# Heuristics on the asymptotic behaviour of the divisor funcion

Let $\sigma(n)$ be the sum of divisor function. We know, by Gronwall's Theorem, that

$${\lim \text{sup}}_{n \to \infty} \frac{\sigma(n)}{n \log \log{n}} =e^\gamma$$

And the Riemann Hypothesis (Robin's Inequality) states that:

$$\frac{\sigma(n)}{n \log \log{n}} <e^\gamma \: \forall n>5040$$

But, regarding the following equality:

$$\sigma(n)-e^\gamma n \log \log{n}= O(f_1(n))$$

• Do we know anything about what is supposed to be $f_1(n)$?

• Do we have any (either formal or heuristic) argument to defend any special function?

• When trying to go beyond that $f_1(n)$, what would be the next step? Finding a function $f_2(n)$ such that

$$\sigma(n)-e^\gamma n \log \log{n}-f_1(n)= O(f_2(n))$$

? If so, do we know/believe anything about it?

• Do we know/believe at which "step" $k$ (if any) we would find something of the form:

$$\sigma(n)-e^\gamma n \log \log{n}-f_1(n)- \cdots -f_k(n)= O(1)$$ ?

Thank you in advantage and apologizes for my bad English.

• The appropriate comparison is not between $\sigma(n)$ and $n \log \log n$ but between $\sigma(n)$ and $e^\gamma n \log \log n$. What text references are you using for your studies? Might I recommend Tenenbaum's Introduction to Analytic and Probabilistic Number Theory? It has many answers you are looking for. – Matthew Conroy Feb 6 '18 at 19:43
• Oops, a typo. I forgot about the factor. Thank you very much for the bibliography, I will take a look at it as soon as I can – user3141592 Feb 6 '18 at 19:48

## 2 Answers

The 1984 article of Robin also contains the unconditional result, for $n \geq 3,$ $$\sigma(n) \leq n e^\gamma \log \log n + \frac{ n \cdot 0.64821364942... }{\log \log n},$$ with the constant giving equality at $n=12.$ The logarithms are natural.

• Nice result there. – Matthew Conroy Feb 6 '18 at 20:44
• @MatthewConroy zakuski.utsa.edu/~jagy/Robin_1984.pdf – Will Jagy Feb 6 '18 at 20:48
• @MatthewConroy I actually prefer the version of his adviser, Nicolas. Most people have trouble programming the superior highly composite numbers and experiments...The primorials are much easier to fiddle with zakuski.utsa.edu/~jagy/Nicolas_1983.pdf – Will Jagy Feb 6 '18 at 20:58
• Thanks: these are nice references. Cheers! – Matthew Conroy Feb 6 '18 at 21:00
• On the other hand, this bound means anything when talking about asymptotic behaviours, since we could also write $\sigma (n) \le n e^{\gamma} \log{\log{n}}+ e^{x}$ – user3141592 Feb 23 '18 at 8:23

One can show $$\sigma(n) \le n e^\gamma \log \log n + O(n).$$ On the other hand, when $n$ is prime, we have $$\sigma(n) = n+1.$$ so, in fact, $$n e^\gamma \log \log n -\sigma(n) = n e^\gamma \log \log n -n -1$$ infinitely often. The sum of divisors function is too erratic to say anything more unless you want to talk about average orders.

Here is a plot of $\sigma(n)-e^\gamma n \log \log n$ for $n<10^5$. Here is a plot of $\frac{\sigma(n)}{e^\gamma n \log \log n}$ for $n<10^5$. • And what about a lim sup instead of an average? And, by the way, is there any paper where I could read a proof of that $O(n)$? Thank you – user3141592 Feb 6 '18 at 20:07
• One reference is Tenenbaum's Introduction to Analytic and Probabilistic Number Theory, , page 86. I don't know what you mean by "what about a lim sup...?". You have Gronwall's Theorem. – Matthew Conroy Feb 6 '18 at 20:13
• Sorry for missexpressing myselft. I meant an asymptotic behaviour of $$\sigma(n) -e^\gamma n \log \log{n} -cn$$ where c is the constant given by $$\text{lim sup}_{n \to \infty} \frac{\sigma(n)-e^\gamma n \log \log{n}}{n}$$ – user3141592 Feb 6 '18 at 20:20
• What do you think the asymptotic behavior is? And what do you think the value of $c$ is? – Matthew Conroy Feb 6 '18 at 20:29
• What analytic approach have you tried? The computations and plotting are, for me, good ways to see what we might want to, and might reasonably try to, prove analytically. What do you think the value of $c$ is? – Matthew Conroy Feb 6 '18 at 20:35