modulo 3 with probability $1/2$. If the sum of two consecutive terms in a sequence is large and divisible by 3, then we also assume that this number is divisible by $3^k$ with probability $1/3^{k-1}$.

How often do we divide by 3 in a 3-free Fibonacci sequence? The following lemma is obvious.

In a 3-free Fibonacci sequence the division happens for every term or for every other term.

In other words, we can not have a subsequence of length 3 such that each term is the sum of the previous two terms. We want to study two polar cases first: stretches where we divide every term and stretches were we divide every other term.

We will call a subsequence of a 3-free Fibonacci sequence where we divide at each step a *division-rich subsequence*. Conversely, we will call a subsequence of a 3-free Fibonacci sequence where we divide at every other step a *division-poor subsequence*.

\[thm:rich\] There exist arbitrary long division-rich subsequences.

The proof is done by explicit construction. Consider the definition of a division-rich subsequence. In this case, we divide by a power of 3 after every addition step, so that $3^{i_n} \cdot a_n = a_{n-1}+a_{n-2}$ for $i_n>0$. Equivalently, $a_{n-2}=3^{i_n} 