Question

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

http://en.wikipedia.org/wiki/Highly_composite_number

Answer 1

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 $$ N_{1/2} = 12, \; N_{1/3} = 2520, \; N_{0.23} = 6983776800, \; N_{0.155} \approx 6.929 \cdot 10^{40}, \; N_{0.1218} \approx 3.309 \cdot 10^{135}.$$

Answer 2

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/classic/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/classic/A061799.