Is there a real number $A$ such that $$\left \lfloor n^{A} \right \rfloor$$ is a prime number (for all natural numbers $n$)? It is obvious that $A>1+\epsilon$ from the prime number theorem.



A more dense analog of the Mills' constant
No, such an $A$ does not exist. First, $A$ cannot be an integer because then $\lfloor n^A\rfloor $ is never a prime for $n\geq 2$. So, assume that $A$ is not an integer. Then by Weyl's equidistribution theorem the fractional parts of $n^{A}/2$ are equidistributed modulo $1$. In particular $\{n^A/2\}\in [0,1/2)$ infinitely often. So $\lfloor n^{A}\rfloor$ is infinitely often an even number.


