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The following question is inspired mostly by this question, answer and the comment by Wojowu there

A naive approach to understanding odd perfect numbers is to make a Dirichlet series where the $n$th odd terms is zero if and only if $n$ is an odd perfect number, namely $$A(s)= \zeta(s)\zeta(s-1) -2 \zeta(s-1).$$

One might hope that one could then use analysis to get some sort of non-trivial statement about when a term can be zero.

One could then look at something like $$\lim_{T \rightarrow \infty} \frac{1}{2T} \int_{-T}^{T} A(s+it)x^{s+it} dt . (1) $$

As long as s is sufficiently large (and in fact , we may take $s>2$), when $x>0$, the limit in (1) is equal to $\sigma(n)-2n$ exactly when $x=n$ and and 0 otherwise. However, there's no content in this statement involving the series $A(s)$ other than that that the Dirichlet series converges if $s>2$. But one might hope that this framework or something like it could get non-trivial statements about when the above integral can be zero with odd $n$.

The most naive thing to start off with here would be to see if one can recover very basic properties of perfect numbers in this analytic context. Here are three statements which are very easy to prove and none of which even require unique prime factorization:

  1. No perfect number is a power of a prime.
  2. No perfect number is congruent to 3 (mod 4).
  3. No perfect number is congruent to 2 (mod 3).

So the question is, can we use this analytic approach to recover any of these statements or any similar statements? This would seem to be a reasonable test that this sort of framework has even a small chance of being productive.

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  • $\begingroup$ @StevenClark Possible that I've made a mistake here. It should be valid for any s in the range of convergence, with a result in x terms of x. But I think this sort of formula should be standard. Checking back, there's an essentially identical formula in Apostol's "Introduction to Analytic Number" (Theorem 11.17). Possible I've made some sort of basic error here though. $\endgroup$
    – JoshuaZ
    Commented Aug 23, 2021 at 18:46
  • $\begingroup$ I don't understand the integral. There are two variables $s$ and $x$ as well as the integration variable $t$. Evaluating the integral using the Dirichlet series for $A(s)$ leads to $-\frac{i}{2 T}\sum\limits_{n=1}^\infty(\sigma_1(n)-2 n)\frac{\left(\frac{x}{n}\right)^{s+i T}-\left(\frac{x}{n}\right)^{s-i T}}{\log\left(\frac{x}{n}\right)}$ and I'm not sure the limit of the term $\underset{T\to\infty}{\text{lim}}\left(-\frac{i}{2 T}\frac{\left(\frac{x}{n}\right)^{s+i T}-\left(\frac{x}{n}\right)^{s-i T}}{\log \left(\frac{x}{n}\right)}\right)$ is determinate. $\endgroup$
    – Steven Clark
    Commented Aug 23, 2021 at 18:47
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    $\begingroup$ I believe I can give you an analytic formula for $\sigma _1(n)-2 n$ (e.g. see the analytic formula for $\sigma_0(n)$ at mathoverflow.net/q/395950 which is based on the answer I posted to one of my own questions at mathoverflow.net/q/395266), but I'm not sure it's going to be useful as I believe the simplification involves the arithmetic function $2\ \phi(n)-n$ (see oeis.org/A083254) which is related to $\sigma_1(n)-2 n$. $\endgroup$
    – Steven Clark
    Commented Aug 23, 2021 at 19:57

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