MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Sloane's 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

share|cite|improve this question
Are you asking to prove that for every $n$ there is an $m$ so that there are exactly $n$ primes in $[m,2m]$? – Gjergji Zaimi Mar 13 '11 at 11:09
@Gjergji, if possible, yes. – Unknown Mar 13 '11 at 11:10
I gave a simple proof that the statement is valid. – GH from MO Mar 13 '11 at 11:29
This seems a bit too easy for MO. But since you got an answer anyway, I see no point in closing... – Gjergji Zaimi Mar 13 '11 at 11:33
It's fine, I don't mean to discourage you from asking further questions, that's why I commented earlier if this was really what you meant to ask. As for your observation, it is still in the easy territory, think about how one constructs arbitrarily large strings of consecutive composites. – Gjergji Zaimi Mar 13 '11 at 12:33
up vote 18 down vote accepted

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 $[m,2m]$.

share|cite|improve this answer
@GH, I understand that $f(m)$ tends to infinity, but I do not see why it might not jump some big numbers. Please clarify. Thanks. – Unknown Mar 13 '11 at 11:31
@Elohemahab, read the proof carefully, especially the part about $f(m)$ never changing by more than 1 when you add one to $m$. – Gerry Myerson Mar 13 '11 at 11:33
The jump $f(m+1)-f(m)$ equals $1$ if $2m+1$ is prime but $m+1$ is not, equals $0$ if both or none of $2m+1$ and $m+1$ are prime, equals $-1$ if $m+1$ is prime but $2m+1$ is not. All cases are covered here. – GH from MO Mar 13 '11 at 11:34
Forgive my laziness. The problem no longer exists. Thanks very much. – Unknown Mar 13 '11 at 11:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.