Define $$f(n):=\frac{\sum_{j=1}^n j^2\cdot j!}{99}$$
Is $f(n)$ prime for only finite many positive integers $n\ge 10$ ?
Approach : If we find a prime number $q>11$ with $q\mid f(q-1)$ , then for every $n\ge q$ , we have $q\mid f(n)$ , therefore only finite many prime numbers of the form $f(n)$. I found however no such prime upto $10^5$.
Motivation : Since $$\sum_{j=1}^n j\cdot j!=(n+1)!-1$$ the search for prime numbers of this form is already covered by the factorial prime search, therefore I chose to try my luck with $$\sum_{j=1}^n j^2\cdot j!$$ for which Wolfram alpha knows no closed form representation. But for $n\ge 10$ , this number is divisible by $99$ , hence the above expression.
$f(n)$ is prime for $n=11,15,37,119,185,221,698,1574,1830$ , the number of digits is $8, 13, 45, 199, 344, 427, 1688, 4355, 5183$ respective. Upto $n=221$ , the primality is proven.