Does the sum of reciprocals of the harmonic divisor numbers converge?
Define the following:
Harmonic divisor number - $n$ such that $\sigma(n) \mid n\sigma_0(n)$. Equivalently, the harmonic mean of divisors of $n$ is an integer.
Deficient, perfect, or abundant number - $n$ such that $2n$ is greater than, equal to, or less than, respectively, $\sigma(n)$, the sum of the divisors of $n$. Every positive integer falls into one of these three categories.
Semiperfect number = $n$ such that $n$ is a sum of distinct proper divisors of $n$.
Practical number - $n$ such that every $k \in [1,n]$ is a sum of distinct divisors of $n$.
We should of course consider whether any known subsets of the harmonic divisor numbers have a divergent sum of reciprocals or if any known supersets have a convergent sum of reciprocals. The perfect numbers are a subset of the harmonic divisor numbers, but have a convergent sum of reciprocals. I don't know of any other subsets to consider. As for supersets, I'm pretty sure every harmonic divisor number is non-deficient, but I can't prove this and it wouldn't be any help anyways because every multiple of an abundant number is also abundant and the sum of reciprocals of the first $r$ multiples of any integer is just a constant times the $r$th harmonic number, thus divergent. Numerical evidence supports the stronger statement that the harmonic divisor numbers are also a subset of the intersection of the practical numbers and semiperfect numbers (same as the practical numbers that aren't powers of two), but the sum of reciprocals of practical semiperfect numbers is divergent still, so this wouldn't help either.
I don't know what other approaches there are to this problem. The harmonic divisor numbers are known to have natural density $0$, so the sum of reciprocals may or may not converge.