Are there infinite many $n\in\mathbb N$ such that $\pi(n)=\sum_{p\leq\sqrt n}p$?

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 $(1)$.

Solution 1:

I think they are finite :

$$\sum_{p\leq\sqrt{x}}p=\int_{2^{+}}^{\sqrt{x}}td\left(\pi(t)\right)= \sqrt{x}\pi(\sqrt{x}) -\int_{2}^{\sqrt‌​{x}}\pi (t)dt$$

suppose we have the equality for infinitely many $x$:

$$\pi(x)= \sqrt{x}\pi(\sqrt{x}) -\int_{2}^{\sqrt‌​{x}}\pi (t)dt$$

$$\sqrt{x}\pi(\sqrt{x}) - \pi(x)= \int_{2}^{\sqrt‌​{x}}\pi (t)dt$$

for $x > 55^{2}$ : $$ \sqrt{x}\frac{\sqrt{x}}{log(\sqrt{x})-4}- \frac{x}{log(x)+2} >\sqrt{x}\pi(\sqrt{x}) - \pi(x)$$

$$ \int_{2}^{\sqrt‌​{x}}\pi (t)dt > \int_{55}^{\sqrt‌​{x}} \frac{t}{log(t)+2}dt +C = e^{-4}\operatorname{Ei}(2(log(\sqrt{x})+2))+C $$

and :

$$ e^{-4}\operatorname{Ei}(2(log(\sqrt{x})+2))+C >> \sqrt{x}\frac{\sqrt{x}}{log(\sqrt{x})-4}- \frac{x}{log(x)+2} $$ so we get a contradiction ..

What do you think ? did I miss something ?

(note : I used the fact that for $x > 55$ :

$$\frac {x}{\ln x + 2} < \pi(x) < \frac {x}{\ln x - 4} $$)

According to my computation the list should stop at 4000 or 5000 ..