Here, according to Wolfram, I assume that $m(n)=1$ for all positive integer $n>1$. Suppose we always get the number that is divisible by $17$. Let $a_0=n$ and $a_i$ be the number we obtain after $i^{th}$ minute passed. If the process stop, which happens only when we get a prime number (which is larger than $17$), then that number is not divisible by $17$, contradiction. So the process must not stop. For any $i\in \mathbb{Z}^+_0$, we get $a_{i+1}=a_i+\frac{a_i}{p_i}-1$ where $p_i$ is the smallest prime divisor of $a_i$. We have $17\mid a_i,a_{i+1}$. If $p_i\neq 17$, then $17\mid \frac{a_i}{p_i}$, which gives $17\mid 1$, impossible. Hence $p_i=17$ for all $i\in \mathbb{Z}^+_0$, and so $a_{i+1}=a_i+\frac{a_i}{17}-1$ for all $i\in \mathbb{Z}^+_0$. Not hard to prove by induction that $a_m=\Big( \frac{18}{17}\Big)^m n-\sum_{j=0}^{m-1}{\Big( \frac{18}{17}\Big)^j } =\frac{18^m(n-17)}{17^m}+17$ for all $m\in \mathbb{Z}^+$. Since $a_m\in \mathbb{Z}$ for all $m\in \mathbb{Z}^+$, we get $n-17=0\Rightarrow n=17$, contradict with $n>1000$, done.