Asymptotic divisor function / primorials

Let $p_n\#\equiv\prod_{k=1}^{n}p_k$ (primorial), and $\sigma(n)=\sum_{d|n}^{}d$ (divisor function).

Could someone please tell me what the general asymptotic of $\dfrac{\sigma(p_n\#)}{p_n\#}$ is? It looks like $O \log(n)$. Is there anything more precise?

Update

Thank you to Daniel Fischer for his answer. Out of interest, I include the plot of $\dfrac{\sigma(p_n\#)}{p_n\#}$ against $\dfrac{6 e^{\gamma}}{\pi^2} \log p_n$. The primorial curve is surprisingly smooth though - why is this?

• FWIW it's usually more straightforward to work in terms of $\sigma_{-1}$, so I'd sooner write $\sigma_{-1}(p_n\#)$ than $\sigma(p_n\#)/p_n\#.$ – Charles May 20 '14 at 19:44
• Yes - that makes sense! – martin May 20 '14 at 19:47

The divisor sum function is multiplicative, and $\sigma(p) = p+1$ for a prime $p$. So

$$\frac{\sigma(p_n\#)}{p_n\#} = \prod_{k=1}^n \left( 1 + \frac{1}{p_k}\right) = \frac{\prod_{k=1}^n \left(1 - \frac{1}{p_k^2}\right)}{\prod_{k=1}^n\left(1 - \frac{1}{p_k}\right)}.$$

By Mertens' third theorem, we have

$$\prod_{k=1}^n \left(1 - \frac{1}{p_k}\right) \sim \frac{e^{-\gamma}}{\log p_n},$$

and by Euler's product formula

$$\lim_{n\to\infty} \prod_{k=1}^n \left(1 - \frac{1}{p_k^2}\right) = \frac{1}{\zeta(2)} = \frac{6}{\pi^2}.$$

Therefore, we have the asymptotics

$$\frac{\sigma(p_n\#)}{p_n\#} \sim \frac{6 e^{\gamma}}{\pi^2} \log p_n.$$

Since $p_n \sim n\log n$, we have $\log p_n \sim \log n$, but $\log n + \log \log n$ is a little more accurate.

• Yes, this looks right. But your $\sim$ is lossy enough that I don't think using $\log n+\log\log n$ will be an improvement over $\log n$. – Charles May 20 '14 at 19:31
• I don't know how good the convergence in Mertens' theorem is. For the $\zeta$ product, the error is less than a factor $1+\frac{1}{p_n}$, so $\log n + \log \log n$ versus $\log n$ makes a bigger difference. It may be an improvement or not, depends on the unknown [to me] quality of the asymptotics in Mertens' theorem. – Daniel Fischer May 20 '14 at 19:41
• Pretty bad, I think. Also dropping the error term $\prod_{k=n+1}^\infty(1-1/p_k^2)$ makes a difference (though not a $\sim$-difference). Still, this is the right answer and I +1'd you. – Charles May 20 '14 at 19:43
• @ Daniel Fischer - beautiful! Thank you very much for your answer :) – martin May 20 '14 at 19:55

Daniel's answer sems to cover it. Note gronwall's theorem, Thm 323 on page 266 of Hardy nad Wright, $$\limsup \frac{\sigma(n)}{n \log \log n} = e^\gamma.$$ H+W use the sequence of numbers $$\operatorname{lcm} \{1,2,3, \ldots, \}$$ to show this, prrof pages 353-354.

Daniel can confirm whether the primorials give a good sequence for Gronwall's.

Meanwhile, the best numbers are not primorials, although they are the product of primorials, meaning that the exponents in the prime factorization decraese, or at least do not increase. The best numbers here are the colossally abundant numbers, by Erdos and Alaoglu, but actually cut out of the original Ramanujan article owing to paper shortages at the time. (Don't ask me, i wasn't there).

Also note that the criterion of Nicolas (1983) is much easier to deal with, as the primorials are the genuine test case there. Answer with proof: Is the Euler phi function bounded below? ... other answer Euler's Phi Function Worst Case

• Yes, - colossally abundant numbers certainly make things a little more difficult in terms of combinatorics though... – martin May 20 '14 at 19:42
• What is the criterion of Nicolas that you refer to? Do you know of any interesting links here? – martin May 20 '14 at 19:43
• @martin, yes, I do, but I have a doctor appointment, allergies. I think you are able to look in my profile and then ask for my favorites. Out of some four pages of those, there are some items about extreme values of Euler's totient function. See those. in any case Nicolas' original paper is available fro his website. – Will Jagy May 20 '14 at 20:05
• Great - thank you :) ... I might add that my French is not good :( – martin May 20 '14 at 20:07
• @martin, the Planat article is in English, arxiv.org/abs/1012.3613 – Will Jagy May 20 '14 at 21:44