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I am interested in a proof that for all $n \in \mathbb{N}$ (with just a few exceptions) there will always be a prime $p$ such that $\frac{n}{3} \lt p \le \frac{n}{2}$. Note that the exact boundaries are important (i.e. we don't allow $\frac{n}{3}$). This seems to be true for all $n \gt 3$ except $n=9$ and $n=21$. Does anyone know how to prove this? I found this paper, showing that there is always a prime in the interval $[2n,3n]$, I'm not sure if there is a way to use that. I was able to prove easily that there will always be a prime with $\frac{n}{2} \lt p \le n$ for all $n \gt 2$ with Bertrand's Postulate. Any ideas?

EDIT: I have found a way to prove it, see the accepted answer below.

For anyone who is interested in this intermediate result, I will leave this here:

I can prove the statement for all numbers that are not of the form $6k+3, k \in \mathbb{N}$. For $n=6k$ it follows directly from the paper showing there is always a prime in the interval $[2k,3k]$. Since $2k$ and $3k$ are clearly not prime for $k>1$, we can exclude them as well (in fact, we are already excluding $2k$ in the case $n=6k$). So if we choose $n_1=6k+1$ and $n_2 = 6k+2$ we get that $$]\frac{n_1}{3},\frac{n_1}{2}] \cap \mathbb{N} = ]\frac{n}{3},\frac{n}{2}] \cap \mathbb{N} \subseteq ]\frac{n_2}{3},\frac{n_2}{2}] \cap \mathbb{N},$$

meaning we have at least the same numbers and therefore there is still a prime $p$ with $\frac{n}{3}\lt p \le \frac{n}{2}$. The same is true for $n_{-1}=6k-1$ and $n_{-2}=6k-2$, where the lower bound goes down, so we certainly don't exclude any numbers on that side and the upper bound goes down only enough to exclude $\frac{n}{2}=3k$, which isn't a prime for $k>1$. This argument doesn't work for $n = 6k \pm 3$.

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    $\begingroup$ We could assume that the boundary values are integers, i.e. $6\mid n$, then writing $n=6k$ gives primes between $2k$ and $3k$. This is proved there. $\endgroup$ – Dietrich Burde Jun 3 '20 at 12:03
  • $\begingroup$ yes, I recognized that too, but I need the proof to hold for all $n \in \mathbb{N}$, not just multiples of 6. In fact I just realized it also follows for numbers of the form $n = 6k+1$ and $n = 6k+2$, since in these cases the lower bound is $2k+\frac{1}{3}$ or $2k+\frac{2}{3}$, so all integers in the interval $[2k,3k]$ are included in the respective intervals as well. (Note that the exceptions mentioned above, 9 and 21, are both of the form $6k+3$) $\endgroup$ – user5615895 Jun 3 '20 at 12:11
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    $\begingroup$ There are stronger forms of Bertrand, certainly strong enough for your purposes. See this $\endgroup$ – lulu Jun 3 '20 at 12:28
  • $\begingroup$ @lulu I don't see how that helps me. Applying Sylvester's Theorem tells me the $\frac{n}{6}$ (more or less, up to rounding errors) integers between $\frac{n}{3}$ and $\frac{n}{2}$ have a prime factor $\gt \frac{n}{6}$. But that doesn't imply that I have a prime there, I could for example have the integer $\frac{2n}{5}$ in that range with $\frac{n}{5}$ being that prime factor. $\endgroup$ – user5615895 Jun 3 '20 at 16:34
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    $\begingroup$ Dusart's result, cited in that link, tells us that for sufficiently large $x$ there is always a prime between $x$ and $\left(1+\frac 1{(\ln x)^3}\right)x$. That is considerably better than the factor of $1.5$ which you need. $\endgroup$ – lulu Jun 3 '20 at 18:59
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This paper claims to prove that there is a prime in $[3n, 4n]$ for all positive integers $n$. That should suffice for your purposes (for sufficiently large $n$).

Of course you could also take your favourite explicit version of the prime number theorem, but that requires a bit more work.

Edit: Let us complete the argument. Take $n$ sufficiently large (how large we will determine at the end). Now take the smallest integer $k$ so that $3k > \frac n 3$. Note that $3k$ is at most $3$ larger than $\frac n3$. By the theorem I quoted above, there is a prime in $[3k, 4k]$. Since $3k \leq \frac n3 + 3$, we have that $4k \leq \frac 49 n + 4$. Now if we take $n \geq 100$ (say) we have that $\frac 49 n + 3 = \frac 12 n + 3 - \frac{1}{18}n < \frac12 n$. Thus the prime that is in the interval $[3k, 4k]$ is also in the interval $\left(\frac n3, \frac n2\right]$.

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  • $\begingroup$ Can you explain how that allows me to conclude there must be a prime between $\frac{n}{3}$ and $\frac{n}{2}$? I don't see it. $\endgroup$ – user5615895 Jun 3 '20 at 18:16
  • $\begingroup$ @user5615895, I've explicitly completed the argument. $\endgroup$ – Mees de Vries Jun 4 '20 at 9:33
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lulu pointed me to a certain part of the Wikipedia page on Bertrand's Postulate, where I found this beauty, where the last Theorem (p. 180) tells us that we can always find a prime $p$, such that for any $x\ge8$ we have $$8 \le x \lt p \lt \frac{3x}{2}$$

So if we have $n\ge24$, then $\frac{n}{3}\ge8$ and the above formula tells us there is a prime $p$ with $8 \le \frac{n}{3} \lt p \lt \frac{3}{2}\cdot\frac{n}{3}=\frac{n}{2}$.

Thanks for everyone who helped.

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