# The action of the unitary divisors group on the set of divisors and odd perfect numbers

Let $$nn$$ be a natural number. Let $$Un={d∈N∣d∣n and gcd(d,n/d)=1}U_n = \{d \in \mathbb{N}\mid d\mid n \text{ and } \gcd(d,n/d)=1 \}$$ be the set of unitary divisors, $$DnD_n$$ be the set of divisors and $$Sn={d∈N∣d2∣n}S_n=\{d \in \mathbb{N}\mid d^2 \mid n\}$$ be the set of square divisors of $$nn$$.

The set $$UnU_n$$ is a group with $$a⊕b:=abgcd(a,b)2a\oplus b := \frac{ab}{\gcd(a,b)^2}$$. It operates on $$DnD_n$$ via:

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

The orbits of this operation “seem” to be

$$Un⊕d=d⋅Und2 for each d∈Sn U_n \oplus d = d \cdot U_{\frac{n}{d^2}} \text{ for each } d \in S_n$$

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

$$σ(n)=∑d∈Sndσ∗(nd2)\sigma(n) = \sum_{d\in S_n} d\sigma^*(\frac{n}{d^2})$$

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

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

$$2n=σ(n)=∑d∈Sndσ∗(nd2)=∑d∈Snd2ω(n/d2)kd2n = \sigma(n) = \sum_{d \in S_n} d \sigma^*(\frac{n}{d^2}) = \sum_{d \in S_n} d 2^{\omega(n/d^2)} k_d$$

where $$kd=σ∗(n/d2)2ω(n/d2)k_d = \frac{\sigma^*(n/d^2)}{2^{\omega(n/d^2)}}$$ are natural numbers.
Let $$ˆd\hat{d}$$ be the largest square divisor of $$nn$$. Then:
$$ω(n/d2)≥ω(n/ˆd2)\omega(n/d^2)\ge \omega(n/\hat{d}^2)$$.

Hence we get:

$$2n=2ω(n/ˆd2)∑d∈Sndld2n = 2^{\omega(n/\hat{d}^2)} \sum_{d \in S_n} d l_d$$
for some natural numbers $$ldl_d$$.

If the prime $$22$$ divides not the prime power $$2ω(n/ˆd2)2^{\omega(n/\hat{d}^2})$$, we must have $$ω(n/ˆd2)=0\omega(n/\hat{d}^2)=0$$ hence $$n=ˆd2n=\hat{d}^2$$ is a square number, which is in contradiction to Eulers theorem on odd perfect numbers.

So the prime $$22$$ must divide the prime power $$2ω(n/ˆd2)2^{\omega(n/\hat{d}^2})$$ and we get:

$$n=2ω(n/ˆd2)−1∑d∈Sndldn = 2^{\omega(n/\hat{d}^2)-1} \sum_{d \in S_n} d l_d$$

with $$ld=σ∗(n/d2)2ω(n/d2)l_d = \frac{\sigma^*(n/d^2)}{2^{\omega(n/d^2)}}$$. Hence the odd perfect number, satisifies:

$$n=∑d2∣ndσ∗(n/d2)2ω(n/d2)=:a(n)n = \sum_{d^2\mid n} d \frac{\sigma^*(n/d^2)}{2^{\omega(n/d^2)}}=:a(n)$$

Hence an odd perfect number satisifies:

$$n=a(n)n = a(n)$$

So my idea was to study the function $$a(n)a(n)$$, 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 $$nn$$ satisfies: $$n=a(n)n=a(n)$$? (checked for $$n=2k+1n=2k+1$$ and $$1≤k≤1071 \le k \le 10^7$$)

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

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

$$a(pk)

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

$$a(p2k+1)=p2(k+1)−12(p−1)a(p^{2k+1})= \frac{p^{2(k+1)}-1}{2(p-1)}$$

and

$$a(p2k)=p2k+1+pk+1−pk−12(p−1)a(p^{2k}) = \frac{p^{2k+1}+p^{k+1}-p^k-1}{2(p-1)}$$

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

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

$$p2k+1>p2(k+1)−12(p−1), for all k≥0p^{2k+1} > \frac{p^{2(k+1)}-1}{2(p-1)}, \text{ for all } k \ge 0$$

and

$$p2k>p2k+1+pk+1−pk−12(p−1), for all k≥1p^{2k} > \frac{p^{2k+1}+p^{k+1}-p^k-1}{2(p-1)}, \text{ for all } k \ge 1$$

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

$$σ(n)=∑d2|ndσ∗(nd2)\sigma(n)=\sum_{d^2|n}d\sigma^{*}(\frac{n}{d^2})$$
2. Let's call $$α(n)\alpha(n)$$ the number of prime divisors of $$nn$$ which appear with an odd exponent in the factorization of $$nn$$. This is what you call $$ω(n/ˆd2)\omega(n/\hat{d}^2)$$. You are right in observing that $$2α(n)2^{\alpha(n)}$$ divides $$σ(n)\sigma(n)$$. This is where Euler's result comes from: If $$nn$$ is an odd perfect number then $$α(n)=1\alpha(n)=1$$.
3. It seems you want to define a new function $$a(n)=σ(n)2α(n)a(n)=\frac{\sigma(n)}{2^{\alpha(n)}}$$, and you conjecture that $$a(n)
for all odd numbers $$nn$$. If true this conjecture would imply that there are no odd perfect numbers. Unfortunately it is false. For example the inequality is reversed at $$n=335272n=3^35^2 7^2$$.