Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have been searching for a bound of the divisor function $d(n)$, meaning the number of divisors of n. So far I have found that it can be bounded by $$ d(n) \le e^{O(\frac{\log n}{\log \log n})}$$ Wigert has proven the constant is $\log 2$ so $$ d(n) \le e^{(\log 2+ o(1)) \frac{\log n}{\log \log n}} $$ However, when I tried to check that bound on a computer, it did not seem right. I have drawn $d(n)$ (in blue), $e^{\frac{\log n}{\log \log n}}$ (in red) and $e^{\log 2 \frac{\log n}{\log \log n}}$ (in green) on the following graph:

enter image description here

Furthermore, when I plot $\frac{\log d(n)}{\log n / \log \log n}$, it does not seem to have $\log 2$ as a limit either. See this other graph:

enter image description here

It appears the constant 1 is a much better fit !

So my question is: is Wigert's bound only true for large $n$ (i.e larger than $10^6$) ?

My confusion comes from the second equation, which is clearly a bound (see this post on Tao's blog) and the result of Wikipedia, which is a limit.

share|cite|improve this question
E. R. Canfield, P. Erdos, C. Pomerance, On a problem of Oppenheim concerning Factorisato Numerorum, J. Number Theory 17 (1983) 1-28, Table 1, column "n". [R. J. Mathar, mathar(AT), Mar 06 2009] has an additional factor of log log log n in the numerator of the exponential. And 10^6 is quite small. – Ross Millikan Sep 11 '11 at 21:40
up vote 2 down vote accepted

If you want to see how the bound is approached for bigger values of $n$ you might want take a look to the last graph in this page about highly composite numbers and the tables linked there. There you see how the ratio $\log(d(n))/\mathrm{Li}(\log(n))$ approaches $\log 2$ as $n$ goes through the highly composite numbers, (ie numbers such that $d(n)>d(m)$ for all $m=1,2,\dots, n-1$).

As $\mathrm{Li}(x) \sim x/\log x$ as $x \to \infty$ the asymptotic is equivalent to the one you state but it seems to work much better, as an example the last entry $n$ in the tables (an integer with 17000 decimal digits) gives $$ \frac{\log d(n)}{\mathrm{Li}(\log n)} = 0.6946\ldots \quad \text{but} \quad \frac{\log d(n)}{\log n/\log\log n}=0.7793\ldots$$

share|cite|improve this answer
Thank you, it is just that I wanted. Before I mark your post as the answer, can you tell me what $Li$ is ? – alex_reader Sep 12 '11 at 18:35
$\mathrm{Li} x$ is the logarithmic integral $=\int_2^x \frac{ dt}{\log t}$ see here. – Esteban Crespi Sep 12 '11 at 19:04

Are you confusing the notions of bound and limit? Consider a function $f(n)$ which is 1 unless $n$ is a power of 2, in which case $f(n)=n$. This function is bounded by $n$; $f(n)\le n$. But it doesn't have $n$ as a limit (or to put it another way, $f(n)/n$ doesn't have 1 as a limit). The function isn't bounded by 1, but the constant 1 looks like a much better fit for the graph than the bound $n$.

$d(n)$, like $f(n)$, is generally quite small, but occasionally very large. The occasional large value pushes the bound up.

share|cite|improve this answer
Thank you for clearing that up. But I understood that the second bound should be true for all n, and it does not look like it. I have edited my question. – alex_reader Sep 12 '11 at 18:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.