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A088689
Jacobsthal numbers modulo 3.
4
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
M. E. Muldoon and A. A. Ungar, Beyond Sin and Cos, Mathematics Magazine, 69,1,(1996).
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
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Oct 06 2003
STATUS
approved