**Conjecture** For arbitrary integers $\ 0 \le k \le m\ $ there exists
integer $\ n\ge m\ $ such that for every natural number $\ s\ $ at
least one of the numbers $\ p(x)+s\ (\text{where}\ k\le x\le n)\ $ is not prime.



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> Here, $\ p(0)=2, p(1)=3,\ldots\ $ is the strictly increasing sequence of all prime numbers.
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A detail oriented prime conjecture
Let $q=p(k)$. Using Dirichlet's theorem on primes in arithmetic progressions, there is $n\geq m$ large enough so that for any $a$ not divisible by $q$, there is some $k\leq x\leq n$ such that $p(x)\equiv a\pmod q$. Also taking $a=0$ and $x=k$, we see this is true for *all* residue classes mod $q$.


Take any natural $s$ (which presumably also requires $s>0$ for you). Pick some $k\leq x\leq n$ such that $p(x)\equiv -s\pmod q$ (which exists by construction). Then $p(x)+s$ is divisible by $q$ and greater than $q$.


