A330080 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.

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, 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, 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

Offset: 1

Author

Andres Cicuttin, Nov 30 2019

Keywords

• nonn

Comments

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

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

Crossrefs

Cf. A068508, A185332, A185341.

Programs

Mathematica

c[1] = 1; c[2] = 1; c[3] = 1;
c[n_] := c[n] = (c[n - 2] + c[n - 1])/c[n - 3];
Table[Floor@c[j], {j, 1, 2^6}]