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By Dirchlet's hyperbola method, one can prove that the average number of divisors of integers $1 \leq n \leq X$ is $\log X$. This question concerns the number of integers $n \leq X$ such that the number of divisors, $d(n)$, is substantially larger than average. Indeed, what is known about the size of the set

$$\displaystyle \{1 \leq n \leq X : d(n) > (\log X)^A \}$$

where $A > 1$ is considered to be a large (but fixed) positive number?

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    $\begingroup$ Karl K. Norton, "On the frequencies of large values of divisor functions", Acta Arithmetica 68:3 (1994), 219-244. See eudml.org/doc/206657 $\endgroup$ Sep 20 at 21:28
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Theorem 1.11 and Theorem 1.22 of the paper by Norton, cited in the comment of Peter Humphries, show that for any fixed $A \ge \log 2$, $$ \frac{X (\log\log X)^{O(1)}}{(\log X)^{B(A)}} \ll_A |\{1\le n\le X:d(n) \ge (\log X)^A\}| \ll_A \frac{X}{(\log X)^{B(A)}}, $$ where $$ B(A):=1+\frac{A}{\log 2}\left(\log\left(\frac{A}{\log 2}\right) -1 \right). $$ Equation (1.37) of the same paper gives the correct order of magnitude: For every fixed $A>\log 2$, $$ |\{1\le n\le X:d(n) \ge (\log X)^A\}| \asymp_A \frac{X}{(\log X)^{B(A)} (\log\log X)^{1/2}}. $$

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The normal order of $\log(d(n))$ is $\log(2)\log\log(n))$. So, for every $\epsilon>0$, $$ \log(d(n))<(1+\epsilon)\log(2)\log(\log(n)) $$ hold for almost all n: that is, if the proportion of $n\le x$ for which this does not hold tends to 0 as $x$ tends to infinity. Thus $$ d(n) < \log(n)^{(1+\epsilon)\log(2)} $$ holds for almost all $n$.

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  • $\begingroup$ Is it possible to give an explicit bound for the number of such $n$? $\endgroup$ Sep 20 at 21:11
  • $\begingroup$ @StanleyYaoXiao 'proportion... tends to 0' does not mean finitely many, so I don't understand your question. $\endgroup$
    – Stopple
    Sep 20 at 21:12
  • $\begingroup$ Sorry, I meant the density of such numbers as a function of $X$. For example can one obtain a bound of $O_A(X(\log X)^{-B(A)})$ for some number $B(A)$ depending on $A$? $\endgroup$ Sep 20 at 21:18
  • $\begingroup$ @StanleyYaoXiao Of course; good question but I don't know. $\endgroup$
    – Stopple
    Sep 20 at 21:54

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