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A120592 G.f. satisfies: 5*A(x) = 4 + 4*x + A(x)^3, starting with [1,2,6].

%I A120592 #13 May 23 2013 21:50:27
%S A120592 1,2,6,40,330,3048,30156,312528,3349170,36809960,412651668,4700098416,
%T A120592 54237852708,632762593680,7450815536280,88435205367456,
%U A120592 1056940049423682,12708927083800296,153636691533864900,1866178021496170800
%N A120592 G.f. satisfies: 5*A(x) = 4 + 4*x + A(x)^3, starting with [1,2,6].
%C A120592 See comments in A120588 for conditions needed for an integer sequence to satisfy a functional equation of the form: r*A(x) = c + b*x + A(x)^n.
%H A120592 Vincenzo Librandi, <a href="/A120592/b120592.txt">Table of n, a(n) for n = 0..70</a>
%F A120592 G.f.: A(x) = 1 + Series_Reversion((1+5*x - (1+x)^3)/4).
%F A120592 G.f.: A(x) = Sum_{n>=0} C(3*n,n)/(2*n+1) * (4+4*x)^(2*n+1) / 5^(3*n+1), due to Lagrange Inversion.
%F A120592 Recurrence: 17*(n-1)*n*a(n) = 108*(n-1)*(2*n-3)*a(n-1) + 12*(3*n-7)*(3*n-5)*a(n-2). - _Vaclav Kotesovec_, Oct 19 2012
%F A120592 a(n) ~ sqrt(250-60*sqrt(15))*((108+30*sqrt(15))/17)^n/(30*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Oct 19 2012
%e A120592 A(x) = 1 + 2*x + 6*x^2 + 40*x^3 + 330*x^4 + 3048*x^5 + 30156*x^6 +...
%e A120592 A(x)^3 = 1 + 6*x + 30*x^2 + 200*x^3 + 1650*x^4 +15240*x^5 +150780*x^6 +...
%t A120592 FullSimplify[Table[SeriesCoefficient[Sum[Binomial[3*k,k]/(2*k+1)*(4+4*x)^(2*k+1)/5^(3*k+1),{k,0,Infinity}],{x,0,n}],{n,0,20}]] (* _Vaclav Kotesovec_, Oct 19 2012 *)
%o A120592 (PARI) {a(n)=local(A=1+2*x+6*x^2+x*O(x^n));for(i=0,n,A=A+(-5*A+4+4*x+A^3)/2);polcoeff(A,n)}
%Y A120592 Cf. A120588 - A120591, A120593 - A120607.
%K A120592 nonn
%O A120592 0,2
%A A120592 _Paul D. Hanna_, Jun 16 2006, Jan 24 2008