I have derived following formula for gcd(n,k), where a $n$ is odd positive integer and a $k$ is integer
$$\gcd(n,k)=\frac{n^2} {\sum_{l=0}^{n-1} \sec^2(\pi k l/n)}$$
I didn't found this formula anywhere. Is it really new? Is it useful for someone or for something?
@sato Let start with this equality $$\sum_{j=1}^\infty f(j)\sum_{l=0}^{n-1}\exp \left(-i 2 \pi\ k l j/n \right) = n \sum_{t=1}^\infty f \left( t \frac n {\gcd(n,k} \right)$$ Using $$f(j)=\frac {x^j} j $$ we receive $$\ln \prod_{l=0}^{n-1} \left(1-x \exp(-i 2 \pi\frac{ k l}n)\right) = \gcd(n,k)\ln\left(1-x^{\frac n {\gcd(n,k)}} \right) {*}$$ where $|x| \leq 1 $ and $x \ne 1 $.
Inserting $x=-1 $ we receive for odd $n$ well known Slavin's formula
$$\gcd(n,k)= \log_2 \prod_{l=0}^{n-1} \left(1+\exp(-i 2 \pi\frac{ k l} n\right) $$
Now we substitute into the equation (*) $x=e^{i \phi} $. After double derivation according $\phi $ we receive
$$\sum_{l=0}^{n-1} \csc^2\left(\frac \phi 2 + \frac {\pi k l} n\right) = \frac {n^2} {\gcd(n,k)} \csc^2\left(\frac {\phi} 2 \frac n {\gcd(n,k)}\right) $$ Inserting $\phi=\pi $ we receive for odd $n$ $$\gcd(n,k)=\frac{n^2} {\sum_{l=0}^{n-1} \sec^2(\pi k l/n)}$$