## Free Fibonacci Sequences

John Conway likes playing with the Fibonacci sequence. He invented many new sequences using the following trick. The next number in the sequence is the sum of the two previous number adjusted in some way. Free Fibonacci sequences were invented this way. Here is the recurrence for an *n*-free Fibonacci sequence: the next number in the sequence is the sum of the previous two numbers divided by the highest possible power of *n*.

Let us calculate a 2-free Fibonacci sequence starting with 5 and 4: 5, 4, 9, 13, 11, 3, 7, 5, 3, 1, 1, 1, …. I leave it to the reader to show that any 2-free sequence ends with a cycle of length one.

Let us try a 3-free Fibonacci sequence starting with 5 and 6: 5, 6, 11, 17, 28, 5, 11, 16, 1, 17, 2, 19, 7, 26, 11, 37, 16, 53, 23, 76, 11, 29, 40, 23, 7, 10, 17, 1, 2, 1, 1, 2, and so on. We are now in the cycle of length 3. Is this always the case? Not quite. If there is a 1-1-2 cycle there should be a 2-2-4 cycle, or any cycle *k*–*k*-2*k*, where *k* is coprime with 3. But the question remains: does it always end in a cycle of length 3?

I published a paper Free Fibonacci Sequences with Brandon Avila. We conjecture that a 3-free Fibonacci sequence always ends in a cycle and support this conjecture with a probabilistic argument. We were amused by how the behavior changes when we move to 4-free Fibonacci sequences. It seems that in this case sequences never cycle. We were even more amused when we moved to 5-free Fibonacci sequences and discovered that the behavior changes again.

When *n* equals 5 there are some sequences that cycle. Can you find the cycles? There are also sequences that grow indefinitely and we do not need a probabilistic argument to prove that. Consider Lucas numbers: 2, 1, 3, 4, 7, 11, and so on. This is a Fibonacci-like sequence that never has a term divisible by 5. Thus Lucas numbers form a 5-free Fibonacci sequence. We made a probabilistic argument that most of the starting terms converge eventually to a Lucas-like sequence that grows indefinitely because there are no terms divisible by 5.

What happens for larger *n*? We didn’t manage to find any cycles there. Would you like to try?

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