Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

if $x=d(n)$ is the number of divisors of $n$, what is the tightest lower-bound for $n$ only given $x$?


share|improve this question
add comment

2 Answers 2

up vote 15 down vote accepted

I will start off with the simplest type, $$ d(n) \leq \sqrt{3 n} $$ and $$ d(n) \leq 24 \left(\frac{n}{315}\right)^{1/3} $$ The first one has equality at $n = 12,$ second at $n =2520.$ Instead of continuing with fractional powers $1/k$ the better results switch to logarithms. Reference is a paper by J. L. Nicolas in a book called Ramanujan Revisited.

With equality at $n = 6983776800 = 2^5 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19$ and $d(n) = 2304,$ $$ d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1.5379398606751... \right)} $$

With equality at a number $n$ near $6.929 \cdot 10^{40},$ $$ d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1 + \frac{1.934850967971...}{\log \log n} \right)} $$

With equality at a number $n$ near $3.309 \cdot 10^{135},$ $$ d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1 + \frac{1}{\log \log n} + \frac{4.762350121177...}{\left(\log \log n \right)^2} \right)} $$

Just to fill in one blank, the special integers $n$ here are "superior highly composite numbers" using Ramanujan's original recipe for prime factorization, which I like to write, with $ \delta > 0,$ as $$ N_\delta = \prod_p \; p^{\left\lfloor \frac{1}{p^\delta - 1} \right\rfloor } $$ So $$ N_{1/2} = 12, \; N_{1/3} = 2520, \; N_{0.23} = 6983776800, \; N_{0.155} \approx 6.929 \cdot 10^{40}, \; N_{0.1218} \approx 3.309 \cdot 10^{135}.$$

share|improve this answer
add comment

It would be nice to have an inequality $n \ge f(x)$. If the poser wants numerical results, here are two:

The least number having exactly x divisors is given by OEIS sequence http://www.oeis.org/A005179. It is a pretty wild function. The nice paper by Grost is recommended.

The least number having x (or more) divisors is given by the OEIS sequence http://www.oeis.org/A061799.

share|improve this answer
add comment

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.