Let $m\geq 1$ an integer and $\sigma(m)=\sum_{d|m}$ the sum of positive divisors function. A positive integer is said to be perfect if and only if $\sigma(n)=2n$. We have that $\sigma(m)$ is a multiplicative function (see this site or [1]), thus we have the known
Fact. If $n$ is an odd perfect number then $\sigma(\sigma(n))=6n$.
Proof. We don't know if there are odd perfect numbers, is an unsolved problem, but assuming the existence then $\sigma(2n)=\sigma(2)\cdot\sigma(n)=3\cdot 2n=6n$, since $gcd(2,n)=1$, and this is the end of the proof.
Hypothesis to work. In this post we assume that $n=n_1$ is the first odd perfect number, if there is one.
Now we define iterated of the sum of divisor function, as is ususal $$\sigma^k(m)=\sigma(\sigma^{k-1}(m))=\sum_{b|\sigma^{k-1}(m)}b$$ for integers $k>1$ and $\sigma^1(m)=\sigma(m)$.
Hypothetical algorithm. And I compute terms of the sequence $$\sigma^k(n)$$ always excluding the possibility that an prime previously computed in the factorization of all previous terms of the sequence $\sigma^t(n)$, are $t<k$, divides $n$. (Please if you want explain my idea in a better sense or words, too the question bellow feel free to edit, thanks.)
Thus my
Question. For the first odd perfect number $n=n_1$ where we can find/estimate the barrier of using this algorithm? Thanks in advance.
I say, $\sigma^3(n)=8\cdot 3\cdot n$ and I assume to add as hypothesis that $3\nmid n$. Now I compute using that the sum of positive divisor function is multiplicative another time... $\sigma^4(n)=\frac{2^4-1}{2-1}\cdot 4\cdot 2n=8\cdot 3\cdot 5\cdot n$. Well Tochard theorem say that perhaps is possible that there are odd perfect numbers of the form $12\lambda+1$, but perhaps I cannot assume yet that $5\nmid n$ (looking the barrier yes I assume and add the hypothesis that $5\nmid n$). Going to the behaviour of the sum of divisor function, and the fact that if $n_1$ exists, then can I apply the Fundamental Theorem of the Arithmetic and Euler's Theorem for odd perfect numbers to bound the barrier of use in previous algorithm?
References:
[1] Wikipedia pages for Divisor function and Odd Perfect Numbers.
Juan, what you need is a variant of factor/sigma chains.
You can check out this MSE link for a naive implementation of factor/sigma chains involving odd perfect numbers $N = {q^k}{n^2}$, where $q = 73$. Note that we still do not know whether $q = 73$, or otherwise.
However, note that we do know that $105 \nmid N$. (See: "Can an odd perfect number be divisible by 105?".)