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I was playing with some Meissel/Lehmer formulas and I found this one. In fact there is a much simpler way to find it when looking closer, so I guess i is well known.

$$\sum\limits_{p\leq x}\pi\left(\frac{x^2}{p}\right)-\sum\limits_{p\gt x}\pi\left(\frac{x^2}{p}\right)=\pi^2(x)$$

There is a similar formula I already explored in an other context (Goldbach).

$$\sum\limits_{p\leq x}\pi(\small 2x-p \normalsize)-\sum\limits_{p\gt x}\pi(\small 2x-p \normalsize)=\pi^2(x)$$

Any paper covering one of them? Were they used anywhere?

Edit: this seems to work also for $$\sum\limits_{p\leq x}\pi\left(\small\sqrt[k]{2 x^k-p^k}\normalsize\right)-\sum\limits_{p\gt x}\pi\left(\small\sqrt[k]{2 x^k-p^k}\normalsize\right)=\pi^2(x)$$ with $k\ge 1$

and for some $k\gt \gamma_x$ we even have

$$\sum\limits_{p}\pi\left(\small\sqrt[k]{2 x^k-p^k}\normalsize\right)=\pi^2(x)$$

To illustrate the influence of $k$, here is a chart with $k=1$ in blue, $k=2$ in purple and $k=16$ in yellow. $x=9$ in this example: wrapping arround $\pi^2(x)$

I guess there are a lot of other working formulas...

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    $\begingroup$ No because $\pi(x)^2 = \sum_{p \le x, q \le x} 1 = \sum_{pq \le x^2} 1 - \sum_{pq \le x^2, p > x} 1-\sum_{pq \le x^2, q > x} 1=\sum_{pq \le x^2} 1 -2\sum_{pq \le x^2, p > x} 1 $ $=\sum_{pq \le x^2,p \le x} 1 -\sum_{pq \le x^2, p > x} 1 =\sum_{ p \le x}\pi(x^2/p)-\sum_{ p>x} \pi(x^2/p) $ is valid for any set of strictly positive integers $A$ and $\pi(x) = \sum_{n \le x, n \in A} 1$ $\endgroup$
    – reuns
    Jan 14, 2019 at 16:22
  • $\begingroup$ That was my guess. It was not very usefull on Goldbach either. $\endgroup$
    – Collag3n
    Jan 14, 2019 at 18:04

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