Let $M$ be the $n\times n$ matrix, known as the *GCD matrix*, of entries $M\_{ij}=\gcd(i,j)$. In the paper


H J S Smith, *On the value of a certain arithmetical determinant*, Proc. London Math. Soc. **7**:208-212 (1875-76)


it is shown that $\det M=\prod\_{k=1}^n\varphi(k)$; where $\varphi(k)$ is the [Euler totient function](https://en.wikipedia.org/wiki/Euler%27s_totient_function).


Lucas introduced the family of sequences defined recursively by $L\_0(s,t)=0, L\_1(s,t)=1$ and
$$L\_n(s,t)=s\cdot L\_{n-1}(s,t)+t\cdot L\_{n-2}(s,t).$$
**Remark.** My interest in these numbers lies in the fact that they are gcd-compatibile:
$$\gcd(L\_i,L\_j)=L\_{\gcd(i,j)}.$$



> 
> **Question 1.** Can Smith's result be extended to the determinantal evaluation
> $$\det\left[\gcd(L\_i(s,t),L\_j(s,t))\right]\_{i,j=1}^n \,\,?$$
> 
> 
> 


**Update.** On the basis of Max Alekseyev's calculations depicted below, I state the following claim.



> 
> **Conjecture.** Define a modified Euler's totient function $\varphi\_L(k):=\sum\_{d\vert n}L\_d(s,t)\cdot\mu\left(\frac{k}d\right)$. Then
> $$\det\left[\gcd(L\_i(s,t),L\_j(s,t))\right]\_{i,j=1}^n=\prod\_{k=1}^n\varphi\_L(k).$$
> 
> 
> 


**Remark.** Observe that $L\_n(2,-1)=n$, therefore **Question 1** is already answered by Smith. In this case, $\varphi\_L=\varphi$.



> 
> **Question 2.** We may start modest. If $s=2, t=1$ then $L\_n(2,1)=P\_n$ is the [Pell sequence](https://en.wikipedia.org/wiki/Pell_number). What is the value of
> $$\det\left[\gcd(P\_i,P\_j)\right]\_{i,j=1}^n\,\,?$$
> 
> 
> 


**POSTSCRIPT.** Following Max Alekseyev's generalized statement, I thought we might still upgrade the theorem slighly.


**Theorem.** For any integer $n>0$ and any variables $v\_1,v\_2,\dots,v\_n$,
$$\det [\,(v\_{\gcd(i,j)})^{\mathbf{\color{red}i}}\,]\_{i,j=1}^n = \prod\_{k=1}^n \sum\_{d|k} \mu(k/d)\cdot (v\_d)^{\mathbf{\color{red}k}}.$$



The GCD-matrix: generalizing a result of Smith?
For Pell numbers, the answer appears to be
$$\det\left[\gcd(P\_i,P\_j)\right]\_{i,j=1}^n = \prod\_{k=1}^n \sum\_{d|k} \mu(\frac{k}{d})\cdot P\_d,$$
i.e. the product of first $n$ terms of the Moebius transform of Pell numbers. At least, this equality holds for all $n\leq 100$.




---


**UPDATE**


A more general statement holds:


**Theorem.** For any integer $n>0$ and any variables $v\_1,v\_2,\dots,v\_n$, 
$$\det [v\_{\gcd(i,j)}]\_{i,j=1}^n = \prod\_{k=1}^n \sum\_{d|k} \mu(k/d)\cdot v\_d.$$


The statement for Lucas sequences directly follows from this theorem, thanks to the property $\gcd(L\_i,L\_j)=L\_{\gcd(i,j)}$.


The theorem can be proved using the following lemma:


**Lemma.** For any integers $0<m<n$,
$$\sum\_{d|n} \mu(n/d)\cdot v\_{\gcd(m,d)} = 0.$$


Indeed, in matrix $V=[v\_{\gcd(i,j)}]\_{i,j=1}^n$, one can replace the row $V\_n$ with the linear combination $\sum\_{d|n} \mu(n/d)\cdot V\_d$ (which does not change the determinant, and nullifies the $n$-th row except for its rightmost element), and then use induction on $n$.


