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
Links
- 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).
Programs
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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
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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 *) -
PARI
a(n)=[0, 1, 1, 0, 2, 2][n%6+1] \\ Charles R Greathouse IV, Oct 16 2015
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
Comments