It’s easy to prove that $d(n)\leq2\sqrt{n}$ for integer $n$, where $d(n)$ is the divisor function. One can prove stronger results, such as $d(n)\leq\sqrt{n}$ for $n\in$ $A035035$, or $d(n)\leq\sqrt[3]{n}$ for $n\in$ $A056758$. What makes these results somewhat useful is that the complements of these sequences are finite. I’m interested in finding these numbers for a given $k$.
The idea I have is to consider the prime factorization of $n$, and use the multiplicativity of $$f(n)=\frac{d(n)}{\sqrt[k]{n}}.$$ Then, for each prime $p$, one can find the maximum value of $f\left(p^\alpha\right)$, and afterwards, the values that are less than the product of all the maximums. This gives finitely many candidates for the exponents of each prime, and we can then check the possible combinations.
For $k=2$ this is somewhat doable, but this gets quite unwieldy even for small $k$. So my question is, is there a better algorithm to find these numbers?
The number that gives the maximum of your expression has, for each prime factor $p,$ exponent $$ \left\lfloor \frac{1}{p^{1/k} - 1} \right\rfloor $$ You just take the product of these, it is a finite product. There are usually a few primes with exponent 1. For fixed $k,$ once $p > 2^k,$ the denominator becomes bigger than $1$ and the floor becomes zero, so such primes do not divide the number.