A question concerning some arithmetic function 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: 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):
$2n-1=945$
$a(2n-1)=1088$
$2n−1+τ(2n−1)/2=953$
A: For Question (1) consider any divisor $d$ of $n$. If $d= 2k$, then we may write $d = p + q$ for any $0 \leq p \leq k-1$ each of which pairs with a $k+1 \leq q \leq 2k$ to satisfy what is counted by $a(n)$. Hence, any even divisor $d$ contributes $d/2$ to $a(n)$. If $d =2k+1$, then we may write $d = p + q$ for any $0 \leq p \leq k$ each of which pairs with a $k+1 \leq p \leq 2k+1$ to satisfy what is counted by $a(n)$. So, any odd divisor $d$ contributes $d/2 + 1/2$ to $a(n)$. It follows Question (1) has a positive answer.
