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$\newcommand{\lcm}{\operatorname{lcm}}$Let $\gcd(i,j)$ and $\lcm(i,j)$ be the greatest common divisor and least common multiple of the pair of positive integers $i$ and $j$. Denote the sum of divisors of $k$ by $\sigma(k)=\sum_{d\vert k}d$.

I noticed that the following is known: if $M_n$ is the $n\times n$ matrix with entries $M_n(i,j)=\sigma(\gcd(i,j))$ then $\det(M_n)=n!$.

By analogy, define the matrix $A_n$ with entries $A_n(i,j)=\sigma(\lcm(i,j))$. I could not find anything about this, so I ask:

QUESTION. Is there an explicit evaluation of the determinant $\det(A_n)$?

Examples. For $n\geq1$, here are few terms of $\det(A_n)$: $$1, -6, 72, -672, 20160, 1451520, -81285120, 1393459200.$$ These numbers do have small primes factors, so it seems reasonable to expect a closed form.

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    $\begingroup$ $1393459200$ turns up as a denominator in the expansion of $(1+x)^{1/x}/e$, oeis.org/A055535 (but the other numbers don't, so never mind). $\endgroup$
    – Gerry Myerson
    Commented Jul 10, 2022 at 1:19
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    $\begingroup$ Actually, the entries $\sigma_k({\rm gcd}(i,j))$ yield the determinant $(n!)^k$. This applies also to $k=0$, where $\sigma_k$ is just the number of divisors. $\endgroup$
    – Denis Serre
    Commented Jul 12, 2022 at 8:56
  • $\begingroup$ Have you any idea of the determinant when the $(i,j)$-entry is just ${\rm lcm}(i,j)$ ? Hand calculation gives $1$, $-2$, $12$, $-48$, $-960$, $11520$, ... $\endgroup$
    – Denis Serre
    Commented Jul 12, 2022 at 12:39
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    $\begingroup$ @DenisSerre, OEIS A060238 $\endgroup$
    – Peter Taylor
    Commented Jul 12, 2022 at 14:09
  • $\begingroup$ First 20 terms of $\det(A_n) / n!$: [1, -3, 12, -28, 168, 2016, -16128, 34560, -112320, -2021760, 24261120, 226437120, -3170119680, -76082872320, -1825988935680, 3773710467072, -67926788407296, -662286186971136, 13245723739422720, 185440132351918080]. First 20 terms of $\det(A_n) / (n! \operatorname{A124052}(n))$: [1, -1, 1, -1, 1, 12, -12, 12, -12, -216, 216, 2016, -2016, -48384, -1161216, 1161216, -1161216, -11321856, 11321856, 158505984] where A124052 is $\sigma(\operatorname{lcm}(1,\ldots,n))$ $\endgroup$
    – Peter Taylor
    Commented Jul 12, 2022 at 14:22

2 Answers 2

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This is only empirical observation, but I was requested to post it as an answer rather than merely a comment.

Define $b(n) = \frac{\det(A_n)}{n! \, \sigma(\operatorname{lcm}(1,\ldots,n))}$ for $n \ge 1$ and $r(n) = \frac{b(n)}{b(n-1)}$ for $n \ge 2$. If the prime factorisation of $n$ is $p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$ with all $e_i > 0$ then empirically (tested up to $n=105$) we have $$r(n) = (-1)^k \begin{cases} 1 & \textrm{if } k = 1\\ \prod_i \frac{p_i^{e_i+1}-1}{p_i^{e_i} - 1} & \textrm{otherwise} \end{cases}$$

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  • $\begingroup$ It may or may not be more useful to write the second case as $\frac{\sigma(n)}{\sigma(n / \operatorname{rad}(n))}$ where the radical $\operatorname{rad}(n)$ is the product of the prime factors of $n$. $\endgroup$
    – Peter Taylor
    Commented Jul 12, 2022 at 15:50
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One can generalize the first problem/result as follows: Let $(a_d)_{d\geq 1}$ be any sequence and $b(m)=\sum_{d|m} a_d$. Then let $M_{i,j}=b( gcd(i,j))$. This matrix has determinant $a_1...a_n$, which can be proved by subtracting the first column from the rest, then the second column from what remains, etc to get a lower diagonal matrix with $a_n$ at the $n$th slot on the diagonal.

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