For all positive integer $n$ we define a finite sequence in the following way: $n_0 = n$, then $n_1\geq n_0$ and has the property that $n_1$ is a multiple of $n_0-1$ such that the difference $n_1 - n_0$ is minimal among all multiple of $n_0 -1 $ that are bigger than $n_0$. More generally $n_k \geq n_{k-1}$ and has the property that is a multiple of $n_0-k$ such that the difference $n_k - n_{k-1}$ is minimal among all multiple of $n_0 -k $ that are bigger than $n_{k-1}$. We stop the procedure when $k=n-1$ so that we define $f$ to be $f(n_0) = n_{n-1} $.
Question: What is the value of $\lim_{n \to \infty} \frac{n^2}{f(n)} $ ?
I'm able to prove that $$ \frac{8}{3} \leq \lim_{n \to \infty} \frac{n^2}{f(n)} \leq 4 $$ but I can't do better. Someone has an idea? I think that the limit is $ \pi $ but don't know how to prove that.
My idea is to prove that $$ f(n) = \frac{n^2}{\pi} + O(n) $$ the general term for $n_k$ is given by $$ n_k = (n-k) \left \lceil \frac{n_{k-1}}{n-k} \right \rceil $$ So that $$ n_k = (n-k) \left \lceil \frac{n-(k-1)}{n-k} \left \lceil \frac{n-(k-2)}{n-(k-1)} \left \lceil \ldots \left \lceil \frac{n-1}{n-2} \left \lceil \frac{n}{n-1} \right \rceil \right \rceil \right \rceil \right \rceil \right \rceil $$