This is a follow-up to an earlier question.
The answer to that question was found on this page. The discussion on OEIS seems to suggest that, for any prime $p$, there should exist a $p$-length arithmetic progression of primes beginning with $p$.
Is this known?
This is only a partial answer for now. Define the $ k $ order configuration of an integer $ n $ as the sequence $ (n\mod p_{i})_{1\leq i\leq k}$. Such a sequence defines a unique residue class modulo $ P_{k} $, with $ P_{k} $ the $ k $ -th primorial.
Now let's consider the primes less than $ p $ and let $C(p) $ denote an integer such that $ p $, $ p+C(p) $ , ..., $ p+(p-1)C(p) $ are all prime. Fix $ k $ large enough and denote with $ C_{i}(p)=C(p)\mod p_{i} $. Necessarily $ p_{i}<p $ implies $ C_{i}(p)=0 $. One can take $ C_{\pi(p)}(p)=1 $ .
Actually it suffices that none of the numbers $ C_{i}(p), 2C_{i}(p),...,(p-1)C_{i}(p) $ be equal to $p_{i}-(p\mod p_{i}) $ , which can be achieved by taking for $C_{i}(p) $ the number $p_{i}-h_{i} $ with $ h_{i} $ greater than 1, less than $ p_{i} $ and coprime with $2(p\mod p_{i}) $. That way you finally obtained a residue class modulo $ P_{k} $ whose less positive representant fulfills your requirements.
