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Let $n$ be a positive number. I want to bound the number of divisors of $n$. Let $d(n)$ is the number of divisors of $n$. Let $n = p_1^{e_1}p_2^{e_2}\cdots p_r^{e_r} $ be the prime factorization of $n$. then

$$d(n) = (e_1 +1)(e_2 +1)\cdots (e_r +1)$$

we can see that $e_i + 1 \le 2^{e_i}$

I don't know how to proceed i.e to prove $d(n) < 2^{(1 + \epsilon) \log n/ \log \log n}$ for $\epsilon$ greater than zero. Is there exist a better bound?

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  • $\begingroup$ Proceed to what? $\endgroup$
    – Hagen von Eitzen
    Commented Feb 4, 2018 at 13:38
  • $\begingroup$ You want to bound the divisors : do you want a bound completely in terms of $n$ and not depending on stray terms? $\endgroup$
    – Sarvesh Ravichandran Iyer
    Commented Feb 4, 2018 at 13:40
  • $\begingroup$ @астон вілла олоф мэллбэрг I have edite the question $\endgroup$
    – user437890
    Commented Feb 4, 2018 at 13:47
  • $\begingroup$ Roughly speaking, we have particularly many divisors when $n$ is the product of $r$ distinct small primes. One can argue that $p_r\sim \log n$ and that there are $\sim\frac{p_r}{\log p_r}$ primes below $p_r$, so $r\sim\frac{log n}{\log \log n}$ ... $\endgroup$
    – Hagen von Eitzen
    Commented Feb 4, 2018 at 13:48
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    $\begingroup$ @RossMillikan reults of Nicolas and/or Robin summarized in math.univ-lyon1.fr/~nicolas/hcnrevisited.pdf $\endgroup$
    – Will Jagy
    Commented Feb 4, 2018 at 19:16

1 Answer 1

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Added: the first inequality is not tied to $\log 2$ well enough. The second one, from Robin's dissertation, does give $$ \frac{\log d(n)}{\log 2} \leq \left( \frac{\log n}{\log \log n} \right) \left( 1 + \frac{1.934850967971...}{\log \log n} \right)$$

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.

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 $3.309 \cdot 10^{135},$ $$ d(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1 + \frac{1}{\log \log n} + \frac{4.762350121177...}{\left(\log \log n \right)^2} \right)} $$

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