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
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
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)
a(n) = binomial(6*n,3*n)*binomial(3*n,n)*binomial(2*n,n). -
Chai Wah Wu, Feb 14 2026