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

1, 2, 6, 40, 330, 3048, 30156, 312528, 3349170, 36809960, 412651668, 4700098416, 54237852708, 632762593680, 7450815536280, 88435205367456, 1056940049423682, 12708927083800296, 153636691533864900, 1866178021496170800

Offset: 0

Author

Paul D. Hanna, Jun 16 2006, Jan 24 2008

Keywords

• nonn

Comments

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.

Examples

			A(x) = 1 + 2*x + 6*x^2 + 40*x^3 + 330*x^4 + 3048*x^5 + 30156*x^6 +...
A(x)^3 = 1 + 6*x + 30*x^2 + 200*x^3 + 1650*x^4 +15240*x^5 +150780*x^6 +...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..70

Crossrefs

Cf. A120588 - A120591, A120593 - A120607.

Programs

Mathematica

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 *)

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)}

Formula

G.f.: A(x) = 1 + Series_Reversion((1+5*x - (1+x)^3)/4).

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.

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

a(n) ~ sqrt(250-60*sqrt(15))*((108+30*sqrt(15))/17)^n/(30*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 19 2012