Numbers with more than n divisors

Question

1. Numbers with more than 4 divisors = multiples of numbers with exactly 4 divisors. This only applies to 4 (and 2, of course): e.g. numbers with more than 3 divisors != multiples of numbers with exactly 3 divisors.
2. Numbers with more than 5 divisors = multiples of OEIS A068993

My quick question is: are these two facts obvious and/or well-known?

I assume that the answer is "yes", but I know pretty much nothing about mathematics, so I wanted to ask this question. I would have asked it in the chat, but I have not enough reputation.

A simple yes/no answer is fine for me, but of course if you want to add an explanation, it's even better. :)

Edit: why I don't think this question was a duplicate. The answer is the same as the answer to Finding out the number of divisors (I have no reason to doubt it, even though I still have to understand the answer to my question!). But, as far as I can understand, "duplicate" applies to questions, not to answers. Otherwise, the following statement from the help wouldn't make sense:

[...] we love (some) dupes. There are many ways to ask the same question, and a user might not be able to find the answer if they're asking it a different way.

Now, my question is not the same as "finding out the number of divisors", so it isn't a duplicate.

Answer 1

The reason that numbers with more than four divisors are multiples of numbers with exactly four divisors are that the numbers exactly four divisors are of the form $pq$ for distinct primes $p,q$ or $p^3$ for prime $p$. To have more than four, the number has to be of the form $p^4, p^2q, \text{ or } pqr$ for primes $p,q,r$ or more complex. All of these are multiples of a number with exactly four factors. Conversely, any multiple of a number $pq$ or $p^3$ will have more than four divisors.

A number with more than five divisors will be a multiple of a number in your OEIS sequence, because this sequence is all numbers of the form $pq$ or $p^4$. As $pqr$ has eight divisors and $p^2q$ and $p^5$ have six, all multiples of the elements of this sequence will have more than five divisors.

Added: for the fact that numbers with more that three divisors are not exactly the numbers that are multiples of numbers with exactly three divisors can be shown by example: $6$ has four divisors, while $1,2,3$ do not have exactly three. The general case is that a number of the form $pq$, with $p,q$ distinct primes, has factors $1,p,q,pq$ and none of $1,p,q$ have exactly three divisors.

Answer 2

If a number is product of distinct primes then it has a power of $2$ number of divisors (it has $2^{\omega(n)}$ divisors, where $\omega(n)$ is the number of prime divisors of $n$).

So if your number is product of distinct primes all of its divisors has a power of $2$ number of divisors.

From here we deduce if $k$ is not a power of $2$ then there is a number $n$ such that $n$ has more than $k$ divisors and no divisor of $n$ has exactly $k$ multiples. We just take $p_1p_2\dots p_r$, where $p_i$ is the $i$'th prime and $2^r>k$.

So now we just have to check $2$ and $4$ are the only such powers of $2$ that work. To see this just now that $p_1^2p_2^2\dots p_r^2$ has $3^r$ divisors, and any divisor of this number that has a power of $2$ number of divisors has at most $2^r$ divisors.

So we must now all we must prove is that given $2^n$ with $n>2$ there is an $r<n$ so that $3^r>2^n$ , so all we must prove is $3^{n-1}>2^{n}\iff \frac{3}{2}^{n-1}>1$, which of course is true when $n>2$. Since $\frac{3}{2}^2=\frac{9}{4}>2$.

Finally, to check $4$ works is easy, if the number is of form $p^k$ and has $4$ or more divisors then $k\geq 3$ and so $p^3$ is the divisor we wanted. Otherwise there are primes $p$ and $q$ dividing $n$ and so $pq$ is the desired divisor.

Proving it works for $2$ should be much easier.

Answer 3

For starters, the $\tau$ function is multiplicative. That means, $\tau(mn)=\tau(m)\tau(n)$ where $(m, n) =1$. But, as you seek intuition, I won't use it in the discussion. Now, your claim is true for $4$. The numbers which have $2$ divisors are prime $p$. Now, if we multiply it by something that the product has more than $4$ integers, then the multiplier must be of the form $p_1 p_2$ where at most one can be $p$ itself. But, we can always represent it as a multiple of $p_1 p_2$ which has exactly $4$ divisors. Now, a similar argument can be stated about integers with exactly $3$ divisors, which must be of the form $p^2$. So, every integer that has more than $4$ divisors can be expressed as a multiple of an integer with exactly $4$ divisors. But, no similar argument can be presented for numbers with higher divisor counts. Hence your claim is indeed true.

For the second case, just use the algebraic form of the $\tau$ function. Although, some simple combinatorics can give you the intuition.