Are there infinite many $n\in\mathbb N$ such that $$\pi(n)=\sum_{p\leq\sqrt n}p,\tag{1}$$ where $\pi(n)$ is the Prime-counting_function?
For example, $n=1,4,11,12,29,30,59,60,179,180,389,390,391,392,\dots$
As I know, $\pi(x)\sim \dfrac{x}{\ln x},\sum_{p\leq x}p\sim \dfrac{x^2}{2\ln x}$, hence $\pi(x)\sim \sum_{p\leq \sqrt x}p$.
It seems that
- 1) it's very often that $\pi(n)>\sum_{p\leq\sqrt n}p$,
- 2) there are infinite many primes $q$ such that $q>\pi(q^2)-\sum_{p<q}p.$
If we can prove 1) and 2) then we get (1), but I can't prove even one of them.
Thanks in advance!
Edit: Use the formula given by Balarka Sen, I get $$\pi(x)\sim \sum_{p\leq\sqrt x}p = \frac{x}{\ln x}(1+\frac{1+o(1)}{\ln x}),$$ but it's not enough to solve our problem.
Edit2: Use the formula given in Dusart's paper and this paper (or this post), I get $$\pi(x)=\frac{x}{\ln x}(1+\frac{1}{\ln x}+\frac{2}{(\ln x)^2}+O(\frac{1}{(\ln x)^3}))\tag 2$$
$$\sum_{p\leq\sqrt x}p =\frac{x}{\ln x}(1+\frac{1}{\ln x}+o(\frac{1}{(\ln x)^2})),\tag 3$$ so 1) is true but 2) is not, and there are only finite many $n$ satisfy $(1)$.