In this post , the function $$f(n):=\sum_{j=1}^n j!^2$$ is mentioned. $f(n)$ seems to be squarefree for every positive integer $n$.
Do we have $n+1\mid f(n)$ for some positive integer $n$ ? The conjecture is that the answer to this question is "no" : For no positive integer $n\le 10^5$ , this divisibility holds.
Motivation : I want to show that the smallest prime factor of $f(n)$ must exceed $n$. If $p$ is a prime number with $p\le n$ and if we have $p\mid f(n)$ , then we can easily conclude $p\mid f(p-1)$ , so this has to be ruled out. It would be enough for this purpose to show the above conjecture for the case that $n+1$ is prime , but I think , if there is a proof then the proof covers all positive integers.
Any ideas ?