OFFSET
0,2
COMMENTS
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.
REFERENCES
Postings to Number Theory List (NMBRTHRY(AT)LISTSERV.NODAK.EDU) by Victor S. Miller, Oct 05 2000.
LINKS
Index entries for linear recurrences with constant coefficients, signature (3,1,-5,-1,1).
FORMULA
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
MATHEMATICA
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 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Oct 05 2000
STATUS
approved