How many divisors does $n!$ have?

270 Views Asked by At

Let $d(n!)$ be the number of divisors of $n!$. I observed that graph of $\log(d(n!)$ is similar to that of $\pi(x)$, the number of prime not exceeding $x$ except for a scaling factor. In other words, the following limit must exist: $$ \lim_{n \to \infty}\frac{\log(d(n!))}{\pi(n)} \approx 1.26 $$

Question: What is known about the number of divisors of $n!$? Can it be proved or disproved?

1

There are 1 best solutions below

3
On BEST ANSWER

A short argument of Erdős et al. (see $\S$2 of the cited article) shows that

$$\log d(n!) \sim \frac{c_0 \log (n!)}{\log^2 \log (n!)} \left[1 + O\left(\frac{\log\log\log (n!)}{\log\log (n!)}\right)\right],$$ where $$\color{#df0000}{c_0 = \int_1^\infty \frac{\log (\lfloor t \rfloor + 1) \,dt}{t^2} = \sum_{k = 2}^\infty \frac{\log k}{k (k - 1)} = 1.25775\!\ldots }.$$ Applying Stirling's Approximation then gives that the leading term of the above series approximation is $$\frac{c_0 n}{\log n} + O\left(\frac{n \log \log n}{(\log n)^2}\right) .$$ On the other hand, the Prime Number Theorem says that $$\pi(n) \sim \frac{n}{\log n} + o \left(\frac{n}{\log n}\right),$$ so $$\color{#df0000}{\boxed{\lim_{n \to \infty} \frac{\log d(n!)}{\pi(n)} = c_0}} .$$

Erdös P., Graham S. W., Ivić A., Pomerance C. (1996) "On the number of divisors of n!." In: Berndt B. C., Diamond H. G., Hildebrand A. J. (eds.) Analytic Number Theory. Progress in Mathematics 138 Birkhäuser Boston.

Remark In fact, we can compute higher asymptotics of the ratio using the formulae of Erdős et al. (and a better bound on the remainder of the above approximation to $\pi(n)$), e.g., $$\frac{\log d(n!)}{\pi(n)} \sim c_0 + \frac{c_1}{\log n} + O\left(\frac{1}{\log^2 n}\right), $$ where $$c_1 := \int_1^\infty \frac{\log(\lfloor t \rfloor + 1) \log t \,dt}{t^2} = 2.11412\!\ldots .$$