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A205882 a(n) = Fibonacci(n)*A109064(n) for n>=1 with a(0)=1.

%I A205882 #27 Dec 14 2017 02:40:54
%S A205882 1,-5,5,20,-45,-25,-80,390,525,-1190,275,-5340,4320,13980,-11310,6100,
%T A205882 -54285,127760,90440,-418100,-101475,-656760,1062660,3152270,-2318400,
%U A205882 -375125,-7283580,19641800,28602990,-77134350,-8320400,-215403040,228722445,422949360
%N A205882 a(n) = Fibonacci(n)*A109064(n) for n>=1 with a(0)=1.
%C A205882 Compare to g.f. of A109064, which is the Lambert series identity:
%C A205882 1 - 5*Sum_{n>=1} L(n,5)*n*x^n/(1-x^n) = eta(x)^5/eta(x^5).
%C A205882 Here L(n,5) is the Legendre symbol given by A080891(n).
%H A205882 G. C. Greubel, <a href="/A205882/b205882.txt">Table of n, a(n) for n = 0..1000</a>
%F A205882 G.f.: 1 - 5*Sum_{n>=1} Fibonacci(n)*L(n,5)*n*x^n / (1 - Lucas(n)*x^n + (-1)^n*x^(2*n)), where L(n,5) is the Legendre symbol and Lucas(n) = A000204(n).
%e A205882 G.f.: A(x) = 1 - 5*x + 5*x^2 + 20*x^3 - 45*x^4 - 25*x^5 - 80*x^6 + 390*x^7 +...
%e A205882 where A(x) = 1 - 1*5*x + 1*5*x^2 + 2*10*x^3 - 3*15*x^4 - 5*5*x^5 - 8*10*x^6 + 13*30*x^7 + 21*25*x^8 +...+ Fibonacci(n)*A109064(n)*x^n +...
%e A205882 The g.f. is illustrated by:
%e A205882 A(x) = 1 - 5*(+1)*1*1*x/(1-x-x^2) - 5*(-1)*2*1*x^2/(1-3*x^2+x^4) - 5*(-1)*3*2*x^3/(1-4*x^3-x^6) - 5*(+1)*4*3*x^4/(1-7*x^4+x^8) - 5*(0)*5*5*x^5/(1-11*x^5-x^10) - 5*(+1)*6*8*x^6/(1-18*x^6+x^12) +...
%e A205882 The values of the Legendre symbol L(n,5) repeat: [1,-1,-1,1,0, ...].
%t A205882 QP = QPochhammer; a[0] = 1; a[n_] := Fibonacci[n] * SeriesCoefficient[ QP[q]^5 / QP[q^5], {q, 0, n}]; Table[a[n], {n, 0, 40}] (* _Jean-François Alcover_, Apr 26 2017 *)
%o A205882 (PARI) {Lucas(n)=fibonacci(n-1)+fibonacci(n+1)}
%o A205882 {a(n)=polcoeff(1-5*sum(m=1,n,kronecker(m,5)*m*fibonacci(m)*x^m/(1-Lucas(m)*x^m+(-1)^m*x^(2*m)+x*O(x^n))),n)}
%o A205882 (PARI) {A109064(n)=local(A); if(n<0, 0, A=x*O(x^n); polcoeff( eta(x+A)^5/eta(x^5+A), n))}
%o A205882 {a(n)=if(n==0,1,fibonacci(n)*A109064(n))}
%Y A205882 Cf. A109064, A080891 (Legendre),A000045 (Fibonacci), A000204 (Lucas), A205884 (variant).
%K A205882 sign
%O A205882 0,2
%A A205882 _Paul D. Hanna_, Feb 01 2012