# Density of primes in the set $\{p_1+2a,…,p_n+2a,…\}$ for every natural $a$, any conjectures?

If we shift the set $$\mathbb P=\{p_1,...,p_n,...\}$$ of all prime numbers by some natural number $$2a$$ to obtain a set $$\mathbb P+2a=\{p_1+2a,...,p_n+2a,...\}$$ then I expect that $$\mathbb P +2a$$ contains an infinite number of prime numbers and that it contains an infinite number of composite numbers.

I would like to know are there any conjectures about density of primes in the set $$\mathbb P+2a$$, that is, what is known about the limit $$\lim_{n \to + \infty} \dfrac{nop\{p_1+2a,...,p_n+2a\}}{n}$$, where $$nop$$ stands for the "number of primes", that is, if $$S$$ is any set then $$nop(S)$$ gives as a number of primes in the set $$S$$.

We can denote $$\lim_{n \to + \infty} \dfrac{nop\{p_1+2a,...,p_n+2a\}}{n}=f(a)$$, and, I would also like to know what is known about $$f$$, for example, is it reasonable to expext that $$f$$ is a constant function?

More particularly, is there any evidence that we could have $$f(a)=0$$ for every $$a \in \mathbb N$$, that is, that shifting of the set of primes by some even number $$2a$$ gives us a set where "almost all" numbers are composite numbers?

• It is known that $f(a)=0$ for all $a$. This can be proven using the same methods as those which establish Brun's theorem. In particular, $\mathbb P+2a$ contains infinitely many composites. It is an open problem whether it contains infinitely many primes for any $a$, and it's open whether it contains a prime for every $a$. Polignac's conjecture is closely related. – Wojowu Jun 7 '19 at 14:58
• @Wojowu If that were an answer , I would accept it. It contains enough information to do a research in this direction. – user141210 Jun 7 '19 at 15:01
• "If we shift a set $\Bbb P = \dots$": is $\Bbb P$ the set of all primes, or a generic infinite set of primes? – Greg Martin Jun 7 '19 at 17:57
• @GregMartin $\mathbb P$ is the set of all prime numbers. – user141210 Jun 7 '19 at 17:58

Fix $$a\geq 1$$. It is elementary to see that $$\mathbb P+2a$$ contains infinitely many composite numbers -- indeed, if this were not the case, then for sufficiently large primes $$p$$, $$p+2a$$ would also be a prime, hence so would be $$p+4a,p+6a,\dots,p+2pa=p(1+2a)$$ which is clearly absurd.
Indeed, it is true that $$f(a)=0$$ - this is more difficult, but can be proven using the same methods as those which prove Brun's theorem. Indeed, those methods show that for some constant $$C$$ there are at most $$C\frac{x}{(\log x)^2}$$ primes $$p such that $$p+2a$$ is a prime, compared to approximately $$\frac{x}{\log x}$$ primes below $$x$$ in total (prime number theorem).
Now, as you can probably guess, existence of primes in $$\mathbb P+2a$$ is a much more difficult problem. The statement that for every $$a$$ this set contains infinitely many primes is known as Polignac's conjecture and it is wide open - indeed, there is no single $$a$$ for which it is known that $$\mathbb P+2a$$ contains infinitely primes (though, by the results of Zhang et al. we know that there are infinitely many such $$a$$, in particular there is such an $$a\leq 123$$), nor is it even known whether $$\mathbb P+2a$$ contains even one prime for every $$a$$.
• Just a remark that while a priori the statement "\mathbb{P} + 2a contains one prime" is much weaker than "$\mathbb{P} + 2a$ contains infinitely many primes", they're essentially equivalent given every method we know in number theory. It would be extraordinary to see a proof of the first statement that does not yield a proof of the second. – Stanley Yao Xiao Jun 7 '19 at 18:10