Primes between $n$ and $2n$ 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.
 A: 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.
A: There is no other such $n$.
For instance,

In 1952, Jitsuro Nagura proved that for $n ≥ 25$, there is always a prime between $n$ and $(1 + 1/5)n$.

This immediately means that for $n \ge 25$, we have one prime between $n$ and $\frac{6}{5}n$, and another prime between $\frac{6}{5}n$ and $\frac65\frac65n = \frac{36}{25}n < 2n$. In fact, $\left(\frac{6}{5}\right)^3 < 2$ as well, so we can be sure that for $n \ge 25$, there are at least three primes between $n$ and $2n$. As you have already checked all $n$ up to $25$ (and more) and found only $2$, $3$, $5$, we can be sure that these are the only ones.

The number of primes between $n$ and $2n$ only gets larger as $n$ increases: it follows from the prime-number theorem that
$$ \lim_{n \to \infty} \frac{\pi(2n) - \pi(n)}{n/\log n} = 2 - 1 = 1,$$ so the number of primes between $n$ and $2n$, which is $\pi(2n) - \pi(n)$, is actually asymptotic to $\frac{n}{\log n}$ which gets arbitrarily large.
