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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

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up vote 19 down vote accepted

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? : $2, 3, 2, 5, 2, 3, 7,...$ for some detail, with computations.

So $$ N_{1/2} = 12, \; N_{1/3} = 2520, \; N_{1/4} = 21621600, $$
$$ N_{0.23} = 6983776800, \; N_{0.155} \approx 6.929 \cdot 10^{40}, \; N_{0.1218} \approx 3.309 \cdot 10^{135}.$$

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For the benefit of future readers, let $x=d(n)$, then inverting the bounds given above, $x^2/3 \leq n, 2520(x/48)^3 \leq n, 21621600(x/576)^4 \leq n$, and soon. The more ambitious can try inverting the bound involving logarithms. Gerhard "Leaves Something For Future Readers" Paseman, 2015.12.04 – Gerhard Paseman Dec 5 '15 at 5:58

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.

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