\cdot a_n - a_{n-1}$. Thus, by choosing $a_n$ and $a_{n-1}$, and selecting a sequence $\{i_m\}$ that satisfies this relationship, the sequence can easily be constructed backwards. Our only requirements are that every term of the sequence is positive, and every step contains a division, so it will suffice to construct a sequence $\{i_m\}$ such that $3^{i_n} \cdot a_n - a_{n-1}>0$ and $i_n>0$ for all $n$.

As an example, let us begin with $a_n=1$, $a_{n-1}=1$. At each step let us choose the smallest possible power for $i_m$. Then $i_n = 1$ satisfies our inequality for the first step, and $a_{n-2}=2$. Continuing, $i_{n-1}=1$ satisfies the inequality for the next step, yielding $a_{n-3}=1$. Next, $i_{n-2}=1$ and $a_{n-3}=1$, followed by $i_{n-3}=2$ and $a_{n-4}=5$. Reading the sequence backwards we get: 1, 1, 2, 1, 5, 4, 11, 1, 32, 49, $\ldots$. This is now sequence A233525 in the OEIS [@OEIS]. when read forward, this sequence is a 3-free sequence containing eight divisions in a row. The process can be continued to arbitrarily many terms for arbitrarily many consecutive divisions.

The growth bound for division-rich subsequences is estimated by the following lemma.

For a division-rich subsequence: $\max\{a_{2k+1},a_{2k+2}\} \leq 2\max\{a_{2k-1},a_{2k}\}/3$.

We can estimate that $a_{2k+1} \leq (a_{2k-1}+a_{2k})/3 