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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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    $\begingroup$ 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). $\endgroup$ – David Schneider-Joseph Mar 20 '17 at 13:00
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    $\begingroup$ @David Schneider-Joseph thanks. $\endgroup$ – Ahmad Mar 20 '17 at 13:20
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    $\begingroup$ 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. $\endgroup$ – Mastrem Mar 20 '17 at 15:23
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    $\begingroup$ 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. $\endgroup$ – Erick Wong Mar 21 '17 at 18:20

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