A131290 1 followed by period 6: repeat [3, 2, 0, -1, 0, 2].

1, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1

Offset: 0

Author

Paul Curtz, Sep 29 2007

Keywords

• sign
• easy

Links

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

Crossrefs

Cf. A079757, A100219, A130869.

Programs

Magma

[1] cat &cat [[3, 2, 0, -1, 0, 2]^^30]; // Wesley Ivan Hurt, Jun 20 2016

Maple

A131290 := proc(n) if n = 0 then 1; else op(((n-1)mod 6)+1,[3,2,0,-1,0,2]) ; fi ; end: seq(A131290(n),n=0..100) ; # R. J. Mathar, Feb 27 2008

Mathematica

PadRight[{1},110,{2,3,2,0,-1,0}] (* or *) Join[{1},LinearRecurrence[ {2,-2,1},{3,2,0},110]] (* Harvey P. Dale, Jun 22 2012 *)

Formula

G.f.: (1+x-2*x^2+x^3)/((1-x)*(1-x+x^2)). - R. J. Mathar, Feb 27 2008

If n mod 6 = 4 then a(n) = (Fibonacci(n-3)*Fibonacci(n+1)) mod 4 -2, else a(n) = (Fibonacci(n-3)*Fibonacci(n+1)) mod 4, n>0. - Gary Detlefs, Dec 12 2010

From Wesley Ivan Hurt, Jun 20 2016: (Start)

a(n) = 2*a(n-1) - 2*a(n-2) + a(n-3) for n>4.

a(0) = 1, a(n) = 1 + cos(n*Pi/3) + sqrt(3)*sin(n*Pi/3) for n>0. (End)

Extensions

More terms from R. J. Mathar, Feb 27 2008