Problem similar to Collatz conjecture

Question

Let $f:\mathbb{Z} \rightarrow \mathbb{Z}$ be given by $f(x) = \frac{n}{2}$ if $n$ is even and $f(n) = n^2 -1$ if $n$ is odd. Consider the sequence $(a_n)$ given by $a_n = f(a_{n-1})$ with $a_0$ a positive integer. Will the sequence eventually reach $1$? ( i.e does there exist an $N$, such that for any $m>N$, $a_m = 1$ no matter what the value of $a_0$ is ?)

At first glance, it does seem to be the case, but taking $a_0 = 13$, it is not clear whether the sequence reaches $1$. Coding it in python, after $1000$ iterations the sequence got exponentially large, to the point that squaring woulnd not be accurate enough in python. I tried looking for patterns in $mod(8)$ but couldn't really make much progress with this approach.

Answer 1

An idea rather than an answer

This can be analysed as a function on odd numbers, $$ F(2k+1)={Od}(k(k+1)),$$ where $Od$ is the odd part function.

Unless $k$ or $k+1$ is a power of $2$, ${Od}(k(k+1))\ge 3k$ and so $F$ is an increasing function unless and until a number of the form $2^n\pm 1$ is reached. From such numbers we immediately have a decreasing chain of numbers of the same form until 1 is reached.

The problem thus becomes one of analysing the rare cases when $ F(2k+1)$ is of the form $2^n\pm 1$.

Example Of the first few odd numbers, $11$ and $13$ are the first not to be of the form $2^n\pm 1$ and $11$ immediately maps to $15=2^4-1.$