I found that the bound $\sqrt{3x}$ appears to work, but I just want to see if there are any tighter bounds, and if not, what a proof for this would look like.
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1Have you looked at https://math.stackexchange.com/questions/1699330/upper-bound-on-number-of-factors – Benjamin Wang Dec 17 '20 at 20:59
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No actually. But I would still like some sort of proof for curiosity sake. – LthMath Dec 17 '20 at 22:30
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https://en.wikipedia.org/wiki/Superior_highly_composite_number – Will Jagy Dec 17 '20 at 23:16
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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 $$ \delta = \frac{\log \left(1 + \frac{1}{k} \right)}{\log p}. $$
So $$ N_{1/2} = 12, \; N_{1/3} = 2520, \; N_{1/4} = 21621600, $$
$$ N_{0.23} = 6983776800, \; N_{0.155} \approx 6.929 \cdot 10^{40}, \; N_{0.1218} \approx 3.309 \cdot 10^{135}.$$
Will Jagy
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