For an integer $n\geq 1$ we denote with $$\sigma(n)=\sum_{d\mid n}d$$ the sum of divisors function. It's well known that is multiplicative. I can to prove the following statement, and I am interesting in a special generalization of this.
Claim.
A) Let $\alpha\geq 2$ an integer and we assume that $N$ is an odd perfect number (see here in this MathWorld's article), then $N$ satisfies $$2N+2\sigma\left(2^{\alpha-2}N\right)=\sigma\left(2^{\alpha-1}N\right).\tag{A}$$
B) Let $\alpha\geq 2$ an integer and we assume that $N$ is an odd integer that satisifies $(A)$, then $N$ is also perfect (and thus $N$ is an odd perfect number).
C) Let $\alpha\geq 2$ an integer and we assume that $N$ is even, thus $N=2^kM$ where $k\geq 1$ and $M\geq 1$ are integers with $\gcd(2,M)=1$. If $N$ satisifies $(A)$ then $$\sigma(M)=2^{k+1}M,$$ and thus our odd integer $M$ is multiperfect.
Now I consider the following procedure (I don't provide an explanation about if my idea to create these conditions is potentitally interesting, or has full mathematical meaning because I don't know it: were only speculative calculations) as a background of my question.
Procedure. Let $\alpha\geq 2$ an integer.
1) We consider for odd integers $N\geq 1$ the condition $$\sigma(N)+2\sigma\left(2^{\alpha-2}N\right)=\sigma\left(2^{\alpha-1}N\right).\tag{2}$$ 2) We multiply the condition $(2)$ by $2^{\beta+1}$, where $\beta\geq 1$ is a fixed integer $$2^{\beta+1}\sigma(N)+2^{\beta+2}\sigma\left(2^{\alpha-2}N\right)=2^{\beta+1}\sigma\left(2^{\alpha-1}N\right).\tag{3}$$ 3) Substracting $\sigma(N)$ and using that the sum of divisor function is multiplicative we deduce $$\sigma\left(2^{\beta}N\right)+2^{\beta+2}\sigma\left(2^{\alpha-2}N\right)=2^{\beta+1}\sigma\left(2^{\alpha-1}N\right)-\sigma(N).\tag{4}$$ 4) Inspired in $(4)$ and the condtion $(A)$ from the Claim we consider for integers $N\geq 1$ (of both parities) $$2N+\sigma\left(2^{\beta}N\right)+2^{\beta+2}\sigma\left(2^{\alpha-2}N\right)=2^{\beta+1}\sigma\left(2^{\alpha-1}N\right)\tag{B}$$ where as was said $\alpha\geq 2$ and $\beta\geq 1$ are integers. $\square$
If we apply one more time previous Procedure we obtain the next condition that we would like to study for integers $N\geq 1$, where $\alpha\geq 2$, $\beta\geq 1$ and $\gamma\geq 1$ are integers $$2N+\sigma\left(2^{\gamma}N\right)+2^{\gamma+1}\sigma\left(2^{\beta}N\right)+2^{\beta+\gamma+3}\sigma\left(2^{\alpha-2}N\right)=2^{\beta+\gamma+2}\sigma\left(2^{\alpha-1}N\right).\tag{C}$$
I want to write the generalization of $(A)$ for $n$ indexes $\alpha\geq 2$, $\beta\geq 1$, $\gamma\geq 1$, $\delta\geq 1\ldots$ integers, and being $N\geq 1$ also an integer, and we want to deduce the corresponding and similar Claim for such condition.
Question. That is, denoting such indexes as $\alpha_1\geq 2$ and for $2\leq i\leq n$ by means of $\alpha_i\geq 1$ all integers, we denote our generalization as $(A_{\alpha_1,\alpha_2,\ldots\alpha_n})$ (see below my attempt to get such condition), I would like to know if it is possible to write a similar Claim for such condition $(A_{\alpha_1,\alpha_2,\ldots\alpha_n})$ for those cases about integers $N\geq 1$ that were stated in such Claim.
I believe that the condition $(A_{\alpha_1,\alpha_2,\ldots\alpha_n})$ is like as (still I can not deduce exactly the right condition) $$2N+\sigma\left(2^{\alpha_n}N\right)+2^{1+\alpha_n}\sigma\left(2^{\alpha_{n-1}}N\right)+2^{\sum_{i=n-1}^n(1+\alpha_i)}\sigma\left(2^{\alpha_{n-2}}N\right)+\ldots+2^{\sum_{i=3}^n(1+\alpha_i)}\sigma\left(2^{\alpha_{2}}N\right)$$ $$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+2\cdot2^{\sum_{i=2}^n(1+\alpha_i)}\sigma\left(2^{\alpha_{1}-2}N\right)=2^{\sum_{i=2}^n(1+\alpha_i)}\sigma\left(2^{\alpha_{1}-1}N\right).$$
I believe that the corresponding and similar Claim that one can deduce for the right condition, and for such cases that were studied, should be very similar than mine. Feel free if you can add in your answer if these conditions are potentially interesting in the study of the unsolved problem about the existence of odd perfect numbers. Many thanks.