Inspired by this quite simple question, I tried to play with the formula.
$(n+1)!+(i+1);\;1\le i\le n$
Which gives at least $n$ consecutive composite numbers. But I've found that very often they are more than $n$.
For $n=3$ we have $3! +2=8; 3!+3=9;3!+4=10$ there are exactly $3$ composite consecutive numbers
But for $n=11$ I've found $15$ consecutive composite numbers. For $n=27$ there are $45$.
I wonder if there is a way to find $n$ such that there are exactly $n$ composite consecutive numbers.
Consider the sequence $$2+(n+1)!\space,\space 3+(n+1)!\space,\space ... \space,\space (n+1)+(n+1)!$$
Then for the first term, you can factor a $2$, for the second a $3$, etc. Thus there are $(n+1)-2+1=n$ terms in the sequence which are all consecutive and composite numbers.