For any non-zero natural integer $n$, we set $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ Show that $$\left(\sum_{\substack{p_1, p_2 \leqslant x \\ p_1 \neq p_2 \text{ prime}}} \operatorname{Card}\left\{n \in \mathbb{N}^* : n \leqslant x, p_1 \mid n \text{ and } p_2 \mid n\right\}\right) - x\ln_2(x)^2 \underset{x \rightarrow +\infty}{=} O\left(x\ln_2(x)\right)$$ One may estimate the cardinality of the set of pairs of prime numbers $(p_1, p_2)$ such that $p_1 p_2 \leqslant x$ when $x$ tends to $+\infty$.
For any non-zero natural integer $n$, we set
$$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$
Show that
$$\left(\sum_{\substack{p_1, p_2 \leqslant x \\ p_1 \neq p_2 \text{ prime}}} \operatorname{Card}\left\{n \in \mathbb{N}^* : n \leqslant x, p_1 \mid n \text{ and } p_2 \mid n\right\}\right) - x\ln_2(x)^2 \underset{x \rightarrow +\infty}{=} O\left(x\ln_2(x)\right)$$
One may estimate the cardinality of the set of pairs of prime numbers $(p_1, p_2)$ such that $p_1 p_2 \leqslant x$ when $x$ tends to $+\infty$.