When does $\sum_{j=1}^n j$ divide $\prod_{j=1}^n j$?

Question

When does $\sum_{j=1}^n j$ divides $\prod_{j=1}^n j$ ?

What I did :

$$ Q=\frac{\prod_{j=1}^n j}{\sum_{j=1}^n j}=\frac{2\times(n-1)!}{n+1}$$

Now, by Wilson theorem, if $n+1$ is prime, $Q \notin \mathbb{Z}$.

My conjecture is that if $n+1$ is composite then $Q \in \mathbb{Z}$, so I write $n+1=ab$ with $1<a,b<n+1$. If $a \ne b$, then it is clear that $Q \in \mathbb{Z}$ as $a, b \leq n-1$ but what about the case $a=b$?

Thank you

Answer 1

If $n+1=a^2$ and $a \ge 3$, $n-1 \gt 2a$ and you get a factor of $a$ in $2(n-1)!$ from $a$ and another one from $2a$ so you have a integer. We just have to check $n=1,3$ by hand, where we get $\frac 11=1$ and $\frac 66$ to conclude it is an integer for all $n+1=a^2$

Answer 2

Suppose that $n+1=m^2$. Then

$$Q=\frac{2(m^2-2)!}{m^2}\;,$$

which will certainly be an integer if $2m\le m^2-2$, so that $m$ and $2m$ are amongst the factors of $(2m^2-2)!$. This will be the case if $m^2-2m-2\ge 0$, i.e., if $(m-1)^2-1\ge 0$, or $(m-1)^2\ge 1$, and hence if $m\ge 2$. Thus, your conjecture is correct. (I assume that you’ve already checked the two small cases not covered by this.)