I need to prove that $\pi _2 (x)=\sum_{p \leq \sqrt{x}} \pi(\frac{x}{p}) + O(\frac{x}{(\log x)^2})$ by using and showing that $\pi _2(x)=\sum_{p_1 \leq \sqrt{x}}\sum _{p_1 < p_2 \leq \frac{x}{p_1}}1$. Where $\pi_2(x)$ denotes the number of integers less than $x$ which is composed of presicely two destinct primes.
And regarding the second sum ($\pi _2(x)=\sum_{p_1 \leq \sqrt{x}}\sum _{p_1 < p_2 \leq \frac{x}{p_1}}1$), maybe someone can explain how it should be understood.
Lastly, maybe someone has encountered the function $\pi_2$ before, and know where I can read about it.