The number of $n$ and $n+2$ that have at most seven prime factors

Question

In Murty and Cojocaru, we have the theorem: As $x\rightarrow\infty$, $$\#\{n\le x: n \text{ and } n+2 \text{ have at most seven prime factors}\}\gg \frac{x}{(\log x)^2}$$ The proof uses Brun's sieve with $A:=\{n(n+2): n\le x\}$ $P$ being the set of primes, and $\omega(2) = 1$, $\omega(p)=2$, so that if $S(A,P,z)$ is the number of elements in $A$ not divisible by any $p<z$, then $$S(A,P,z)>\frac{1}{2}x\prod_{2<p<z}\left(1-\frac{2}{p}\right)\left(1-F(\lambda)\right) + O(z^{G(\lambda)})$$ for specific functions $F$ and $G$ that are unimportant here. The book then picks an appropriate $\lambda$ such that $F(\lambda) > 1 + O(1/\log z)$ and $G(z) < 8$. Then the book says this implies $$S(A,P,z) \gg \frac{x}{\log^2x} + O(z^\theta)$$ with $\theta < 8$

1. My first question is how do we get this? I can only get $$S(A,P,z)\gg \frac{x}{\log x}\prod_{2<p<z}\left(1-\frac{2}{p}\right) + O(z^\theta)$$ I imagine it has something to do with Merten's Theorem, but I'm not sure.

The book then continues to say that if we choose $z:=x^{1/u}$ with $u<8$, we get the result.

2. I'm not able to draw the connection between $S(A,P,x^{1/u})$ and $\#\{n\le x: n \text{ and } n+2 \text{ have at most seven prime factors}\}$. Does anyone have insight as to why this implication is true?

Any help is greatly appreciated. Thanks in advance!

Answer 1

1. Yes, by Merten's we have $\prod_{2 < p < z} \left(1 - \frac{2}{p}\right) \geq \prod_{2 < p < z} \left(1 - \frac{1}{p}\right)^2 \gg \left(\frac{1}{\log z}\right)^2$. Then following the book's calculation,

$$ S(\mathcal{A}, \mathcal{P}, z) \gg \frac{x}{\log^2 z}\left(1 - F(\lambda)G(\lambda, z)\right) + O\left(z^{1 + \frac{2.01}{e^{\lambda} - 1}}\right) $$

Where $F(\lambda) = \frac{2\lambda^2 e^{2\lambda}}{1 - \lambda^2 e^{2 + 2\lambda}}$ and $G(\lambda, z) = \exp\left(\frac{4c_1}{\lambda \log z}\right)$. Now, note that as $z \to \infty$, $G(\lambda, z) \ll 1 + O\left(\frac{1}{\log z}\right)$. Also, letting $\lambda' = \lambda e^{\lambda}$ we see that $F'(\lambda') = \frac{2\lambda'^2}{1 - e^2\lambda'^2}$, and we can find that $\lambda' < \sqrt{\frac{1}{2 + e^2}}$ will give us $F'(\lambda') < 1$. This corresponds to $\lambda < W\left(\sqrt{\frac{1}{2 + e^2}}\right) = 0.2533\ldots$. With this constant, $1 + \frac{2.01}{e^{\lambda} - 1} = 7.97\ldots < 8$. Picking $z = x^{1 / (\theta + \epsilon)} < x^{1 / 8}$ and combining everything gives

$$ S(\mathcal{A}, \mathcal{P}, z) \gg \frac{x}{\log^2 z}\left(1 - F(\lambda)\left(1 + \frac{1}{\log z}\right)\right) + O(z^{\theta}) \gg \frac{x}{\log^2 x} $$

2. By definition, $S(A, \mathcal{P}, z)$ is the number of $n \leq x$ such that $p < z \implies p \not \mid n(n + 2) \implies (p \not \mid n \, \land p \not \mid n + 2)$. If we write out the factorization of $n = p_1p_2 \cdots p_k$, we note that $p_i \geq z$, so $n \geq z^k = x^{k / u}$. Since $n \leq x$, $k < u < 8$ and $n$ has at most seven prime factors. Same for $n + 2$.

Hope this helps! I am studying sieve theory as well :)