Let $$f(n)=\sum_{j=1}^n (j!)^{j!}$$ for integer $n\ge 1$
The first few values are $f(1)=1 , f(2)=5 , f(3)=46661 , f(4)=1333735776850284124449081472890437$
- Is $f(n)$ a prime number for any $n>2$ ?
The smallest prime factors of $f(n)$ for $n=3,\cdots, 13$ , except $n=10$ are
3 29
4 7
5 41
6 241
7 43
8 3593
9 17
11 61
12 13
13 13
For $n\ge 12$ , $13$ is a prime factor, so the last possible prime number is $f(10)$. It has $$23\ 804\ 069$$ digits.I found no prime factor upto $3\cdot 10^{10}$ (doublecheck would be fine!). $f(10)$ is prime with probability about $1$ to $1$ million, but a prime factor would be very welcome !
- Are the $f(n)$ squarefree except $f(21$) ?
The only non-squarefree case I found so far is $$83^2\mid f(21)$$ Is there any other pair $(n,p)$ with prime $p$ and $p^2\mid f(n)$ ?
Is $f(n)$ a semiprime for any $n>3$ ? The next possible case (apart from $f(10)$) is $n=34$ (which is almost surely NO semiprime) and for $n\ge 72$ , we have $13\cdot 73\mid f(n)$ so a semiprime is not possible anymore.
The complete factorization of $f(5)$ would be very welome.here is the partial factorization. The remaining composite has $193$ digits.