Consider finite sequence $x_j^k, j=0,...,k-1$:
$0, csc^2(\frac{\pi}{k}), csc^2(\frac{2\pi}{k}),...,csc^2(\frac{(k-1)\pi}{k})$
Suppose its discrete Fourier transform be
$X_j^k=\Sigma_{l=0}^{k-1}x_l^ke^{-\frac{2\pi lj}{k}}$
Interestingly $X_j^k$ has simple patterns:
$X_j^2 =1,-1$
$X_j^3 = 8/3, -4/3, -4/3$
$X_j^4 = 5, -1, -3, -1$
$X_j^5 = 8, 0, -4, -4, 0$
$X_j^6 = 35/3, 5/3, -13/3, -19/3, -13/3, 5/3$
$X_j^7=16, 4, -4, -8, -8, -4, 4$
...
Is there a way to calculate the formula of $X_j^k$ ?
A step towards the solution
I will use in the sequel capital $K,L$ instead of $k,l$ (lowercase $l$ is too similar with number $1$).
Let $f(t):=\csc(\pi t)^2=\dfrac{1}{\sin(\pi t)^2}$ which is computed on discrete values
$$\forall L, \ 1\leq L \leq K \ \ \ \ \ \ t=\dfrac{L}{K}.$$
Displaying the values of the Discrete Fourier Transform as curves, we observe discrete parabolas (see graphics below for values of $k$ between $4$ and $25$.)
After some computations, one obtains, in fact, quadratic expressions:
$$\tag{1}\forall L, \ 1\leq L \leq K \ \ \ \ \ \ \hat{f}(L)=2\left(L-\frac{K}{2} - 1\right)^2 - \dfrac{K^2}{6}-\dfrac{1}{3},$$
which take exactly the same values as the Discrete Fourier Transform of the given sequence, but without any proof for the moment.
Edit: as the result is a quadratic function + shifts, which are mapped to second derivation and multiplications by complex exponentials, it may be interesting to use Inverse Discrete Fourier Transform.
Another track: take advantage of the following identity:
$$\left(\dfrac{\sin \pi t}{\pi t}\right)^2 \times \dfrac{1}{(\sin \pi t)^2}=\dfrac{1}{t^2}$$
transformed, by Discrete Fourier Transform, into:
$$\underbrace{\Lambda(u)}_{triangle} \underbrace{\star}_{conv.} DFT(csc^2)(u) =DFT(\dfrac{1}{t^2})(u)$$