What positive integer $n$ maximizes the function $f(n) = \sigma_0(n)/n$? A friend and I were having a discussion on which base would be best. I argued that 12 would be the best because it has the most divisors relative to its size. However, I'm not sure that 12 is actually the number that maximizes this ratio. To investigate, I formalized my observation by claiming that 12 maximizes the function $f(z) = \sigma_0(z)/z$ where $\sigma_0(n) = \sum_{d|n} d^0$ is the function which counts the divisors of $n$. I found some articles and some interesting properties of $\sigma_0$ but nothing which I was able to use to prove this property. I'm not too familiar with this kind of thing so I wasn't sure exactly how to go about it.
Does anyone have any idea as to how one could prove this? Right now, it seems like the formula which would be most useful would be that $$\sigma_o(n) = \Pi_{i = 1}^{\omega(n)}(1 - a_i)$$ where $\omega(n)$ is the number of distinct prime factors of $b$ so that $n = \Pi_{i = 1}^{\omega(n)}p_i^{a_i}$.
Thank you in advance!
EDIT: On thinking about it a little more, it seems like 12 definitely does not maximize this. For example, 6 has 4 divisors whereas 12 has 6 of them. As a commenter also pointed out, 3 has 2 divisors. The best does seem to be 2, though, with two divisors. If $\sigma_0(n) = n$, then for all $m \leq n$, we would have that $m|n$. That would imply that every prime less than $n$ would be included in the prime factorization of $n$. This is a fairly strong property that I suspect only 2 holds.
 A: First note that $\displaystyle \frac{\sigma_0(n)}{n} = \prod_p \frac{\alpha_p+1}{p^{\alpha_p}}$ where $\alpha_p \ge 0$.
But $\displaystyle\frac{\alpha_p+1}{p^{\alpha_p}} < 1$ for all prime $p$ and all $\alpha >0$ with the only exception of $p = 2$ and $\alpha = 1$, which means that the maximum is achieved when all the $\alpha$'s are $0$ ($n = 1$) or when all but $\alpha_2$ are $0$ and $\alpha_2 = 1$ (n = 2).
A: From this answer, we know that
$$\sigma_0(n)\leq n^{\frac{1.0660186782977...}{\log \log n}}<n^{ \frac{2}{\log \log n}}$$
(with equality at $n=6983776800$). This then implies that
$$\frac{\sigma_0(n)}{n}<n^{ \frac{2}{\log \log n}-1}$$
Now, it is easy to see that for $n\geq 1619$ we have
$$\frac{2}{\log \log n}-1<0$$
Then for $n\geq 1619$ we know
$$\frac{\sigma_0(n)}{n}<n^{ \frac{2}{\log \log n}-1}<n^0=1$$
But
$$\frac{\sigma_0(1)}{1}=\frac{\sigma_0(2)}{2}=1$$
Now, we only have to check all integers $3\leq n\leq 1618$. These are easily checked and we conclude that the function is maximized at $n\in\{1,2\}$.

EDIT: If you wanted the case $n\geq 3$, then in much the same manner we see that for $n\geq 2880$ we have
$$n^{\frac{2}{\log \log n}-1}<\frac{3}{4}$$
Then after checking all integers $5\leq n\leq 2879$ we may conclude that the function is maximized at $n=4$ where
$$\frac{\sigma_0(4)}{4}=\frac{3}{4}$$
