For the rest of this chapter, we will focus on \(\sigma_1=\sigma\) itself, since the sum of divisors function has a deep richness of its own. This short section asks a particularly interesting question.

We can ask about things like evenness or other patterns you may have encountered, of course. Let's look at a different area for exploration.

As you manipulate the Sagelet above, try to see what possibilities there are for the relative size of \(\sigma(n)\) with respect to \(n\) itself. (Or, one can ask what the constant \(C_n\) is such that \(\sigma(n)=C_n\cdot n\).)

Quite a few certainly are above and below \(2\). If you look carefully, you will see that only one of the numbers above has a sum of divisors without 1 or 2 as the integer part. What is it?

Another thing we can do is take big powers of numbers with lots of small prime divisors (and hence lots of factors). Such numbers might get us big ratios.

You'll notice that although we quickly go above 3, but don't seem to get much further. Why?

A helpful thing to think about with this is the following rewrite, using the formula for \(\sigma(n)\): \[\frac{\sigma(n)}{n}=\frac{\prod_{i=1}^k \left(\frac{p_i^{e_i+1}-1}{p_i-1}\right)}{\prod_{i=1}^k p_i^{e_i}} = \prod_{i=1}^k\frac{p_i-1/p_i^{e_i}}{p_i-1} \approx \prod_{i=1}^k\frac{p_i}{p_i-1}\, ,\] and we should expect this approximation to be very close when \(e_i\) are all quite large. So the idea is that since \(\frac{p}{p-1}>1\), if we multiply by enough of these we will get very large numbers and so \(\sigma(n)/n\) will be greater than any given \(C\), and then \(\sigma(n)>Cn\).

Of course, \(p=2\) is the best for this since \(\frac{2}{2-1}=2\), but the others will hopefully be useful. For instance, \(n=2^{10} 3^{10}\) will have \[\sigma(n)/n=\frac{2-1/2^{10}}{2-1}\frac{3-1/3^{10}}{3-1} \approx \frac{2}{2-1}\frac{3}{3-1}=3\] so certainly \(\sigma(6^{10})\) will be nearly \(3\cdot 6^{10}\).

If we multiply it by 5 as well that should do it, and that gives the results we saw in the previous cell. \[\frac{2-1/2^{10}}{2-1}\frac{3-1/3^{10}}{3-1}\frac{5-1/5}{5-1}\approx\frac{2}{2-1}\frac{3}{3-1}\frac{5}{5-1}=2\cdot \frac{3}{2}\cdot \frac{5}{4}=\frac{15}{4}=3.75\; .\]

Continuing this for more primes suggests the following fact.

The argument above not completely rigorous, but good enough for now. It turns out that trying to prove it this way could bring the distribution of primes to the table. Later we might see a more formal proof.