For every positive integer $n$, $\tau(n)$ is the number of divisors of $n$. If we list the ratio of each positive integer $n$ to $\tau(n)$，they form a rational sequence
1,1,3/2,4/3,5/2,3/2,…
Because $\tau(n)$ is not greater than $\sqrt{3n}$, the ratio becomes not less than $\sqrt{n/3}$, every term in the sequence will appear finitely often.
If those which are not integers are deleted from the sequence, then this sequence becomes A036762 in OEIS.
1,1,2,3,2,3,3,4,5,7,…
If an integer $m$ appears exactly $k$ times in this integer sequence, it may be described that the frequency of $m$ in the sequence is $k$. And then there will be a frequency sequence, the $n$th terms is $k$ indicates $n$ appears exactly $k$ times in A036762. This sequence is A051521 in OEIS.
2,2,3,1,2,1,2,2,1,1,…
0 firstly appears in the 18th term，which means there is not a positive integer $n$ equals to $18\tau(n)$.
Question: Does every integer greater than 3 also appear in the sequence?
The 11264th term is 4, the 64000th term is 5, and the 82944th is 6, which is from A217125, A217126 and A217127.
Motivation: It has been proved that 0 appears infinitely often in A051521, and I have a proof for that if a positive integer appears in A051521, then it will appear infinitely often.
(1) For a positive integer $c$, if every positive integer $m$ which satisfy that $m$ equals to the product of $\tau(m)$ and $\tau(c)$, is coprime to $c$, and there are $t$ different such $m$, $t>0$,
then there will be exactly $t$ different positive integers $n$，which satisfy that $\tau(n)$ is coprime to $c$, and that $n$ equals to the product of $\tau(n)$ and $c$.
(2) For an integer $c$ greater than 1, if every prime factor $p$ of $c$ satisfy that $p-1$ is not less than the product of $(a+2)/(a+1)$ and $\tau(c)$，where $a$ is the exponent of $p$ in the prime factorization of $c$,
then every positive integer $n$ satisfy that $n$ equals to the product of $\tau(n)$ and $c$, will also satisfy that $\tau(n)$ is coprime to $c$.
This can be proved with inequality and prime factorization.
(3) For every positive integer $k$, there will be infinitely many positive integers $c$ satisfy that $\tau(c)$ equals $k$, And we can choose those $c$ which also satisfy that every prime factor of $c$ is coprime to every solution $m$ to equation $m$ equals $\tau(m)$ multiplied by $k$, and at the same time these prime factors satisfy the condition in (2), In short, these prime factors are large enough.
Then use (1) and (2), it can be showed that the numbers of solution to these two equations are equal. One is that $m$ equals to the product of $\tau(m)$ and $k$, and the other is that $n$ equals to the product of $\tau(n)$ and $c$.
So the number of solution to one of these equation, also the frequency of some integer ratio of $n$ to $\tau(n)$, also the $k$-th term in A051521, will appear one more time in the $c$-th term, thus it will appear infinitely often in A051521.
That's to say that the frequency of a nonnegative integer $m$ in the frequency sequence is either 0 or infinity. And if the answer of the question above is true, it will be proved that every nonnegative integer $m$ appears infinitely often in A051521.
