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A057592
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a(n) = Fibonacci(n+1)^2 + 4*Fibonacci(n).
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0
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1, 5, 8, 17, 37, 84, 201, 493, 1240, 3161, 8141, 21092, 54865, 143061, 373608, 976609, 2554357, 6683444, 17491097, 45781949, 119841976, 313723305, 821294493, 2150106052, 5628936097, 14736560549, 38580516296, 101004617393, 264432735685, 692292618516
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OFFSET
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0,2
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COMMENTS
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Theorem: only the first term is a square. Proof from Don Coppersmith: (F[n+1] + 2)^2 = F[n+1]^2 + 4*F[n+1] + 4 > F[n+1]^2 + 4*F[n]. But (F[n+1] + 1)^2 -(F[n+1]^2 + 4*F[n])= 2*F[n+1] + 1 - 4*F[n] is odd and positive, so can't be 0. Thus our number is trapped between 2 successive squares.
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REFERENCES
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Postings to Number Theory List (NMBRTHRY(AT)LISTSERV.NODAK.EDU) by Victor S. Miller, Oct 05 2000.
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LINKS
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FORMULA
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G.f.: -(4*x^4-7*x^3-8*x^2+2*x+1)/(x^5-x^4-5*x^3+x^2+3*x-1). - Alois P. Heinz, May 22 2021
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MATHEMATICA
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Table[Fibonacci[n + 1]^2 + 4*Fibonacci[n], {n, 0, 200}] (* and *) CoefficientList[Series[(-4 z^4 + 7 z^3 + 8 z^2 - 2 z - 1)/(z^5 - z^4 - 5 z^3 + z^2 + 3 z - 1), {z, 0, 100}], z] (* Vladimir Joseph Stephan Orlovsky, Jun 30 2011 *)
4#[[1]]+#[[2]]^2&/@Partition[Fibonacci[Range[0, 30]], 2, 1] (* Harvey P. Dale, May 21 2021 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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