TITLE: Numbers that are divisible by the number of primes smaller than them
QUESTION [14 upvotes]: Let $\pi(n)$ denote the number of primes less than or equal to $n$ (a.k.a the prime-counting function).
For certain values of $n$, the value of $\frac{n}{\pi(n)}$ is integer.
Here are the first few examples:

$n=  8,\pi(n)= 4,\frac{n}{\pi(n)}=2$
$n= 27,\pi(n)= 9,\frac{n}{\pi(n)}=3$
$n= 30,\pi(n)=10,\frac{n}{\pi(n)}=3$
$n= 33,\pi(n)=11,\frac{n}{\pi(n)}=3$
$n= 96,\pi(n)=24,\frac{n}{\pi(n)}=4$
$n=100,\pi(n)=25,\frac{n}{\pi(n)}=4$
$n=120,\pi(n)=30,\frac{n}{\pi(n)}=4$
$n=330,\pi(n)=66,\frac{n}{\pi(n)}=5$
$n=335,\pi(n)=67,\frac{n}{\pi(n)}=5$
$n=340,\pi(n)=68,\frac{n}{\pi(n)}=5$
$n=350,\pi(n)=70,\frac{n}{\pi(n)}=5$
$n=355,\pi(n)=71,\frac{n}{\pi(n)}=5$
$n=360,\pi(n)=72,\frac{n}{\pi(n)}=5$

$\textbf{Has it been proved that }\mathbf{\forall{k>1},\exists{n}:\frac{n}{\pi(n)}=k}$?
Two aspects which "intuitively" support this statement are:

The prime-number theorem, which implies $\frac{n}{\pi(n)}\approx\ln{n}$.
There seem to be several such values of $n$ for each value of $k$.

But I'm not sure how either one of them can be used in order to establish a proof.

REPLY [2 votes]: Yes this is proved by S.W.Golomb.
I cannot resist to show a piece of my work:http://arxiv.org/abs/1311.1398