<s>[INST] Prove that $\dfrac{\sigma_1(n)}{n} = \sigma_{-1}(n)$ where $\sigma_x(n)$ is the sum of the $x$th powers of the positive divisors of $n$. I computed $\dfrac{\sigma_1(n)}{n}$ and $\sigma_{-1}(n)$ on a good hundred values of $n$, and they seem to always match.
For example:
$\dfrac{\sigma_1(6)}{6} = \dfrac{1 + 2 + 3 + 6}{6} = \dfrac{12}{6} = 2$
$\sigma_{-1}(6) = \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{6}{6} + \dfrac{3}{6} + \dfrac{2}{6} + \dfrac{1}{6} = \dfrac{12}{6} = 2$
Humm... Now that I'm actually writing all of this down, I think I have an idea how to prove this...
For every positive integer $n$, all the $m$ divisors of $n$ can be grouped into $\dfrac{m}{2}$ pairs {$d_k$ ; $d_{m+1-k}$} such that $d_k \times d_{m+1-k} = n$ with $1 \leq k \leq m$
$d_k = \dfrac{n}{d_{m+1-k}}$
$\dfrac{\sigma_1(n)}{n} = \dfrac{d_1}{n} + \dfrac{d_2}{n} + \dfrac{d_3}{n} +$ $... + \dfrac{d_{m-2}}{n} + \dfrac{d_{m-1}}{n} + \dfrac{d_m}{n} $
$\sigma_{-1}(n) = \dfrac{1}{d_1} + \dfrac{1}{d_2} + \dfrac{1}{d_3} +$ $... + \dfrac{1}{d_{m-2}} + \dfrac{1}{d_{m-1}} + \dfrac{1}{d_m} + $
$\sigma_{-1}(n) = \dfrac{d_m}{n} + \dfrac{d_{m-1}}{n} + \dfrac{d_{m-2}}{n} +$ $... + \dfrac{d_3}{n} + \dfrac{d_2}{n} + \dfrac{d_1}{n}$
$\sigma_{-1}(n) = \dfrac{\sigma_1(n)}{n}$
And voila!
I'm not totally sure it's perfectly rigorous but it seems to be correct.
If $n$ is a square there's $d_{\frac{m+1}{2}}$ that isn't in a pair but we have: $d_{\frac{m+1}{2}}^2 = n$ so $d_{\frac{m+1}{2}} = \dfrac{n}{d_{\frac{m+1}{2}}}$
Well, how about you try to find some different proofs for my equality then? Maybe someone can find an even simpler and more elegant proof.
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Hallowed is 120.
$\dfrac{\sigma_1(120)}{120} = \dfrac{360}{120} = 3$
$\sigma_{-1}(120) = 3$
 [/INST] $\sigma_k(p^a)=\dfrac{1-p^{k(a+1)}}{1-p^k}=p^{ka}\dfrac{p^{-k(a+1)}-1}{p^{-k}-1}=p^{ka}\sigma_{-k}(p^a)$
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