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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. –  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
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    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
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    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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    Very interesting question. Just as an observation, if we took $n$ to be a prime and a finite list of values less than $n$ as stated in the problem and if the list of values is congruent to successive primes mod $n$, then we could encode that entire list using just the initial prime's index and deterministically recover the elements of the list with no additional information (except the size of the list). Perhaps, this is intricately linked to Shannon Entropy. – vvg Oct 05 '22 at 06:28
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    For those seeking Shiu's paper, here is the citation: Shiu, D.K.L. (2000), Strings of Congruent Primes. Journal of the London Mathematical Society, 61: 359-373. https://doi.org/10.1112/S0024610799007863 – vvg Oct 05 '22 at 06:47

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