We set, for all real $t \geqslant 2$, $$R(t) = \sum_{\substack{p \leqslant t \\ p \text{ prime}}} \frac{\ln(p)}{p} - \ln(t)$$ Show, using the result from question 16, that $$\sum_{\substack{p \leqslant n \\ p \text{ prime}}} \frac{1}{p} = 1 + \ln_2(n) - \ln_2(2) + \frac{R(n)}{\ln(n)} + \int_2^n \frac{R(t)}{t(\ln(t))^2} \, \mathrm{d}t$$

Let $x$ be a positive real number greater than or equal to 1 and $q \in \mathbb{N}^{*}$. Justify that the quantity $$\operatorname{Card}\{n \in \mathbb{N} \cap [1,x] : n \equiv 0 \pmod{q}\} - \frac{x}{q}$$ is bounded in absolute value by a real number independent of $x$ and $q$.

For any non-zero natural integer $n$, we set $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ Prove, using a summation interchange, that $\frac{1}{x} \sum_{n \leqslant x} \omega(n) \underset{x \rightarrow +\infty}{=} \ln_{2}(x) + O(1)$.

Let $x$ be a positive real number greater than or equal to 1 and $q \in \mathbb{N}^*$. Justify that the quantity $$\operatorname{Card}\{n \in \mathbb{N} \cap [1,x] : n \equiv 0 \pmod{q}\} - \frac{x}{q}$$ is bounded in absolute value by a real number independent of $x$ and $q$.

For any non-zero natural integer $n$, we set $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ Prove, using a change of order of summation, that $\frac{1}{x} \sum_{n \leqslant x} \omega(n) \underset{x \rightarrow +\infty}{=} \ln_2(x) + O(1)$.

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 $$\frac{1}{x} \sum_{n \leqslant x} \left(\omega(n) - \ln_{2}(x)\right)^{2} \underset{x \rightarrow +\infty}{=} \frac{1}{x}\left(\sum_{n \leqslant x} \omega(n)^{2}\right) - \ln_{2}(x)^{2} + O\left(\ln_{2}(x)\right).$$

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 $$\sum_{n \leqslant x} \omega(n)^{2} = \sum_{\substack{p_{1} \leqslant x \\ p_{1} \text{ prime}}} \sum_{\substack{p_{2} \leqslant x \\ p_{2} \text{ prime}}} \operatorname{Card}\left\{n \in \mathbb{N}^{*} : n \leqslant x, p_{1} \mid n \text{ and } p_{2} \mid n\right\}$$

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 $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ Conclude that $\frac{1}{x}\left(\sum_{n \leqslant x} \left(\omega(n) - \ln_{2}(x)\right)^{2}\right) \underset{x \rightarrow +\infty}{=} O\left(\ln_{2}(x)\right)$.

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 $$\frac{1}{x} \sum_{n \leqslant x} \left(\omega(n) - \ln_2(x)\right)^2 \underset{x \rightarrow +\infty}{=} \frac{1}{x}\left(\sum_{n \leqslant x} \omega(n)^2\right) - \ln_2(x)^2 + O\left(\ln_2(x)\right)$$

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 $$\sum_{n \leqslant x} \omega(n)^2 = \sum_{\substack{p_1 \leqslant x \\ p_1 \text{ prime}}} \sum_{\substack{p_2 \leqslant x \\ p_2 \text{ prime}}} \operatorname{Card}\left\{n \in \mathbb{N}^* : n \leqslant x, p_1 \mid n \text{ and } p_2 \mid n\right\}$$

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 $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ Conclude that $\frac{1}{x}\left(\sum_{n \leqslant x} \left(\omega(n) - \ln_2(x)\right)^2\right) \underset{x \rightarrow +\infty}{=} O\left(\ln_2(x)\right)$.

Show that Theorem 1 is equivalent to the following statement:
Let $f(x) \in \mathbf{Q}\llbracket x \rrbracket$ be an exponential polynomial such that $f(1) = \sum_{i=1}^{s} P_i(1) e^{c_i}$ vanishes. Then $f(x)/(x-1)$ is still an exponential polynomial.
(An exponential polynomial is any power series with rational coefficients of the form $f(x) = \sum_{i=1}^{s} P_i(x) e^{c_i x}$, where $c_1, \ldots, c_s \in \mathbf{Q}$ are rationals and $P_1, \ldots, P_s \in \mathbf{Q}[x]$ are polynomials.)

For any non-zero natural integer $n$, we set $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ We define $\mathscr{S} = \left\{n \geqslant 3 : \left|\frac{\omega(n) - \ln_{2}(n)}{\sqrt{\ln_{2}(n)}}\right| \geqslant \left(\ln_{2}(n)\right)^{1/4}\right\}$. Show that $$\lim_{x \rightarrow +\infty} \frac{1}{x} \operatorname{Card}\{n \leqslant x : n \in \mathscr{S}\} = 0$$ One may begin by writing $\operatorname{Card}(\mathscr{S} \cap [1,x]) \underset{x \rightarrow +\infty}{=} \operatorname{Card}(\mathscr{S} \cap [\sqrt{x}, x]) + O(\sqrt{x})$ and note that in the sum on the right-hand side, the difference $\left|\ln_{2}(n) - \ln_{2}(x)\right|$ remains bounded.

For any non-zero natural integer $n$, we set $$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$ We set $\mathscr{S} = \left\{n \geqslant 3 : \left|\frac{\omega(n) - \ln_2(n)}{\sqrt{\ln_2(n)}}\right| \geqslant \left(\ln_2(n)\right)^{1/4}\right\}$. Show that $$\lim_{x \rightarrow +\infty} \frac{1}{x} \operatorname{Card}\{n \leqslant x : n \in \mathscr{S}\} = 0$$ One may begin by writing $\operatorname{Card}(\mathscr{S} \cap [1,x]) \underset{x \rightarrow +\infty}{=} \operatorname{Card}(\mathscr{S} \cap [\sqrt{x}, x]) + O(\sqrt{x})$ and note that in the sum on the right-hand side, the difference $\left|\ln_2(n) - \ln_2(x)\right|$ remains bounded.

Let $D$ be a common denominator of the rational numbers $a_1, \ldots, a_r$ and let $A = \max(1, |a_1|, \ldots, |a_r|)$. Show that $D^n u_n \in \mathbf{Z}$ for all $n \geq 0$.

Let $C = 1 + |s_1| + \cdots + |s_r|$. Show that $D^n v_n(k) \in \mathbf{Z}$ and that there exists a real number $c_2 > 0$ such that $$|v_n(k)| \leq c_2 A^n C^k \text{ for all } n \geq kr.$$

Deduce that $k!$ divides $D^n v_n(k)$ for all $n$ and $k$ such that $n \geq kr$.

Let $n \in \mathbb { N } ^ { * }$. Show that
$$\prod _ { \substack { n + 2 \leqslant p \leqslant 2 n + 1 \\ p \text { prime } } } p \leqslant \binom { 2 n + 1 } { n } \leqslant 4 ^ { n } .$$

Show that, for all $n \in \mathbb { N } ^ { * }$,
$$\prod _ { \substack { p \leqslant n \\ p \text { prime } } } p < 4 ^ { n } .$$
One may proceed by induction and perform the inductive step by discussing according to the parity of $n$.

Deduce that, for all real $x \geqslant 1$,
$$\prod _ { \substack { p \leqslant x \\ p \text { prime } } } p < 4 ^ { x } .$$

Let $p$ be a prime number. Show that, for all $n \in \mathbb { N }$,
$$v _ { p } ( n ! ) = \sum _ { k = 1 } ^ { + \infty } \left\lfloor \frac { n } { p ^ { k } } \right\rfloor$$

Deduce that, for all $n \in \mathbb { N } ^ { * } , k \in \mathbb { N }$ and $p$ prime number: if $p ^ { k }$ divides $\binom { 2 n } { n }$, then $p ^ { k } \leqslant 2 n$.

Let $n \in \mathbb { N } ^ { * }$. Justify that
$$\prod _ { \substack { p \leqslant n \\ p \text { prime } } } p \geqslant \prod _ { \substack { \sqrt { n } < p \leqslant n \\ p \text { prime } } } p$$