9
$\begingroup$

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

$\endgroup$
  • $\begingroup$ One can denote the floor function at $x$ by $\lfloor x\rfloor$, coded as \lfloor x\rfloor. That seems to prevail today. I suspect the $[x]$ notation was formerly completely standard simply because typesetters didn't have little pieces of steel in a shape appropriate for anything like the notation usually favored now. $\endgroup$ – Michael Hardy May 17 '13 at 18:39
  • $\begingroup$ I think when you say "like $n$", you meant to give an example? $n=3$ would work. $\endgroup$ – Calvin Lin May 17 '13 at 18:56
  • 1
    $\begingroup$ It seems so. Experimentally I got a little less than $\frac {n}{\log(n)^2}$ primes for $n\le 10^8$ but proving it is another matter... $\endgroup$ – Raymond Manzoni May 17 '13 at 18:58
  • 1
    $\begingroup$ It's A062009 , with no other comment. $\endgroup$ – ama May 17 '13 at 19:14
6
$\begingroup$

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 $x/\log^2x$.)

$\endgroup$
  • $\begingroup$ In my comment I was considering the primes $p\le n$ (and not the $n$ first primes as asked...) getting the following table for $n=10^1$ to $10^{10}$ : $2, 7, 23, 125, 723, 4865, 34444, 254132, 1951646, 15466581$. So that your $x\sim \frac n{\log n}$ explaining the discrepancy and comforting your fine statistical analysis (+1). $\endgroup$ – Raymond Manzoni May 19 '13 at 8:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.