2
$\begingroup$

It is known that $$ \exp\left\{k(1/1.71\cdots+o(1))\right\} < H(k) < \exp\left\{k(1/1.13862\cdots+o(1))\right\}, $$ where $H(k)$ is the $k^{th}$ highly composite number.

Question: Does the number of divisors of the highly composite numbers $d(H(k))$ achieve the growth rate given by $$ \lim_{n \rightarrow \infty} \sup \frac{\log d(n)}{\frac{ \log n}{\log \log n}}=\log 2? $$ To be more precise let $n_k=H(k).$ Does $$ \lim_{k \rightarrow \infty} \sup \frac{\log d(n_k)}{\frac{ \log n_k}{\log \log n_k}}=\log 2? $$

$\endgroup$
10
  • $\begingroup$ highly composite number? $\endgroup$
    – Wlod AA
    Jan 4, 2020 at 3:06
  • 1
    $\begingroup$ @WlodAA see this: en.wikipedia.org/wiki/Highly_composite_number $\endgroup$ Jan 4, 2020 at 3:26
  • 1
    $\begingroup$ In brief: "A highly composite number is a positive integer with more divisors than any smaller positive integer has." $\endgroup$ Jan 4, 2020 at 4:00
  • 1
    $\begingroup$ @WlodAA yes. OEIS A002182. $\endgroup$
    – kodlu
    Jan 4, 2020 at 4:51
  • 3
    $\begingroup$ The wikipedia link has a link to Erdos paper. The Lemma 1 of that paper implies that the limsup in your question is positive. $\endgroup$ Jan 4, 2020 at 5:04

1 Answer 1

4
$\begingroup$

Applying Lemma 1 from Erdos paper, we are able to obtain $$ \limsup_{k\rightarrow\infty} \frac{\log d(n_k)}{\frac{ \log n_k}{\log \log n_k}}\geq \frac16\log 2. $$ Here's the proof. Let $N$ be a highly composite number other than 4, 36, we have the prime factorization of $N$ as $$ N=2^{a_2}3^{a_3}\cdots p^{a_p}, \ \ a_2\geq \cdots\geq a_p=1. $$ Then $$ d(N)=(a_2+1)(a_3+1)\cdots(a_p+1) \geq 2^{\pi(p)} $$ Taking logarithms, $$ \log d(N)\geq \pi(p) \log 2. $$ Lemma 1 from Erdos paper states $p>c\log N$ for some positive $c$. But, by following his argument there, we can see that $c$ can be taken as $1/6-\epsilon$. (see Proof of Lemma 1)

Then it follows that $$ \log d(N) \geq (1/6-\epsilon)\frac{\log N \log 2}{\log \log N}. $$ Therefore, we have the result.

Remark

We actually have $$ \lim_{k\rightarrow\infty} \frac{\log d(n_k)}{\frac{\log n_k}{\log\log n_k}} = \log 2. $$ The proof is given in this 66 page paper by Ramanujan. More specifically, see section 32 to 39.

$\endgroup$
3
  • $\begingroup$ It seems that the final Remark is a complete answer to the question in the OP, and the rest is probably best deleted. $\endgroup$ Jan 5, 2020 at 19:53
  • $\begingroup$ @GregMartin It might make sense to have it after the Remark, but it seems relevant for giving a quick and easy bound that is almost as good. $\endgroup$
    – JoshuaZ
    Jan 5, 2020 at 20:05
  • $\begingroup$ @GregMartin, Thanks for your comment. I actually find the simple argument quite nice, so I'll accept the answer as it is. $\endgroup$
    – kodlu
    Jan 6, 2020 at 5:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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