I was thinking about consecutive integers and I wondered if anyone had done work exploring whether a sequence of $2n$ consecutive integers (i.e. 101,102,103,...,100+2n) always contains at least one integer with a least prime factor $p > n$.

I had been thinking about it for some time when I noticed this post: question in prime numbers

Using complete residue systems for $p$# and the Chinese Remainder Theorem, I noticed that it is possible to find a sequence of $38$ (2*19) consecutive integers that has only one least prime factor greater than $19$. It is also true of a sequence of $62$ (2*31) that you can find a sequence with only one least prime factor greater than $31$.

For any sequence where $n \le 23$ with the exception of $19$, there are at least two integers where the least prime factor is greater than $2$.

The fact that any sequence of $46$ (2*23) integers always has at least $2$ such integers while a sequence of $38$ (2*19) integers might just have $1$ suggested to me that there might be an elementary argument there.

I did a Google search and could find very little on least prime factors. As you can probably tell from this write up, I am an amateur very interested in learning more about number theory.

I would greatly appreciate any help that can be provided in showing me a proof or an example that there exists integers $x,n$ such that $x+1, x+2,$ ... $, x+2n$ are all divisible by some prime $p \le n$

Thanks very much

-Larry