Question

Let $\theta_1 = \sqrt[3]{2}$ and

$$\theta_2 = 2 + \sqrt[3]{4} + \sqrt[3]{2} = {\theta_1}\left(1 + \theta_1 + {\theta_1}^2\right).$$

Now, if $N = {q^k}{n^2}$ is an Odd Perfect Number (OPN) given in Eulerian form (where $q$ is prime, $q \equiv k \equiv 1 \pmod 4$, and $\gcd(q, n) = 1$), then (please do allow me to publicize this article here, albeit unsolicited) in the paper titled The Abundancy Index of Divisors of Odd Perfect Numbers, I proved that $I(q^k) < \theta_1 < I(n)$, where $I(x) = \frac{\sigma_1(x)}{x}$ is the abundancy index of the positive integer $x$ and $\sigma_1(x)$ is the sum of the divisors of $x$.

Now, note that $\theta_2 \approx 4.8473221018630726395189162465505366109617447926013609$ (click here for the WolframAlpha computation). Notice that $q \geq 5 > \theta_2$.

But we can rewrite $\theta_2$ as:

$$\theta_2 = {\theta_1}\left(1 + \theta_1 + {\theta_1}^2\right) = {\theta_1}\left({\frac{{\theta_1}^3 - 1}{\theta_1 - 1}}\right) = \frac{\theta_1}{\theta_1 - 1}.$$

Since $q > \theta_2 = \frac{\theta_1}{\theta_1 - 1}$, we have:

$$\frac{q}{q - 1} < \theta_1.$$

Consequently:

$$q > {\theta_1}^3 + {\theta_1}^2 + \theta_1 > {\left(\frac{q}{q - 1}\right)}^3 + {\left(\frac{q}{q - 1}\right)}^2 + \frac{q}{q - 1}.$$

This implies that:

$$I(q) = \frac{\sigma_1(q)}{q} = \frac{q + 1}{q} = 1 + \frac{1}{q} < 1 + \frac{1}{q}{\left[\frac{1}{{\left(\frac{q}{q - 1}\right)}^3 - 1}\right]}$$

An explicit form for the upper bound is given by this WolframAlpha computation as:

$$I(q) < \frac{4q^3 - 6q^2 + 4q - 1}{3q^3 - 3q^2 + q}$$

The ubiquity of the $\frac{1}{2}$'s in the real root and complex roots of the upper bound for $I(q)$ is striking:

Real root: $$q = \frac{1}{2}$$

Complex roots: $$q = \frac{1}{2} - \frac{i}{2}$$ $$q = \frac{1}{2} + \frac{i}{2}.$$

Must this possibly imply a relationship between the OPN Conjecture and the Riemann Hypothesis (RH)? Perhaps somebody with more experience on and greater expertise in (algebraic and/or analytic) number theory could shed some light on how and why we have this "coincidence".

Any ideas/comments/suggestions are most welcome!

Answer 1

The polite thing would be to let Greg Martin's comment stand. However, you asked for suggestions: I think the following will help, although the tone will be less polite. I intend full respect.

1) Take a cynical (or disbelieving) view. (With your own work, that can be hard, especially after you have sweated out some result. I know. I've been there.) One such application of cynicism boils away the post down to: I(q)= 1 + 1/q < 1 +R(q)/q, where R(q) is some rational function of q such that R(q)> 1 when q>= 5. Since R(q) has magical properties, does that mean I(q) does too?

Even if R(q) gave you the exponent n that is in a nontrivial solution to Fermat's Most Famous Equation ( and tempting you to say "In Your Face!" to Andrew Wiles), it is an arbitrary choice which will not serve as an informative bound and not tell you anything more than what you have already assumed, which is that q is at least 5 and that I(q)= 1 + 1/q. If you want to link up Riemann with odd perfect numbers, using R(q) as a not provably strong upper bound won't do it. R(q) may be crucial in some other work, but your post does not show this.

If you have trouble applying cynicism to your work, get some private help. MathOverflow can help resolve technical tricky issues sometimes, but this posting is not like that.

2) Self promotion can be done in a subtle fashion on MathOverflow, and the community will not run you out on a rail. Instead of the paragraphs above referencing your work, say "In my work [1], I proved this and was moved to consider that...", followed by the technical question of interest, and a brief hyperlinked bibitem resolving [1] . I suggest editing your post to tone that aspect down.

Gerhard "Self Promotion? Who Me? Pshaw!" Paseman, 2013.02.20