a(n) = 2*(4*n+1)! / ((n+1)!*(3*n+2)!) = binomial(4*n+1, n+1) - 9*binomial(4*n+1, n-1).
G.f.: hypergeom([1,1/2,3/4,5/4],[2,4/3,5/3],256*x/27) = 1 + 120*x/(Q(0)-120*x); Q(k) = 8*x*(2*k+1)*(4*k+3)*(4*k+5) + 3*(k+2)*(3*k+4)*(3*k+5) - 24*x*(k+2)*(2*k+3)*(3*k+4)*(3*k+5)*(4*k+7)*(4*k+9)/Q(k+1); (continued fraction). -
Sergei N. Gladkovskii, Nov 25 2011
a(n) = binomial(4*n + 2, n + 1) / ((2*n + 1) * (3*n + 2)). -
Michael Somos, Mar 28 2012
0 = F(a(n), a(n+1), ..., a(n+8)) for all n in Z where a(-1) = 3/4 and F() is a polynomial of degree 2 with integer coefficients and 29 monomials. -
Michael Somos, Dec 23 2014
D-finite with recurrence: 3*(3*n+2)*(3*n+1)*(n+1)*a(n) - 8*(4*n+1)*(2*n-1)*(4*n-1)*a(n-1) = 0. -
R. J. Mathar, Oct 21 2015
a(n) ~ 2^(8*n+7/2) / (sqrt(Pi) * n^(5/2) * 3^(3*n+5/2)). -
Vaclav Kotesovec, Feb 26 2016
E.g.f.: 3F3(1/2,3/4,5/4; 4/3,5/3,2; 256*x/27). -
Ilya Gutkovskiy, Feb 01 2017
G.f. y(x) satisfies:
0 = x^3*y^4 + 3*x^2*y^3 + x*(8*x+3)*y^2 - (20*x-1)*y + 16*x-1.
0 = x*(256*x - 27)*deriv(y,x) - 8*x^2*y^3 - 25*x*y^2 + 4*(24*x-11)*y + 44.
(End)
a(n) = Integral_{x=0...256/27} x^n*W(x), where
W(x) = (sqrt(2)/Pi)*(h1(x) - h2(x) + h3(x)) and
h1(x) = 3F2([-6/12,-2/12, 2/12], [ 3/12, 9/12], (27*x)/256)/((x/2)^(1/2));
h2(x) = 3F2([-3/12, 1/12, 5/12], [ 6/12, 15/12], (27*x)/256)/(x^(1/4));
h3(x) = 3F2([ 3/12, 7/12, 11/12], [18/12, 21/12], (27*x)/256)/(x^(-1/4)*32).
This integral representation is unique as the solution of n-th Hausdorff power moment of the function W. Using only the definition of a(n), W(x) can be proven to be positive. W(x) is singular at x = 0 and for x > 0 is monotonically decreasing to zero at x = 256/27. (End)
a(n) = (1/27^n) * Product_{1 <= i <= j <= 3*n} (3*i + j + 3)/(3*i + j - 1). Cf.
A006013. -
Peter Bala, Feb 21 2023
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