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%I #12 May 22 2021 19:59:14
%S 1,5,8,17,37,84,201,493,1240,3161,8141,21092,54865,143061,373608,
%T 976609,2554357,6683444,17491097,45781949,119841976,313723305,
%U 821294493,2150106052,5628936097,14736560549,38580516296,101004617393,264432735685,692292618516
%N a(n) = Fibonacci(n+1)^2 + 4*Fibonacci(n).
%C 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.
%D Postings to Number Theory List (NMBRTHRY(AT)LISTSERV.NODAK.EDU) by Victor S. Miller, Oct 05 2000.
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (3,1,-5,-1,1).
%F 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
%t 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 *)
%t 4#[[1]]+#[[2]]^2&/@Partition[Fibonacci[Range[0,30]],2,1] (* _Harvey P. Dale_, May 21 2021 *)
%Y Cf. A000045.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_, Oct 05 2000
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