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Let $P$ be a finite set of primes of cardinality $k$. Consider the set $$\Phi(P)=\{n\in\mathbb{Z}~:~\exists{p}\in{P} \text{ such that }p\mid{n}\}.$$

Let $\Lambda(P)$ be the largest number of consecutive integers that all lie in $\Phi(P)$. Can the function $$\Lambda(k)=\sup_{|P|=k}\Lambda(P)$$ be bounded in terms of $k$ alone?

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The answer is yes, and Iwaniec (1978) proved that $\Lambda(k)\ll k^2\log^2 k$.

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  • $\begingroup$ Let $P$ in an arbitrary finite subset of $\mathbb{Z}_{\geq 2}$, and $P'$ is the set of prime divisors of the elements of $P$, then $\Phi(P)\subset\Phi(P')$. Hence $\Lambda(P)\leq\Lambda(P')$, and for $\Lambda(P')$ you can apply Iwaniec's bound. BTW, if you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$ Commented Jan 23 at 19:12
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    $\begingroup$ @user1868607 I suggest that you post ask your new question in a separate post, with precise notation etc. as in the present post. $\endgroup$ Commented Jan 23 at 20:41
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    $\begingroup$ The absolutely trivial bound for the initial problem is $N\le 2k\cdot 4^{2k}$. Indeed, for any $P$, the product of primes less than $P$ is $M\le 4^P$ (Chebushev), so considering only numbers congruent to $1$ modulo $M$, we get $N/M\le k+(N/M)\sum_p 1/p\le k+(N/M)(k/P)$ where $p$ runs over the first $k$ primes after $P$. Now just take $P=2k$. This bound trivially extends to the setting of arbitrary numbers $\ge 2$ (replace each number by some its prime divisor) and to the arithmetic progression setting when the divisor set consists of primes. Of course, one can do much better in that case too. $\endgroup$ Commented Jan 23 at 22:27
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    $\begingroup$ @fedja Please encourage the OP to ask the arithmetic progression version in a separate post rather than discussing the problem here. $\endgroup$ Commented Jan 24 at 0:22
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    $\begingroup$ @user1868607 New questions belong in new posts, not comments. $\endgroup$ Commented Jan 26 at 0:04

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