Question: Let $d(n)=\sum_{d|n}1$, be the divisor function. Estimate the asymptotic behaver $$D(N):=\max\{d(n):1\leq n\leq N\}$$ when $N$ is large.
I know $$\sum_{n=1}^N d(n)=\sum_{n=1}^N [\frac{N}{n}]=N\sum_{n=1}^N\frac{1}{n} + O(N)\approx N{\rm log} N.$$ So $D(N)\geq \frac{1}{N}\sum_{n=1}^N d(n)\approx {\rm log}N$. But I can't determine if this is optimal, i.e. $D(N)=O({\rm log}N)$.
Wigert showed the bound $$ d(n) \leq e^{(\log 2 + o(1))(\log n / \log \log n)}.$$ There are other ways of approaching this bound, but I think it's not too hard to show bounds of this shape. We'll do this in two ways.
Let $P_m = \prod_{j \leq m} p_j$ denote the $m$th primorial, the product of the first $m$ primes. One way to get numbers with lots of divisors would be to consider the primorials. These aren't necessarily the best, but the number of divisors of $P_m$ is $2^m$, a very easy number to understand.
The prime number theorem implies that $P_m \sim e^{O(1) m / \log m}$. Thus $P_m \lesssim n$ if $m \lesssim \frac{\log n}{\log \log n}$. Such a $P_m$ will have at least $$ 2^m \sim 2^{O(1) \log n / \log \log n}$$ divisors, which is of the claimed size.
Instead of using primorials as approximations for numbers with maximal numbers of divisors (which they aren't, actually), we can be more precise with a bit of extra work. I repeat this analysis from Terry Tao, though it appears in various places in the literature.
Suppose $$ n = \prod p_j^{\alpha_j}.$$ Then $$ d(n) = \prod (1 + \alpha_j). $$ Consider the ratio $$ \frac{d(n)}{n^\epsilon} = \prod \frac{1 + \alpha_j}{p_j^{\alpha_j \epsilon}} $$ for an $\epsilon > 0$ that we will optimize later. For each prime $p_j$, the factor $\frac{1 + \alpha_j}{p_j^{\alpha_j \epsilon}}$ will be small when $p_j$ is large (in comparison to $\epsilon$). When $p_j > O(1 / \epsilon)$, for example, we find that $$ p_j^{\alpha_j \epsilon} > e^{\alpha_j} \geq 1 + \alpha_j, $$ and thus for such primes $$ \frac{1 + \alpha_j}{p_j^{\alpha_j \epsilon}} \leq 1. $$
For small primes, when $p_j < O(1 / \epsilon)$, Taylor expansion show that $$ p_j^{\alpha_j \epsilon} = \exp(\epsilon \alpha_j \log p_j) \geq 1 + \epsilon \alpha_j \log p_j. $$ Thus for such primes, $$ \frac{1 + \alpha_j}{p_j^{\alpha_j \epsilon}} \leq \frac{1 + \alpha_j}{1 + \alpha_j \epsilon \log p_j} \leq O\left( \frac{1}{\epsilon \log p_j}\right),$$ where the implicit constant does not depend on $\alpha_j$.
Thus in total, we find that $$ \frac{d(n)}{n^\epsilon} \ll \prod_{p < O(1/\epsilon)} O\left(\frac{1}{\epsilon \log p_j}\right) \ll O\left( \frac{1}{\epsilon} \right)^{\exp(1/\epsilon)} = \exp(\exp(O(1/\epsilon))). $$ (We bound $\log p_j$ by a constant $\log 2$ trivially, and the number of primes trivially). Therefore we find that $$ d(n) \ll n^{\epsilon} \exp (\exp (O(1/\epsilon))). $$ Choosing $\epsilon = C / \log \log n$ for sufficiently large $C$ causes $n^{\epsilon}$ to dominate, giving the claimed bound.
This shows an upper bound.
This answer is intended to prove the result that
$$ \limsup_{n\to\infty}{\log d(n)\log\log n\over\log n}=\log2.\tag1 $$
Let $n_k$ be the product of first $k$ primes, so that $d(n_k)=2^k$. By Chebyshev's inequality (i.e. the weak prime number theorem), we know there exists $c>0$ such that for large $k$,
$$ \log n_k=\vartheta(p_k)=\sum_{p\le p_k}\log p>cp_k. $$
This indicates that
$$ \log\log n_k>(1+o(1))\log p_k. $$
Moreover, since $\vartheta(p_k)\le\pi(p_k)\log p_k=k\log p_k$, we have
$$ {\log d(n_k)\over\log2}=k={k\log p_k\over\log p_k}>(1+o(1)){\log n_k\over\log\log n_k}. $$
This inequality indicates that
$$ \limsup_{n\to\infty}{\log d(n)\log\log n\over\log n}\ge\log2. $$
To turn this one-sided result into an equality, we need to obtain a general upper bound for $d(n)$:
Suppose $n=\prod_{p|n}p^{\alpha_p}$. Then by definition, we can choose $2\le t\le n$ so that
\begin{align} d(n) &=\prod_{p|n}(1+\alpha_p)=\prod_{\substack{p|n\\p\le t}}(1+\alpha_p)\prod_{\substack{p|n\\p>t}}(1+\alpha_p) \end{align}
For all $p|n$, we know for certain that $\alpha_p\le\log n/\log2$ with equality iff $p=2$. As a result, we have
$$ \prod_{\substack{p|n\\p\le t}}(1+\alpha_p)\le\left(1+{\log n\over\log2}\right)^t $$
When product over $p|n\wedge p>t$, we have
\begin{aligned} \prod_{\substack{p|n\\p>t}}(1+\alpha_p) &\le\prod_{\substack{p|n\\p>t}}2^{\alpha_p}=\prod_{\substack{p|n\\p>t}}p^{\alpha_p\log2/\log p} \\ &<\left(\prod_{p|n}p^{\alpha_p}\right)^{\log2/\log t}=\exp\left(\log2{\log n\over\log t}\right) \end{aligned}
Putting them all together gives
$$ \log d(n)<t\log\left(1+{\log n\over\log2}\right)+\log2{\log n\over\log t}<\log2{\log n\over\log t}+O(t\log\log n) $$
To continue, set $t=e^{-u}\log n$ so that
$$ \log d(n)<\log2{\log n\over\log\log n-u}+O(e^{-u}\log n\log\log n) $$
To continue, set $u=3\log\log\log n$, so we conclude that
$$ {\log d(n)\over\log2}<{\log n\over\log\log n}\left\{1+O\left(\log\log\log n\over\log\log n\right)\right\} $$
Combining this with the aforementioned inequality, we finish the proof for (1).
