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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 $n!-1$

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    If I'm not missing something, isn't $n=3$ a simple example, where you get $2$? Maybe you've underspecified your question (perhaps you mean "is it a prime infinitely often?" or something similar). – 2017-02-17
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    I think it's probably rare, maybe never - but a computation as far as $n=30$ is not a lot of evidence for never. – 2017-02-17
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    yes, the only is n=3 , just a wrong i go to edit it, by the way i meant the primes of this form are rare – 2017-02-17
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    For $n \leqslant 1000$: $3,11,14,53,110,216,322,364,389$, where primality is certain for $3,11,14,53$, and probable for the others (the numbers passed a BPSW test). – 2017-02-17
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    Yes, because $\lfloor\sqrt{110!}\rfloor$ is prime! – 2017-02-17
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    Cf. https://oeis.org/A273869 – 2017-06-04

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$\lfloor {\sqrt{53!}}\rfloor=65382591597917144387816492317568177$ 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.

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    Is there any good heuristic for $\lfloor \sqrt{n!} \rfloor$ to be more or less "random"? – 2017-02-18
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    @Ian, there's no evidence of Mersenne numbers, $n! \pm 1$, or $\lfloor \sqrt{n!} \rfloor$ is random. – 2017-02-18