# Consecutive integers each of which has a large prime factor

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 $$n$$ ?

• Ooops, I'd better delete that comment :) Commented Feb 25, 2019 at 16:00
• 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. Commented Feb 25, 2019 at 16:37
• 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. Commented Feb 25, 2019 at 16:38
• @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. Commented Feb 25, 2019 at 16:41
• 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. Commented Feb 25, 2019 at 17:59