A001421 a(n) = (6*n)!/((n!)^3*(3*n)!).
1, 120, 83160, 81681600, 93699005400, 117386113965120, 155667030019300800, 214804163196079142400, 305240072216678400087000, 443655767845074392936328000, 656486312795713480715743268160, 985646873056680684690542988249600, 1497786250388951255453847206769124800
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + 120*x + 83160*x^2 + 81681600*x^3 + ... A(x)^(1/2) = 1 + 60*x + 39780*x^2 + 38454000*x^3 + ... + A092870(n)*x^n + ...
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..310 (terms 0..75 from Vincenzo Librandi)
- Timothy Huber, Daniel Schultz, and Dongxi Ye, Ramanujan-Sato series for 1/pi, Acta Arith. (2023) Vol. 207, 121-160. See p. 11.
- 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.
Crossrefs
Programs
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Magma
[Factorial(6*n)/(Factorial(n)^3*Factorial(3*n)): n in [0..15]]; // Vincenzo Librandi, Oct 26 2011
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Maple
f := n->(6*n)!/( (n!)^3*(3*n)!);
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Mathematica
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 *) -
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
Formula
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.: 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.
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)
a(n) = (8^n/n!^3)*Product_{k = 0..3*n-1} (2*k + 1). - Peter Bala, Feb 26 2023
a(n) = 24*(6*n - 1)*(2*n - 1)*(6*n - 5)*a(n-1)/n^3. - Neven Sajko, Jul 19 2023
From Karol A. Penson, Jan 21 2025: (Start)
a(n) = Integral_{x=0..1728} x^n*W(x), with W(x) = W1(x) + W2(x) + W3(x), where
W1(x) = hypergeometric([1/6, 1/6, 1/6], [1/3, 2/3], x/1728)/(6*sqrt(Pi)*x^(5/6)*Gamma(5/6)^3),
W2(x) = - hypergeometric([1/2, 1/2, 1/2], [2/3, 4/3], x/1728)/(24*Pi^2*sqrt(x)), and
W3(x) = hypergeometric([5/6, 5/6, 5/6], [4/3, 5/3], x/1728)*Gamma(5/6)^3/(1536*Pi^(7/2)*x^(1/6)). This integral representation is unique as W(x) is the solution of the Hausdorff power moment problem on x = (0, 1728). Using only the definition of a(n), W(x) can be proven to be positive. W(x) is singular at x = 0, with singularity x^(-1/6), and for x > 0 is monotonically decreasing to zero at x = 1728. (End)
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