A227462 E.g.f. equals the series reversion of x - x^2*exp(x^2).
1, 2, 12, 144, 2400, 50760, 1310400, 39984000, 1407490560, 56143130400, 2502730137600, 123302540148480, 6653043673436160, 390182090889951360, 24713127865269043200, 1681180295452142284800, 122252619549518954496000, 9463442426016057083404800, 776944497581062575154790400
Offset: 1
Keywords
Examples
E.g.f.: A(x) = x + 2*x^2/2! + 12*x^3/3! + 144*x^4/4! + 2400*x^5/5! +... where A(x) = x + A(x)^2*exp(A(x)^2). The e.g.f. satisfies: (3) A(x) = x + x^2*exp(A(x)^2) + 2*x^3*exp(2*A(x)^2) + 5*x^4*exp(3*A(x)^2) + 14*x^5*exp(4*A(x)^2) + 42*x^6*exp(5*A(x)^2) +... (4) log(A(x)/x) = x*exp(A(x)^2) + 3*x^2*exp(2*A(x)^2)/2 + 10*x^3*exp(3*A(x)^2)/3 + 35*x^4*exp(4*A(x)^2)/4 + 126*x^5*exp(5*A(x)^2)/5 +... (5) A(x) = x + x^2/2*exp(x^2) + d/dx x^4/4*exp(2*x^2)/2! + d^2/dx^2 x^6/8*exp(3*x^2)/3! + d^3/dx^3 x^8/16*exp(4*x^2)/4! +... (6) log(A(x)/x) = x*exp(x^2)/2 + d/dx x*exp(2*x^2)/2! + d^2/dx^2 x^2*exp(3*x^2)/3! + d^3/dx^3 x^3*exp(4*x^2)/4! +...
Programs
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Mathematica
Table[(n-1)!*SeriesCoefficient[(x/(x-x^2*E^(x^2)))^n,{x,0,n-1}],{n,1,20}] (* Vaclav Kotesovec, Jul 28 2013 *) -
PARI
{a(n)=n!*polcoeff(serreverse(x-x^2*exp(x^2 +x*O(x^n))), n)} for(n=1,25,print1(a(n),", ")) -
PARI
/* E.g.f. A(x) = x*Catalan( x*exp(A(x)^2) ): */ {a(n)=local(A=x); for(i=1,n,A=(1-sqrt(1-4*x*exp(A^2 +x^2*O(x^n)) ))/2*exp(-A^2 +x*O(x^n)) ); n!*polcoeff(A,n)} for(n=1,25,print1(a(n),", ")) -
PARI
{Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D} {a(n)=local(A=x); A=x+sum(m=1, n, Dx(m-1, x^(2*m)*exp(x^2+x*O(x^n))^m/m!)); n!*polcoeff(A, n)} for(n=1,25,print1(a(n),", ")) -
PARI
{Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D} {a(n)=local(A=x+x^2+x*O(x^n)); A=x*exp(sum(m=1, n, Dx(m-1, x^(2*m-1)*exp(x^2+x*O(x^n))^m/m!)+x*O(x^n))); n!*polcoeff(A, n)} for(n=1,25,print1(a(n),", "))
Formula
E.g.f. A(x) satisfies:
(1) A(x) = x + A(x)^2*exp(A(x)^2).
(2) A(x) = x*Catalan( x*exp(A(x)^2) ) where Catalan(x) = (1-sqrt(1-4*x))/(2*x).
(3) A(x) = x*Sum_{n>=0} binomial(2*n+1,n)/(2*n+1) * x^n * exp(n*A(x)^2).
(4) A(x) = x*exp( Sum_{n>=1} binomial(2*n-1,n) * x^n/n * exp(n*A(x)^2) ).
(5) A(x) = x + Sum_{n>=1} d^(n-1)/dx^(n-1) x^(2*n) * exp(n*x^2) / n!.
(6) A(x) = x*exp( Sum_{n>=1} d^(n-1)/dx^(n-1) x^(2*n-1) * exp(n*x^2) / n! ).
a(n) ~ n^(n-1) * s * sqrt((1+2*s^2)/(2+10*s^2+4*s^4)) / (exp(n) * ((s*(1+2*s^2))/(2*(1+s^2)))^n), where s = 0.3788063540000847107637564... is the root of the equation 2*s*(1+s^2)*exp(s^2) = 1. - Vaclav Kotesovec, Jan 07 2014
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