Sum of square root of primes

Question

I was playing around with prime numbers and a question came into my mind:
Let $S(n)$ denote the sum of square roots of primes from $2$ to the $n$th prime number. Are there infinitely many numbers $n$ so that $\left\lfloor S(n) \right\rfloor$ is prime itself? (Where $\left\lfloor X \right\rfloor$ denotes the floor function.) Please tell me if you had any ideas about it. I actually could do nothing. xD

Answer 1

If this problem is solvable, it would require advanced tools from analytic number theory. As for a heuristic: $S(n)$ is asymptotically $\sum_{k=1}^n \sqrt{k\log k} \sim \frac23n^{3/2}\sqrt{\log n}$; and I don't see any reason why $\lfloor S(n)\rfloor$ is more or less likely to be even, a multiple of $3$, or more generally divisible by any fixed prime. (This independence is borne out by numerical experiments.) Thus the prediction would be that $\lfloor S(n)\rfloor$ is just as likely to be prime as a random integer of the same size, which is about $1/\log S(n) \sim 2/(3\log n)$. In other words, I would predict that the number of $n\le x$ for which $\lfloor S(n)\rfloor$ is prime should be asymptotically $\frac23x/\log x$. (Numerical experiments also make this appear more likely than a constant times $x/\log^2x$.)