|
%I #11 Jun 07 2016 13:51:45
%S 1,1,5,35,355,4585,72485,1353275,29150275,711535825,19409915525,
%T 585181872275,19321831403875,693431767653625,26876449852377125,
%U 1118833620294264875,49786969727179559875,2358371859530852790625,118483568521991801253125,6292649927564072269071875
%N E.g.f.: Series_Reversion(x - x^2/2 - x^3/3).
%F E.g.f. satisfies: A(x) = Integral 1/(1 - A(x) - A(x)^2) dx.
%F a(n) ~ 12^(n-1/2) * n^(n-1) / (5^(1/4) * exp(n) * (5*sqrt(5)-7)^(n-1/2)). - _Vaclav Kotesovec_, Jun 15 2015
%F Conjecture: +19*a(n) +21*(-2*n+3)*a(n-1) -4*(3*n-5)*(3*n-7)*a(n-2)=0. - _R. J. Mathar_, Jun 07 2016
%e E.g.f.: A(x) = x + x^2/2! + 5*x^3/3! + 35*x^4/4! + 355*x^5/5! + 4585*x^6/6! +...
%e where A(x - x^2/2 - x^3/3) = x.
%t Rest[CoefficientList[InverseSeries[Series[x - x^2/2 - x^3/3, {x, 0, 20}], x],x] * Range[0, 20]!] (* _Vaclav Kotesovec_, Jun 15 2015 *)
%o (PARI) {a(n) = local(A=x); A = serreverse(x - x^2/2 - x^3/3 + x*O(x^n)); n!*polcoeff(A,n)}
%o for(n=1,25,print1(a(n),", "))
%o (PARI) {a(n) = local(A=x); for(i=1,n, A = intformal(1/(1 - A - A^2 + x*O(x^n)))); n!*polcoeff(A,n)}
%o for(n=1,25,print1(a(n),", "))
%K nonn
%O 1,3
%A _Paul D. Hanna_, Jun 14 2015
|
