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A001421
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a(n) = (6n)!/((n!)^3*(3n)!).
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10
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1, 120, 83160, 81681600, 93699005400, 117386113965120, 155667030019300800, 214804163196079142400, 305240072216678400087000, 443655767845074392936328000, 656486312795713480715743268160
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OFFSET
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0,2
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COMMENTS
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Self-convolution of A092870, where A092870(n) = (12^n/n!^2) * Product_{k=0..n-1} (12k+1)*(12k+5). - Paul D. Hanna, Jan 25 2011
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LINKS
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Seiichi Manyama, Table of n, a(n) for n = 0..310 (terms 0..75 from Vincenzo Librandi)
M. Kaneko and D. Zagier, Supersingular j-invariants, hypergeometric series and Atkin's orthogonal polynomials, pp. 97-126 of D. A. Buell and J. T. Teitelbaum, eds., Computational Perspectives on Number Theory, Amer. Math. Soc., 1998 (See Eq. 31.)
R. S. Maier, Nonlinear differential equations satisfied by certain classical modular forms, arXiv:0807.1081 [math.NT], 2008_2010, p. 34 equation (7.29b).
R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
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FORMULA
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o.g.f.: Hypergeometric2F1(5/12, 1/12; 1; 1728x)^2. - Jacob Lewis (jacobml(AT)uw.edu), Jul 28 2009
a(n) = binomial(2n,n) * (12^n/n!^2) * Product_{k=0..n-1} (6k+1)*(6k+5). - Paul D. Hanna, Jan 25 2011
G.f.: A(x) = 1 + 120*x + 83160*x^2 + 81681600*x^3 + ... - Paul D. Hanna, Jan 25 2011
A(x)^(1/2) = 1 + 60*x + 39780*x^2 + 38454000*x^3 + ... + A092870(n)*x^n + ... - Paul D. Hanna, Jan 25 2011
G.f.: F(1/6, 1/2, 5/6; 1, 1; 1728*x), a hypergeometric series. - Michael Somos, Feb 28 2011
0 = y^3*z^3 - 360*y^4*z^2 + 43200*y^5*z - 1728000*y^6 - 16632*x*y^2*z^3 + 7691328*x*y^3*z^2 - 1738520064*x*y^4*z + 176027074560*x*y^5 + 92207808*x^2*y*z^3 - 69176553984*x^2*y^2*z^2 + 23624298528768*x^2*y^3*z - 2853152143441920*x^2*y^4 - 170400029184*x^3*z^3 + 224945232150528*x^3*y*z^2 - 92759146352345088*x^3*y^2*z + 11686511179538104320*x^3*y^3 where x = a(n), y = a(n+1), z = a(n+2) for all n in z. - Michael Somos, Sep 21 2014
a(n) ~ 2^(6*n - 1) * 3^(3*n) / (Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Apr 07 2018
From Peter Bala, Feb 14 2020: (Start)
a(n) = binomial(6*n,n)*binomial(5*n,n)*binomial(4*n,n) = ( [x^n](1 + x)^(6*n) ) * ( [x^n](1 + x)^(5*n) ) * ( [x^n](1 + x)^(4*n) ) = [x^n](F(x)^(120*n)), where F(x) = 1 + x + 227*x^2 + 123980*x^3 + 92940839*x^4 + 82527556542*x^5 + 81459995686401*x^6 + ...
appears to have integer coefficients. For similar results see A008979.
Supercongruences: a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k - apply Mestrovic, equation 39, p. 12.
a(n) = [(x*y*z)^n] (1 + x + y + z)^(6*n). (End)
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MAPLE
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f := n->(6*n)!/( (n!)^3*(3*n)!);
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MATHEMATICA
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Factorial[6 n]/(Factorial[3n] Factorial[n]^3) (* Jacob Lewis (jacobml(AT)uw.edu), Jul 28 2009 *)
a[ n_] := SeriesCoefficient[ HypergeometricPFQ[ {1/6, 1/2, 5/6}, {1, 1}, 1728 x], {x, 0, n}] (* Michael Somos, Jul 11 2011 *)
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PROG
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(PARI) {a(n)=(2*n)!/n!^2*(12^n/n!^2)*prod(k=0, n-1, (6*k+1)*(6*k+5))} \\ Paul D. Hanna, Jan 25 2011
(MAGMA) [Factorial(6*n)/(Factorial(n)^3*Factorial(3*n)): n in [0..15]]; // Vincenzo Librandi, Oct 26 2011
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CROSSREFS
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Cf. A092870; variants: A184423, A008977, A184892, A184896, A184898. - Paul D. Hanna, Jan 25 2011
Cf. A289292.
Cf. A145492, A145493, A145494, A008979.
Sequence in context: A065961 A333043 A058528 * A107446 A184887 A279579
Adjacent sequences: A001418 A001419 A001420 * A001422 A001423 A001424
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KEYWORD
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nonn
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AUTHOR
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N. J. A. Sloane, Glenn K Painter (KUPK78A(AT)prodigy.com)
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STATUS
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approved
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