Does there exist a positive real $k$ such that for all real $r\geq k$, "$\sigma_r(m)<\sigma_r(n)$ for every $m<n$" for infinitely many odd positive integers $n$? $\sigma_r(n)$ is the sum of the $r$th powers of the divisors of $n$.
I did some computational experiments a while back involving highly abundant numbers - positive integers $n$ such that $\sigma_1(m)<\sigma_1(n)$ for every $m<n$ - and it seemed as if $3$ and $10$ might be the only terms that aren't practical numbers, i.e., their canonical prime factorizations do not satisfy "$p_i\leq 1+\sigma(p_1^{a_1}p_2^{a_2}...p_{i-1}^{a_{i-1}})$ for every $i\in[1,\omega(n)]$," where $\omega(n)$ is the numbers of distinct prime factors of $n$ (clearly every practical number, excluding $1$, which is practical, is even).
It's also (relatively) well-known that all highly composite numbers - positive integers $n$ such that $\sigma_0(m)<\sigma_0(n)$ for every $m<n$ - are practical numbers (a consequence of the fact that all products of primorials, which include highly composite numbers, are practical numbers).
I considered that perhaps all sufficiently large terms of all sets of this type are practical numbers, even though for larger exponents there come to be many odd or otherwise non-practical numbers in the corresponding set. It seems intuitive that, at the very least, there could never be infinitely many primes for a given $r$.
I wondered if someone could produce a proof that infinitely many odd terms or infinitely many non-practical terms exist for sufficiently large $r$. Considering that practical numbers have $0$ natural density, this could be done by showing that these sets have positive natural density for sufficiently large $r$, but I'm doubtful that it's true, because I suspect that each set eventually consists of only practical numbers.