A065885 a(n)-1, a(n) and a(n)+1 form three consecutive integers that can be factored into Fibonacci numbers.

2, 3, 4, 5, 9, 25, 26, 64, 169, 441, 1156, 3025, 7921, 20736, 54289, 142129, 372100, 974169, 2550409, 6677056, 17480761, 45765225, 119814916, 313679521, 821223649, 2149991424, 5628750625, 14736260449, 38580030724, 101003831721, 264431464441, 692290561600, 1812440220361

Offset: 1

Author

John W. Layman, Nov 28 2001

Keywords

• nonn
• easy

Comments

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]

Examples

			440 = 8*55, 441 = 21^2, 442 = 13*34, so 441 is a term of the sequence.
		

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Crossrefs

Cf. A000045, A007598, A065108.

Programs

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

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

Formula

Except for n = 1, 2, 4 and 7, a(n) is the square of a Fibonacci number.

From Colin Barker, Sep 30 2016: (Start) (based on the signature given in the link)

a(n) = 2*a(n-1)+2*a(n-2)-a(n-3) for n>10.

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)).

(End)

a(n) = 3*a(n-1) - a(n-2) - 2*(-1)^n for n >= 10. - Greg Dresden, May 18 2020