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%I #29 Dec 14 2023 05:15:51
%S 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,
%T 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,
%U 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
%N Jacobsthal numbers modulo 3.
%C Period 6 = A175286(3).
%H M. E. Muldoon and A. A. Ungar, <a href="http://www.jstor.org/stable/2691389">Beyond Sin and Cos</a>, Mathematics Magazine, 69,1,(1996).
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,-1,1,-1,1).
%F E.g.f.: exp(x) - exp(-x/2)*cos(sqrt(3)*x/2) - 3*exp(x/2)*sin(sqrt(3)*x/2)/sqrt(3);
%F E.g.f.: F(1, 3, 1, x) + F(1, 3, 2, x) + F(1, 6, 4, x) + F(1, 6, 5, x);
%F a(n) = a(n-6), with a(0)=0, a(1)=a(2)=1, a(3)=0, a(4)=a(5)=2;
%F a(n) = 1 - cos(2*Pi*n/3) - 3*sin(Pi*n/3)/3.
%F a(n) = A001045(n) mod 3.
%F 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
%F a(n) = ( n * floor( 3(n+1)/2 ) - 2n ) mod 3. - _Wesley Ivan Hurt_, Oct 13 2013
%p A088689:=n->(n*floor(3*(n+1)/2) - 2*n) mod 3; seq(A088689(k), k=0..70); # _Wesley Ivan Hurt_, Oct 13 2013
%t Table[Mod[n*Floor[3(n+1)/2] - 2n, 3], {n,0,100}] (* _Wesley Ivan Hurt_, Oct 13 2013 *)
%t LinearRecurrence[{1,-1,1,-1,1},{0,1,1,0,2},120] (* _Harvey P. Dale_, Apr 09 2020 *)
%o (PARI) a(n)=[0, 1, 1, 0, 2, 2][n%6+1] \\ _Charles R Greathouse IV_, Oct 16 2015
%K easy,nonn
%O 0,5
%A _Paul Barry_, Oct 06 2003
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