a(n) = Sum_{m=0..n} binomial(n, m) *
A000172(m). [Barrucand]
D-finite with recurrence: (n+1)^2 a(n+1) = (10*n^2+10*n+3) * a(n) - 9*n^2 * a(n-1). -
Matthijs Coster, Apr 28 2004
Sum_{n>=0} a(n)*x^n/n!^2 = BesselI(0, 2*sqrt(x))^3. -
Vladeta Jovovic, Mar 11 2003
a(n) = Sum_{p+q+r=n} (n!/(p!*q!*r!))^2 with p, q, r >= 0. -
Michael Somos, Jul 25 2007
a(n) = hypergeom([1/2, -n, -n], [1, 1], 4). -
Mark van Hoeij, Jun 02 2010
G.f.: 2*sqrt(2)/Pi/sqrt(1-6*z-3*z^2+sqrt((1-z)^3*(1-9*z))) * EllipticK(8*z^(3/2)/(1-6*z-3*z^2+sqrt((1-z)^3*(1-9*z)))). -
Sergey Perepechko, Feb 16 2011
G.f.: Sum_{n>=0} (3*n)!/n!^3 * x^(2*n)*(1-x)^n / (1-3*x)^(3*n+1). -
Paul D. Hanna, Feb 26 2012
G.f.: 1/(1-3*x)*(1-6*x^2*(1-x)/(Q(0)+6*x^2*(1-x))), where Q(k) = (54*x^3 - 54*x^2 + 9*x -1)*k^2 + (81*x^3 - 81*x^2 + 18*x -2)*k + 33*x^3 - 33*x^2 +9*x - 1 - 3*x^2*(1-x)*(1-3*x)^3*(k+1)^2*(3*k+4)*(3*k+5)/Q(k+1); (continued fraction). -
Sergei N. Gladkovskii, Jul 16 2013
G.f.: G(0)/(2*(1-9*x)^(2/3)), where G(k) = 1 + 1/(1 - 3*(3*k+1)^2*x*(1-x)^2/(3*(3*k+1)^2*x*(1-x)^2 - (k+1)^2*(1-9*x)^2/G(k+1) )); (continued fraction). -
Sergei N. Gladkovskii, Jul 31 2013
a(n) = [x^(2n)] 1/agm(sqrt((1-3*x)*(1+x)^3), sqrt((1+3*x)*(1-x)^3)). -
Gheorghe Coserea, Aug 17 2016
0 = +a(n)*(+a(n+1)*(+729*a(n+2) -1539*a(n+3) +243*a(n+4)) +a(n+2)*(-567*a(n+2) +1665*a(n+3) -297*a(n+4)) +a(n+3)*(-117*a(n+3) +27*a(n+4))) +a(n+1)*(+a(n+1)*(-324*a(n+2) +720*a(n+3) -117*a(n+4)) +a(n+2)*(+315*a(n+2) -1000*a(n+3) +185*a(n+4)) +a(n+3)*(+80*a(n+3) -19*a(n+4))) +a(n+2)*(+a(n+2)*(-9*a(n+2) +35*a(n+3) -7*a(n+4)) +a(n+3)*(-4*a(n+3) +a(n+4))) for all n in Z. -
Michael Somos, Oct 30 2017
G.f. y=A(x) satisfies: 0 = x*(x - 1)*(9*x - 1)*y'' + (27*x^2 - 20*x + 1)*y' + 3*(3*x - 1)*y. -
Gheorghe Coserea, Jul 01 2018
Sum_{k>=0} binomial(2*k,k) * a(k) / 6^(2*k) =
A086231 = (sqrt(3)-1) * (Gamma(1/24) * Gamma(11/24))^2 / (32*Pi^3). -
Vaclav Kotesovec, Apr 23 2023
The g.f. T(x) obeys a period-annihilating ODE:
0=3*(-1 + 3*x)*T(x) + (1 - 20*x + 27*x^2)*T'(x) + x*(-1 + x)*(-1 + 9*x)*T''(x).
The periods ODE can be derived from the following Weierstrass data:
g2 = (3/64)*(1 + 3*x)*(1 - 15*x + 75*x^2 + 3*x^3);
g3 = -(1/512)*(-1 + 6*x + 3*x^2)*(1 - 12*x + 30*x^2 - 540*x^3 + 9*x^4);
which determine an elliptic surface with four singular fibers. (End)
a(n) = Sum_{k = 0..n} binomial(n, k)^2 * binomial(3*k, 2*n) (Almkvist, p. 16). -
Peter Bala, May 22 2025
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