Let $A$ be the $n$ square matrix with as entries the greatest common divisors of the respective indices: $(A)_{i,j}=\gcd(i,j)$, $$A=\begin{Bmatrix}\gcd(1,1) & \gcd(1,2) & ... & \gcd(1,n) \\ \gcd(2,1) & \gcd(2,2) & ... & \gcd(2,n) \\ \vdots & \vdots & \ddots & \vdots \\\gcd(n-1,1) & \gcd(n-1,2) & ... & \gcd(n-1,n)\\\gcd(n,1) & \gcd(n,2) & ... & \gcd(n,n)\end{Bmatrix}$$ ok so $\phi(n)$ denotes Euler's totient function, then $$\det(A)=\phi(1)\phi(2)...\phi(n)=\prod_{i=1}^n\phi(i)\quad\quad(1)$$ this can permit us to determine $\phi(n)$ if we know the values of $\phi$ for the numbers less than $n$ because $$\underbrace{\det(A) = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) \prod_{i=1}^n a_{i,\sigma_i}}_{\text{Leibniz formula}}\to\dfrac{\det(A)}{\phi(1)...\phi(n-1)}=\dfrac{\sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) \prod_{i=1}^n a_{i,\sigma_i}}{\phi(1)...\phi(n-1)}=\phi(n)$$ but idk if this is useful
- so (1) is a surprising result, but how to prove it?
- are there similar examples where $(A)_{i,j}=f(i,j)$ and $$\det(A)=\prod_{i=1}^n g(i)?$$
