Let $\tau^\*(n)=$ number of odd divisors of $n$, and let $a(n) = \#\{(p,q)| 0 \le p < q, p+q|n\}$, $\sigma(n)=$ sum of divisors of $n$, $\tau(n)=$ number of divisors of $n$.


1) Question: Is it true that $a(n) = 1/2(\sigma(n)+\tau^\*(n))$?


2) Question: From this one it would follow that no odd perfect numbers exist, so I suppose this question, might be interesting, but out of reach:
Is $a(2n-1) < 2n-1+\tau(2n-1)/2, \forall n \in \mathbb{N}$?


Why would it follow? Becaus if $n$ is an odd perfect number, then $\sigma(n)=2n$, $\tau^\*(n)=\tau(n)$ and from 1), it follows that:


$$ a(n) = n + \tau(n)/2$$


in contradiction to 2).


The question 1) could be proved through a cyclic sieving phenomenon, which is described here:


<https://math.stackexchange.com/questions/3223033/a-question-on-a-possible-cyclic-sieving-phenomenon>


But maybe there is also another proof, more direct?


Thanks for your help!



A question concerning some arithmetic function
There's a direct proof for 1):


$\text{#}\{(p,q)|0≤p<q,p+q=k\}$ is $k/2$ for $k$ even, and $k/2+1/2$ for $k$ odd.


So $\text{#}\{(p,q)|0≤p<q,p+q|n\}$


$=\sum\_{k|n, k \text{ even}}{\text{#}\{(p,q)|0≤p<q,p+q=k\}} + \sum\_{k|n, k \text{ odd}}{\text{#}\{(p,q)|0≤p<q,p+q=k\}}$


$=\sum\_{k|n, k \text{ even}}{k/2} + \sum\_{k|n, k \text{ odd}}{k/2+1/2}$


$=\sum\_{k|n}{k/2} + \sum\_{k|n, k \text{ odd}}{1/2}$


$=1/2(σ(n)+τ^∗(n))$.


**EDIT:** $n=473$ is a counterexample for 2):


$2n-1=945$


$a(2n-1)=1088$


$2n−1+τ(2n−1)/2=953$


