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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?

edited Dec 2 2010 at 20:19

asked Dec 2 2010 at 19:59

Alex Botros
225●1●5

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

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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

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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

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@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

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Ergenheim · Answer 1 · 2011-02-04 05:49:47Z UTC

A powerful lead toward creating a proof can be found using the Croft Spiral Sieve:

Note, referencing the Croft Spiral Sieve, that all squared primes are either mod30,1 or mod30,19. both of which have primes positioned in p-2 (at mod30,29 and mod30,17, respectively). To see why this is the case, first explore this: http://www.primesdemystified.com, in order to understand this: http://primesdemystified.com/ChordSequenceFactorizations.html

Generating Chen primes.

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