twice less probable, so the average growth is at least the cube root of $1.51023^2\cdot 0.453822 = 1.03507$, which is greater than 1.

The result is greater than 1, which means that our model sequence does not cycle all the time. Therefore, extending the argument to 4-free sequences, we can safely say that 4-free sequences do not cycle all the time. Taking our computational experiments and our intuition into account we are comfortable with the following conjecture:

With probability 1 a 4-free Fibonacci sequence does not cycle.

So 4-free Fibonacci sequences do not cycle. If $n$ grows will it mean that $n$-free Fibonacci sequences for $n > 4$ will not cycle either?

5-free Fibonacci sequences {#sec:5free}
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Let us look at the Lucas sequence modulo 5: $2$, $1$, $3$, $4$, $2$, 1 $\ldots$ and see that no term is divisible by 5. Clearly, no term in the Lucas sequence will require that we factor out a power of 5, and the terms will grow indefinitely. Thus, the Lucas sequence is itself a 5-free Fibonacci sequence. This is something new. We do not need a probabilistic argument to show that there are 5-free Fibonacci sequences that do not cycle.

On the 