Bounding the number of semiprimes of the form $2p\pm1$

Question

I would like an upper bound on the number of semiprimes of the form $2p+1$ (and the same with $2p-1$), where $p$ is prime. Is there a general result I can apply? I have not studied sieve theory (but am willing to learn if pointed appropriately).

Trivially there are $O(x/\log x)$ up to $x$ using only the fact that $p$ must be prime. The goal would be to show that there are $$ O\left(\frac{x\log\log x}{\log^2x}\right). $$

I don't need a good constant, an O-bound is fine (and as I understand, the expected constant is not attainable).

Answer 1

The most relevant result I can find is the following from Halberstam & Richart "Sieve Methods", Corollary 5.8.4 in page 179.

Let $q$ and $h$ be integers, and let $y$ and $x$ be real numbers satisfying $$ h\neq 0, \ \ (q,h)=1, \ \ 2|qh, \ \ 1\leq q<y\leq x. $$ Then $$ |\{p: x-y<p\leq x, (p-h)/q = p' \}| $$ $$ \leq 16 \prod_{p>2}\left(1-\frac1{(p-1)^2}\right)\prod_{2<p|qh}\frac{p-1}{p-2}\frac{y/q}{\log^2(y/q)}\left(1+O\left(\frac{\log\log 3|h|y}{\log(y/q)}\right)\right). $$

In here, $p$ and $p'$ are primes, and the implied constants in $O$-term is absolute.

This is not exactly the one you will need. But, you would have to modify it to work with $(2p\pm 1)/q$ instead of $(p-h)/q$ (so that $h=\pm 1$). This modification ultimately has to come from you by studying this book up to this point.

Once this is done, then sum over odd primes $q$ up to $\sqrt x$. This will give the bound $O((x\log\log x)/\log^2 x)$ as you desired.