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%I #17 Aug 21 2016 02:21:15
%S 1,2,14,170,2942,66122,1828094,59999690,2279713982,98416537802,
%T 4758057790334,254660698294730,14948094619714622,954799973383805642,
%U 65929683232042222334,4893783473730361785290,388585728666772137187262,32866836871725315639639242,2950043008427035823210644094
%N E.g.f. satisfies: A(x) = 1/(1 - Integral A(x) + A(x)^2 dx).
%C Compare to: G(x) = 1/(1 - Integral G(x) dx) holds when G(x) is the e.g.f. of odd double factorials (A001147).
%C Compare to: G(x) = 1/(1 - Integral G(x)^2 dx) holds when G(x) is the e.g.f. of triple factorial numbers (A007559).
%C Compare to: G(x) = 1 + Integral G(x)+G(x)^2 dx holds when G(x) is the e.g.f. of A000629 (number of necklaces of partitions of n+1 labeled beads).
%H Vaclav Kotesovec, <a href="/A235369/b235369.txt">Table of n, a(n) for n = 0..200</a>
%F E.g.f. A(x) satisfies:
%F (1) A(x) = 1/(1 - 2*Series_Reversion( Integral (1-2*x)^2/(1-x) dx )).
%F (2) A(x) = 1/sqrt(1 - 4*Series_Reversion( Integral sqrt(1-4*x)*C(x) dx )), where C(x) = 1 + x*C(x)^2 = (1 - sqrt(1-4*x))/(2*x) is the Catalan function (A000108).
%F (3) A(x) = exp( Integral A(x)^2 + A(x)^3 dx ). - _Paul D. Hanna_, Aug 21 2016
%F (4) A(x) = 1 + Integral A(x)^3 + A(x)^4 dx. - _Paul D. Hanna_, Aug 21 2016
%F a(n) ~ n! / (GAMMA(1/3) * 3^(1/3) * (log(2)-1/2)^(n+1/3) * n^(2/3)). - _Vaclav Kotesovec_, Jan 27 2014
%e E.g.f.: A(x) = 1 + 2*x + 14*x^2/2! + 170*x^3/3! + 2942*x^4/4! + 66122*x^5/5! +...
%e Related series.
%e A(x)^2 = 1 + 4*x + 36*x^2/2! + 508*x^3/3! + 9780*x^4/4! + 238684*x^5/5! +...
%e The series 1 - 1/A(x) = Integral A(x) + A(x)^2 dx and begins:
%e 1 - 1/A(x) = 2*x + 6*x^2/2! + 50*x^3/3! + 678*x^4/4! + 12722*x^5/5! +...
%e where the series reversion of (1 - 1/A(x))/2 equals Integral (1-2*x)^2/(1-x) dx.
%e The series 1 - 1/A(x)^2 begins:
%e 1 - 1/A(x)^2 = 4*x + 4*x^2/2! + 28*x^3/3! + 340*x^4/4! + 5884*x^5/5! +...
%e where the series reversion of (1 - 1/A(x)^2)/4 begins:
%e x - x^2/2 - 2*x^3/3 - 5*x^4/4 - 14*x^5/5 - 42*x^6/6 - 132*x^7/7 - 429*x^8/8 - 1430*x^9/9 +...+ -A000108(n)*x^(n+1)/(n+1) +...
%e Also, we have
%e A(x) = exp( Integral A(x)^2 + A(x)^3 dx ), and thus
%e A'(x) = A(x)^3 + A(x)^4, where
%e A(x)^3 = 1 + 6*x + 66*x^2/2! + 1062*x^3/3! + 22530*x^4/4! + 593766*x^5/5! + 18705666*x^6/6! + 685717542*x^7/7! + 28674765570*x^8/8! +...
%e A(x)^4 = 1 + 8*x + 104*x^2/2! + 1880*x^3/3! + 43592*x^4/4! + 1234328*x^5/5! + 41294024*x^6/6! + 1593996440*x^7/7! + 69741772232*x^8/8! +...
%t CoefficientList[1/(1-2*InverseSeries[Series[-2*x^2-Log[-1+x]+I*Pi,{x,0,20}],x]),x]*Range[0,20]! (* _Vaclav Kotesovec_, Jan 27 2014 *)
%o (PARI) {a(n) = my(A=1); for(i=1,n, A = 1/(1 - intformal(A + A^2+x*O(x^n))));n!*polcoeff(A,n)}
%o for(n=0,20,print1(a(n),", "))
%o (PARI) {a(n) = my(A = 1/(1 - 2*serreverse( intformal((1-2*x)^2/(1-x +x*O(x^n)) ))));n!*polcoeff(A,n)}
%o for(n=0,20,print1(a(n),", "))
%o (PARI) {a(n) = my(A=1); for(i=1, n, A = exp( intformal(A^2+A^3 +x*O(x^n)) ) ); n!*polcoeff(A, n)}
%o for(n=0, 20, print1(a(n), ", ")) \\ _Paul D. Hanna_, Aug 21 2016
%o (PARI) {a(n) = my(A=1+x); for(i=1, n, A = 1 + intformal(A^3 + A^4 +x*O(x^n))); n!*polcoeff(A, n)}
%o for(n=0, 20, print1(a(n), ", ")) \\ _Paul D. Hanna_, Aug 21 2016
%Y Cf. A228608.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Jan 08 2014
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