Evaluate $\lfloor n / \lfloor n / \lfloor \sqrt n \rfloor \rfloor \rfloor$ for positive integers $n$.

Question

I'm considering:

$$ \left\lfloor {n \over \lfloor n / \lfloor \sqrt n \rfloor\rfloor} \right\rfloor \:\:\:\: \forall n \in \mathbf N^+ $$

which seems to be $\lfloor \sqrt n \rfloor$.

Is there any way to prove or disprove?

Answer 1

Let $n$ be a real number such that $n\geq 1$, and $m:=\left\lfloor\sqrt{n}\right\rfloor$. Then, we have $$m^2\leq n< (m+1)^2=m^2+2m+1\leq m^2+3m\,.$$ Thus, $$m\leq \frac{n}{m}< m+3\,.$$ Set $k:=\left\lfloor\dfrac{n}{m}\right\rfloor$, so that $k\in\{m,m+1,m+2\}$. If $k=m$, then $m^2\leq n<m(m+1)$. That is, $$m\le\frac{n}{k}<m+1\,,\text{ whence }\left\lfloor\frac{n}{k}\right\rfloor=m\,.$$ If $k=m+1$, then $m(m+1)\leq n<m(m+2)$. Thus, $$m\leq \frac{n}{k}<\frac{m(m+2)}{m+1}=\frac{(m+1)^2-1}{m+1}<m+1\,,$$ so $\left\lfloor\dfrac{n}{k}\right\rfloor=m$. If $k=m+2$, then $m(m+2)\leq n<(m+1)^2$. Hence, $$m\leq \frac{n}{k}<m+\frac{1}{m+2}<m+1\,,\text{ making }\left\lfloor\frac{n}{k}\right\rfloor=m\,.$$ In other words, $$\left\lfloor\frac{n}{\left\lfloor\frac{n}{\left\lfloor\sqrt{n}\right\rfloor}\right\rfloor}\right\rfloor=\left\lfloor\sqrt{n}\right\rfloor$$ for all real numbers $n\geq 1$.