I'm studying the Brun's sieve and I'm stuck with the proof of the following statement:
Let $f(x) \in \mathbb{Z}[x]$ of degree $k \geq 1$, such that the first coefficient $a_k>0$. Let $m,M \in \mathbb{N}$ such that $m$ is odd and $\mu(M)\neq 0$.
Let $\mathcal{A} := f(\mathbb{N})\cap \mathbb{N}^*$. Then, for $x\rightarrow+\infty$,
$$\mathcal{S}(\mathcal{A}, f(x), M) \leq x \prod_{\substack{p \vert M}} \Big( 1- \frac{\rho(p)}{p} \Big) + O \Bigg(x \displaystyle\sum_{r=m}^{\omega(M)}\Big( \frac{ek\Sigma(M)}{r} \Big)^r \Bigg) + O \Big( (k\omega(M))^{m-1} \Big).$$
where, for $n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$, $\mu (n)$ is defined as
\begin{cases}
1 \qquad \qquad \text{if} \ n=1 \\
0 \qquad \qquad \text{if} \ \alpha_i = 2 \ \text{for some} \ i \in \{1,...,k\} \ \ \ ; \\
(-1)^k \qquad \text{if} \ \alpha_i = 1 \ \forall i \in \{1,...,k\}
\end{cases}
$\rho(n)=\vert \{ h \ (\text{mod} \ n) : f(h)\equiv 0 \ (\text{mod} \ n) \} \vert$;
$\Sigma(n)=\sum_{p|M}p^{-1}$;
$\mathcal{S}(\mathcal{A},y,M) := \vert \{ a \in \mathcal{A} \cap [1,y] : (a,M)=1 \} \vert $
and $e$ is the neper number.
I managed to prove the following inequality: $$\mathcal{S}(\mathcal{A}, f(x), M) \leq x \sum_{\substack{d \vert M \\ \omega(d) < m}} \frac{\mu(d) \rho(d)}{d} + O \Big( \sum_{\substack{d \vert M \\ \omega(d) < m}} \rho(d) \Big) = x \Bigg[ \prod_{\substack{p \vert M}} \Big( 1-\frac{\rho(p)}{p} \Big) + \displaystyle\sum_{r=m}^{\omega(M)} \sum_{\substack{ d \vert M \\ \omega(d)=r}} \frac{\rho(d) \mu(d)}{d} \Bigg] + O \Big( \sum_{\substack{d \vert M \\ \omega(d) < m}} \rho(d) \Big)$$
Which seems very likely what I have to prove, if I take into consideration the following lemma:
Let $M \in \mathbb{N}^*$ such that $\mu(M)\neq 0$ and let $f(x) \in \mathbb{Z}[x]$ of degree $k$. Then $$\sum_{\substack{d \vert M \\ \omega(d)\leq m } } \rho (d) \leq e^{(k \omega (M))^{m-1}}.$$
I think that, by this, I can conclude $$O \Big( \sum_{\substack{d \vert M \\ \omega(d) < m}} \rho(d) \Big) = O \Big( e^{(k\omega(M))^{m-1}} \Big)$$ Which is still incorrect, but very much similar to what I need.
Whilst for the big O inside the squared bracket, since $\mu$ can only assume the value $0$ or $\pm1$, it seems reasonable that $$ \displaystyle\sum_{r=m}^{\omega(M)} \sum_{\substack{ d \vert M \\ \omega(d)=r}} \frac{\rho(d) \mu(d)}{d}=O \Bigg( \displaystyle\sum_{r=m}^{\omega(M)} \sum_{\substack{ d \vert M \\ \omega(d)=r}} \frac{\rho(d)}{d} \Bigg)$$
And if this is true, I think I can conclude that $$\displaystyle\sum_{r=m}^{\omega(M)} \sum_{\substack{ d \vert M \\ \omega(d)=r}} \frac{\rho(d) \mu(d)}{d} = O \Bigg(\displaystyle\sum_{r=m}^{\omega(M)}\Big( \frac{ek\Sigma(M)}{r} \Big)^r \Bigg)$$
Since, from a previuos result, I know that $$\displaystyle\sum_{r=m}^{\omega(M)} \sum_{\substack{ d \vert M \\ \omega(d)=r}} \frac{\rho(d) \mu(d)}{d} \leq \displaystyle\sum_{r=m}^{\omega(M)}\Big( \frac{ek\Sigma(M)}{r} \Big)^r $$
In my attempt I used the fact that, if $f(z) = O (g(z)) $ and $g(z) \leq h(z)$, than $f(z) = O(h(z))$. Is this assumption correct?
Thanks in advance for your help.
Edit: I solved the problem, it was just a bad notation. The author defines $e(A(x))=e^{A(x)}$ for a certain function $A(x)$. But in the case of the lemma I was using, the $e$ was the Neper number, this means that $$\sum_{\substack{d \vert M \\ \omega(d)\leq m } } \rho (d) \leq e(k \omega (M))^{m-1}$$ which leads me to the thesis.