x)^(3*n) ) = [x^n]( F(x)^(12*n) ), where [x^n] is the coefficient extraction operator and where F(x) = 1 + x + 11*x^2 + 350*x^3 + 15293*x^4 + 794433*x^5 + 45958617*x^6 + ... appears to have integral coefficients. Cf. A000897 and A001451. - Peter Bala, Dec 30 2019

a(n) = A006480(n) * A000984(n). - F. Chapoton, Sep 14 2025

From Karol A. Penson, May 05 2026: (Start)

G.f.: hypergeom([1/3,1/2,2/3],[1,1],108*x).

a(n) = Integral_{x=0..108} x^n*W(x), where W(x) = sqrt(3)*2^(2/3)*hypergeom([1/3, 1/3, 1/3], [2/3, 5/6], x/108)/(6*Gamma(5/6)^2*Gamma(2/3)^2*x^(2/3)) - hypergeom([1/2, 1/2, 1/2], [5/6, 7/6], x/108)/(Pi^2*sqrt(x)) + (3*sqrt(3)*2^(1/3)*Gamma(5/6)^2*Gamma(2/3)^2*hypergeom([2/3, 2/3, 2/3], [7/6, 4/3], x/108))/(16*Pi^4*x^(1/3)). (End)

EXAMPLE

G.f.: A(x) = 1 + 12*x + 540*x^2 + 33600*x^3 + 2425500*x^4 +...

G.f. of A184424 equals A(x)^(1/2):

A(x)^(1/2) = 1 + 6*x + 252*x^2 + 15288*x^3 + 1089270*x^4 + 84963060*x^5 +...+ [(3^n/n!^2)*Product_{k=1..n} (6*k-4)*(6*k-5)]*x^n +...

MATHEMATICA

Table[((2n)!(3n)!)/(n!)^5, {n, 0, 20}] (* Harvey P. Dale, Dec 18 2018 *)

PROG

(PARI) {a(n)=(3*n)!*(2*n)!/n!^5}

(PARI) {a(n)=polcoeff(sum(m=0, n, 3^m*prod(k=1, m, (6*k-4)*(6*k-5))/m!^2*x^m+x*O(x^n))^2, n)}

(Python)

from math import comb

def A184423(n): return comb(3*n, n)*comb(2*n, n)**2 # Chai Wah Wu, Feb 17 2026