5
$\begingroup$

Is it theoretically possible for there to always be a twin prime pair between $n$ and $2n$ for all sufficiently large $n$ (assuming of course that there are infinitely many twin primes) or would this contradict Brun's Theorem? Thanks.

$\endgroup$
3
  • $\begingroup$ I think the question can be rephrased as: let $S\subseteq\Bbb N$ contain a natural number between $n$ and $2n$ for every $n$. Is it true that $\sum_{s\in S} \frac1s$ diverges? $\endgroup$
    – Myself
    Commented Dec 21, 2014 at 0:37
  • $\begingroup$ ... in which case the answer would be no, because we could take $S = \{2^i\mid i\in\Bbb N\}$. In other words if by accident all twin prime pairs were (roughly) $2^i\pm 1$, maybe twice as much just to be safe, there would be a pair between every number and its double and Bruns theorem would still hold. $\endgroup$
    – Myself
    Commented Dec 21, 2014 at 0:42
  • $\begingroup$ I agree with your above analysis, but I was also wondering if Brun's constant is too small to allow for this. $\endgroup$
    – Ari
    Commented Dec 21, 2014 at 0:46

1 Answer 1

5
$\begingroup$

Brun's theorem follows from the bound $$ \pi_2(x) = O\left(\frac{x(\log\log x)^2}{(\log x)^2}\right), $$ where $\pi_2(x)$ is the number of twin primes less than $x$. On the other hand, the existence of twin primes between each $n$ and $2n$ only implies $\pi_2(x) = \Omega(\log x)$ (as Myself notes in the comments), so there is no contradiction.

Moreover, it is conjectured that $\pi_2(x) \sim 2C_2 \displaystyle\frac{x}{(\log x)^2}$ for some constant $C_2$. This implies that for large enough $n$, there exist many twin primes between each $n$ and $2n$ (in fact, roughly $C_2 \displaystyle\frac{n}{(\log n)^2}$).

$\endgroup$

You must log in to answer this question.

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