For any integer $n$, the factorial $n!$ is the product of all positive integers up to and including $n$. Then in the sequence
$$n!+2,n!+3,… ,n!+n$$
the first term is divisible by $2$, the second term is divisible by $3$, and so on. Thus, this is a sequence of $(n − 1)$ consecutive composite integers, which definitely not coprime with $n!$.
Question: Is this the longest sequence of consecutive integers which are not coprime with $n!$ (less than $n!$)?
On the other words, is $(n-1)$ is the length of the longest sequence of consecutive integers which less than $n$ factorial and not coprime with it? Or can we find longer?
Best Answer
Starting with $x=7$ and noting that $8,9,10$ all have factors in common with $7!$, we quickly get the counterexample $$n=7\quad \&\quad \{7!-2,\;7!-3,\;7!-4,\;7!-5,\;7!-6,\;7!-7,\;7!-8,\;7!-9,\;7!-10\}$$
which is a string of length $9$ (and all terms are less than $7!$)