Starting point. Let $P\subseteq\mathbb{N}$ be the set of primes. By just "looking at" how the first primes are distribute, there is a lot going for the following statement makes a lot of sense:

For all integers $n,k \geq 2$ we have $$\Big|[2, 2+n]\cap P\Big| \geq \Big|[k, k+n]\cap P\Big|.$$

Interestingly, it is not known whether this statement is true. This inspired the following considerations.

Taking this to finite arithmetical progressions. For $a, n\in\mathbb{N}$, let $A^{\leq n}_a=\{j\cdot a: j\in\mathbb{N}, 0\leq j\leq n\}$. If $n\in \mathbb{N}$ and $B\subseteq \mathbb{N}$, let $n+B:= \{n+b: b\in B\}$. Is there a counterexample for the following statement?

For all integers $a, k, n \geq 2$ with $a$ odd we have $$\Big|[2+A^{\leq n}_a\cap P\Big| \geq \Big|[k+A^{\leq n}_a\cap P\Big|.$$

Starting point. Let $P\subseteq\mathbb{N}$ be the set of primes. Looking at $P$ and seeing how $P$ thins out as we progress to higher numbers is enough to make the following statement plausible:

For all integers $n,k \geq 2$ we have $$\Big|[2, 2+n]\cap P\Big| \geq \Big|[k, k+n]\cap P\Big|.$$

Interestingly, it is not known whether this statement is true, and the consensus seems to be that the statement is likely false. This inspired the following considerations.

Taking this to finite arithmetical progressions. For $a, n\in\mathbb{N}$, let $A^{\leq n}_a=\{j\cdot a: j\in\mathbb{N}, 0\leq j\leq n\}$. If $n\in \mathbb{N}$ and $B\subseteq \mathbb{N}$, let $n+B:= \{n+b: b\in B\}$. Is there a counterexample for the following statement?

For all integers $a, k, n \geq 2$ with $a$ odd we have $$\Big|[2+A^{\leq n}_a\cap P\Big| \geq \Big|[k+A^{\leq n}_a\cap P\Big|.$$