%I #11 Jan 04 2014 03:45:42
%S 1,1,6,41,456,6301,108576,2207981,52012416,1390239481,41593598976,
%T 1376769180401,49955931795456,1971671764875541,84095262825824256,
%U 3854514200269774901,188942180401957502976,9863099585213327293681,546266997049408050364416,31993839349571172423492281
%N E.g.f. 1/(1 - sin(5*x))^(1/5).
%F E.g.f. A(x) satisfies: A(x) = (cos(5*x/2) - sin(5*x/2))^(-2/5).
%F O.g.f.: 1/G(0) where G(k) = 1 - (5*k+1)*x - 5*(k+1)*(5*k+2)/2*x^2/G(k+1) [continued fraction formula from A144015 due to _Sergei N. Gladkovskii_].
%F a(n) ~ n! * sqrt(5+sqrt(5)) * GAMMA(3/5) * 2^(n-9/10) * 5^n / (n^(3/5) * Pi^(n+7/5)). - _Vaclav Kotesovec_, Jan 03 2014
%e E.g.f.: A(x) = 1 + x + 6*x^2/2! + 41*x^3/3! + 456*x^4/4! + 6301*x^5/5! +...
%e O.g.f.: 1/(1-x - 5*1*2/2*x^2/(1-6*x - 5*2*7/2*x^2/(1-11*x - 5*3*12/2*x^2/(1-16*x - 5*4*17/2*x^2/(1-21*x - 5*5*22/2*x^2/(1-...)))))), a continued fraction.
%t CoefficientList[Series[1/(1-Sin[5*x])^(1/5), {x, 0, 20}], x]* Range[0, 20]! (* _Vaclav Kotesovec_, Jan 03 2014 *)
%o (PARI) {a(n)=local(X=x+x*O(x^n)); n!*polcoeff((1-sin(5*X))^(-1/5), n)}
%o for(n=0, 20, print1(a(n), ", "))
%o (PARI) {a(n)=local(A=1+x+x*O(x^n)); for(i=0, n, A=exp(intformal(A^(5/2)/subst(A^(5/2), x, -x)))); n!*polcoeff(A, n)}
%o for(n=0, 20, print1(a(n), ", "))
%Y Cf. A001586, A007788, A144015, A227544, A230114.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Dec 20 2013
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