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?

More generally, I am also interested in every conjecture about this topic that you know of.

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    $\begingroup$ 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. $\endgroup$ – Wojowu Jun 7 '19 at 14:58
  • $\begingroup$ @Wojowu If that were an answer , I would accept it. It contains enough information to do a research in this direction. $\endgroup$ – user141210 Jun 7 '19 at 15:01
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    $\begingroup$ "If we shift a set $\Bbb P = \dots$": is $\Bbb P$ the set of all primes, or a generic infinite set of primes? $\endgroup$ – Greg Martin Jun 7 '19 at 17:57
  • $\begingroup$ @GregMartin $\mathbb P$ is the set of all prime numbers. $\endgroup$ – 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 $p<x$ 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 $a$.

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    $\begingroup$ 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. $\endgroup$ – Stanley Yao Xiao Jun 7 '19 at 18:10

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