Extraordinary Numbers

Question

Can you please explain what are Extraordinary Numbers in detail? At the same time, I would also like to confirm whether the equivalent problem of Riemann Hypothesis mentioned here is correct (like it's accepted by the mathematical community or not). As far as I know:

Take a function $G$ for integer $n > 1$ such that

$G(n) = σ(n)/(n(ln(ln(n))))$

where $σ(n)$ is the divisors sum function i.e. the sum of the positive divisors of $n$ and $ln$ denotes natural logarithm.

A positive composite integer $N$ is said to be an extraordinary number if its

$G(N) ≥ max[G(N/p), G(aN)]$

for all prime factor $p$ of $n$ and for all multiples $a*N$ of $N$.

It seems like it has been proved that the Riemann Hypothesis is true if and only if $4$ is the only extraordinary number. But I am unable to find the proof in some trusted peer-reviewed journal so am not sure.

Source https://planetmath.org/extraordinarynumber#:~:text=A%20positive%20integer%20N%20N%20is%20said%20to,true%20iff%204%20is%20the%20only%20extraordinary%20number.

EDIT: Thanks @amateurdotcounter for providing the link to the journal where the proof for this equivalent problem is published. Now I seek the answer to my first question only i.e. if anybody can explain the problem in simple, clear and concise way.

Answer 1

The page you linked references an arXiv preprint (which corresponds to this publication).

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If you read the arXiv preprint you'll find the following definition:

Definition 1. A positive integer $N$ is extraordinary if $N$ is composite and satisfies:
(i). $G(N ) ≥ G(N/p)$ for all prime factors $p$ of $N$ , and
(ii). $G(N ) ≥ G(aN )$ for all multiples $aN$ of $N$ .

(Where $G(n):=σ(n)/n \ln(\ln(n))$ (for $n>1$) is shown earlier in Theorem 1.)

And, from the line setting up this definition, and the lack of other references to other "extraordinary number" papers, it seems that this publication is the first source of this coined phrase "extraordinary number".

That's all it is... an "extraordinary number" is a (composite) number that satisfies (i) and (ii) from above.

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And finally this is why your question is difficult to answer, not because papers on "extraordinary numbers" are so rare, but because you haven't really "shown your work" in any way that allows folks to guess what point(s) in the above definition is confusing you.

(A) Do you know what a composite number is?
(B) Do you know how to calculate $\sigma (n)$ for $n$ needed to calculate $G(n)$?
(C) Do you know how to calculate $\ln (n)$ for $n$ needed to calculate $G(n)$?
(D) Do you know how to calculate G(n) for $n$ based on it's form in Theorem 1?

If "yes" to all the above, then what more of the definition needs to be explained "in detail"?

Or are you having trouble more-so with applying that later on?

(E) Are you having issues calculating $G(N/p)$ for all prime factors or $N$?
(F) Are you having issues calculating $G(aN)$ for all integer multiples $a$?

If you were playing around with the numbers and got stuck at point (E) then someone could answer that pretty simply, or if you got stuck at (F) they could likely help walk you through that... but having said nothing, nobody knows whether you're stuck at (A) or (B) or (C) or (D) or (E) or (F) or somewhere else entirely.

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So, let's actually do the work and play around with one of these things.

Let's see if $N=6$ is extraordinary.

Is it composite?
Is (i) satisfied?
Is (ii) satisfied?

The number $N=6$ has prime factors $\{2,3\}$ and so it is composite.

Furthermore, for (i) that means calculating $G(N/p)$ fro all possible prime factors requires calculating both $\{G(3),G(2)\}$ values. $$ G(3)=\frac{\sigma(3)}{3\ln(\ln(3))}=\frac{(1+3)}{3\ln(\ln(3))}\approx 14.17718... $$ $$ G(2)=\frac{\sigma(2)}{2\ln(\ln(2))}=\frac{(1+2)}{2\ln(\ln(2))}\approx -4.09263... $$ And of course we'll also need $G(6)$ to make those $G(N ) ≥ G(N/p)$ comparisons. $$ G(6)=\frac{\sigma(6)}{6\ln(\ln(6))}=\frac{(1+2+3+6)}{6\ln(\ln(6))}\approx 3.42937... $$ And so, we can see that $G(6)\geq G(2)$ is true, but unfortunately $G(6)<G(3)$ and so (i) is not satisfied for all possible prime factors $p$ and therefore $N=6$ is not an "extraordinary number".

We could (should?) stop there, but maybe working through an example of (ii) would also be illustrative.

This time, to prove (ii) we need to show whether $G(N)≥G(aN)$ for all integer multiples $a$ (or more simply, disprove it by finding a single counterexample $a$ value).

Obviously, $G(N)\geq G(aN)$ works for $a=1$ because they'd be exactly equal.

So, let's test whether $G(N)≥G(aN)$ for the $a=2$ case of our $N=6$ problem:

$$ G(12)=\frac{\sigma(12)}{12\ln(\ln(12))}=\frac{(1+2+3+4+6+12)}{12\ln(\ln(12))}\approx 2.56344... $$

So, it looks like $a=2$ works (since this number is smaller that the $G(6)\approx 3.42..$ calculated earlier).

What about $a=3$? $G(18)\approx 2.04136...$ is smaller than $G(6)$, confirmed.

What about $a=4$? $G(24)\approx 2.16213...$ is smaller than $G(6)$, confirmed.

We obviously can't manually verify this for every single infinite possible $a$ value... so we'd ideally need to find some relationship that does prove $G(N)\geq G(AN)$ for some big $A$ values beyond a certain point. This is where Robin's inequality gets involved in the paper: $$ G(n) < e^\gamma+\frac{0.6483}{(\ln(\ln(n)))^2} $$ for $(n>1)$. This gives us a maximum upper bound (overestimate) of how big $G(n)$ could be for a given value of $n$. So, if we calculate the Robin upper bound for $n=12$ we see that: $$ G(12) < e^\gamma+\frac{0.6483}{(\ln(\ln(12)))^2} \approx 2.99067... $$ which is quite a bit above the calculated value we found earlier, but it very importantly is still well below the $G(6)$ we caluclated. And then, because Robin's upper bound is monotonically decreasing for large enough values we can say confidently that $G(6)>e^\gamma+\frac{0.6483}{(\ln(\ln(12)))^2}\geq e^\gamma+\frac{0.6483}{(\ln(\ln(6a)))^2}\geq G(6a)$ (for $a\geq 2$) will always apply and so $G(6)\geq G(6a)$ will always apply, and so (ii) is satisfied.

So, for $N=6$ we have:

Is it composite? YES.
Is (i) satisfied? NO, $p=2$ provides a counterexample where $G(N)\geq G(N/p)$ fails.
Is (ii) satisfied? YES.

And therefore $N=6$ is NOT extraordinary.

Similar work can be done to prove whether another specifically chosen number (e.g $N=234,879,230,902$) is or isn't extraordinary.