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%I #29 May 20 2020 15:35:41
%S 2,3,4,5,9,25,26,64,169,441,1156,3025,7921,20736,54289,142129,372100,
%T 974169,2550409,6677056,17480761,45765225,119814916,313679521,
%U 821223649,2149991424,5628750625,14736260449,38580030724,101003831721,264431464441,692290561600,1812440220361
%N a(n)-1, a(n) and a(n)+1 form three consecutive integers that can be factored into Fibonacci numbers.
%C In general it can be shown that F(n-1)F(n+1), F(n)^2, F(n-2)F(n+2) form three consecutive increasing integers when n is odd and F(n-2)F(n+2), F(n)^2, F(n-1)F(n+1) for three consecutive increasing integers when n is even. Thus the sequence is infinite. [Corrected by _Charles R Greathouse IV_, Jul 17 2012]
%H Colin Barker, <a href="/A065885/b065885.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).
%F Except for n = 1, 2, 4 and 7, a(n) is the square of a Fibonacci number.
%F From _Colin Barker_, Sep 30 2016: (Start) (based on the signature given in the link}
%F a(n) = 2*a(n-1)+2*a(n-2)-a(n-3) for n>10.
%F G.f.: x*(2-x-6*x^2-7*x^3-6*x^4+x^5-37*x^6-29*x^7+14*x^8+x^9) / ((1+x)*(1-3*x+x^2)).
%F (End)
%F a(n) = 3*a(n-1) - a(n-2) - 2*(-1)^n for n >= 10. - _Greg Dresden_, May 18 2020
%e 440 = 8*55, 441 = 21^2, 442 = 13*34, so 441 is a term of the sequence.
%o (PARI) a(n)=if(n>7,fibonacci(n-2)^2,[2,3,4,5,9,25,26][n]) \\ _Charles R Greathouse IV_, Jul 17 2012
%o (PARI) Vec(x*(2-x-6*x^2-7*x^3-6*x^4+x^5-37*x^6-29*x^7+14*x^8+x^9)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ _Colin Barker_, Sep 30 2016
%Y Cf. A000045, A007598, A065108.
%K nonn,easy
%O 1,1
%A _John W. Layman_, Nov 28 2001
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