%I #20 Mar 29 2019 21:34:57
%S 1,162,480330,10273699314,1847922279733674,2953111740267791663874,
%T 42388685550188054819159463162,5474356451772724705359722799690452754,
%U 6362678346319775964713654070623548738367762634,66555927069904206114784802924703047820915991846751358882,6265787358184263065518163906632414087797095676323706723479988918042
%N G.f. A(x) satisfies: 1 + 3*x = Sum_{n>=0} (3^n + q*sqrt(A(x)))^n * x^n / (1 + 3^n*q*x*sqrt(A(x)))^(n+1), where q = 9/2.
%C For a given integer k > 1, there exists an integer series G(x,k)^2 that satisfies: 1 + k*x = Sum_{n>=0} (k^n + q*G(x,k))^n * x^n / (1 + k^n*q*x*G(x,k))^(n+1) iff q^2 = k^4/(2*k-2). In that case, G(x,k)^2 = 1 + k^3*(k^2-3)*x + k^4*(2*k^8 - 18*k^4 + 21*k^2 + 2*k + 1)*x^2/2 + ...; the g.f. for this sequence is A(x) = G(x,k=3)^2.
%H Paul D. Hanna, <a href="/A325063/b325063.txt">Table of n, a(n) for n = 0..50</a>
%F Let q = 9/2, then g.f. A(x) satisfies:
%F (1) 1 + 3*x = Sum_{n>=0} (3^n + q * sqrt(A(x)))^n * x^n / (1 + 3^n * q * x*sqrt(A(x)))^(n+1).
%F (2) 1 + 3*x = Sum_{n>=0} (3^n - q * sqrt(A(x)))^n * x^n / (1 - 3^n * q * x*sqrt(A(x)))^(n+1).
%e G.f.: A(x) = 1 + 162*x + 480330*x^2 + 10273699314*x^3 + 1847922279733674*x^4 + 2953111740267791663874*x^5 + 42388685550188054819159463162*x^6 + ...
%e Let q = 9/2 and B = A(x)^(1/2), then
%e 1 + 3*x = 1/(1 + x*q*B) + (3 + q*B)*x/(1 + 3*x*q*B)^2 + (3^2 + q*B)^2*x^2/(1 + 3^2*x*q*B)^3 + (3^3 + q*B)^3*x^3/(1 + 3^3*x*q*B)^4 + (3^4 + q*B)^4*x^4/(1 + 3^4*x*q*B)^5 + (3^5 + q*B)^5*x^5/(1 + 3^5*x*q*B)^6 + (3^6 + q*B)^6*x^6/(1 + 3^6*x*q*B)^7 + ...
%e and also
%e 1 + 3*x = 1/(1 - x*q*B) + (3 - q*B)*x/(1 - 3*x*q*B)^2 + (3^2 - q*B)^2*x^2/(1 - 3^2*x*q*B)^3 + (3^3 - q*B)^3*x^3/(1 - 3^3*x*q*B)^4 + (3^4 - q*B)^4*x^4/(1 - 3^4*x*q*B)^5 + (3^5 - q*B)^5*x^5/(1 - 3^5*x*q*B)^6 + (3^6 - q*B)^6*x^6/(1 - 3^6*x*q*B)^7 + ...
%o (PARI) /* Set k = 3 to generate this sequence (requires high precision) */
%o {a(n,k) = my(q = k^2/sqrt(2*k-2), A=[1, k^3*(k^2-3), 0]); for(i=0, n,
%o A=concat(A, 0); A[#A-1] = round( polcoeff( sum(m=0, #A, (k^m + q * Ser(A)^(1/2))^m * x^m / (1 + k^m * q * x*Ser(A)^(1/2))^(m+1) ), #A)/k^4)); A[n+1]}
%o for(n=0, 20, print1(a(n,k=3), ", "))
%Y Cf. A325062, A325064, A325065, A325066.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Mar 26 2019