Finding $m∈\mathbb N$ such that $d^n(m)$ is not a perfect square for any $n\geq1$

Question

Let , for $k ∈\mathbb N$ , $d(k)$ denote the number of positive divisors of $k$ ; define $d^n (k)$ recursively as $d^1(k)=d(k)$ , for $n\geq1$ , $d^{n+1}(k)=d(d^n(k))$ , how do we find those $m∈\mathbb N$ such that $d^n(m)$ is never a perfect square for any $n\geq1$ ?

Answer 1

Start at the bottom, and work your way up.

2 is such a number, since $d^n(2)=2$ for all $n$.

Every prime $p$ is such a number, since $d(p)=2$.

Every number of the form $q^{p-1}$ is such a number, where $p$ and $q$ are prime, since $d(q^{p-1})=p$.

Then you can find the numbers $n$ for which $d(n)=q^{p-1}$, and so on.

http://oeis.org/A009087 is related.