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

1,1

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]

Links

Colin Barker, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

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

Example

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

Prog

(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

Crossrefs

Cf. A000045, A007598, A065108.

Sequence in context: A323289 A355374 A357601 * A274331 A177064 A092233

Adjacent sequences: A065882 A065883 A065884 * A065886 A065887 A065888

Keyword

nonn,easy

Author

John W. Layman, Nov 28 2001

Status

approved