## Generating Chen primes.

Let

$A_p(n)=\#\{q<n \vert \ qp-2 \ is \ prime\}$

Where p,q are prime, n is an integer. My question is, it seems fairly reasonable to assume that for a fixed $n$, $A_p(n)$ can be bounded by a decreasing function, but can I prove it. Moreover, can it be shown that $\lim_{p \rightarrow \infty} A_p(n)=0$? Does anyone know of any work done on the subject?

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Of course notice that if p=1, then A_p(n) is about n/ln(n), and that for any p, this MUST be the largest possible value of A_p(n) for a fixed n. – Alex Botros Dec 2 2010 at 20:09
The limit equaling zero contradicts a few well known and believed conjectures - specifically $3p-2$ should be prime for infinitely many primes $p$. – Dror Speiser Dec 2 2010 at 20:59
So the conjecture is: If $n$ is fixed and $p$ is sufficiently large (in terms of $n$), then none of $2p-2,3p-2, \dots, qp-2$ (with $q$ going through the primes) are themselves prime. If I understand the question correctly, then I suspect that's false, as I believe that there are infinitely many prime pairs $p, 3p-2$. – Kevin O'Bryant Dec 2 2010 at 21:03
In the light of the comments by Dror and Kevin, what makes it seem reasonable to assume that $A_p(n)$ decreases as $p$ increases? – Gerry Myerson Dec 2 2010 at 22:14
@Alex: Consider $A_p(4)$. If $3p-2$ is prime, then $A_p(4)=1$. So $A_p(4)$ does not go to zero as $p$ goes to infinity (if the standard conjectures are true). Understood? I vote to close on the grounds that if the poster doesn't understand the question, the rest of us can't be expected to, either. – Gerry Myerson Dec 3 2010 at 5:06