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The groundbreaking work of Maynard and Tao showed the following fundamental result:

For any integer $m$, there exists a number $k(m)$ such that for any admissible set $H$ of $k$ integers (with $k$ at least $k(m)$), there are infinitely many integers $n$ for which the set $n+H$ contains at least $m$ primes.

My question is about a natural and stronger version. Suppose we want to force primes to appear in different parts of the set simultaneously. More precisely, suppose we are given numbers $r$, and $m_1, m_2, \ldots, m_r$.

Question. Does there exist a constant $C$ (depending on $r$ and all the $m_j$) such that for any admissible set $H$ of $k$ integers (with $k$ at least $C$), and for any way of splitting $H$ into $r$ smaller groups $H_1$, $H_2$,$\ldots$, $H_r$, there are infinitely many integers $n$ with the following property?

For every single group $H_j$ (where $j=1$ to $r$), the set $n+H_j$ contains at least $m_j$ primes.

In other words, can we guarantee that we can find infinitely many $n$ such that each part of the partition contains many primes at the same time?

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    $\begingroup$ I doubt it, since this would imply the twin prime conjecture: take $r=2$ and $m_1=2$ and take a very large admissible set $H$ containing $0$ and $2$, and $H_1=\{0,2\}$. You should probably add a condition like each $H_j$ is sufficiently large depending on $m_j$. $\endgroup$ Commented Nov 18 at 10:08
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    $\begingroup$ This is obviously false if $m_j<|H_j|$, but as Thomas says, you probably want to ask for $H_j$ large enough relative to $m_j$. I think an equivalent formulation would be: for any $m,r$ there some $C$ such that given $r$ admissible tuples $H_i$ of size $k$, there are infinitely many $n$ such that $n+H_i$ contains $m$ primes for each $i$. That is, we ask for Maynard-Tao for a number of admissible tuples simultaneously. $\endgroup$ Commented Nov 18 at 11:38
  • $\begingroup$ Thank you, Thomas Bloom and Wojowu; $H_j$ needs to be sufficiently large relative to $m_j$. $\endgroup$ Commented Nov 19 at 0:22

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I don't think this can be done even for $H_j$ arbitrarily large in terms of $m_j$. The method of Maynard's work, and prior work in the same direction, proceeds roughly by finding a set $S$ of $n$, expressed in terms of arithmetic progressions, such that $n+h$ has no small prime factors for $n \in S$ and $h\in H$, then applying results counting primes in arithmetic progressions to show that many numbers $n+h$ must be prime, which implies by a pigeonhole argument that for some $n$ we must have $n+h$ prime for many $h$.

The problem is that even if for each $j$ and each $h \in H_j$ we know that $n+h$ is prime for many $n \in S$, the pigeonhole doesn't tell us that we simultaneously have $n+h_1$ and $n+h_2$ prime for $h_1 \in H_1$ and $h_2\in H_2$. Rather, we would need some control on the number of $n\in S$ with $n+h_1$ prime and $n+h_2$ prime, which would be stronger than the twin prime conjecture instead of merely a variant of the Bombieri-Vinogradov theorem.

So it seems to me that, if this could be done, it would require a completely new method rather than a small variation on existing techniques.

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  • $\begingroup$ Thank you, Will Sawin $\endgroup$ Commented Nov 19 at 0:23

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