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Writing a little better the previous question: is it true that if we let $a$ and $b$ be coprime integers, then the arithmetic progression : $a + bh: h\in {\mathbb Z}$, contains a sequence of $k$ "consecutive" primes: $a + nb,\,a + (n+1)b,\,\ldots,a + (n+k-1)b$, for possibly all integer $k$?

I wrote "possibly for all $k$ " because there are some $k$ for which the elements of the progression cannot be prime, for example if $a>1$ and if $k>a$ then at least one of $n,(n+1),\ldots,(n+k-1)$ is a multiple of $a$. On the other hand, in the case $a=1$, $b>2$ letting $h=b-2$ gives $1+bh=1+b(b-2)$ that is not prime (if $b=3$ then one of: $1+3n$, $1+3(n+1)$ is even), so $k$ is bounded by $b$, whereas in the case $a=1$ and $b=1$: one of $1+n,1+(n+1)$ is even, therefore $k\leq1$. If we require that $a$ and $b$ are large enough, then it’s not immediate that $k$ needs to be small.

informally: in each arithmetic progression, there are "arbitrarily" long sequences of primes.

This question comes after read the Green–Tao theorem on arithmetic progressions but I understand this as: there exist arithmetic progressions of primes, with $k$ terms, where $k$ can be any natural number, which essentially is not the same as the question before, even more: Green–Tao is a consequence of the previous one.

summarizing questions:

  • Is this a conjecture?
  • can be found references about the problem in order to try solve it?

I apologize for my poor English. I just speak Spanish. So please excuse me if occasionally the translation is not perfect.

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  • $\begingroup$ What exactly is your question? $\endgroup$ – Moishe Kohan Nov 13 '14 at 0:52
  • $\begingroup$ @Jean-ClaudeArbaut Your first sentence is on OP! Except that they assert it only for $k>a$ and not $k \ge a$. This is not really an answer to the intended question. Though admittedly it is not quite clear what is the intended question. But it is definitely not what you make it to be as this is addressed in OP.. $\endgroup$ – quid Nov 13 '14 at 9:08
  • $\begingroup$ @Jean-ClaudeArbaut I think a reading close to the intent is to interpret the "possibly" as "where this is/could be possible" so for all $k$ where there is no simple reason why it cannot be prime, technically where there is no local obstruction. This is essentially what Charles answer is doing. $\endgroup$ – quid Nov 13 '14 at 12:37
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    $\begingroup$ @quit You are probably right, but the OP should clarify this I think. Then, only the case $a=1$ is interesting, since for $a>1$ he already showed $k$ must be bounded. Thank you for your comments, I realize I didn't understand the question. $\endgroup$ – Jean-Claude Arbaut Nov 13 '14 at 12:52
  • $\begingroup$ Also posted to MO, mathoverflow.net/questions/187096/… $\endgroup$ – Gerry Myerson Nov 13 '14 at 22:11
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Let $a=3$ and $b=2$. Then the conjecture fails, for one of $3+2n$, $3+2n+2$, and $3+2n+4$ is divisible by $3$.

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  • $\begingroup$ This is mentioned in OP (with k> a instead of k>=a) and thus likely does not actually respond to the inquery. $\endgroup$ – quid Nov 13 '14 at 9:09
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First of all, $b$ must be a multiple of $k\#$ for such a progression to exist (well, apart from one progression using some of the primes up to $k$). Once you have that necessary condition you can derive the result from Dickson's conjecture, which is still unproved:

Conjecture: For any admissible collection of pairs $(a_1,b_1),\ldots,(a_n,b_n)$ there is some $n$ (hence infinitely many $n$) such that $a_i+b_in$ are prime for all $1\le i\le n.$

Th definition of "admissible" is slightly complicated but it comes down to not having any fixed prime divisors.

Proof sketch of your conjecture from Dickson's: find a bunch of large primes and require that each one, in sequence, divides $a+bn+1,a+bn+2,\ldots,a+b-1+bn,a+b+1+bn,\ldots,a-1+(n+k-1)b$ which you can do by picking residue classes and combining them with the CRT.

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It is not even known if for some specified $b$ there are infinitely many $a$ such that $a$ and $a+b$ are prime. So, your question is wide open.

(By recent results by Zhang and others it is however known that there exists such a $b$.)

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  • $\begingroup$ While your question is open, the OP's one has a crystal clear answer. $\endgroup$ – Jean-Claude Arbaut Nov 13 '14 at 7:27
  • $\begingroup$ This is debatable as the question is a bit unclear, and thus I should perhaps not have answered it. For details see the other comment. $\endgroup$ – quid Nov 13 '14 at 9:05

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