# Does every prime $p$ appear in a $p$-term arithmetic progression of primes? [duplicate]

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?

• I believe that is not known. – GH from MO Aug 10 '18 at 2:58
• A closely related question here in MO at which you may want to take a look later on: mathoverflow.net/q/260783/1593 – José Hdz. Stgo. Aug 10 '18 at 3:32
• 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. – Sylvain JULIEN Aug 10 '18 at 11:04
• This could be closed as a duplicate of the thread mentioned by José? – Jeppe Stig Nielsen Aug 12 '18 at 0:03

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.
• 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 $p$. – Sylvain JULIEN Aug 12 '18 at 13:26
• 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 $(\frac{a}{\log^{2}(p+pC(p))})^{p-2}>\frac{1}{p+pC(p)}$. – Sylvain JULIEN Aug 12 '18 at 13:49