I worked this out while working on the Lagarias/Robin/Nicolas inequalities:

$$\sigma(n) = \frac{n^2}{\phi(n)}\cdot\prod_{p|n}{\left(1-\frac{1}{p^{k+1}}\right)}$$

where $k$ is the largest exponent of $p$ which divides $n$. I haven't seen any relationship as explicit as this stated anywhere that I've been reading, although it makes the $\frac{n^2}{\phi(n)}$ upper bound on $\sigma(n)$ very obvious. Does anyone happen to know offhand if this is a well known identity and know where I can read more about it?

According to the Robin inequality, the product term there is what would be responsible for keeping this inequality true for n > 5040 if RH is true:

$$\sigma(n) < e^\gamma\cdot n\cdot \log(\log(n))$$

According to the Nicolas inequality, the lack of the product term makes this inequality true for all primorial n if RH is true:

$$\frac{\sigma(n)}{\prod_{p|n}{\left(1-\frac{1}{p^{k+1}}\right)}} > e^\gamma\cdot n\cdot\log(\log(n))$$

or, in a simplified fashion:

$$\prod_{p|n}{\left(\frac{p}{p-1}\right)} > e^\gamma\cdot \log(\log(n))$$

It comes from manipulation of Mertens' 3rd theorem that LHS $\sim e^\gamma\cdot \log(p)$, where $p$ is the largest prime dividing $n$. As it's sufficient only to show this for primorials, the RHS actually turns out to be:

$$e^\gamma\cdot \log(\theta(p)) \sim e^\gamma\cdot \log(p)$$

Side question, but is it sufficient to prove this inequality to note that LHS approaches the value asymptotically from above, and RHS approaches the value asymptotically from below? Numerically, LHS actually seems to approach $e^\gamma\cdot\log(q)$ from above, where $q$ would be the next prime that's larger than $p$, but I haven't determined how to prove that.

  • $\begingroup$ seems very similar to proof of Theorem 329 of Hardy and Wright, on page 267 of my edition (fifth). The outcome is, for a positive constant $A,$ that $$ A < \frac{\sigma(n) \phi(n)}{n^2} < 1. $$ It turns out that $A = \frac{6}{\pi^2},$ that is in a footnote on the same page. Refers to Theorem 280 $\endgroup$ – Will Jagy Nov 17 '16 at 17:52
  • $\begingroup$ I guess that would imply the lower bound of $\frac{6}{\pi^2}$ for $\prod_{p|n}{(1-\frac{1}{p^{k+1}})}$ which seems to hold for primorials. $\endgroup$ – J. Whidden Nov 17 '16 at 20:38
  • $\begingroup$ Does this imply that $\sigma(n) \cdot \frac{\pi^2}{6} > e^\gamma \cdot n \cdot $log$($log$(n))$? $\endgroup$ – J. Whidden Nov 17 '16 at 20:51
  • $\begingroup$ As to your last question, I don't know. J.-L. Nicolas is still active and posts most of his articles math.univ-lyon1.fr/~nicolas/publications.html $\endgroup$ – Will Jagy Nov 18 '16 at 1:14

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