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>



What is the lower bound for highly composite numbers?
I will start off with the simplest type, $$ d(n) \leq \sqrt{3 n} $$ and $$ d(n) \leq 48 \left(\frac{n}{2520}\right)^{1/3} $$ and
$$ d(n) \leq 576 \left(\frac{n}{21621600}\right)^{1/4}. $$
The first one has equality only at $n = 12,$ second only at $n =2520,$ third only at $n= 21621600.$ Instead of continuing with fractional powers $1/k$ the better results switch to logarithms. Reference is a 1988 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 $d(n) = 2304,$
$$ d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1.5379398606751... \right)} = n^{ \left( \frac{1.0660186782977...}{\log \log n} \right) }. $$ Full details of the proof appear in J.-L. Nicolas et G. Robin. *Majorations explicites pour le nombre de diviseurs de n*, Canad. Math. Bull., 26, 1983, 485--492. The next two appear in the dissertation of Robin, are repeated in the 1988 Nicolas survey article indicated.


With equality at a number $n$ near $6.929 \cdot 10^{40},$
$$ d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1 + \frac{1.934850967971...}{\log \log n} \right)}. $$ Compare this one with Theorem 317 in Hardy and Wright, attributed to Wigert (1907),
$$ \limsup \frac{\log d(n) \log \log n}{\log n} = \log 2. $$


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
$$ N\_\delta = \prod\_p \; p^{\left\lfloor \frac{1}{p^\delta - 1} \right\rfloor } $$
The first (largest) $\delta$ that assigns an exponent $k$ to a prime $p$ is 
$$ \delta = \frac{\log \left(1 + \frac{1}{k} \right)}{\log p}. $$
See [Is there a formula that can predict the primes in the sequence of ratios of consecutive superior highly composite numbers? : $2, 3, 2, 5, 2, 3, 7,...$](https://mathoverflow.net/questions/185493/is-there-a-formula-that-can-predict-the-primes-in-the-sequence-of-ratios-of-cons/185529#185529) for some detail, with computations. 


So $$ N\_{1/2} = 12, \; N\_{1/3} = 2520, \; N\_{1/4} = 21621600, $$  

$$ N\_{0.23} = 6983776800, \; N\_{0.155} \approx 6.929 \cdot 10^{40}, \; N\_{0.1218} \approx 3.309 \cdot 10^{135}.$$


