# every arithmetic progression contains a sequence of $k$ “consecutive” primes for possibly all natural numbers $k$?

I ask the same question here:https://math.stackexchange.com/q/1019404/192097

writing a little better the previous question: it´s true that if we let $a$ and $b$ be coprime integers, then the arithmetic progesion : $a + bh: h\in Z$, contains a sequence of $k$ "consecutive" primes : $a + nb,a + (n+1)b,...,a + (n+k-1)b$, for possibly all integer $k$?.

I wrote "possibly for all $k$ " because there are some $k$ for which the elements of the progression can not be prime, for example if $a>1$ and if $k>a$ then at least one of $n,(n+1),...,(n+k-1)$ is a multiple of $a$. On the other hand, in the case $a=1$, $b>2$ letting $h=b-2$ is immediate that $1+bh=1+b(b-2)$ 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 is not immediate that $k$ need 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, wich essentially is not the same as the question before, even more: Green- Tao is a consecuence 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.

• This has been conjectured but is open. In various generalities: Hardy-Littlewood, Schinzel hypothesis H, Bateman-Horn,... – Felipe Voloch Nov 13 '14 at 21:44
• Specifically if you require every prime $p$ at most $k$ divides $b$, you should be good by Hardy-Littlewood. – Will Sawin Nov 13 '14 at 23:10

In 2000, Shiu proved that if $(a,q)=1$, then there are arbitrarily long strings of consecutive primes all equivalent to $a$ modulo $q$. See: http://jlms.oxfordjournals.org/content/61/2/359