Generalisation of Bertrand's postulate: not just one but two primes Just a sunday's question. Bertrand's postulate states that for every $n>1$, there is always at least one prime $p$ such that $n \lt p \lt 2n$.
Playing around with some small numbers suggests the question if there is possibly not just one but there are at least two primes in that interval.
From the following numerical example we can see that the answer seems to be affirmative for $n\ge6$
\begin{array}{ccc}
n&\{n-1,2n-1\}&\text{list of primes }p\\
 1 & \{0,1\} & \{\} \\
 2 & \{1,3\} & \{3\} \\
 3 & \{2,5\} & \{5\} \\
 4 & \{3,7\} & \{5,7\} \\
 5 & \{4,9\} & \{7\} \\
 6 & \{5,11\} & \{7,11\} \\
 7 & \{6,13\} & \{11,13\} \\
 8 & \{7,15\} & \{11,13\} \\
 9 & \{8,17\} & \{11,13,17\} \\
 10 & \{9,19\} & \{11,13,17,19\} \\
\end{array}
Question
I would greatly appreciate some references for this most probably well known result.
EDIT
Here I add some plots which show the number of primes between $n+1$ and $2n-1$ over different ranges of $n$.
I have fitted the curves with the function
$$b(n) = \frac{A n }{\log(1+B n)}$$
The parameters are shown in the plots.



Discussion

*

*These graphs illustrate the proof of Erdös: "for any $k$ and sufficiently large $n$ there are always at least $k$ primes between $n$ and $2n$", as was pointed out in a comment of @lulu and in the solution of @Thomas Preu


*The number $b(n)$ is not monotonous on a small scale.


*The shape of the fitted functions is just what was to be expected from the prime counting function $\pi(n)$. Indeed we have exactly just $b(n) = \pi(2n) - \pi(n)$.

Here is a precise form of what you are looking for. Theorem 3.6 of Das and Paul's 2019 paper shows that

For any integer $n$ and $k$, there are at least $k-1$ primes between
$n$ and $kn$ when $n \ge 1.1\ln(2.5k) $

