6
$\begingroup$

One motivation for this question is a paper by Erdos and Hall "Values of the divisor function on short intervals", in which the authors obtain the leading asymptotics

$$x(\log x)^{2(\sqrt 2-1)-\epsilon}\leq\sum_{n\leq x}\min{(d(n),d(n+1))}\leq x(\log x)^{2(\sqrt 2-1)}$$ and

$$\sum_{n\leq x}\max{(d(n),d(n+1),...,d(n+k-1)})\sim kx\log x.$$ In particular, it is possible that the answer to this question will lead to refinements of these bounds. It is also interesting simply as a curiosity.

Let $r(n)=d(n+1)/d(n)$. Heath-Brown proved that $r(n)=1$ for infinity many $n$, but I don't know how often this is expected to happen. Recent advances in knowledge of the gaps between primes suggests that small and large values are common.

  1. What is known about the distribution of values of $r(n)$?

  2. In particular, is it known that $r(n)$ is neither bounded above or below?

EDIT: $r(n)$ is neither bounded above or below - the proof is easy so I suppose that's not news though.

$\endgroup$
  • $\begingroup$ It might be interesting to study the ratio $ \log d(n+1)/\log d(n) $ instead of $ r(n) $ . $\endgroup$ – Sylvain JULIEN Feb 13 '17 at 18:19
4
$\begingroup$

Under GRH, Titchmarsh showed 1931 that $\sum_{p\leq x}\tau(p+a)\sim C(a)x$, where summation runs over primes only. 1963 Linnik proved the same unconditionally, hence there are many $n$ such that $d(n)=2$, $d(n+1)\gg\log n$ and similarly for $n-1$. Hence $r(n)$ is quite often as big as $\log n$ and as small as $\frac{1}{\log n}$.

In general $r(n)$ is much closer to 1. Using some standard techniques in probabilistic number theory you can show that the joint distribution of $(\omega(n), \omega(n+1))$ is normal with mean and variance $\log\log x$, hence the distribution of $\log r(n)$ is normal with variance $2\log\log n$.

You can construct extreme values of $r(n)$ by taking an integer $n$ with many divisors, and look for a prime $p\equiv \pm 1\pmod{n}$. By Heath-Browns version of Linnik's theorem, for $n>n_0$ such a prime $p$ exists satisfying $p\leq n^{5.5}$, hence $\max_{p\leq x} d(p+1)\geq \max_{n\leq x^{2/11}}d(n)$, thus for some small $c>0$ we have that $r(n)>e^{c\log n/\log\log n}$ has infinitely many solutions.

$\endgroup$
  • $\begingroup$ When you say $r(n)$ is much closer to $1$, do you know of some estimate that describes how much closer? I'm considering the average value of $e^{-|\log r(n)|/2}$, and my guess is also that this is $>1/2$, perhaps even $\sim \gamma$, based on calculations up to $10^8$. $\endgroup$ – Kevin Smith Feb 19 '17 at 21:15
  • $\begingroup$ $\log r(n)$ has a normal distribution with mean $2\log\log n$, so if $\omega(n)\rightarrow\infty$, then $|r(n)|<\omega(n)\sqrt{\log\log n}$ for a set of density 1. Note that the events $p|n$ and $p|n+1$ are negatively correlated, so as long as $\log\log n$ is not much larger than 1, $\omega(n)$ and $omega(n+1)$ do not look independent. Also $e^{-\sqrt{\log\log n}/2}$ looks pretty constant for all $n$ accessible by a computer. $\endgroup$ – Jan-Christoph Schlage-Puchta Feb 20 '17 at 21:41
  • $\begingroup$ Thank you , Jan-Christoph. Do you mean "$|\log r(n)|<\omega(n)\sqrt{\log\log n}$ for a set of density $1$? Would you clarify what you mean by "negatively correlated" and "do not look independent", please. $\endgroup$ – Kevin Smith Feb 21 '17 at 15:37
  • $\begingroup$ Of course I understand that the events $p|n$ and $p|(n+1)$ cannot both happen. But the number of primes less than $n^{1/\loglog n}$ is still large compared to $\log\log n$. $\endgroup$ – Kevin Smith Feb 21 '17 at 15:50
  • $\begingroup$ The expectation of $\omega(n)\omega(n+1)$ is $\sum_{p, q: p|n, q|n+1}1$ which is smaller than the expectation of $\omega(n)$ squared, since the latter contains an additional diagonal sum. However, the difference is smaller order than the non-diagonal main term, so the joint distribution of $\omega(n)$ and $\omega(n+1)$ is asymptotically independent with normal marginal distributions (see en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Kac_theorem ). However, for small $x$ the sum of magnitude $\sum_p\frac{1}{p^2}$ is not negligible compared to $\log\log x$, so numerics can be misleading. $\endgroup$ – Jan-Christoph Schlage-Puchta Feb 21 '17 at 17:18

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.