I'm looking for "long" arithmetic sequences that contain only prime (positive) numbers. For example: $$7,37,67,97,127,157$$

Is there any known bound for the length of these kind of sequences? Is it known that there exist arbitrarily long sequences?


I'm aware of Dirichlet's theorem. I also know that if $a_1$ is the first term and $d$ the difference of the sequence, then it should be $\gcd(a_1,d)=1$ and the length of the sequence is $\le a_1$ (because $a_1+a_1d$ is not prime).

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    $\begingroup$ Yes , this is known as the Green-Tao theorem : en.wikipedia.org/wiki/Green%E2%80%93Tao_theorem $\endgroup$ – user252450 Jan 20 '16 at 16:50
  • $\begingroup$ If the primes must be consecutive, I do not know whether it is only conjectured or proven. If the primes are arbitary, you can construct arbitary long sequences. $\endgroup$ – Peter Jan 20 '16 at 17:01
  • $\begingroup$ @Peter The example included is not made of consecutive primes ;) $\endgroup$ – N. S. Jan 20 '16 at 17:05
  • $\begingroup$ @Peter Is the variant about consecutive primes really conjectured to be true ? I didn't heard about it but it seems very unlikely that for some $n$ we will have : $$g_n=g_{n+1}=\ldots=g_{n+10^{100}}$$ where $g_n=p_{n+1}-p_n$ is the prime gap . But , who knows...maybe it's true (that would be mind-blowing ) $\endgroup$ – user252450 Jan 20 '16 at 17:13
  • $\begingroup$ The gap should be a primorial. It does not work for an arbitary gap. $\endgroup$ – Peter Jan 20 '16 at 17:17

The Green-Tao theorem guarantees that you can find such sequences as long as you want.

If I remember right, the proof also contains some bound on how far one needs to go to be guaranteed to find such a progression, but the bound is insanely high..

As of few days ago, the longest known such sequence is consisting of 26 numbers, but the above theorem guarantees that there are sequences as long as we want...

  • $\begingroup$ Ok, I see that the theorem is very "young" (12 years old). Thank you and ComplexPhi. $\endgroup$ – ajotatxe Jan 20 '16 at 16:54
  • $\begingroup$ @ajotatxe Yes, it is. For mathematics this is basically yesterday :) $\endgroup$ – N. S. Jan 20 '16 at 16:58

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