TITLE: Do all members of this sequence have $8$ divisors?
QUESTION [5 upvotes]: Define the set $A$ as a subset of $\mathbb{N}$ as follows: Take all of the divisor pairs of a natural number $n$ (except $n$ and $1$), divide the larger of each pair by the smaller, and then take the resultant product. If this number $=n,$ it is a member of the set.
For example, the divisor pairs of $24$ (excluding $1$ and itself) are $(2,12),(3,8),(4,6).$ Divide the larger by the smaller: $6,8/3,3/2$ and multiply them together $=24.$ Hence $24$ is a member of the set.
The sequence starts: $1, 24, 30, 40, 56, 64, 70, 105, 135, 154\dots$
With the exception of $1$ and $p^6$ (where $p$ is prime), all members of $A$ appear to have $8$ divisors. That is, $\sigma_0(n)=8,$ for all $n\in A.$ Is this true?
There are other ways to formulate the sequence, eg: the product of the largest of each divisor pair excluding $n$ ifself $=n^2.$ (If $n$ is a square, include $\sqrt{n}$.)
eg for $n=24,$ take the larger of each divisor pair: $6,8,12.$ Their product $=24^2.$ 
I have seached oeis, but the sequence doen't feature.

REPLY [5 votes]: Using your alternative definition, it is quite obvious that the assertion is true.
Indeed, if $d_i$ are the largest of each divisor pair, then (excluding the possibility of $n$ being a square) $\sqrt n<d_i<n$. Hence the product of four such terms is too big for $n^2$ and the product of two is too small. This only leaves the option of three $d_i$, which mean four divisor pairs if we count $(1,n)$, which in turn means eight divisors.
With $n$ being a square, the same line of thought leads to four divisor pairs one of which is $(\sqrt n,\sqrt n)$. This means seven divisors, which is only possible for $n=p^6$.
As for the possible prime signature of the non-square $n$, eight divisors are possible in the numbers of the form $p^7,\,p^3q,\text{ or }pqr$. By writing out the divisors and plugging them in the formula, we immediately see that:

$pqr$ with $p<q<r$ works if $r<pq$;
$p^3q$ works if $p<q<p^3$;
$p^7$ never works.