# Is any minimal covering number other than 12 a primitive abundant number?

Recall, a collection of congruences with distinct moduli, each greater than 1, such that each integer satisfies at least one of the congruences, is said to be a covering system (or a covering system of distinct congruences). Given a covering system, we will define $$m$$ to be the LCM of the moduli, and we will by slight abuse of notation refer to $$m$$ as the LCM of the moduli.

It is well known and an easy exercise that for any covering system, $$m$$ must be abundant, that is that $$\sigma(n)>2n$$ where $$\sigma(n)$$ is the sum of the positive divisors of $$n$$.

Note that if there exists a covering system $$S$$ with modulus $$m$$, then there is a covering system with modulus $$km$$ by just taking the $$S$$ throwing in a redundant restriction mod $$km$$. We can thus talk about minimal covering numbers, where by a minimal covering number we mean a number $$m$$ such that there is a covering system where $$m$$ is the LCM of the system, but where no proper divisor of $$m$$ is if the LCM of a system. The sequence of minimal covering numbers is A160559 and begins $$12$$, $$80$$, $$90$$, $$210$$... .

This sequence is infinite. In particular, for any odd prime $$p$$, $$2^{p-1}p$$ will be in this sequence.

A number is said to be a primitive abundant number if $$\sigma(n)>2n$$ and no proper divisor of $$n$$ is an abundant number. The motivation for this is very similar to that of the minimal covering number, namely noting that any multiple of an abundant number is abundant. A minor note of terminology: some authors define a number to be primitive abundant if $$\sigma(n) \geq 2n$$ and $$n$$ has no proper divisor satisfying that inequality(which would allow one to include perfect numbers as primitive abundant). We will not be doing so here, although doing so would allow us to delete the "other than 12" from the title question.

The question in the title has two motivations which both suggest that the answer is "no."

First, the fact that minimal covering numbers must be abundant is in practice a very weak statement. Generally, the smallest abundant divisor of a minimal covering number is much smaller than the number. For example, 80 is a minimal covering number, and its smallest abundant divisor is 20.

More generally, if $$p$$ is an odd prime, then $$2^ap$$ is either a perfect or primitive abundant number where $$a=\lfloor \log_2 p \rfloor$$. So if $$m$$ is a minimal covering number of the form $$2^{p-1}p$$ then $$m$$'s smallest primitive abundant divisor is of order $$\log m$$.

Second, there is an open problem of Erdos and Selfridge asking whether there is any covering system with all moduli odd. This is equivalent to asking whether the minimal covering numbers are always even. Here, the size of such an $$m$$ must be very large under some reasonable assumptions. For example, $$(3)(5)(7)(11)(13)$$ is an odd primitive abundant number but there's a paper by Song Guo, Zhi-Wei Sun that shows that an odd square free $$m$$ must have at least 22 distinct prime divisors. So, it seems like in the hypothetical case of odd $$m$$, in at least the squarefree situation, we'd expect such an $$m$$ to be large compared to the smallest squarefree primitive odd abundant numbers.

This motivates the title question of whether 12 is the only minimal covering number which is a primitive abundant number.

One thing to note which leads to a related question: One can have minimal covering systems with $$\frac{\sigma(m)}{m}$$ arbitrarily close to 2, and the subsequence of $$2^{p-1}p$$ is an example of such. This leads to the second question: Is there some non-trivial improvement of this sort of inequality for minimal covering systems? In particular, can we find a slow growing increasing function $$f(x)$$ such that if $$m$$ is a minimal covering then $$\sigma(m) \geq 2m\left(1+\frac{1}{f(m)}\right)$$ and where that same inequality is not satisfied by all sufficiently large primitive abundant numbers? There are inequalities in the Sun and Guo paper that are almost of this form, but there one seems to need a function $$f(x)$$ which takes in input about the prime factorization of $$m$$ and so isn't an increasing function. This question may also be relevant.