For example the list $(2, 1, 2, 1)$ is congruent $\pmod 3$ to the consecutive primes $(5, 7, 11, 13)$. But how about the list $(1,1,1,1,1,1,1,1,2,3,4,3,2,3,1) \mod 5$?

More generally, we are given some integer $n \geq 2$ and a finite list of integers that are coprime to and less than $n$. Is it always possible to produce the same list by consecutive primes $\pmod n$?

Formally: given $n \geq 2$ and $(a_0,a_1,\cdots \,a_k)$ such that for all $i$, $GCD(a_i, n) = 1$, is there a list of consecutive primes such that each $p_i \equiv a_i \pmod n$?

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    No, for example $(2, 2) \pmod 4$ would require two even primes. The question might be more interesting if you require that entries in the list are coprime with $n$ (a generalization of disallowing 0). – David Schneider-Joseph Mar 20 '17 at 13:00
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    @David Schneider-Joseph thanks. – Ahmad Mar 20 '17 at 13:20
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    Very interesting question. Altough I would be surprised if this wasn't true, I would be even more surprised if there was an easy proof. – Mastrem Mar 20 '17 at 15:23
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    This is an open problem except in the case where all $a_i$ are equal, which is a famous theorem of Shiu. There are a few other cases that are consequently true for simple combinatorial reasons, such as $(1,1,1,1,1,1,1,1,2)$ mod $3$, but essentially Shiu's theorem is the best we have, AFAIK. – Erick Wong Mar 21 '17 at 18:20
  • For the record, you can find $\{1,1,1,1,1,1,1,1,2,3,4\} \pmod {5}$ starting at $1313286451$, which looks about in line with what I'd expect from the random model. So yeah, it certainly seems likely to me that arbitrarily long sequences of this sort exist. – Trevor Aug 11 '20 at 01:19
  • Is there a Wikipedia link to the Shiu Theorem? I wan't able to find any statement of the Theorem via google search. – cr001 Aug 16 '20 at 10:09
  • nobody intrested in my bounty ?? – mick Aug 17 '20 at 11:04
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    @cr001 Can't find a wiki link. TLDR, it was proven by Daniel Shiu in the 90s, and the statement is as follows: for any coprime $a,b\in\mathbb N$, there exists arbitrarily long sequences of consecutive primes, all congruent to $a\bmod b$. Similar but different than Greenwood-Tao, which doesn't require consecutiveness of the primes. – Rushabh Mehta Aug 17 '20 at 17:23
  • What about using a generalized Dirichlet series but with multiple entries, i.e. characters $\chi: (\mathbb{Z}/n\mathbb{Z} )^* \times \ldots \times (\mathbb{Z}/n\mathbb{Z} )^* \to \mathbb{C}^*$ ? I just don't know how to implement the "consecutive primes" constraint.. – Andrea Marino Mar 21 '21 at 16:19
  • I Think the question is more interesting if we consider the case where $n$ is a prime itself. – mick Mar 29 '21 at 20:05
  • Well lets first note, that even residues modulo odd $n$ aren't possible to be prime unless the multiplier on $n$ is odd. Likewise, odd residues aren't representing primes unless the multiplier on $n$ is even in this case. This at least forces gap size constraints (1,4,2,3) mod 5 won't happen below gap sizes 8,8,6 with gaps 3 mod 5, 3 mod 5, and 1 mod 5 Your long one needs gaps of 10,10,10,10,10,10,10,6,6,6,4,4,6,8 – Roddy MacPhee Mar 21 '21 at 12:09
  • @RushabhMehta How is that Greenwood-Tao result an advance on Dirichlet's theorem that every AP of integers $a, a+d,\dots$ with $d>0$ coprime to $a$ contains arbitrarily many primes? – Rosie F May 09 '22 at 09:45
  • It's not, and I never said it was... – Rushabh Mehta May 09 '22 at 13:53

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