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Let $\gcd(a,b)$ denote the greatest common divisor function. H. J. S. Smith proved that $$\det\left[\gcd(i,j)\right]_{i,j=1}^n=\prod_{k=1}^n\varphi(k),$$ where $\varphi(k)$ denotes Euler's totient function. More generally, it is known that if $\ell\in\Bbb{P}$ then $$\det\left[\gcd(i,j)^{\ell}\right]_{i,j=1}^n=\prod_{k=1}^n\varphi_{\ell}(k),$$ where $\varphi_{\ell}(k):=\sum_{d\vert k}d^{\ell}\mu\left(\frac{k}d\right)$ denotes Jordan's totient function and $\mu$ is the Mobius function.

Let $L_n(s,t)=s\cdot L_{n-1}(s,t)+t\cdot L_{n-2}(s,t)$, or simply $L_n$, be the Lucas polynomials with the convention that $L_0=0, L_1=1$.

Experimental results suggest the below computation which I like to state our by introducing a "generalized" Jordan totient function.

Question: Let $p$ stand for a prime number. If $n, s, t, \ell\in\Bbb{N}, s\neq1$ and $\gcd(s,t)=1$, then is the following true? $$\det\left[\gcd(L_i,L_j)^{\ell}\right]_{i,j=1}^n=\prod_{k=1}^n\sum_{d\vert k}L_d^{\ell}\cdot \mu\left(\frac{k}d\right) =(L_n!)^{\ell}\prod_{k=1}^n\prod_{p\vert k}\left(1-\frac1{p^{\ell}}\right).$$

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  • $\begingroup$ Do the $L_i$ form a gcd-sequence? That is, do we have $\gcd\left(L_i, L_j\right) = L_{\gcd\left(i,j\right)}$ ? And does your $\mathbb{N}$ contain $0$ ? $\endgroup$ Commented Aug 25, 2018 at 23:29
  • $\begingroup$ Yes to the first. No to the second. $\endgroup$ Commented Aug 26, 2018 at 0:17
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    $\begingroup$ Then, $\gcd\left(L_i, L_j\right)^{\ell} = \gcd\left(L_i^{\ell}, L_j^{\ell}\right)$, so your left hand side is the gcd-determinant for the $n$-tuple $\left(L_1^{\ell}, L_2^{\ell}, \ldots, L_n^{\ell}\right)$, which has a well-known formula (see, e.g., Lemma 7.1 in Chris Godsil, An Introduction to the Moebius Function, arXiv:1803.06664v1). $\endgroup$ Commented Aug 26, 2018 at 0:25
  • $\begingroup$ @darijgrinberg, yes that was mentioned in a previous question as well: mathoverflow.net/questions/262153/… . I think the second equality is the main part of the question. $\endgroup$ Commented Aug 26, 2018 at 0:39
  • $\begingroup$ Indeed. E.g., the formula in question directly follows from the one in my answer by setting $v_i=L_i^{\ell}$. $\endgroup$ Commented Aug 26, 2018 at 12:14

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