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%I #24 Jan 17 2017 05:52:20
%S 1,1,3,16,126,1333,17895,293461,5721390,129948787,3384796695,
%T 99848190706,3301868304168,121369298328835,4923587573624940,
%U 219090125559917698,10637377855875861600,560928617456424367993,31993928581562975604588,1966682218962058310721178
%N E.g.f. A(x) satisfies: A'(x) = exp(A(A(x))).
%C The e.g.f A(x) is divergent according to the answer by Pietro Majer to the MathOverflow question linked below. - _Tom Copeland_, Jan 16 2017
%H Paul D. Hanna, <a href="/A214645/b214645.txt">Table of n, a(n) for n = 1..160</a>
%H MathOverflow, <a href="http://mathoverflow.net/questions/258525/how-do-i-solve-this-displaystyle-f-ef-1">How do I solve this: df/dx = exp[f^{(-1)}(x)]</a>, answer by Tom Copeland to a MathOverflow question by Zeraoulia Rafik, 2017.
%H MathOverflow, <a href="http://mathoverflow.net/questions/258611/f-ef-1-again/258991#258991">df/dx = exp[f^{(-1)}(x)] again</a>, answer by Pietro Majer to a MathOverflow question by Fan Zheng, 2017.
%F E.g.f. A(x) satisfies:
%F (1) A''(x) = exp( 2*A(A(x)) + A(A(A(x))) ).
%F (2) exp(-A(x)) = d/dx Series_Reversion(A(x)).
%F (3) A(x) = Series_Reversion( Integral exp(-A(x)) dx ).
%F (4) A(x) = log(F(x)) where F(x) satisfies F( Integral 1/F(x) dx ) = exp(x) and equals the e.g.f. of A233335.
%e E.g.f.: A(x) = x + x^2/2! + 3*x^3/3! + 16*x^4/4! + 126*x^5/5! + 1333*x^6/6! + ...
%e Related expansions:
%e A'(x) = 1 + x + 3*x^2/2! + 16*x^3/3! + 126*x^4/4! + 1333*x^5/5! + ...
%e A(A(x)) = log(A'(x)) = x + 2*x^2/2! + 9*x^3/3! + 65*x^4/4! + 657*x^5/5! + 8627*x^6/6! + 140433*x^7/7! + 2744360*x^8/8! + 62894577*x^9/9! + ...
%e The exponential of e.g.f. A(x) equals the e.g.f. of A233335:
%e exp(A(x)) = 1 + x + 2*x^2/2! + 7*x^3/3! + 38*x^4/4! + 292*x^5/5! + 2975*x^6/6! + 38350*x^7/7! + 604433*x^8/8! + 11351659*x^9/9! + ... + A233335(n)*x^n/n! + ...
%o (PARI) {a(n)=local(A=x+x^2);for(i=0,n,A=intformal(exp(subst(A,x,A+x*O(x^n)))));n!*polcoeff(A, n)}
%o for(n=0, 25, print1(a(n), ", "))
%Y Cf. A233335 (exp), A259267.
%K nonn
%O 1,3
%A _Paul D. Hanna_, Jul 24 2012
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