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4

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

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

Thanks, as always

flag	asked Mar 13 2011 at 10:58 To be cont'd 1,699●15●30

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 2011 at 11:09

@Gjergji, if possible, yes. – To be cont'd Mar 13 2011 at 11:10

I gave a simple proof that the statement is valid. – GH Mar 13 2011 at 11:29

3

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 2011 at 11:33

1

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 2011 at 12:33

show 3 more comments

GH · Accepted Answer · 2011-03-13 11:12:48Z UTC

16

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 $[m,2m]$.

answered Mar 13 2011 at 11:12

GH
15.1k●1●31●53

	@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. – To be cont'd Mar 13 2011 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 2011 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 Mar 13 2011 at 11:34
	Forgive my laziness. The problem no longer exists. Thanks very much. – To be cont'd Mar 13 2011 at 11:50

A partial converse to Bertrand’s Postulate

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