3
$\begingroup$

  • Using my computer, I found that in the interval $[1, N]$ the probability of finding a prime number between $x-\ln(x)$ and $x+\ln(x)$ is greater than constant $c$ where $N=10^2, 10^3,...,10^{9}$, $x$ is the positive integer number in $[1, N]$ and $c > 0.8$.

  • Question: Does the result hold for $N \to +\infty $?

Improve this question
$\endgroup$
1
  • 13
    $\begingroup$ The number of primes in such an interval is expected to be Poisson with parameter $2$. Thus the probability is $1-e^{-2} \approx 0.86$. We don't know how to prove that. $\endgroup$
    – Lucia
    Commented May 4 at 4:04

1 Answer 1

Reset to default
13
$\begingroup$

It is known that the probability in question exceeds a positive constant for $N>N_0$, and it is conjectured that it tends to $1-e^{-2}\approx 0.864665$ as $N\to\infty$. I suggest to read the introduction of this paper by Teräväinen for an overview of this research area.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Is this specific to this interval $[x-\ln x,x+\ln x]$ or is there something similar for any interval, say, $[x-c\ln x,x+c\ln x]$ for $c>0$? $\endgroup$
    – YCor
    Commented May 4 at 7:46
  • 4
    $\begingroup$ @YCor The introduction of that paper answers that too. It is known that the probability is positive for all large enough $N$, and is conjectured to tend to $1-e^{-2c}$. $\endgroup$
    – Wojowu
    Commented May 4 at 8:24
  • 1
    $\begingroup$ @YCor A very simple observation that helped me see what's going on: The prime number theorem says that, near $x$, the density of primes is approximately $x/(\ln x)$. So in your interval of length $2c\ln x$, we can expect approximately $2c$ primes. That "explains" the factor $2$ in the comments by Lucia and Wojowu. That the distribution should be Poisson (in Lucia's comment) seems to be another instance of the distribution of primes being random unless one can find a reason against randomness. $\endgroup$
    – Andreas Blass
    Commented May 4 at 17:29
  • 1
    $\begingroup$ @AndreasBlass The density is not $x/(\ln x)$ but $1/\ln x$. $\endgroup$
    – GH from MO
    Commented May 4 at 18:12
  • 2
    $\begingroup$ @GHfromMO Thanks for the correction, which fortunately also makes the next sentence correct. $\endgroup$
    – Andreas Blass
    Commented May 4 at 18:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .