elementary number theory - sum of factorials dividing the factorial of the sum

The coefficients of the multinomial formula are integers where the numerator is the factorial of a sum of integers and the denominator is the product of the factorials of those same integers.

Suppose instead of taking a product of those factorials in the denominator we took the sum of those factorials. When would that expression be an integer?

Based on some calculations, I suspect, but I cannot prove in general, that if the integers are consecutive forming the sum, then there exists a non-negative integer $m$ depending on the number of integers $k$ in the sum such that if $n \ge m$ then the expression is always an integer.

In other words, there exists $m$ depending on $k$ such that for all $n \ge m$ the following is always an integer:

$$\frac{\left(\sum_{i=n}^{n+k-1}i\right)!}{\sum_{i=n}^{n+k-1}i!}$$

For $k = 1, m = 0$. For $k = 2, m = 1$. Based on some calculations, I conjecture, but I have no proof, for $k = 3, m = 1$ and for $k = 4, m = 4$.

Is this type of problem familiar to anyone? Perhaps someone has a solution to it or suggestions where I could go for more information.

Answer

In the $k=4$ case,
$$\frac{\left(\sum_{i=n}^{n+k-1}i\right)!}{\sum_{i=n}^{n+k-1}i!} = \frac{(4n+6)!}{n!(n+2)(n^2+5n+5)}.$$

For a lot of (likely infinitely many) values of $n$, $n^2+5n+5$ is a prime larger than $4n+6$, and so the expression is not an integer for those $n$.

(In the $k=3$ case, we get

$$\frac{(3n+3)!}{n!(n+2)^2}$$
and since $2(n+2)<3n+3$, this simplifies to an integer for $n\ge1$. Similarly, with $k=2$, we have
$$\frac{(2n+1)!}{n!(n+2)}$$
which again simplifies to an integer for $n\ge1$.)