A224788 E.g.f. satisfies: A(x) = exp( Integral A(x)/(1 - x*A(x)^2) dx ).
1, 1, 3, 18, 168, 2142, 34704, 682740, 15810372, 421339176, 12702393792, 427435993512, 15881634963216, 645804320863680, 28527455317884336, 1360332028008819360, 69645942884911181184, 3810436222004101378656, 221867131720533800409216, 13698420738298341356760768
Offset: 0
Keywords
Examples
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! +...
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 0..300
- Eric Weisstein, MathWorld: Exponential Integral
Programs
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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 *) -
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),", "))
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
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