OFFSET
0,3
COMMENTS
Compare to: C(x) = exp( Integral C(x)^2/(1 - x*C(x)^2) dx ), which is satisfied by: C(x) = (1-sqrt(1-4*x))/(2*x) (Catalan numbers, A000108).
Compare to: W(x) = exp( Integral W(x)/(1 - x*W(x)) dx ), which is satisfied by: W(x) = LambertW(-x)/(-x) = Sum_{n>=0} (n+1)^(n-1)*x^n/n!.
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 0..300
Eric Weisstein, MathWorld: Exponential Integral
FORMULA
E.g.f. derivative: A'(x) = A(x)^2 / (1-x*A(x)^2). - Vaclav Kotesovec, Feb 19 2014
a(n) ~ n^(n-1) / (sqrt(2) * exp(n) * r^(n+1/4)), where r = 0.28396034297... is the root of the equation Ei(1/sqrt(r)) - Ei(1) + exp(1) = (r+sqrt(r)) * exp(1/sqrt(r)), where Ei is the Exponential Integral. - Vaclav Kotesovec, Feb 19 2014
EXAMPLE
E.g.f.: A(x) = 1 + x + 3*x^2/2! + 18*x^3/3! + 168*x^4/4! + 2142*x^5/5! +...
where
log(A(x)) = x + 2*x^2/2! + 11*x^3/3! + 99*x^4/4! + 1236*x^5/5! + 19752*x^6/6! +...
A(x)/(1-x*A(x)^2) = 1 + 2*x + 11*x^2/2! + 99*x^3/3! + 1236*x^4/4! + 19752*x^5/5! +...
MATHEMATICA
a = ConstantArray[0, 21]; a[[1]]=1; a[[2]]=1; Do[a[[n+2]] = n!*Sum[a[[i+1]]*a[[n-i+1]]/i!/(n-i)!, {i, 0, n}] + n!*Sum[a[[j+1]]/(j-1)!*Sum[a[[i+1]]*a[[n-j-i+1]]/i!/(n-j-i)!, {i, 0, n}], {j, 1, n}], {n, 1, 18}]; a (* Vaclav Kotesovec, Feb 19 2014 *)
FindRoot[ExpIntegralEi[1/Sqrt[r]] - ExpIntegralEi[1] + E == (r+Sqrt[r]) * E^(1/Sqrt[r]), {r, 1/2}, WorkingPrecision->50] (* program for numerical value of the radius of convergence r, Vaclav Kotesovec, Feb 19 2014 *)
PROG
(PARI) {a(n)=local(A=1+x); for(i=1, n, A=exp(intformal(A/(1-x*A^2 +x*O(x^n))))); n!*polcoeff(A, n)}
for(n=0, 20, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Apr 28 2013
STATUS
approved