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Let v(1)=v(2)=v(3)=1, v(n)=(-1)^n*sign(v(n-1)-v(n-2))*v(n-3), then a(n) =1+v(n).
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%I #8 Oct 13 2022 13:36:02

%S 2,2,2,1,2,2,1,0,2,1,0,0,1,0,0,1,2,0,1,2,2,1,2,2,1,0,2,1,0,0,1,0,0,1,

%T 2,0,1,2,2,1,2,2,1,0,2,1,0,0,1,0,0,1,2,0,1,2,2,1,2,2,1,0,2,1,0,0,1,0,

%U 0,1,2,0,1,2,2,1,2,2,1,0,2,1,0,0,1,0,0,1,2,0,1,2,2,1,2,2,1,0,2,1,0,0,1,0,0

%N Let v(1)=v(2)=v(3)=1, v(n)=(-1)^n*sign(v(n-1)-v(n-2))*v(n-3), then a(n) =1+v(n).

%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,0,0,0,-1,1).

%F a(n) is a period-8 sequence with period (1, 0, 2, 1, 0, 0, 1, 0, 0, 1, 2, 0, 1, 2, 2, 1, 2, 2)

%K nonn,easy

%O 1,1

%A _Benoit Cloitre_, Nov 24 2002