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%I #15 Oct 26 2017 09:37:29
%S 1,1,2,10,143,5959,904224,696895088,3563009122225,144004257475683137,
%T 52273888783668336094726,189699379891906830471022186526,
%U 7572226826806850232281722700245568807,3627110408773444347271222282038547230122245455,22586092882428159778440302586299616247303225297287979548,1969016989037466758104728399066094312610056241493227691736998060636,2574833047387344521023398134994106823445574761658761070132072595536874966252691
%N G.f.: exp( Sum_{n>=1} L(n) * x^n/n ), where Sum_{n>=1} L(n) * x^n/n! = Series_Reversion( log(1 + x/A(x)) ).
%C This sequence is motivated by the following conjectures:
%C (C1) Given integer series G(x) such that G(0) = G'(0) = 1, define L(n) by
%C Sum_{n>=1} L(n) * x^n/n! = Series_Reversion( log(G(x)) )
%C then exp( Sum_{n>=1} L(n) * x^n/n ) is also an integer series;
%C (C2) Given G(x) = 1 + x*G(x)^m, define L(n) by
%C Sum_{n>=1} L(n) * x^n/n! = Series_Reversion( log(G(x)) )
%C then exp( Sum_{n>=1} L(n) * x^n/n ) = (1 + m*x)/(1 + (m-1)*x).
%H Paul D. Hanna, <a href="/A294115/b294115.txt">Table of n, a(n) for n = 0..50</a>
%e G.f.: A(x) = 1 + x + 2*x^2 + 10*x^3 + 143*x^4 + 5959*x^5 + 904224*x^6 + 696895088*x^7 + 3563009122225*x^8 + 144004257475683137*x^9 +...
%e The logarithm of the g.f. begins
%e log(A(x)) = x + 3*x^2/2 + 25*x^3/3 + 531*x^4/4 + 29041*x^5/5 + 5388603*x^6/6 + 4871887945*x^7/7 + 28498490189571*x^8/8 + 1296006243863566561*x^9/9 +...+ L(n)*x^n/n +...
%e such that the same coefficients L(n) are also found in
%e Series_Reversion( log(1 + x/A(x)) ) = x + 3*x^2/2! + 25*x^3/3! + 531*x^4/4! + 29041*x^5/5! + 5388603*x^6/6! + 4871887945*x^7/7! + 28498490189571*x^8/8! +...+ L(n)*x^n/n! +...
%o (PARI) {a(n) = my(A=1+x, L); for(i=0,n, L = x*serlaplace( 1/x*serreverse( log(1 + x/A +O(x^(n+2))) ) ); A = exp(L);); polcoeff(A, n)}
%o for(n=0, 30, print1(a(n), ", "))
%Y Cf. A293379.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Oct 22 2017
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