I was thinking about an elementary question
For some given $k\in \mathbb N$, find $k$ consecutive composite integers.
The answer that is generally expected is
The following $k$ integers are consecutive and composite $$(k+1)!+2\\(k+1)!+3\\(k+1)!+4\\.\\.\\.\\(k+1)!+(k+1)$$
While this is a perfectly fine answer, I was thinking about how big these numbers will get even for very small values of $k$. For example, the first three consecutive numbers (found by trial and error) are $$8,9,10$$ and the first four consecutive composite numbers are $$24,25,26,27$$ but the sets we obtain through taking factorials are $26,27,28$ and $122,123,124,125$ respectively. So I was thinking of trying to make these numbers smaller.
One way I thought was to use the $\operatorname{L.C.M.}$ of these numbers instead, i.e., saying
The following $k$ integers are consecutive and composite $$\operatorname{L.C.M.}\{1,2,\dots ,k+1\}+2\\\operatorname{L.C.M.}\{1,2,\dots ,k+1\}+3\\\operatorname{L.C.M.}\{1,2,\dots ,k+1\}+4\\.\\.\\.\\\operatorname{L.C.M.}\{1,2,\dots ,k+1\}+(k+1)$$
While this bound gives us the correct answer for finding the first three consecutive composite integers, for four, it generates the set $62,63,64,65$. It is quite clear that even this one will shoot farther away from the first few numbers with increasing $k$.
So, the question is, is it possible to point out exactly the first few composite numbers? If not, then how close can we get to it?
