Fraction Transforms

Question

Here's a number theory problem I'm having some difficulty with:

Say we transform a fraction by the following rule: we start with some fraction $\frac{m}{n}$ with $m > n$ and then convert it to $\frac{fr}{n}$, where $f = \lfloor m/n \rfloor$ and $r = m \bmod{n}$. We repeat the process until our fraction is less than 1 or an integer. Given an $n$ is there an $m$ so that we end up with a fraction less than 1?

For $n = 2$ it isn't that hard, 3, 7, 15, 31, etc. work, but what about other values of $n$?

Answer 1

One can merely work backwards. Suppose we want to end up at $1/n$. Then the preceding term could be $1 + \frac{1}{n} = (n+1)/n$. Then the preceding term could be $(n+1) + \frac{1}{n} = (n^2+n+1)/n$. By inspection, any improper fraction of the form $(n^k+n^{k-1}+\cdots+n^2+n+1)/n$ will end up at $1/n$.