this is a conjecture or a result? every arithmetic progression contains a sequence of $k$ "consecutive" primes for possibly all natural numbers $k$?

Question

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.

Answer 1

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 $b$.)

Answer 2

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 $3$.

Answer 3

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.