Finite Factors of Refactorable Numbers This was just a question that I came up with while learning about refactorable numbers.
While looking through the sequence of refactorable numbers (1, 2, 8, 9, 12, 18....), I decided to look at the number of factors that each refactorable number had. This gave me the sequence (1, 2, 4, 3, 6, 6....). I saw that while refactorable numbers were all distinct, my numbers repeated. My question is that will all the numbers in my sequence repeat infinitely, or will some of these stop showing up after a finite value? Either way how do I prove this?
Since this is just a problem that I came up with I don´t have much work on it, but here is the wikipedia page that I  reading about refactorable numbers from:
https://en.wikipedia.org/wiki/Refactorable_number
 A: Each prime $p$ appears exactly once in your list.  Proof:  $p^{p-1}$ is a refactorable number with $p$ divisors, and it is the only one because every number with exactly $p$ divisors has the form $q^{p-1}$ for some prime $q$.
A: For an odd prime $p$ the number $8p$ has $4 \cdot 2 = 8$  divisors, which divides $8p.$  This is repeated infinitely often
For an odd prime $q > 3$ the number $12q$ has $6 \cdot 2 = 12$  divisors, which divides $12q.$
For an odd prime $r > 5$ the number $80r$ has $10 \cdot 2 = 20$  divisors, which divides $80r.$
For any prime $s \neq 3$ the number $9s^2$ has $3 \cdot 3 = 9$  divisors, which divides $9s^2.$
For any prime $t \neq 5$ the number $625 t^4$ has $5 \cdot 5 = 25$  divisors, which divides $625 t^4.$
For any prime $u \neq 7$ the number $117649 u^6$ has $7 \cdot 7 = 49$  divisors, which divides $117649 u^6$
There are also slowly growing families. For any prime $v > 2$ take
$$   v^{v-1} $$ which has exactly $v$ divisors.
Any $w $  take $$  2^{\left( 2^w -1\right)}  $$
$$  3^{\left( 3^w -1\right)}  $$
$$  5^{\left( 5^w -1\right)}  $$
$$  7^{\left( 7^w -1\right)}  $$
or
$$ \color{red}{ p^{\left( p^w -1\right)} } $$
A: To fully generalize bits from the two existing answers:

Every square-free $n$ appears $k!$ times where $k$ is the number of prime divisors of $n$. Every other number appears infinitely often.

The two crucial bits to solving this are seeing that the number
$$p_1^{n_1}p_2^{n_2}\ldots p_k^{n_k}$$
has $(n_1+1)(n_2+1)\cdots (n_k+1)$ factors and that
$$p_1^{n_1}p_2^{n_2}\ldots p_k^{n_k} \text { divides } q_1^{m_1}q_2^{m_2}\ldots q_k^{m_k}$$
if and only if every prime on the left appears on the right with an equal or greater exponent. Using these facts lets us easily constrain both which numbers have $n$ divisors and which are divisible by $n$ knowing the prime factorization of $n$.
First, for the square free case, suppose that
$$n=p_1p_2p_3\cdots p_k$$
for distinct primes $p_1,\ldots,p_k$. Note that, in order for $n$ to divide some $m$, that $m$ must have at least these same prime factors - but, for a number with $k$ distinct prime factors to have $n$ factors total, it must be of the form
$$q_1^{p_1-1}q_2^{p_2-1}\cdots q_k^{p_k-1}$$
for distinct primes $q_1,\ldots,q_k$. Thus, any number divisible by $n$ and having $n$ factors must be of the form $$p_1^{p_{\pi(1)}-1}p_2^{p_{\pi(2)}-1}\cdots p_k^{p_{\pi(k)}-1}$$ for some permutation $\pi$ of the indices $\{1,2,\ldots,k\}$. There are thus $k!$ multiples of $n$ with $n$ factors.
If $n$ is not square free, let us write
$$n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$$
and assume $a_1>1$. Then, consider, for any prime $q$:
$$m=p_1^{p_1^{a_1-1} - 1}\cdot p_2^{p_2^{a_2}-1}\cdot p_3^{p_3^{a_3}-1} \cdots p_k^{p_k^{a_k}-1}\cdot q^{p_1-1}$$
where we reduce the power of $p_1$ to "make room" for a power of $q$. This number has $n$ factors for every choice of prime $q$, hence $n$ appears infinitely often in your sequence.
