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 $p_{1} p_{2} \leqslant x$ as $x$ 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 $p_{1} p_{2} \leqslant x$ as $x$ tends to $+\infty$.