login
Let u(1) = u(2) = u(3) = 1; u(n) = sign(u(n-1)-u(n-2))*u(n-3), then a(n) = 1+u(n).
1

%I #9 Nov 30 2018 19:40:23

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

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

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

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

%F a(n) is a 9-periodic sequence with period (1, 0, 0, 1, 0, 2, 1, 2, 2)

%t PadRight[{2},120,{1,2,2,1,0,0,1,0,2}] (* _Harvey P. Dale_, Nov 30 2018 *)

%K nonn

%O 1,1

%A _Benoit Cloitre_, Nov 24 2002