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A205966
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a(n) = Fibonacci(n)*A004016(n) for n>=1, with a(0)=1, where A004016(n) is the number of integer solutions (x,y) to x^2 + x*y + y^2 = n.
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7
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1, 6, 0, 12, 18, 0, 0, 156, 0, 204, 0, 0, 864, 2796, 0, 0, 5922, 0, 0, 50172, 0, 131352, 0, 0, 0, 450150, 0, 1178508, 3813732, 0, 0, 16155228, 0, 0, 0, 0, 89582112, 289893804, 0, 758951832, 0, 0, 0, 5201933244, 0, 0, 0, 0, 28845161856, 140017356882, 0, 0
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OFFSET
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0,2
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COMMENTS
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Compare g.f. to the Lambert series of A004016: 1 + 6*Sum_{n>=1} Kronecker(n,3)*x^n/(1 - x^n).
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LINKS
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FORMULA
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G.f.: 1 + 6*Sum_{n>=1} Fibonacci(n)*Kronecker(n,3)*x^n/(1 - Lucas(n)*x^n + (-1)^n*x^(2*n)).
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EXAMPLE
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G.f.: A(x) = 1 + 6*x + 12*x^3 + 18*x^4 + 156*x^7 + 204*x^9 + 864*x^12 +...
where A(x) = 1 + 1*6*x + 2*6*x^3 + 3*6*x^4 + 13*12*x^7 + 34*6*x^9 + 144*6*x^12 +...+ Fibonacci(n)*A004016(n)*x^n +...
The g.f. is also given by the identity:
A(x) = 1 + 6*( 1*x/(1-x-x^2) - 1*x^2/(1-3*x^2+x^4) + 3*x^4/(1-7*x^4+x^8) - 5*x^5/(1-11*x^5-x^10) + 13*x^7/(1-29*x^7-x^14) - 21*x^8/(1-47*x^8-x^16) +...).
The values of the symbol Kronecker(n,3) repeat [1,-1,0, ...].
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MATHEMATICA
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A004016[n_] := SeriesCoefficient[(QPochhammer[q]^3 + 9 q QPochhammer[q^9]^3)/QPochhammer[q^3], {q, 0, n}]; Join[{1}, Table[Fibonacci[n]*b[n], {n, 1, 50}]] (* G. C. Greubel, Mar 05 2017 *)
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PROG
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(PARI) {Lucas(n)=fibonacci(n-1)+fibonacci(n+1)}
{a(n)=polcoeff(1 + 6*sum(m=1, n, kronecker(m, 3)*fibonacci(m)*x^m/(1-Lucas(m)*x^m+(-1)^m*x^(2*m) +x*O(x^n))), n)}
for(n=0, 60, print1(a(n), ", "))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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