A001421 a(n) = (6*n)!/((n!)^3*(3*n)!).

1, 120, 83160, 81681600, 93699005400, 117386113965120, 155667030019300800, 214804163196079142400, 305240072216678400087000, 443655767845074392936328000, 656486312795713480715743268160, 985646873056680684690542988249600, 1497786250388951255453847206769124800

Offset

0,2

Comments

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

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.

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

Example

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 + ...

Maple

f := n->(6*n)!/( (n!)^3*(3*n)!);

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 *)

Prog

(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

Crossrefs

Cf. A092870; variants: A184423, A008977, A184892, A184896, A184898. - Paul D. Hanna, Jan 25 2011

Cf. A289292.

Cf. A145492, A145493, A145494, A008979.

Sequence in context: A364512 A333043 A058528 * A107446 A184887 A279579

Adjacent sequences: A001418 A001419 A001420 * A001422 A001423 A001424

Keyword

nonn

Author

N. J. A. Sloane, Glenn K Painter (KUPK78A(AT)prodigy.com)

Status

approved