# Does there exist a positive $k$ s.t. for all $r\geq k$, “$\sigma_r(m)<\sigma_r(n)$ for every $m<n$” for infinitely many odd positive integers $n$?

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.

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Can you find any $r$ such that there are infinitely many odd $n$ for which $\sigma_r(m)<\sigma_r(n)$ for all $m<n$? –  Matthew Conroy Jan 15 '14 at 1:50
@MatthewConroy I haven't. I assumed that if it were true for some $r$ it would be true for all greater $r$. I don't know if this is intended to be a question or a correction. –  Jaycob Coleman Jan 15 '14 at 6:10
It's a question. What I see when I calculate $\sigma_r$ is that there are "record breaking" odd $n$, but only small ones, for each $r$. As $r$ increases, there are more/larger ones, but always finite (i.e., in calculations, the odd record-breakers just stop at some point, while the even record-breakers continue). My hunch is that there are never an infinite number of record-breaking odd $n$. A good place to start is to show that, for $r=1$, there are only finitely many odd record-breaking $n$. Then try $r=2$, and see if you can extend the proof to all $r$. –  Matthew Conroy Jan 15 '14 at 18:45
@MatthewConroy Yes that's what I was getting at. The practical numbers (excluding 1) are a subset of the even numbers, with natural density 0, and it seems plausible that all sufficiently large terms for each $r$ are practical numbers. But on the weak question, now that I think about it, Daniel Fischer proved that $3$ is the only odd term for $r=1$ here. I will try to adapt that proof for $r=2$. –  Jaycob Coleman Jan 16 '14 at 3:04
I'm not sure i understood your question correctly, but if true, does it mean that $\sigma(2n+1)$ has a lower bound for large enough natural number $n$? –  CODE Jan 16 '14 at 15:31