%I #10 Jan 13 2019 01:02:02
%S 1,3,54,1500,51450,2000376,84523824,3789772272,177645575250,
%T 8619977394600,429995817814992,21940447761898848,1140809521021467024,
%U 60271001744583000000,3228065652622114800000,174953984892890573016000,9580981919014895332205250,529506723911785149640077000,29503054954798945147262250000,1655865088781640962375927700000,93546915429941104997312052606000,5316303408243224817579669328104000
%N Coefficients a(n) of x^n*y^n*z^n in function A = A(x,y,z) such that A = 1 + x*B*C, B = 1 + y*C*A, and C = 1 + z*A*B, for n >= 0.
%H Paul D. Hanna, <a href="/A323325/b323325.txt">Table of n, a(n) for n = 0..50</a>
%F a(n) = 3/4 * binomial(2*n, n)^3 / (n+1) for n >= 1, with a(0) = 1.
%F a(n) = 3/4 * (2*n)!^3 / (n!^6 * (n+1)) for n >= 1.
%o (PARI) {a(n) = my(A=1, B=1, C=1); for(i=0, 2*n,
%o A = 1 + x*B*C +x*O(x^n);
%o B = 1 + y*A*C +y*O(y^n);
%o C = 1 + z*A*B +z*O(z^n));
%o polcoeff(polcoeff(polcoeff(A, n, x), n, y), n, z)}
%o for(n=0, 30, print1(a(n), ", "))
%o (PARI) /* Using binomial formula */
%o {a(n) = 0^n/4 + 3/4 * binomial(2*n, n)^3 / (n+1)}
%o for(n=0, 30, print1(a(n), ", "))
%Y Cf. A323324.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Jan 12 2019