A088689 Jacobsthal numbers modulo 3.

0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1, 0, 2, 2, 0, 1, 1

Offset

0,5

Comments

Period 6 = A175286(3).

Links

Table of n, a(n) for n=0..104.

M. E. Muldoon and A. A. Ungar, Beyond Sin and Cos, Mathematics Magazine, 69,1,(1996).

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

Formula

E.g.f.: exp(x) - exp(-x/2)*cos(sqrt(3)*x/2) - 3*exp(x/2)*sin(sqrt(3)*x/2)/sqrt(3);

E.g.f.: F(1, 3, 1, x) + F(1, 3, 2, x) + F(1, 6, 4, x) + F(1, 6, 5, x);

a(n) = a(n-6), with a(0)=0, a(1)=a(2)=1, a(3)=0, a(4)=a(5)=2;

a(n) = 1 - cos(2*Pi*n/3) - 3*sin(Pi*n/3)/3.

a(n) = A001045(n) mod 3.

G.f.: x*(1+2*x^3)/(1-x+x^2-x^3+x^4-x^5); a(n)=a(n-1)-a(n-2)+a(n-3)-a(n-4)+a(n-5). - Paul Barry, Jul 27 2005

a(n) = ( n * floor( 3(n+1)/2 ) - 2n ) mod 3. - Wesley Ivan Hurt, Oct 13 2013

Maple

A088689:=n->(n*floor(3*(n+1)/2) - 2*n) mod 3; seq(A088689(k), k=0..70); # Wesley Ivan Hurt, Oct 13 2013

Mathematica

Table[Mod[n*Floor[3(n+1)/2] - 2n, 3], {n, 0, 100}] (* Wesley Ivan Hurt, Oct 13 2013 *)
LinearRecurrence[{1, -1, 1, -1, 1}, {0, 1, 1, 0, 2}, 120] (* Harvey P. Dale, Apr 09 2020 *)

Prog

(PARI) a(n)=[0, 1, 1, 0, 2, 2][n%6+1] \\ Charles R Greathouse IV, Oct 16 2015

Crossrefs

Sequence in context: A124210 A287447 A110568 * A076898 A174294 A089385

Adjacent sequences: A088686 A088687 A088688 * A088690 A088691 A088692

Keyword

easy,nonn

Author

Paul Barry, Oct 06 2003

Status

approved