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G.f. satisfies: 4*A(x) = 3 + x + A(x)^3, starting with [1,1,3].
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%I #21 Aug 20 2017 11:44:47

%S 1,1,3,19,150,1326,12558,124590,1278189,13449205,144342627,1573990275,

%T 17389407984,194228357568,2189610888840,24881753664840,

%U 284708154606318,3277578288381318,37934510719585350,441152315040444150

%N G.f. satisfies: 4*A(x) = 3 + x + A(x)^3, starting with [1,1,3].

%C 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 Vincenzo Librandi, <a href="/A120590/b120590.txt">Table of n, a(n) for n = 0..70</a>

%H Thomas M. Richardson, <a href="http://arxiv.org/abs/1609.01193">The three 'R's and Dual Riordan Arrays</a>, arXiv:1609.01193 [math.CO], 2016.

%F G.f.: A(x) = 1 + Series_Reversion(1+4*x - (1+x)^3).

%F G.f.: A(x) = Sum_{n>=0} C(3*n,n)/(2*n+1) * (3+x)^(2*n+1) / 4^(3*n+1), due to Lagrange Inversion.

%F Recurrence: 13*(n-1)*n*a(n) = 81*(n-1)*(2*n-3)*a(n-1) + 3*(3*n-7)*(3*n-5)*a(n-2). - _Vaclav Kotesovec_, Oct 19 2012

%F a(n) ~ sqrt(32-18*sqrt(3))*((81+48*sqrt(3))/13)^n/(12*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Oct 19 2012

%F G.f.: 4 * sin( arcsin(3 * sqrt(3) * (3 + x) / 16) / 3) / sqrt(3). - _Benedict W. J. Irwin_, Oct 19 2016

%e A(x) = 1 + x + 3*x^2 + 19*x^3 + 150*x^4 + 1326*x^5 + 12558*x^6 +...

%e A(x)^3 = 1 + 3*x + 12*x^2 + 76*x^3 + 600*x^4 + 5304*x^5 + 50232*x^6 +...

%t FullSimplify[Table[SeriesCoefficient[Sum[Binomial[3*k,k]/(2*k+1)*(3+x)^(2*k+1)/4^(3*k+1),{k,0,Infinity}],{x,0,n}] ,{n,0,20}]] (* _Vaclav Kotesovec_, Oct 19 2012 *)

%o (PARI) {a(n)=local(A=1+x+3*x^2+x*O(x^n));for(i=0,n,A=A-4*A+3+x+A^3);polcoeff(A,n)}

%Y Cf. A120591 (A(x)^3); A244594, A120588, A120592 - A120607.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Jun 16 2006, Jan 24 2008