Combinatorial Interpretation of a Certain Product of Factorials

Solution 1:

This is an idea of approach, and it does not completely answer to the question. Define $$ \mathcal{R}_n(x) = \{r \le x: (r,n) = 1\}$$ and note that, by Inclusion-Exclusion principle $$ | \mathcal{R}_n(x) | = \sum_{d \mid n} \mu(d) \left \lfloor \frac{x}{d} \right \rfloor $$ Let $S_n$ be the sum in the text. Now, for every prime $p \le n$, consider $ V_p(S_n) $.
Recalling Polignac identity on $V_p(k!)$, we have $$ V_p(S_n) = \sum_{d \mid n} \mu(n/d) V_p(d!) = \sum_{d \mid n} \mu(n/d) \sum_{k=1}^{\infty} \left \lfloor \frac{d}{p^k} \right \rfloor = \sum_{k=1}^{\infty} \sum_{d \mid n} \mu(d) \left \lfloor \frac{n}{dp^k} \right \rfloor = \sum_{k=1}^{\infty} |\mathcal{R}_n(n/p^k) | $$ Take $r \in \mathcal{R}_n(n)$: the number of times it is counted in the last infinite sum is the number of positive integer solutions $k$ of $n/p^k \ge r$, i.e. $ \lfloor \log_p n/r \rfloor $. So $$ \sum_{k=1}^{\infty} |\mathcal{R}_n(n/p^k)| = \sum_{r \in \mathcal{R}_n(n)} \lfloor \log_p (n/r) \rfloor $$ Define $q_p(x) := \max \{ p^n \ | \ p^n \le x\}$, and note that $ q_p(x) =p^{ \lfloor \log_p x \rfloor} $. So, in conclusion: $$ S_n = \prod_{p \le n} p^{V_p(S_n)} = \prod_{\substack{p \le n \\ r \in \mathcal{R}_n(n) }} p^{ \lfloor \log_p (n/r) \rfloor } = \prod_{\substack{r,p \le n \\(r,n)=1 }} q_p(n/r) $$

I don't know, actually, if this can be useful; I think that it gives a combinatorial meaning to involved quantities, even if not much to the formula itself.

Hope it helps, Andrea

EDIT: I found an error in an inequality I used; so I can't actually state two bounds (upper and lower) of the same order. I would like to know more about this: Second-order asymptotics for $\pi(n), \theta(n)$