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There are many results about consecutive integers all having small prime factors. But what about consecutive integers each of which has a large prime factor?

More precisely, let $P(n)$ be the greatest prime factor of $n$.

Is it true that for each $k \geq 1$ there exists $n \geq 1$ such that $P(n + i) > \sqrt{n + i}$ for all $i=1,\dots,k$ ?

If yes, what is an upper bound in terms of $k$ for the least possible $n$ ?

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  • $\begingroup$ Ooops, I'd better delete that comment :) $\endgroup$
    – Chris Wuthrich
    Commented Feb 25, 2019 at 16:00
  • $\begingroup$ fine for $k=2$ with Chinese Remainder Theorem. For $k \geq3,$ need to get a better search for nearly consecutive $p_j$ such that $ n \equiv -1 \pmod p_1, $ $ n \equiv -2 \pmod p_2, $ $ n \equiv -3 \pmod p_3.$ but $n$ is especially small. I'd say it is already worth trying $k=3$ by computer, see how difficult it might be. $\endgroup$
    – Will Jagy
    Commented Feb 25, 2019 at 16:37
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    $\begingroup$ You might check out mathoverflow.net/questions/255269 (of which this question is almost a duplicate) and similar questions. My belief is that not much is known, and that your upper bound on n will be exponential in k, and that proving this is hard. Gerhard "But Go Check For Yourself" Paseman, 2019.02.25. $\endgroup$
    – Gerhard Paseman
    Commented Feb 25, 2019 at 16:38
  • $\begingroup$ @Will, for k=3 my computer says n=4. For k=5 my computer says n=18. Gerhard "Check OEIS For Bigger Numbers" Paseman, 2019.02.25. $\endgroup$
    – Gerhard Paseman
    Commented Feb 25, 2019 at 16:41
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    $\begingroup$ I see now in a comment to 255269 that I ran such a program and provided partial results (so replace (650 660) by (364 374)) with an interval of 40 rough numbers somewhere below 27 million. Gerhard "Is K Big Enough Now?" Paseman, 2019.02.25. $\endgroup$
    – Gerhard Paseman
    Commented Feb 25, 2019 at 17:59

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