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%I #44 Sep 26 2023 09:56:32
%S 1,12,540,33600,2425500,190702512,15849497664,1369618398720,
%T 121821136479900,11079206239530000,1025579963180813040,
%U 96310511463483233280,9152842704012278107200,878622906816654279840000
%N a(n) = (2*n)!*(3*n)!/n!^5.
%C 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
%C Diagonal of the rational function R(x,y,z,w) = 1/(1-(w*y+w*z+x+y+z)). - _Gheorghe Coserea_, Jul 15 2016
%C From _Peter Bala_, Jun 28 2023: (Start)
%C 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.
%C 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)
%H Gheorghe Coserea, <a href="/A184423/b184423.txt">Table of n, a(n) for n = 0..200</a>
%H Timothy Huber, Daniel Schultz, and Dongxi Ye, <a href="https://doi.org/10.4064/aa220621-19-12">Ramanujan-Sato series for 1/pi</a>, Acta Arith. (2023) Vol. 207, 121-160. See p. 11.
%H Romeo Meštrović, <a href="https://arxiv.org/abs/1111.3057">Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012)</a>, arXiv:1111.3057 [math.NT], 2011.
%H Titus Piezas III, <a href="https://sites.google.com/view/tpiezas/0013-article-3-pi-formulas-and-the-monster">0013: Article 3 (Pi Formulas and the Monster Group)</a>.
%F Self-convolution of A184424:
%F 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).
%F 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
%F 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
%F a(n) ~ 3^(1/2)/(2*Pi^(3/2)) * n^(-3/2) * 108^n. - _Ilya Gutkovskiy_, Jul 15 2016
%F 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
%e G.f.: A(x) = 1 + 12*x + 540*x^2 + 33600*x^3 + 2425500*x^4 +...
%e G.f. of A184424 equals A(x)^(1/2):
%e 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 +...
%t Table[((2n)!(3n)!)/(n!)^5,{n,0,20}] (* _Harvey P. Dale_, Dec 18 2018 *)
%o (PARI) {a(n)=(3*n)!*(2*n)!/n!^5}
%o (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)}
%Y Cf. A184424, A275047.
%Y Related to diagonal of rational functions: A268545-A268555.
%K nonn,easy
%O 0,2
%A _Paul D. Hanna_, Jan 13 2011
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