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%I #13 Oct 29 2019 13:05:37
%S 1,1,3,25,433,12501,529531,30495613,2272643745,211761416233,
%T 24055076979091,3267213865097601,522451410607362193,
%U 97120159467079471165,20765771676360919883403,5060640084128464622069221
%N E.g.f. satisfies: A(x) = exp( x/(1 - x*A'(x)/A(x)) ).
%H Vaclav Kotesovec, <a href="/A182962/b182962.txt">Table of n, a(n) for n = 0..250</a>
%F E.g.f.: A(x) = exp(x*F(x)) where F(x) = 1 + x*F(x)*d/dx[x*F(x)] is the o.g.f. of A088716.
%F E.g.f. satisfies: [x^n/n!] A(x)^n = n^2*[x^(n-1)/(n-1)!] A(x)^n for n>=1.
%F E.g.f. satisfies: [x^n/n!] A(x)^(n+1) = (n+1)*A156326(n) for n>=0.
%F E.g.f.: A(x) = x/Series_Reversion(x*G(x)) where A(x*G(x)) = G(x) is the e.g.f. of A156326, which satisfies:
%F . G(x) = exp( Sum_{n>=1} n^2 * A156326(n-1)*x^n/n! ).
%F a(n) ~ c * (n!)^2 * n, where c = 0.21795078944715106549... (see A238223). - _Vaclav Kotesovec_, Feb 22 2014
%e E.g.f.: A(x) = 1 + x + 3*x^2/2! + 25*x^3/3! + 433*x^4/4! +...
%e The logarithm of the e.g.f. is the integer series:
%e log(A(x)) = x + x^2 + 3*x^3 + 14*x^4 + 85*x^5 + 621*x^6 + 5236*x^7 + 49680*x^8 +...+ A088716(n)*x^(n+1) +...
%e ...
%e The coefficients of [x^n/n!] in the powers of e.g.f. A(x) begin:
%e A^1: [(1),(1), 3, 25, 433, 12501, 529531, 30495613, ...];
%e A^2: [1,(2),(8), 68, 1120, 30832, 1260544, 70737536, ...];
%e A^3: [1, 3,(15),(135), 2169, 57303, 2261439, 123523515, ...];
%e A^4: [1, 4, 24,(232),(3712), 94944, 3622336, 192461056, ...];
%e A^5: [1, 5, 35, 365, (5905),(147625), 5460475, 282185825, ...];
%e A^6: [1, 6, 48, 540, 8928, (220176),(7926336), 398625408, ...];
%e A^7: [1, 7, 63, 763, 12985, 318507,(11210479),(549313471), ...];
%e A^8: [1, 8, 80, 1040, 18304, 449728, 15551104,(743759360), ...];
%e ...
%e In the above table, the coefficients in parenthesis are related by:
%e 1*1 = 1; 8 = 2^2*2; 135 = 3^2*15; 3712 = 4^2*232; 147625 = 5^2*5905;
%e this illustrates: [x^n/n!] A(x)^n = n^2*[x^(n-1)/(n-1)!] A(x)^n.
%e ...
%e Also note that the main diagonal in the above table begins:
%e [1*1, 2*1, 3*5, 4*58, 5*1181, 6*36696, 7*1601497, 8*92969920, ...];
%e this illustrates: [x^n/n!] A(x)^(n+1) = (n+1)*A156326(n).
%e ...
%e Let G(x) denote the e.g.f. of A156326:
%e G(x) = 1 + x + 5*x^2/2! + 58*x^3/3! + 1181*x^4/4! + 36696*x^5/5! +...
%e then G(x) satisfies: G(x) = A(x*G(x)) and A(x) = G(x/A(x)) where
%e G(x) = exp( Sum_{n>=1} n^2 * A156326(n-1)*x^n/n! ).
%e ...
%t m = 16; A[_] = 1;
%t Do[A[x_] = Exp[x/(1 - x A'[x]/A[x])] + O[x]^m, {m}];
%t CoefficientList[A[x], x] Range[0, m-1]! (* _Jean-François Alcover_, Oct 29 2019 *)
%o (PARI) {a(n)=local(A=1+x);for(i=1,n,A=exp(x/(1 - x*deriv(A)/A+x*O(x^n))));n!*polcoeff(A,n)}
%o (PARI) {a(n)=local(A=[1,1]);for(i=2,n,A=concat(A,0);A[#A]=((#A-1)*Vec(Ser(A)^(#A-1))[#A-1]-Vec(Ser(A)^(#A-1))[#A])/(#A-1));n!*A[n+1]}
%Y Cf. A088716, A300735, A300986, A300988, A300990, A300992.
%Y Cf. A156326, A238223.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jan 01 2011
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