3-free Fibonacci sequences. Consider an example of 3-free Fibonacci numbers: 0, 1, 1, 2, 1, 1, 2, and so on. The sequence turns into a cycle of length 3. There are only 3 different 3-free Fibonacci numbers.

We can multiply a 3-free sequence by a number not divisible by 3 to get another 3-free sequence. Thus, in general we can get cycles of the form $k$, $k$, $2k$, where $k$ is not divisible by 3.

Any cycle of length 3 in a 3-free Fibonacci sequence is of the form $k$, $k$, $2k$.

Consider the length 3 cycle $a$, $b$, $c$. From the definition of 3-free Fibonacci sequences, we know the following relations:

$$\begin{aligned}
a+b=3^xc \nonumber \\
b+c=3^ya \label{eq:b+c}\\
c+a=3^zb. \label{eq:c+a}\end{aligned}$$

Furthermore, no term in the sequence is divisible by 3. Then, by the pigeon-hole principle, at least two of the terms $a,b,c$ must be congruent modulo 3. Without loss of generality, take $a \equiv b$ (mod 3). Then $a+b \not\equiv 0$ (mod 3), so we have that $x=0$ and $a+b=c$. Now substitute for $c$ and add equations (\[eq:b+c\]) and (\[eq:c+a\]) to get that $a+b=3^{y-1}a+3^{z-1}b$. Since $3\nmid a+b$, either $y=1$ or $z=1$. If $y=1$, then $b=3^{z-1}b$, hence $z=1$. Similarly, $z=1$ implies 