It is known for any $n \gt 1$ there will be some prime in the range $(n/2,n]$ which will only occur once in the factorization of $n!$ by Bertrand's Postulate, my question here is to know more about primiality test of the integer part of square root of n!
Question Could :$ \lfloor {\sqrt{n!}}\rfloor $ a prime number ?
Note: according to some computations here which i did i don't got an example only $n=3$ ,and i think it's a rare to get primes of the titled form
Note: The motivation of this question is to check the prime factorization of $n!-1$
$\lfloor {\sqrt{53!}}\rfloor=65382591597917144387816492317568177$ is prime. Check it here! http://factordb.com/index.php?query=65382591597917144387816492317568177
Edit: According to the Prime Number Theorem, the probability of random number $N$ being a prime is approximately $1/\log{N}$. If we assume the sequence of number $\lfloor {\sqrt{n!}}\rfloor$ is random, then probability it being a prime is$$\frac1{\log\lfloor {\sqrt{n!}}\rfloor}\approx \frac1{\log\sqrt{n!}}=\frac2{\log{n!}}=\frac2{n\log n-n+O(\log n)}>\frac2{n\log n}$$ Since (expected total number of primes in the sequence) $\ge\sum_1^{\infty}\frac2{n\log n}$ diverges, there is likely to be infinitely many prime numbers in the sequence $\lfloor {\sqrt{n!}}\rfloor$, and by exactly same argument, there are likely to be infinitely many Mersenne primes, primes of the form $n!\pm1$, etc.