Yesterday I wrote this equation involving the sum of divisors function $\sigma(l)=\sum_{d\mid l}d$, $$\sigma(2n)=2\left(n+\sigma(n)\right).\tag{1}$$ Due to its very simple form I don't know if it is in the literature. I try to explore it in the context of odd perfect numbers - in which I've created it.
Claim 1. If there exists an odd perfect number $n$, then it satisfies $(1)$.
Also, one can prove the next claim.
Claim 2. Let an integer $n=2^{\alpha}m$, where $\alpha\geq 0$ is integer and the positive integer $m$ satisfies $\gcd(2,m)=1$. If our integer $n$ satisfies $(1)$ with $\alpha=0$ then is an odd perfect number. If our integer $n$ satisfies $(1)$ with $\alpha\geq 1$ then $m$ satisfies $\sigma(m)=2^{\alpha+1}m$, that is $m$ should be perfect or multiperfect.
Proof of Case $\alpha\geq 1$. One writes from our assumptions and equation $(1)$ that $$\sigma(2^{\alpha+1}m)=2(2^{\alpha}m+\sigma(2^{\alpha}m)).\tag{2}$$ And using that the sum of divisors function is multiplicative then $$(2^{\alpha+2}-1-2(2^{\alpha+1}-1))\sigma(m)=2^{\alpha+1}m,\tag{3}$$ concluding $\sigma(m)=2^{\alpha+1}m.\square$
Question. Is it possible to do more work on this equation? Many thanks.
I hope that my question is clear. I would like to rule out some case, or deduce a stronger result about the form of the solutions of $(1)$. Please if this equation or question was in the literature refer to it and I will try to search and read those statements. Thus I am looking for feedback about the previous equation, since it is an open problem I am waiting for some answer from your knowledges on odd perfect numbers about if you can get some statement in the exploration of the solutions of $(1)$, and I am going to accept the answer with more merit.
Computational fact. I've tested the solutions of $(1)$ using this code (this is a line)
for (i = 1, 1000000,if(sigma(2*i)==2*(i+sigma(i)),print(i)))
using this Sage Cell Server, you need to choice GP as Language and press Evaluate.