TITLE: Primes between $n$ and $2n$
QUESTION [8 upvotes]: I know that there exists a prime between $n$ and $2n$ for all $2\leq n \in \mathbb{N}$ . Which number is the fourth number that has just one prime in its gap? First three numbers are $2$ , $3$ and $5$ . I checked with computer until $15000$ and couldn't find next one. Maybe, you can prove that there is no other number with this condition?
Also, when I say, a number $n$ has one prime in its gap it means the set $X = \{x: x$ is prime and $n<x<2n\}$ has only one element.
Thanks for any help.

REPLY [5 votes]: There is a nice and rather easy proof of Bertrand's postulate (due to Erdős), which can be found here for example. At some point in the proof, one obtains:
$$4^{\frac{1}{3}n}\leq (2n)^{\sqrt{2n}+1}\cdot\prod_{n<p<2n}p$$
To prove Betrand's postulate, one would assume to the contrary, that this product is empty and conclude a contradiction for $n$ large enough. 
But, more precise:
$$\prod_{n<p<2n}p\geq \frac{4^{\frac{1}{3}n}}{(2n)^{\sqrt{2n}+1}}\quad\Rightarrow\quad|X|\geq \log_{2n}\left(\frac{4^{\frac{1}{3}n}}{(2n)^{\sqrt{2n}+1}}\right)$$
I leave further simplifications to you. I am sure, that you can show, that this is bigger than $2$ for $n$ large enough.