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%I #11 Jul 04 2015 12:20:59
%S 1,1,2,7,38,292,2975,38350,604433,11351659,249042701,6283114723,
%T 179995680530,5794486077958,207806806310354,8241414107222095,
%U 359171801820266717,17107537203463252273,886296777786378900077,49732564234138336160086,3011177123882906437153214,196063383282648338166793297
%N E.g.f. A(x) satisfies: A( Integral 1/A(x) dx ) = exp(x).
%H Paul D. Hanna, <a href="/A233335/b233335.txt">Table of n, a(n) for n = 0..160</a>
%F E.g.f. satisfies: A(x) = exp( Series_Reversion( Integral 1/A(x) dx ) ).
%F E.g.f.: exp(G(x)) where G(x) = exp(G(G(x))) is the e.g.f. of A214645.
%e E.g.f.: A(x) = 1 + x + 2*x^2/2! + 7*x^3/3! + 38*x^4/4! + 292*x^5/5! + 2975*x^6/6! +...
%e Related expansions.
%e Integral 1/A(x) dx = x - x^2/2! - x^4/4! - 6*x^5/5! - 52*x^6/6! - 591*x^7/7! - 8404*x^8/8! +...
%e The series reversion of Integral 1/A(x) dx equals log(A(x)) and begins:
%e log(A(x)) = x + x^2/2! + 3*x^3/3! + 16*x^4/4! + 126*x^5/5! + 1333*x^6/6! + 17895*x^7/7! + 293461*x^8/8! +...+ A214645(n)*x^n/n! +...
%e and equals the e.g.f. of A214645.
%o (PARI) {a(n)=local(A=1+x+x*O(x^n));for(i=1,n,A=exp(serreverse(intformal(1/A+x*O(x^n)))));n!*polcoeff(A,n)}
%o for(n=0,30,print1(a(n),", "))
%Y Cf. A233336, A214645 (log).
%K nonn
%O 0,3
%A _Paul D. Hanna_, Dec 07 2013
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