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NOTATION $\ p(0)\!=\!2\quad p(1)\!=\!3\quad\ldots\ $ -- the strictly increasing sequence $\ \mathbb P\ $ of all primes.

Conjecture $$\forall_{k\in\mathbb Z_{>4}}\,\exists_{m\,n\in\mathbb P}\quad (\ k<m<n<k+p(k)\quad \&\quad p(m)\equiv p(n)\!\!\!\mod p(k)\ ) $$

Example Prime $\ p(5)=11\ $ supports the above conjecture:

  • $\quad p(5)=11\ <\ p(8)=19\ <\ p(13)=41 $
  • $\quad 19\equiv 41\mod 11 $
  • $\quad 5\ <\ 8\ <\ 13\ <\ 5+p(5) = 16.$

A related simple observation (that I made at the beginning of starting the topic of consecutive primes):

Theorem Let primes $\ q\ $ and $\ p(m) \ldots p(n)\ $ be such that the sequence does not represent all residua mod $q$, but the sequence extended by either one of the primes $\ p(m-1)\ $ and $\ p(n+1)\ $ does in both cases. Then $\ p(m-1)\equiv p(n+1)\mod q.$

(This observation also relates to the equally obvious remark made by @EmilJerabek).

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    $\begingroup$ If you asked only for $k<m<n\le k+p(k)$, this would follow immediately from the pigeonhole principle, as there are $p(k)$ numbers of the form $p(n)$ for $k<n\le k+p(k)$, but only $p(k)-1$ nonzero residues modulo $p(k)$. $\endgroup$
    – Emil Jeřábek
    Commented Sep 30, 2020 at 14:09
  • $\begingroup$ There are exactly $\ q-1\ $ non-zero residua mod q possible, for q:= p(k). There is a chance, as it happened for $\ q=2\ 3\ 7\ $ that all of them occur sharply between $\ q\ $ and $\ p(k+p(k)). But it looks like the above conjecture is a sure thing, and much more should hold. ### Of course, residue 0 mod q cannot occur for any prime r > q (or different from q). It's beside the point anyway. It seems that this time, @EmilJeřábek, you're confused for a change (instead of me being so so many times). $\endgroup$
    – Wlod AA
    Commented Sep 30, 2020 at 14:29
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    $\begingroup$ Nothing you say contradict anything I said. So how does that make me confused? $\endgroup$
    – Emil Jeřábek
    Commented Sep 30, 2020 at 14:49
  • $\begingroup$ Indeed, I got confused one more time, but this time only meta-confused. I assumed that you are contradicting the mathematical correctness of my post - I rushed before I read your comment carefully, sorry. $\endgroup$
    – Wlod AA
    Commented Sep 30, 2020 at 17:04
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    $\begingroup$ To be clear, I’m not contradicting your post; I’m saying that if we weaken the conjecture just a little bit by raising the upper bound by $1$, it becomes easily provable. $\endgroup$
    – Emil Jeřábek
    Commented Sep 30, 2020 at 17:55

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