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 function1. 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? $$\det\left[\gcd(L_i,L_j)^{\ell}\right]_{i,j=1}^n=\prod_{k=1}^n\sum_{d\vert k}L_d^{\ell}\cdot \mu\left(\frac{k}d\right) =(L_n!)^{\ell}\prod_{k=1}^n\prod_{p\vert k}\left(1-\frac1{p^{\ell}}\right).$$ \rm

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? $$\det\left[\gcd(L_i,L_j)^{\ell}\right]_{i,j=1}^n=\prod_{k=1}^n\sum_{d\vert k}L_d^{\ell}\cdot \mu\left(\frac{k}d\right) =(L_n!)^{\ell}\prod_{k=1}^n\prod_{p\vert k}\left(1-\frac1{p^{\ell}}\right).$$