maximum decreases: $m_{k+1} < m_k$. Hence, the sequence has to stabilize.

It follows from the proof that for a sequence starting with $a_1$, $a_2$ the number of steps until the cycle is reached is not more than $\max\{a_1,a_2\}$. On the other hand, the subsequence before the cycle can be arbitrary long. It follows from the following lemma.

For any two odd numbers $a_1$, $a_2$, a preceding odd number $a_0$ can be found so that $a_0$, $a_1$, and $a_2$ form a 2-free Fibonacci sequence.

Pick a positive integer $k$ so that $2^k a_2 > a_1$ and set $a_0$ to be equal to $2^k a_2 - a_1$.

There are many ways to build predecessors to a given 2-free Fibonacci sequence. The minimal such sequence is built when we choose the smallest power of 2 that still allows us to have positive members in the sequence. We explicitly build such an example starting with $a_1=3$, and $a_2=1$. Reversing the indexing direction we get: 1, 3, 1, 5, 3, 7, 5, 9, 1, $\ldots$, which is now sequence A233526.

Next, we want to continue with 3-free Fibonacci sequences. Are they as simple as 2-free sequences?

3-free Fibonacci Sequences {#sec:3free}
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Let us look at 