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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.

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  • $\begingroup$ I don't know if it is in literature, feel free to made feedback, you are welcome to improve previous post, tittle, grammar... $\endgroup$
    – user243301
    Nov 5, 2015 at 10:20

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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?".)

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  • $\begingroup$ Very thanks much @ArnieDris, I will take notes and neccesarly I take the decission to choose now your answer. $\endgroup$
    – user243301
    Nov 10, 2015 at 13:36
  • $\begingroup$ You're welcome Juan! =) $\endgroup$ Nov 10, 2015 at 13:38
  • $\begingroup$ I ensure that I will read your notes, and previous references, I am interesting in perfect numbers too, this morning I am eliminate $\sqrt{1+8n}$ from two previous of my posts giving that if $n$ if an even perfect number then (if there are no mistakes): $2\phi(n)\sigma(2n)=2\phi(n)+\sigma(n)\cdot(2\phi(n)+(n-1))$. Perhaps it isn't useful to study qualitively even perfect numbers, but your references and answers encourage to me to continue. $\endgroup$
    – user243301
    Nov 10, 2015 at 13:43
  • $\begingroup$ If previous relation has not mistakes (I run a program until $10^4$ and only finds even perfect numbers), then it is easy to prove that if $n$ is a even perfect number with its Euler's form, $n$ satisfies previous cited equation. Remains the other implication $\to$, we put one $2$ inside the factor $\sigma(2n)$, but we need to pay in the use of Euler's totient function @ArnieDris $\endgroup$
    – user243301
    Nov 10, 2015 at 13:54

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