TITLE: Could this conjecture be proved ? (sum of even powers of cotangents in arithmetic progression )
QUESTION [11 upvotes]: Having tried (in vain) to answer this question, I worked the explicit formulae of
$$\color{blue}{S_k=\sum _{n=1}^m \Big[\cot \left(\frac{n \,\pi  }{2 m+1}\right)\Big]^k}$$ where $k$ is an even integer.
To my surprise, the CAS I used were able to produce explicit results only for $k=2$ and $k=4$. All the remaining was done by hand using pen and paper.
What I noticed is that apparently
$$\color{red}{S_{2k}=\frac{m (2 m-1)}{a_k} \left( \sum_{n=0}^{2(k-1)} b_n\, m^n \right)}$$ where the  $a_k$'s
$$\color{blue}{\{3,45,945,14175,467775\}}$$ are the first terms of sequence $A171080$ in $OEIS$:
$$a_k = \prod_{q\;\text{odd prime}\\ \;\;\leq 2k+1}q^{\lfloor 2k/(q-1) \rfloor}$$
Concerning the coefficients $b_n$, they are given in the following table
$$\color{blue}{\left(
\begin{array}{cc}
 k & \text{list of the } b_n \\
 2 & \{1\} \\
 4 & \{-9,10,4\} \\
 6 & \{135,-252,8,112,32\} \\
 8 & \{-1575,3834,-1388,-2248,496,864,192\} \\
 10 & \{42525,-122760,77040,70400,-57920,-38720,14720,14080,2560\}
\end{array}
\right)}$$

My questions are :

Why cannot we compute explicit forms for odd values of $k$ ?
Could this conjecture be proved ?
What are these polynomials ?
Could we find their generic formula ?

REPLY [2 votes]: In an article by Ejsmont and Lehner a method using operator trace is applied to evaluate finite sums involving the cotangent. The paper gives numerous references concerning finite trigonometric sums which seem to have a rich recent history. We use here a result obtained by Berndt and Yeap by a contour integration technique. It reads
\begin{equation}
\sum_{n=1}^{q-1}\cot^{2k}\frac{n\pi}{q}=(-1)^kq-(-1)^k2^{2k}\sum_{\substack{j_0,j_1,\ldots,j_{2k}\ge0 \\j_0+j_1+\ldots+j_{2k}=k}}q^{2j_0}\prod_{p=0}^{2k}\frac{B_{2j_p}}{(2j_p)!}\tag{BY}\label{BY}
\end{equation}
where $B_s$ are the Bernoulli numbers. It seems that no similar results with odd powers of the function were found.
To evaluate
\begin{equation}
 S_{2k}=\sum _{n=1}^m \left[\cot \left(\frac{n \,\pi  }{2 m+1}\right)\right]^{2k}
\end{equation}
we remark that, by symmetry,
\begin{equation}
 S_{2k}=\frac{1}{2}\sum _{n=1}^{2m} \left[\cot \left(\frac{n \,\pi  }{2 m+1}\right)\right]^{2k}
\end{equation}
then, by (\ref{BY}),
\begin{equation}
 (-1)^kS_{2k}=m+\frac{1}{2}-2^{2k-1}\sum_{\substack{j_0,j_1,\ldots,j_{2k}\ge0 \\j_0+j_1+\ldots+j_{2k}=k}}(2m+1)^{2j_0}\prod_{p=0}^{2k}\frac{B_{2j_p}}{(2j_p)!}
\end{equation}
Under this form, it appears that the sum is polynomial in $m$.
The summation is however difficult to perform. It can be transformed by the use of the series expansion of $\coth Z$
\begin{equation}
 Z\coth Z=1+\sum_{s=1}^\infty\frac{B_{2s}}{(2s)!}(2Z)^{2s}
\end{equation}
One has then
\begin{align}
 S_{2k}&=(-1)^k\left( m+\frac{1}{2}\right)\left[ 1-\left[Z^{-1}\right]\left( \coth\left( (2m+1)Z \right)\coth^{2k}(Z) \right) \right]\\
 &=(-1)^k\left( m+\frac{1}{2}\right)\left[ 1-\frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon} \coth\left( (2m+1)z \right)\coth^{2k}(z)\,dz \right]\label{eq:Cauchy}
\end{align}
where $\left[z^r\right]\left( F(z) \right)$ represents the coefficient of $z^r$ in the Laurent series of $F(z)$ at $z=0$.
The latter expression for $S_{2k}$ vanishes when $m=0$, as $\operatorname{Res}\left( \coth^{2k+1}z,z=0 \right)=1$. Moreover, from the identity $\coth 2z=(\coth z+(\coth z)^{-1})/2$, it also vanishes for $m=1/2$ as proposed. Then,  polynomials $Q_k(m)$ exist such that
\begin{equation}
 S_{2k}=m(2m-1)Q_k(m)
\end{equation}
where
\begin{align}
 Q_k(m)&=\frac{(-1)^k(2m+1)}{2m(2m-1)}\left[ 1-\frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon} \coth\left( (2m+1)z \right)\coth^{2k}(z)\,dz \right]\\
 &=\frac{(-1)^k(2m+1)}{2m(2m-1)}\left[ 1-\left[Z^{-1}\right]\left( \coth\left( (2m+1)Z \right)\coth^{2k}(Z) \right) \right]
\end{align}
Evaluated with a CAS, the latter formula gives the results which are given in the OP up to $k=5$. Higher values of $k$ were numerically tested up to $S_{20}$ for several values of $m$.
Some values of $m$ are remarkable. For instance, it is not difficult to show that $Q_k(-1)=(-1)^{k+1}/3$ and $Q_k\left( -\frac{3}{2} \right)=\frac{(-1)^{k+1}}{3}$. Moreover $Q_k\left( \frac{3}{2} \right)=\frac{1}{3}$. Indeed, for $m=3/2$,
\begin{align}
 \coth\left( (2m+1)z \right)&=\coth(4z)\\
 &=\frac{1}{2}\left( \coth 2z+\frac{1}{\coth 2z} \right)\\
 &=\frac{1}{4}\left( \coth z+\frac{1}{\coth z} \right)+\frac{\coth z}{1+\coth^2z}
\end{align}
the integral reads
\begin{align}
 \frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon} \coth\left( 4z \right)\coth^{2k}(z)\,dz &=\frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon} \frac14\left( \coth^{2k+1}(z)+\coth^{2k-1}(z) \right)\,dz+I_k\\
 &=\frac{1}{2}+I_k
\end{align}
where
\begin{equation}
 I_k=\frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon} \frac{\coth^{2k+1}(z)}{1+\coth^2z}\,dz
\end{equation}
We have the recurrence relation
\begin{equation}
 I_k=1-I_{k-1}
\end{equation}
and
\begin{align}
 I_0&=\frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon} \frac{\coth(z)}{1+\coth^2z}\,dz\\
 &=\frac{1}{2}\frac{1}{2i\pi}\int_{\left|z\right|=\varepsilon}\tanh(2z)\,dz\\
 &=0
\end{align}
Thus $I_{2s}=0$ and $I_{2s+1}=1$. Finally,
\begin{equation}
Q_k\left( \frac32 \right)=
\begin{cases}
\frac23\left( 1-\frac{1}{2}-0\right)=\frac{1}{3}&\text{ when } k=2s\\
-\frac23\left( 1-\frac{1}{2}-1 \right)=\frac{1}{3}&\text{ when } k=2s+1
                                \end{cases} 
\end{equation}
Thus $Q_k\left( \frac{3}{2} \right)=\frac{1}{3}$.
This also shows, by adapting the signs, that, for $m=-5/2$, the values of the polynomials oscillate between $1/3$ and $-1/5$. Adapting to the decomposition of $\coth 3z$, should also make it possible to show (as suggested by @Blue in a comment) that $Q_k(1)=3^{-k}$, which is numerically observed.