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E.g.f. satisfies: A'(x) = A(x)^3 / A(-x) with A(0) = 1.
3

%I #11 Jul 03 2014 05:44:43

%S 1,1,4,22,184,1888,24352,364336,6372352,125098624,2765195776,

%T 67161837568,1795080211456,51946830487552,1628857441189888,

%U 54705106541123584,1968709261466042368,75262309701303402496,3057220808668673081344,131069643224297960046592

%N E.g.f. satisfies: A'(x) = A(x)^3 / A(-x) with A(0) = 1.

%F E.g.f.: 1/(1 - 3*Series_Reversion( Integral 1/(1-9*x^2)^(1/3) dx ))^(1/3).

%F Limit n->infinity (a(n)/n!)^(1/n) = Pi*2^(5/3)/(sqrt(3)*GAMMA(2/3)^3) = 2.3191905339278567... - _Vaclav Kotesovec_, Jan 28 2014

%e E.g.f.: A(x) = 1 + x + 4*x^2/2! + 22*x^3/3! + 184*x^4/4! + 1888*x^5/5! +...

%e Related series.

%e A(x)^3 = 1 + 3*x + 18*x^2/2! + 144*x^3/3! + 1512*x^4/4! + 19224*x^5/5! +...

%e Note that 1 - 1/A(x)^3 is an odd function:

%e 1 - 1/A(x)^3 = 3*x - 18*x^3/3! - 216*x^5/5! - 18144*x^7/7! - 3483648*x^9/9! +...

%e where Series_Reversion((1 - 1/A(x)^3)/3) = Integral 1/(1-9*x^2)^(1/3) dx.

%t CoefficientList[1/(1 - 3*InverseSeries[Series[Integrate[1/(1-9*x^2)^(1/3),x],{x,0,20}],x])^(1/3),x] * Range[0,20]! (* _Vaclav Kotesovec_, Jan 28 2014 *)

%o (PARI) /* By definition A'(x) = A(x)^3 / A(-x): */

%o {a(n)=local(A=1); for(i=0, n, A=1+intformal(A^3/subst(A, x, -x) +x*O(x^n) )); n!*polcoeff(A, n)}

%o for(n=0, 25, print1(a(n), ", "))

%o (PARI) /* E.g.f. 1/(1 - 3*Series_Reversion(Integral (1-9*x)^(1/3) dx))^(1/3): */

%o {a(n)=local(A=1);A=1/(1-3*serreverse(intformal(1/(1-9*x^2 +x*O(x^n))^(1/3))))^(1/3);n!*polcoeff(A, n)}

%o for(n=0,25,print1(a(n),", "))

%Y Cf. A235322, A235345.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Jan 05 2014