Are there infinitely many $x$ for which $\pi(x) \mid x$?

Let $\pi(x)$ denote the Prime Counting Function.

• One observes that, $\pi(6) \mid 6$, $\pi(8) \mid 8$. Does $\pi(x) \mid x$ for only finitely many $x$, or is this fact true for infinitely many $x$.
• I think you switched 'finitely' and 'infinitely' in you question. I have no idea, but look forward to hearing some great responses. – Brandon Carter Oct 7 '10 at 18:07
• @Brandon: Yes, i did that because finitely and finitely doesn't sound realistic – anonymous Oct 7 '10 at 18:09
• @Chandru1: I don't understand your comment. Could you please clarify what didn't sound realistic? – Jonas Meyer Nov 21 '10 at 8:23

Well, it's not hard to show that for every natural $k>2$ equation $x=k\pi(x)$ has a positive solution.

Proof by contradiction: Imagine, that $x\ne k\pi(x)$ for every natural $x$. But then - for $x=2$:$x-k\pi(x)=2-k<0$ and for very large $x$: $x-k\pi(x)\sim x(1-\frac{k}{\ln x})>0$.

So there should be such $t$, that $t-k\pi(t)<0$ and $(t+1)-k\pi(t+1)>0$. But then from one hand:

$$t+1-k\pi(t+1)-(t-k\pi(t))\ge 2$$ as a difference of positive integer and negative integer, but from the other hand:

$$t+1-k\pi(t+1)-(t-k\pi(t))=1-k(\pi(t+1)-\pi(t))\le 1$$

• How did you get from $x-k\pi(x)$ to $x(1-\frac{k}{\ln x})$ ? – configurator Oct 8 '10 at 2:36
• In other words, how do you know that pi(x) is 1/ln(x)? – configurator Oct 8 '10 at 2:38
• @configurator: $\pi(x)$ is asymptotic to $x/\ln x$ by the prime number theorem: en.wikipedia.org/wiki/Prime_number_theorem. Hence, for sufficiently large $x$, those expressions have the same sign. – Jonas Meyer Oct 8 '10 at 4:40
• @Jonas: Thanks for the link – configurator Oct 8 '10 at 13:46
• Am I right if the only properties of $\pi(x)$ you used are that it is 1) monotonically increasing 2) $\pi(x)\ll x$ 3) $x<k\pi(x)$ for at least one $x\in\Bbb Z$ ? – punctured dusk Sep 1 '15 at 17:40

This is Sloane's A057809, but unfortunately there's not much information there.

Heuristically, the chance that $\pi(x)\mid x$ is $\displaystyle\frac{\log{x}}{x}$ which suggests that there are $0.5\log^2 x$ such numbers up to $x$, that is, infinitely many.

• A comment in the sequence suggests that there are log x 'clumps' of numbers up to x. – Charles Oct 12 '10 at 19:31
• The clumping makes sense -- there's a clump for $x/\pi(x) = 1$, a clump for $x/\pi(x) = 2$, and so on. – Michael Lugo Mar 22 '11 at 23:43