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a(n) = floor(b(n)), where b(1) = b(2) = b(3) = 1 and b(n) = (b(n-1) + b(n-2))/b(n-3) for n > 3.
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%I #18 Dec 10 2023 17:41:24

%S 1,1,1,2,3,5,4,3,1,1,0,1,2,4,4,4,2,1,0,1,1,2,3,5,3,2,1,1,0,1,2,4,4,3,

%T 1,1,0,1,1,3,4,4,2,1,0,1,1,2,3,5,3,2,1,1,0,1,2,4,4,3,1,1,0,1,1,3,4,4,

%U 2,1,0,1,1,2,3,5,3,2,1,1,0,1,2,4,4,3,1,1,0,1,1,3,4,4,2,1,0,1,1,2,3,5

%N a(n) = floor(b(n)), where b(1) = b(2) = b(3) = 1 and b(n) = (b(n-1) + b(n-2))/b(n-3) for n > 3.

%C This sequence seems quasiperiodic with the same quasiperiod of A185332(n)/A185341(n) (see related comments of Michael Somos in A068508).

%C Conjecture: 0 <= a(n) <= 5.

%t c[1] = 1; c[2] = 1; c[3] = 1;

%t c[n_] := c[n] = (c[n - 2] + c[n - 1])/c[n - 3];

%t Table[Floor@c[j], {j, 1, 2^6}]

%Y Cf. A068508, A185332, A185341.

%K nonn

%O 1,4

%A _Andres Cicuttin_, Nov 30 2019