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There are multiple results on the sieve method, and I wanted to ask about the following variant (to know if it is trivial by one of the current versions of the sieve method, or seems a challenging problem)

Let $\varepsilon>0$ be fixed. Then there exists $0<\delta<\varepsilon$, such that for all sufficiently large $k$ the following is true.

Let the primes strictly between $\sqrt k$ and $k$ be written as $\sqrt k <p_1\le \ldots \le p_r <k$. Let $M= \prod_{j=1}^r p_j$. For every $1 \le j \le r$, let $I_j= \left \{1,2,\ldots, \left \lceil p_j^{1-\delta} \right \rceil \right \}$. Then among every $M^\varepsilon$ consecutive integers, there is at least one value $x$ for which $ x \bmod p_j$ belongs to $I_j$ for every $1 \le j \le r$.

Note: By the prime number theorem, $M=k^{\Theta( k/ \log{k} )}$. The average size of intersection is at least $M^{\varepsilon-\delta}$, which is large for $k$ large. One can e.g. assume $(\varepsilon-\delta) k^{0.01}> \log{k}$. So maybe this is easy for the experts.

Thanks in advance.

ps: if the problem is solved, with almost all k instead of all k, I would also be happy. But then the proof may use different ideas.

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  • $\begingroup$ what does "choose intervals $I_j$ with the first $p_j^{1-\delta}$ numbers" mean? $\endgroup$ Commented Sep 22, 2024 at 18:26
  • $\begingroup$ You need to define $I_j$ and $M$ more explicitly. The way it's currently written the question is hard to parse. $\endgroup$ Commented Sep 22, 2024 at 19:11
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    $\begingroup$ @AlexeiEntin I believe that $M=\prod_{j=1}^r p_j$, so probably $M$ is the product of the primes in $[\sqrt{k},k]$. Also, I believe that $I_j=\mathbb{N}\cap[1,p_j^{1-\delta}]$. $\endgroup$ Commented Sep 22, 2024 at 19:15
  • $\begingroup$ The above (reply by GH) is correct. Should have written first $p_j^{1-\delta}$ natural numbers to be clearer. Alternatively, one can also aim more general; for every $I_j\subset [p_j]$ for which $\vert I_j \rvert = p_j^{1-\delta}$. $\endgroup$ Commented Sep 22, 2024 at 20:01
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    $\begingroup$ Question has been edited to clarify all the issues. $\endgroup$ Commented Sep 23, 2024 at 5:53

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As an update; it is now part of https://arxiv.org/pdf/2410.09138

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  • $\begingroup$ It says there, "For example when $x$ is even and $y > x$ is odd and not much larger than $x$, we may have that ${\rm lcm}(x, x + 1, x + 2) = x(x + 1)(x + 2) >y(y+1)(y+2)/2 = {\rm lcm}(y, y + 1, y + 2)$." But when $x$ is even and $y$ is odd, then ${\rm lcm}(x, x + 1, x + 2) = x(x + 1)(x + 2)/2$ and $y(y+1)(y+2) = {\rm lcm}(y, y + 1, y + 2)$ $\endgroup$ Commented Jul 28 at 7:53

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