number-theory / Number Theory, Divisors
How large must $y(\varepsilon, n)$ be so that every interval $(x, x+y)$ contains at most $\varepsilon y$ integers having a divisor in $(n, 2n)$? The candidate proof gives the sharp fixed-$\varepsilon$ order $y = \Theta_\varepsilon(n)$, uniformly in the translate.
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How large must $y(\varepsilon, n)$ be so that every interval $(x, x+y)$ contains at most $\varepsilon y$ integers having a divisor in $(n, 2n)$? The candidate proof gives the sharp fixed-$\varepsilon$ order $y = \Theta_\varepsilon(n)$, uniformly in the translate.
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How large must $y(\varepsilon, n)$ be so that every interval $(x, x+y)$ contains at most $\varepsilon y$ integers having a divisor in $(n, 2n)$? The candidate proof gives the sharp fixed-$\varepsilon$ order $y = \Theta_\varepsilon(n)$, uniformly in the translate.
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