any sequence, regardless of how many factors of 3 are divided out from each term, will have the same underlying structure in its parity. Since we have reduced the sequence to the point where there exists at least one odd number, we know that the sequence reduced modulo 2 must be congruent to 1, 1, 0, 1, 1, 0. Only cycles of length $3n$ are permitted in this structure, and therefore permitted in 3-free sequences.

We checked all the starting pairs of numbers from 1 to 1000, and all these sequences end in a 3-cycle.

The 3-free Fibonacci sequences are in many ways similar to the notorious Collatz sequences [@La10], for which it is still not known if every sequence eventually cycles. We do not expect that it is easy to prove or disprove that ever 3-free Fibonacci sequence ends in a cycle. On the other hand, it is possible to make probabilistic arguments to support different claims about free Fibonacci sequences.

Here is the base of the argument. Suppose we encounter a number in a sequence that is divisible by 3, then after removing all powers of 3, let us assume that the resulting number has the remainder 1 