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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 $$\Big|[2+A^{\leq n}_a\cap P\Big| \geq \Big|[k+A^{\leq n}_a\cap P\Big|.$$

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    $\begingroup$ I do not understand what "there is a lot going for the following statement makes a lot of sense" means. $\endgroup$ Commented 13 hours ago
  • $\begingroup$ You are right @DanielAsimov - I changed that part in the post and hope it makes more sense $\endgroup$ Commented 5 hours ago

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A valid counterexample is $a=7,k=3,n=2$, since $A^{\le2}_7$ contains $7$ and $14$ (and possibly $0$, depending on the convention for $\mathbb N$), so $2+A^{\le2}_7$ has at most the prime $2$, while $3+A^{\le2}_7$ contains the primes $3$ and $17$, making the right-hand side larger.

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    $\begingroup$ This looks correct to me. $2+A_7^{\leq 2}=\{2,9,16\}$ and $3+A_7^{\leq 2}=\{3,10,17\}$. Since the range of $j$ in the question is shown to include $0$, then $0\in \mathbb{N}$. $\endgroup$ Commented 19 hours ago
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    $\begingroup$ I wonder why the answer was downvoted? $\endgroup$ Commented 18 hours ago
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Consider the 3-tuple $(a, n, k)$ then we can find a lot of counter-example. By a simple brute-force python script, we can find : $$(7, 2, 3), (7, 2, 5), (7, 4, 3), (7, 8, 3), (7, 8, 5), (7, 9, 3), (7, 9, 5), (7, 9, 10), (7, 10, 3), (7, 10, 5), (7, 10, 10), (7, 11, 3), (11, 6, 7), (7, 12, 3), (7, 12, 5), (7, 13, 3), (7, 13, 5), (7, 13, 10), (7, 13, 12), (11, 13, 6), (13, 2, 3), (13, 2, 5), (13, 2, 11),...$$

And a lot of others (I had to stop the script because it was finding to much solutions and made my computer bug). Here is the code if you want to test yourself :


def is_prime(num):
    if num <= 1: return False
    if num <= 3: return True
    if num % 2 == 0 or num % 3 == 0: return False
    i = 5
    while i * i <= num:
        if num % i == 0 or num % (i + 2) == 0: return False
        i += 6
    return True

def find_tuples():
    max_val = 2
    try:
        while True:
            for a in range(3, max_val + 1, 2):
                for n in range(2, max_val + 1):
                    ref = sum(1 for j in range(n + 1) if is_prime(2 + j * a))
                    for k in range(2, max_val + 1):
                        if max(a, n, k) == max_val:
                            res = sum(1 for j in range(n + 1) if is_prime(k + j * a))
                            if res > ref:
                                print(f"({a}, {n}, {k}),", end=" ", flush=True)
            max_val += 1
    except KeyboardInterrupt:
        pass

if __name__ == "__main__":
    find_tuples()
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