In Sept. 2013, I investigated the determinant $$D_n=\det[\gcd(i-j,n)]_{1\le i,j\le n}$$ and computed the values $D_1,\ldots,D_{100}$ (cf. http://oeis.org/A228884). To my surprise, they are all positive!

Question. Does $D_n>0$ hold for all $n=1,2,3,\ldots$?

I believe that $D_n$ is always positive. How to prove this?

It is easy to see that $D_n$ is divisible by $\sum_{k=1}^n\gcd(k,n)=\sum_{d\mid n}\varphi(d)\frac nd$. It seems that $\varphi(n)^{\varphi(n)}\sum_{k=1}^n\gcd(k,n)$ divides $D_n$. Maybe there is a simple explanation for this.

Your comments are welcome!


1 Answer 1


Denote $f(k)={\rm gcd}(k,n)$. Clearly $f$ is an $n$-periodic function. $D_n$ is the circulant $D_n=\det(f(i-j):0\leqslant i,j\leqslant n-1)$ which equals $\prod_{k=0}^{n-1}h(\omega^k)$ where $\omega=e^{2\pi i/n}$ and $h(t)=f(0)+f(1)t+\ldots+f(n-1)t^{n-1}$. We have $$ h(t)=\sum_{d|n} \varphi(d)(1+t^d+t^{2d}+\ldots+t^{(n/d-1)d}). $$ If $t=\omega^k$, the sum $1+t^d+t^{2d}+\ldots+t^{(n/d-1)d}$ either equals $n/d$ (if $n$ divides $kd$) or equals $(1-\omega^{kn})/(1-\omega^{kd})=0$ otherwise. In any case it is a non-negative integer, and for $d=n$ it is strictly positive.


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