From $\prod_{d\mid n}d=n^{\sigma_0(n)/2}$ to $n!=\operatorname{lcm}(1,\ldots,n)^{e(n)}$, where $\sigma_0(n)$ is the number of divisors

229 Views Asked by At

We know that $$\prod_{d\mid n}d=n^{\sigma_{0}(n)/2}$$ for every integer $n\geq 1$, where $\sigma_{0}(n)$ is the number of positive divisors of $n$, see for example [1] (exercise 10, page 47). And for this sequence of divisors $1=d_{1}<d_{2}<\cdots<d_{\sigma_{0}(n)}=n$, we have that $$\operatorname{lcm}(d_{1},d_{2},\cdots,d_{\sigma_{0}(n)})=n.$$ It is know that the average order of $\sigma_0(n)$ is $\log n$ for example by this formula $$\frac{1}{x}\sum_{n\leq x}\sigma_{0}(n)=\log x +O(1)$$ (for another, see for example Theorem 3.3 in page 57 of [1] and more about the growth in Theorem 13.12).

Now we consider $e(n)$ defined for every integer $n>1$ by $$n!=\operatorname{lcm}(1,2,\cdots,n)^{e(n)}$$

Question What about the growth of $e(n)$? Can you improve my computations or give more details?

My attempt was from the relationship between the second Chebyshev function, defined for $x>0$, as (really in terms of von Mangoldt function) $\psi(n)=\sum_{p^a\leq x}\log p$, where the sum is extended over all prime powers least or equal than $x$ (see this site Math Stack Exchange or the page of Wikipedia corresponding Chebyshev function), satisfies $$\operatorname{lcm}(1,2,\cdots,n)=e^{\psi(n)}$$ and too satisfies an equivalence with Prime Number Theorem, as this form $\psi(x)\sim x$ (Theorem 4.4, page 79 of [1]). I write from this $\psi(n)\sim n$ for large values of integers $n$, and will use Stirling equivalence.

Question (Solved by an user in comments) Can I use $\psi(x)\sim x$, as I said, this is I made a substitution from a real $x$ to the variable in integers $n$ (and compute the limit as I show)? I believe that is 'yes' since both sequences reals and integeres are distributed as same manner, mod 1, is this?

We take logarithms in the equation that defines the exponent $e(n)$, for $n>1$ and we take limits

$$\lim_{n\to\infty}e(n)=\lim_{n\to\infty}\frac{\log \sqrt{2\pi n}+n(\log n-\log e)}{n}$$

Thus dividing by $n$, we compute $e(n)\sim \log n$, or in this form $\lim_{n\to\infty}\frac{e(n)}{\log n}=1$.

I don't know if this problem is in the literature, my only goal is learn and edit the best post. Thanks in advance.

References:

[1] Apostol, Introduction to Analytic Number Theory, Springer.

[2] This Mathematics Stack Exchange, second Chebyshev's function, Stirling equivalence, Prime Number Theorem.