What is the lower bound for highly composite numbers? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T03:57:06Z http://mathoverflow.net/feeds/question/43103 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43103/what-is-the-lower-bound-for-highly-composite-numbers What is the lower bound for highly composite numbers? mna 2010-10-21T22:04:18Z 2010-11-05T16:57:20Z <p>if $x=d(n)$ is the number of divisors of $n$, what is the tightest lower-bound for $n$ only given $x$?</p> <p><a href="http://en.wikipedia.org/wiki/Highly_composite_number" rel="nofollow">http://en.wikipedia.org/wiki/Highly_composite_number</a></p> http://mathoverflow.net/questions/43103/what-is-the-lower-bound-for-highly-composite-numbers/43105#43105 Answer by Will Jagy for What is the lower bound for highly composite numbers? Will Jagy 2010-10-21T22:17:31Z 2010-10-22T19:46:54Z <p>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.</p> <p>With equality at $n = 6983776800 = 2^5 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19$ and $d(n) = 2304,$ $$d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1.5379398606751... \right)}$$</p> <p>With equality at a number $n$ near $6.929 \cdot 10^{40},$ $$d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1 + \frac{1.934850967971...}{\log \log n} \right)}$$</p> <p>With equality at a number $n$ near $3.309 \cdot 10^{135},$ $$d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1 + \frac{1}{\log \log n} + \frac{4.762350121177...}{\left(\log \log n \right)^2} \right)}$$ </p> <p>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 $$N_{1/2} = 12, \; N_{1/3} = 2520, \; N_{0.23} = 6983776800, \; N_{0.155} \approx 6.929 \cdot 10^{40}, \; N_{0.1218} \approx 3.309 \cdot 10^{135}.$$</p> http://mathoverflow.net/questions/43103/what-is-the-lower-bound-for-highly-composite-numbers/44894#44894 Answer by tdnoe for What is the lower bound for highly composite numbers? tdnoe 2010-11-04T23:27:02Z 2010-11-05T16:57:20Z <p>It would be nice to have an inequality $n \ge f(x)$. If the poser wants numerical results, here are two:</p> <p>The least number having exactly x divisors is given by OEIS sequence <a href="http://www.oeis.org/classic/A005179" rel="nofollow">http://www.oeis.org/classic/A005179</a>. It is a pretty wild function. The nice paper by Grost is recommended.</p> <p>The least number having x (or more) divisors is given by the OEIS sequence <a href="http://www.oeis.org/classic/A061799" rel="nofollow">http://www.oeis.org/classic/A061799</a>.</p>