A235369 E.g.f. satisfies: A(x) = 1/(1 - Integral A(x) + A(x)^2 dx).
1, 2, 14, 170, 2942, 66122, 1828094, 59999690, 2279713982, 98416537802, 4758057790334, 254660698294730, 14948094619714622, 954799973383805642, 65929683232042222334, 4893783473730361785290, 388585728666772137187262, 32866836871725315639639242, 2950043008427035823210644094
Offset: 0
Keywords
Examples
E.g.f.: A(x) = 1 + 2*x + 14*x^2/2! + 170*x^3/3! + 2942*x^4/4! + 66122*x^5/5! +... Related series. A(x)^2 = 1 + 4*x + 36*x^2/2! + 508*x^3/3! + 9780*x^4/4! + 238684*x^5/5! +... The series 1 - 1/A(x) = Integral A(x) + A(x)^2 dx and begins: 1 - 1/A(x) = 2*x + 6*x^2/2! + 50*x^3/3! + 678*x^4/4! + 12722*x^5/5! +... where the series reversion of (1 - 1/A(x))/2 equals Integral (1-2*x)^2/(1-x) dx. The series 1 - 1/A(x)^2 begins: 1 - 1/A(x)^2 = 4*x + 4*x^2/2! + 28*x^3/3! + 340*x^4/4! + 5884*x^5/5! +... where the series reversion of (1 - 1/A(x)^2)/4 begins: 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) +... Also, we have A(x) = exp( Integral A(x)^2 + A(x)^3 dx ), and thus A'(x) = A(x)^3 + A(x)^4, where 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! +... 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! +...
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 0..200
Crossrefs
Cf. A228608.
Programs
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Mathematica
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 *) -
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)} for(n=0,20,print1(a(n),", ")) -
PARI
{a(n) = my(A = 1/(1 - 2*serreverse( intformal((1-2*x)^2/(1-x +x*O(x^n)) ))));n!*polcoeff(A,n)} for(n=0,20,print1(a(n),", ")) -
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)} for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Aug 21 2016 -
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)} for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Aug 21 2016
Formula
E.g.f. A(x) satisfies:
(1) A(x) = 1/(1 - 2*Series_Reversion( Integral (1-2*x)^2/(1-x) dx )).
(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).
(3) A(x) = exp( Integral A(x)^2 + A(x)^3 dx ). - Paul D. Hanna, Aug 21 2016
(4) A(x) = 1 + Integral A(x)^3 + A(x)^4 dx. - Paul D. Hanna, Aug 21 2016
a(n) ~ n! / (GAMMA(1/3) * 3^(1/3) * (log(2)-1/2)^(n+1/3) * n^(2/3)). - Vaclav Kotesovec, Jan 27 2014
Comments