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This is a follow-up to an earlier question.

The answer to that question was found on this page. The discussion on OEIS seems to suggest that, for any prime $p$, there should exist a $p$-length arithmetic progression of primes beginning with $p$.

Is this known?

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    $\begingroup$ I believe that is not known. $\endgroup$ – GH from MO Aug 10 '18 at 2:58
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    $\begingroup$ A closely related question here in MO at which you may want to take a look later on: mathoverflow.net/q/260783/1593 $\endgroup$ – José Hdz. Stgo. Aug 10 '18 at 3:32
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    $\begingroup$ I think such a $ p $ -term arithmetic progression should always exist. Maybe you can try to use the notion of configuration I defined in my question "About Goldbach's conjecture" to prove it. I will have 6 days off work in a row from tonight so I'll do my best to think about it seriously and give you some news. $\endgroup$ – Sylvain JULIEN Aug 10 '18 at 11:04
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    $\begingroup$ This could be closed as a duplicate of the thread mentioned by José? $\endgroup$ – Jeppe Stig Nielsen Aug 12 '18 at 0:03
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This is only a partial answer for now. Define the $ k $ order configuration of an integer $ n $ as the sequence $ (n\mod p_{i})_{1\leq i\leq k}$. Such a sequence defines a unique residue class modulo $ P_{k} $, with $ P_{k} $ the $ k $ -th primorial.

Now let's consider the primes less than $ p $ and let $C(p) $ denote an integer such that $ p $, $ p+C(p) $ , ..., $ p+(p-1)C(p) $ are all prime. Fix $ k $ large enough and denote with $ C_{i}(p)=C(p)\mod p_{i} $. Necessarily $ p_{i}<p $ implies $ C_{i}(p)=0 $. One can take $ C_{\pi(p)}(p)=1 $ .

Actually it suffices that none of the numbers $ C_{i}(p), 2C_{i}(p),...,(p-1)C_{i}(p) $ be equal to $p_{i}-(p\mod p_{i}) $ , which can be achieved by taking for $C_{i}(p) $ the number $p_{i}-h_{i} $ with $ h_{i} $ greater than 1, less than $ p_{i} $ and coprime with $2(p\mod p_{i}) $. That way you finally obtained a residue class modulo $ P_{k} $ whose less positive representant fulfills your requirements.

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  • $\begingroup$ Why the downvote ? $\endgroup$ – Sylvain JULIEN Aug 12 '18 at 10:45
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    $\begingroup$ Probably because the method doesn't work (you ensure indivisibility by small primes, but the numbers could still have a large prime factor. There is no suitable choice of k.) $\endgroup$ – Will Sawin Aug 12 '18 at 13:12
  • $\begingroup$ The sequence $ C(p), 2C(p),\cdots \frac{p-1}{2}C(p) $ forms an arithmetic sequence of primality radii of $ p+\frac{p-1}{2}C(p) $. This may help to bound $ k $ in terms of $ p $. $\endgroup$ – Sylvain JULIEN Aug 12 '18 at 13:26
  • $\begingroup$ In other words, $ C(p) $ is a primality radius of $ p+C(p), p+2C(p),\cdots p+(p-2)C(p) $. Note also that the proportion of primality radii of $ n $ is greater than $ \frac{a}{\log^{2} n }$ for some absolute $ a>0 $. So that if should suffice to require $ (\frac{a}{\log^{2}(p+pC(p))})^{p-2}>\frac{1}{p+pC(p)} $. $\endgroup$ – Sylvain JULIEN Aug 12 '18 at 13:49

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