[Sloane's A077463](http://oeis.org/A077463/) obviously suggests that for any positive integer $n$ there exist $n$ consecutive primes and only them in between $m$ and $2m$ for some natural number $m$.


For instance, for


$n=1$, take $m=2$; $\hspace{.2in}$$2<3<4$;


$n=2$, take $m=7$; $\hspace{.2in}$$7<11,13<14$;


$n=3$, take $m=9$; $\hspace{.2in}$$9<11,13,17<18$;


$n=4$, take $m=15$; $\hspace{.1in}$$15<17,19,23,29<30$,$\hspace{.1in}$ etc.


This problem offers a partial converse(given the number of primes, one seeks an exact interval $[m,2m]$) to Bertrand's Postulate(given an interval, one seeks at least one prime in it).


I would like to know


1. whether this problem is solved, or
2. whether there are stronger known conjectures to which it is a consequence.


Thanks, as always 



A partial converse to Bertrand's Postulate
Consider the function $f(m):=\pi(2m)-\pi(m)$ which counts the number of primes $m< p \leq 2m$. It is easy to see that $f(m+1)-f(m)$ equals $-1$ or $0$ or $1$ depending whether $m+1$ and $2m+1$ are primes. On the other hand, by simple estimates for prime numbers, it can be seen that $f(m)$ tends to infinity. Therefore $f(m)$ takes all positive integer values.


EDIT: Of course the same holds for $\pi(2m)-\pi(m-1)$ which counts the primes in $[m,2m]$.


