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A205882
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a(n) = Fibonacci(n)*A109064(n) for n>=1 with a(0)=1.
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4
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1, -5, 5, 20, -45, -25, -80, 390, 525, -1190, 275, -5340, 4320, 13980, -11310, 6100, -54285, 127760, 90440, -418100, -101475, -656760, 1062660, 3152270, -2318400, -375125, -7283580, 19641800, 28602990, -77134350, -8320400, -215403040, 228722445, 422949360
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OFFSET
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0,2
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COMMENTS
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Compare to g.f. of A109064, which is the Lambert series identity:
1 - 5*Sum_{n>=1} L(n,5)*n*x^n/(1-x^n) = eta(x)^5/eta(x^5).
Here L(n,5) is the Legendre symbol given by A080891(n).
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LINKS
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FORMULA
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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).
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EXAMPLE
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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 +...
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 +...
The g.f. is illustrated by:
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) +...
The values of the Legendre symbol L(n,5) repeat: [1,-1,-1,1,0, ...].
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MATHEMATICA
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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 *)
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PROG
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(PARI) {Lucas(n)=fibonacci(n-1)+fibonacci(n+1)}
{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)}
(PARI) {A109064(n)=local(A); if(n<0, 0, A=x*O(x^n); polcoeff( eta(x+A)^5/eta(x^5+A), n))}
{a(n)=if(n==0, 1, fibonacci(n)*A109064(n))}
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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