probabilistic argument that any 3-free Fibonacci sequence ends in a cycle.

According to our probabilistic assumptions we divide either every term or every other term with the same probability $1/2$. So on average we divide on every $1.5$ step. But for how much we divide on average?

\[thm:average3\] On average we divide by $3^{3/2}$.

Now we want to use the fact that when we divide by a power of 3 we on average divide by more than 3. If the number is large, we divide by 3 with probability $2/3$, by 9 with probability $2/9$ and so on. So the average division is by $$3^{2/3} \cdot 9^{2/9} \cdot 27^{2/27} \cdot \ldots.$$ We can say it differently. We can say that we divide by 3 with probability 1 and additionally we divide by 3 more with probability $1/3$, and by 3 more with probability $1/9$, so the result is 3 to the power $$1+1/3+1/9+1/27 + 1/81 + \ldots = 3/2.$$ Since the above sum is equal to $3/2$, every time we divide, we on average divide by $3^{3/2}$.

Notice that the average number we divide by is approximately 5.2 which is more than 5.

Let us build a probabilistic sequence that 