A positive integer $n$ is called $k$-rough if all of the prime factors of $n$ strictly exceed $k$. For $k$ fixed and $n$ large, what's the shortest interval $[n,n+t]$ that contains a $k$-rough number ?
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Being $k$-rough is the same as being relatively prime to the product of primes up to $k$. Hence if $P(k)$ denotes that product, and $j$ denotes the Jacobsthal function, then $t=j(P(k))-1$ works for every $n$, while $t=j(P(k))-2$ fails for infinitely many $n$'s. In particular, it follows from a result of Iwaniec (1971) that $t=ck^2$ works for some absolute constant $c>0$.
answered Feb 3, 2022 at 22:49
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2$\begingroup$ It looks like Maier and Pomerance (1990) conjectured that $j(P(k)) = k(\log k)^{2+o(1)}$. $\endgroup$ Commented Feb 4, 2022 at 0:48
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$\begingroup$ After checking the paper of Iwaniec, $t=c log^2(k)k^2$. $\endgroup$– khattabCommented Feb 19, 2022 at 23:50
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1$\begingroup$ @khattab No. Iwaniec talks about the first $r$ primes, while you talk about the primes up to $k$. That is, Iwaniec's $r$ equals $\pi(k)\sim k/\log k$ here, and Iwaniec's bound $O(r^2\log^2 r)$ is really $O(k^2)$ here. $\endgroup$ Commented Feb 20, 2022 at 0:58
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