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%I #16 Oct 23 2017 13:05:25
%S 1,1,4,20,126,972,8876,93580,1119328,14986944,222184136,3614288272,
%T 64022264176,1226914925840,25295189791296,558317369479616,
%U 13136590271813856,328243850207690432,8680766764223956416,242245419192494844096,7113910552105144027136,219304957649505551899136,7081169542830272102170752,238996807468258679150596352
%N E.g.f. A(x) satisfies: A(x) = exp( Integral B(x) dx ) such that B(x) = exp(2*x) * exp( Integral A(x) dx ), where the constant of integration is zero.
%C Compare to: G(x) = exp( Integral G(x) dx ) when G(x) = 1/(1-x).
%H Vaclav Kotesovec, <a href="/A266490/b266490.txt">Table of n, a(n) for n = 0..200</a>
%F E.g.f. A(x) satisfies:
%F (1) A(x) = exp( Integral A(x) + 2*log(A(x)) dx ).
%F (2) A(x) = A'(x)/A(x) - 2*log(A(x)).
%F (3) log(A(x)) = exp(2*x) * Integral exp(-2*x)*A(x) dx.
%F (4) A(x) = exp( Series_Reversion( Integral 1/(exp(x) + 2*x) dx ) ).
%F a(n) ~ c^(n+1) * n!, where c = 1/Integral_{x=0..infinity} 1/(2*x + exp(x)) dx = 1.4650202775490107369040248583790383461628786237838809798971... - _Vaclav Kotesovec_, Aug 21 2017
%e E.g.f.: A(x) = 1 + x + 4*x^2/2! + 20*x^3/3! + 126*x^4/4! + 972*x^5/5! + 8876*x^6/6! + 93580*x^7/7! + 1119328*x^8/8! + 14986944*x^9/9! + 222184136*x^10/10! +...
%e such that log(A(x)) = Integral B(x) dx
%e where B(x) = 1 + 3*x + 10*x^2/2! + 40*x^3/3! + 206*x^4/4! + 1384*x^5/5! + 11644*x^6/6! + 116868*x^7/7! + 1353064*x^8/8! + 17693072*x^9/9! + 257570280*x^10/10! +...
%e and A(x) and B(x) satisfy:
%e (1) A(x) = B'(x)/B(x) - 2,
%e (2) B(x) = A'(x)/A(x),
%e (3) B(x) = A(x) + 2*log(A(x)),
%e (4) log(A(x)) = Integral B(x) dx,
%e (5) log(B(x)) = Integral A(x) dx + 2*x.
%e The Series Reversion of log(A(x)) equals Integral 1/(exp(x) + 2*x) dx:
%e Integral 1/(exp(x) + 2*x) dx = x - 3*x^2/2! + 17*x^3/3! - 145*x^4/4! + 1649*x^5/5! - 23441*x^6/6! + 399865*x^7/7! - 7957881*x^8/8! + 180997857*x^9/9! - 4631289697*x^10/10! +...
%e so that A( Integral 1/(exp(x) + 2*x) dx ) = exp(x).
%t a[ n_] := a[n] = If[ n < 1, Boole[n == 0], Sum[ Binomial[n - 1, k - 1] a[n - k] Sum[ 2^(j - 1) a[k - j], {j, k}], {k, n}]]; (* _Michael Somos_, Aug 08 2017 *)
%o (PARI) {a(n) = my(A=1+x, B=1+x); for(i=0, n, A = exp( intformal( B + x*O(x^n) ) ); B = exp( intformal( 2 + A ) ) ); n!*polcoeff(A, n)}
%o for(n=0, 30, print1(a(n), ", "))
%o (PARI) {a(n) = n! * polcoeff( exp( serreverse( intformal( 1/(exp(x +x*O(x^n)) + 2*x) ) )), n)}
%o for(n=0, 30, print1(a(n), ", "))
%Y Cf. A266328, A266329.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jan 27 2016
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