Method to find colossally abundant numbers? Colossally abundant numbers are positive integers $n$ for which there exists a positive exponent $\epsilon$ such that
$$\frac{\sigma(n)}{n^{1+\epsilon}}>\frac{\sigma(m)}{m^{1+\epsilon}}$$
for all integers $m>1,m\ne n$. Here, $\sigma(n)$ denotes the sum-of-divisors function $\sum_{d|n}d$.
The first few colossally abundant numbers are $2,6,12,60,120,360...$ (from Wolfram Mathworld here).
My question is, how does one go about discovering such numbers, or proving that a number is colossally abundant? One can't test individually for all combinations of $n,m,\epsilon$, so there must be an algebraic method. What is it? (Google is no help.)
UPDATE
In light of @Mindlack's and @John Omielan's helpful comments below, and in order not to end up with an extended comment section, I thought it might be good to elaborate on my original question here.


*

*@John: Yes, I take your point, but it still sounds a lot like
searching for a needle in a haystack. Maybe that's what you're
trying to say?

*@Mindlack:


*

*OK, so setting $n=2$ gives $\frac{\sigma(n)}{n^{1+\epsilon}}=\frac{\sigma(2)}{2^{1+\epsilon}}=\frac{3}{2^{1+\epsilon}}$, with you so far

*But where does $\frac{\sigma(n)}{n}=\sum_{d|n}\frac{n/d}{n}$ come from? It seems to me that we have $\frac{\sigma(n)}{n}=\frac{\sum_{d|n}d}{n}$. So you are suggesting that $\frac{\sum_{d|n}d}{n}=\sum_{d|n}\frac{n/d}{n}$... How so? Surely we should have $\frac{\sigma(n)}{n}=\frac{\sum_{d|n}d}{n}=\sum_{d|n}\frac{d}{n}$

*And... well, there I lose you. I can't follow the rest because I can't really get past this one issue



I'm 100% sure it's me being stupid - I'm teaching myself all this stuff for the first time, and on my own. I realise that no one on MathStackExchange signs up to hold the hands of newbies, but if you have the time (or anyone else does) I'd really appreciate some clarification.
BTW: aren't we all, as a community and as a species, incredibly that such sites exist? Wow.
 A: I would suggest trying it backwards: given some $\epsilon >0$, it is easy to find an $N$ such that $\frac{\sigma(n)}{n^{1+\epsilon}} < 3/2^{1+\epsilon}$ for all $n \geq N$ (because $\sigma(n) \leq n\ln{n}+n$). 
Then you just optimize $\frac{\sigma(n)}{n^{1+\epsilon}}$ over $2 \leq n \leq N$. 
A: Given some $\delta > 0,$ the correct exponent (to build a Colossally Abundant Number by prime factorization) for some prime $p$ is
$$ \left\lfloor \frac{\log (p^{1 + \delta} - 1) - \log(p^\delta - 1)}{\log p} \right\rfloor  \; - \; 1.  $$
This is Theorem 10 on page 455 of Alaoglu and Erdos (1944). For a fixed $\delta,$ the exponents either stay the same or decrease for increasing $p,$ and eventually the exponent 0 is reached, so there is your complete number. For a fixed $p,$ the exponent either stays the same or increases with decreasing $\delta.$ 
I'm not seeing any lists that show $\delta$ and the result, so here, if I call $f(\delta)$ the corresponding colossally abundant number for $\delta,$ I calculate $$ f(1) = 1, \; f(1/2) = 2, \; f(1/4) = 6, \; f(1/6) = 12, \; f(1/10) = 60, \; f(1/12) = 120,$$
then
$$ f(1/14) = 360, \; f(1/17) = 2520, \; f(1/25) = 5040, \; f(1/31) = 55440, \; f(1/39) = 720720,$$
 and so on as $\delta$ decreases. 
If you want the first (largest) $\delta$ for which a favorite prime $p$ gets assigned exponent $k,$ let
$$  \delta = \frac{\log(p^{k+1} - 1) - \log(p^{k+1} - p)}{\log p}                        $$ 
A: Briggs outlines an approach in his paper Abundant Numbers and the Riemann Hypothesis. Another method would be to multiply the primes found in the integer sequence A073751. Additional methods can be found in the appendix of Schwabhäuser.
