Let $n$ be a natural number. Let $U\_n = \{d \in \mathbb{N}\mid d\mid n \text{ and } \gcd(d,n/d)=1 \}$ be the set of unitary divisors, $D\_n$ be the set of divisors and $S\_n=\{d \in \mathbb{N}\mid d^2 \mid n\}$ be the set of square divisors of $n$.


The set $U\_n$ is a group with $a\oplus b := \frac{ab}{\gcd(a,b)^2}$. It operates on $D\_n$ via:


$$ u \oplus d := \frac{ud}{\gcd(u,d)^2}$$


The orbits of this operation "seem" to be


$$ U\_n \oplus d = d \cdot U\_{\frac{n}{d^2}} \text{ for each } d \in S\_n$$


From this conjecture it follows (also one can prove this directly since both sides are multiplicative and equal on prime powers):


$$\sigma(n) = \sum\_{d\in S\_n} d\sigma^\*(\frac{n}{d^2})$$


where $\sigma^\*$ denotes the sum of unitary divisors.


Since $\sigma^\*(k)$ is divisible by $2^{\omega(k)}$ if $k$ is odd, where $\omega=$ counts the number of distinct prime divisors of $k$, for an odd perfect number $n$ we get (Let now $n$ be an odd perfect number):


$$2n = \sigma(n) = \sum\_{d \in S\_n} d \sigma^\*(\frac{n}{d^2}) = \sum\_{d \in S\_n} d 2^{\omega(n/d^2)} k\_d $$


where $k\_d = \frac{\sigma^\*(n/d^2)}{2^{\omega(n/d^2)}}$ are natural numbers.
Let $\hat{d}$ be the largest square divisor of $n$. Then:
$\omega(n/d^2)\ge \omega(n/\hat{d}^2)$.


Hence we get:


$$2n = 2^{\omega(n/\hat{d}^2)} \sum\_{d \in S\_n} d l\_d$$
for some natural numbers $l\_d$.


If the prime $2$ divides not the prime power $2^{\omega(n/\hat{d}^2})$, we must have $\omega(n/\hat{d}^2)=0$ hence $n=\hat{d}^2$ is a square number, which is in contradiction to Eulers theorem on odd perfect numbers.


So the prime $2$ must divide the prime power $2^{\omega(n/\hat{d}^2})$ and we get:


$$n = 2^{\omega(n/\hat{d}^2)-1} \sum\_{d \in S\_n} d l\_d$$


with $l\_d = \frac{\sigma^\*(n/d^2)}{2^{\omega(n/d^2)}}$. Hence the odd perfect number, satisifies:


$$n = \sum\_{d^2\mid n} d \frac{\sigma^\*(n/d^2)}{2^{\omega(n/d^2)}}=:a(n)$$


Hence an odd perfect number satisifies:


$$n = a(n)$$


So my idea was to study the function $a(n)$, which is multiplicative on odd numbers, on the right hand side and what properties it has to maybe derive insights into odd perfect numbers.


**The question is if it ever can happen that an odd number $n$ satisfies: $n=a(n)$?** (checked for $n=2k+1$ and $1 \le k \le 10^7$)


**Edit**:
Conjecture: For all odd $n \ge 3$ we have $a(n)<n$. This would prove that there exists no odd perfect number.


This conjecture could be proved as follows:
Since $a(n)$ is multiplicative, it is enough to show that for an odd prime power $p^k$ we have


$$a(p^k) < p^k$$


The values of $a$ at prime powers are not difficult to compute and they are:


$$a(p^{2k+1})= \frac{p^{2(k+1)}-1}{2(p-1)}$$


and


$$a(p^{2k}) = \frac{p^{2k+1}+p^{k+1}-p^k-1}{2(p-1)}$$


However, I am not very good at proving inequalities, so:


**If someone has an idea how to prove the following inequalities for odd primes $p$ that would be very nice:**


$$p^{2k+1} > \frac{p^{2(k+1)}-1}{2(p-1)}, \text{ for all } k \ge 0$$


and


$$p^{2k} > \frac{p^{2k+1}+p^{k+1}-p^k-1}{2(p-1)}, \text{ for all } k \ge 1$$


Thanks for your help!


**The inequalities have been proved here:
<https://math.stackexchange.com/questions/3807399/two-inequalities-for-proving-that-there-are-no-odd-perfect-numbers>**



The action of the unitary divisors group on the set of divisors and odd perfect numbers
Here are some general comments:


1. You don't need to bring these actions of abelian groups on various sets of divisors. The identity
$$\sigma(n)=\sum\_{d^2|n}d\sigma^{\*}(\frac{n}{d^2})$$
is easy to check directly, without appeal to anything fancy.
2. Let's call $\alpha(n)$ the number of prime divisors of $n$ which appear with an odd exponent in the factorization of $n$. This is what you call $\omega(n/\hat{d}^2)$. You are right in observing that $2^{\alpha(n)}$ divides $\sigma(n)$. This is where Euler's result comes from: If $n$ is an odd perfect number then $\alpha(n)=1$.
3. It seems you want to define a new function $a(n)=\frac{\sigma(n)}{2^{\alpha(n)}}$, and you conjecture that $$a(n)<n$$
for all odd numbers $n$. If true this conjecture would imply that there are no odd perfect numbers. Unfortunately it is false. For example the inequality is reversed at $n=3^35^2 7^2$.


