On integer values which are attained by $n/\pi(n)$ only once

Question

Let $\pi (n)$ denote the prime counting function. I can prove that $\mathbb N \setminus \{1\} \subseteq \{n/ \pi(n) : n \in \mathbb N \}$ . Now for every integer $m>1$ , define $s(m) := \{ n \in \mathbb N : n>1 , n/ \pi(n)=m \}$ . Does there exist any integer $m$ such that $|s(m)|=1$ ? If there exists such integers , then are there infinitely many of them ?

Answer 1

Not a complete answer, but a table showing the frequences using the primes upto $10^{10}$. The first column is the value of $\large \frac{n}{\pi(n)}$ and the second column shows how often it appeared :

2   4
3   3
4   3
5   6
6   7
7   6
8   6
9   3
10   9
11   1
12   18
13   11
14   12
15   21
16   3
17   10
18   33
19   31
20   32
21   24

Apparently, the frequences do not strictly increase. Value $16$ is particular suprising. It appeared only $3$ times. In total, only $243$ times, an integer-value appears.

UPDATE : You are right, misao. I didn't understand the argument, but finally I could prove your claim. Sorry for my doubt ...

Moreover, the following can also be proven : If $k>2$ and $f(p)<k$, then there exists $N>p$ with $f(N)=k$. This implies misao's claim.

Proof : Suppose , $q>p$ is the smallest number with $f(q)\ge k$. If $f(q)=k$, we are done. Otherwise, we have $p<\pi(p)\cdot k\le \pi(q)\cdot k<q$. If there is no prime $P$ with $p<P<q$, then we can choose $N:=\pi(p)\cdot k$ and we have $\frac{N}{\pi(N)}=\frac{N}{\pi(p)}=k$. If there is at least a prime $P$ with $p<P<q$, assume $P$ is the largest of those. Now, we have either $f(P)=k$ and are done, or we have $P<\pi(P)\cdot k\le \pi(q)\cdot k<q$ and $N:=\pi(P)\cdot k$ does the job.

So, if $s$ is the smallest number with $f(s)=k$ ($k>2$), then there is another number $t$ with $f(t)=k$, if and only if for some positive integer $m$, the interval $[s,s+mk]$ contains at least $m$ primes.