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.](https://en.wikipedia.org/wiki/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
$$
\lim\_{k \rightarrow \infty} \sup \frac{\log d(n\_k)}{\frac{ \log n\_k}{\log \log n\_k}}=\log 2?
$$



Divisor Function of Highly Composite Numbers
Applying Lemma 1 from Erdos [paper](https://www.renyi.hu/~p_erdos/1944-04.pdf), 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](http://ramanujan.sirinudi.org/Volumes/published/ram15.pdf) by Ramanujan. 
More specifically, see section 32 to 39. 


