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%I #12 Jul 17 2015 10:03:11
%S 1,1,3,15,81,561,4683,44415,479241,5793921,77332563,1130944815,
%T 17984844801,308888337681,5698762943643,112401325405215,
%U 2360158641832761,52564270139375841,1237645528139173923,30717272450961249615,801500394828539219121
%N E.g.f. satisfies: A'(x) = (1 + A(x))*(1 + A(x)^2).
%F E.g.f.: Series_Reversion( Integral 1/((1+x)*(1+x^2)) dx ).
%F E.g.f.: Series_Reversion( (log((1+x)^2/(1+x^2)) + 2*atan(x))/4 ).
%F E.g.f. A(x) satisfies: (1 + A(x))^2/(1 + A(x)^2) = exp(4*x) / exp(2*atan(A(x))).
%F a(n) ~ 2^(2*n+1) * n^n / (exp(n) * Pi^(n+1/2)). - _Vaclav Kotesovec_, Jul 17 2015
%e E.g.f.: A(x) = x + x^2/2! + 3*x^3/3! + 15*x^4/4! + 81*x^5/5! + 561*x^6/6! +..
%e where
%e (1 + A(x))*(1 + A(x)^2) = 1 + x + 3*x^2/2! + 15*x^3/3! + 81*x^4/4! + 561*x^5/5! +...+ a(n+1)*x^n/n! +...
%e The series reversion of the e.g.f. equals Integral 1/(1+x+x^2+x^3) dx:
%e Series_Reversion(A(x)) = x - x^2/2 + x^5/5 - x^6/6 + x^9/9 - x^10/10 + x^13/13 - x^14/16 + x^17/17 - x^18/18 +...
%e which equals (log((1+x)^2/(1+x^2)) + 2*atan(x))/4.
%t Rest[CoefficientList[InverseSeries[Series[(Log[(1+x)^2/(1+x^2)] + 2*ArcTan[x])/4, {x, 0, 20}], x],x] * Range[0, 20]!] (* _Vaclav Kotesovec_, Jul 17 2015 *)
%o (PARI) {a(n) = local(A=x); for(i=1,n, A = intformal( (1+A)*(1+A^2) +x*O(x^n))); n!*polcoeff(A,n)}
%o for(n=1,30,print1(a(n),", "))
%o (PARI) {a(n) = local(A=x); A = serreverse( intformal( 1/((1+x)*(1+x^2) +x*O(x^n)) )); n!*polcoeff(A,n)}
%o for(n=1,30,print1(a(n),", "))
%K nonn
%O 1,3
%A _Paul D. Hanna_, Jul 10 2015
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