TITLE: an integer sum of products of tangents
QUESTION [22 upvotes]: This question arose from my initial attempts at answering this question. I later found a way to transform the desired sum into a sum of squares of tangents, but before I did, I found numerically that apparently
$$
\sum_{l=1}^n\tan\frac{jl\pi}{2n+1}\tan\frac{kl\pi}{2n+1}=m_{jkn}(2n+1)
$$
with integer factors $m_{jkn}$, for which I haven't been able to find an explanation. If $j$ or $k$ is coprime to $2n+1$, we can sum over $jl$ or $kl$ instead, so most cases (in particular all for $2n+1$ prime) can be reduced to the case $j=1$. Here are the numerically determined factors $m_{1kn}$ for $n\le18$ (with $n$ increasing downward and $k$ increasing to the right):
$$
\begin{array}{r|rr}
&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18\\\hline1&1\\
2&2&0\\
3&3&-1&1\\
4&4&0&1&0\\
5&5&-1&1&1&1\\
6&6&0&2&-2&0&0\\
7&7&-1&2&-1&1&0&1\\
8&8&0&2&0&2&2&2&0\\
9&9&-1&3&1&1&-3&1&-1&1\\
10&10&0&3&-2&2&-1&1&0&1&0\\
11&11&-1&3&-1&1&-1&1&3&-1&1&1\\
12&12&0&4&0&2&0&0&-4&0&0&0&0\\
13&13&-1&4&1&3&0&1&-1&1&1&3&0&1\\
14&14&0&4&-2&2&2&2&0&2&4&0&0&2&0\\
15&15&-1&5&-1&3&-3&3&-1&3&-5&1&1&1&-1&1\\
16&16&0&5&0&2&-1&2&0&1&-2&1&1&-2&-2&1&0\\
17&17&-1&5&1&3&-1&2&-1&1&-1&1&5&1&0&1&1&1\\
18&18&0&6&-2&4&0&2&0&2&-2&2&-6&0&0&4&2&0&0\\
\end{array}
$$
(See also the table in this answer to the other question, which shows the case $j=k+1$; in that case the rows of the table sum to $0$ because of the identity that's the subject of the other question.)
The values $m_{11n}=n$ reflect the sum of squares of tangents that I determined in my answer to the other question. I have no explanation for the remaining values. I've tried using the product formula for the tangent; multiplying by a third tangent to use the triple tangent product formula; and finding a polynomial whose roots are the products being summed; but none of that worked out.
This vaguely reminds me of character theory; the values $\tan\frac{kl\pi}{2n+1}$ for fixed $k$ are like characters, and their dot products are integer multiples of the "group order" $2n+1$; though if they were characters the dot products couldn't be negative.
I'd appreciate any insight into this phenomenon, and of course ideally a way to calculate the $m_{jkn}$.
[Update:]
I've verified the periodicities that Brian observed in comments up to $n=250$:
$$m_{1,k,n+k} = m_{1kn}+[k \text{ odd}]\;,$$
$$m_{1,k+4d+2,k+4d+2+d}=m_{1,k,k+d}\;,$$
where the bracket is the Iverson bracket.

REPLY [8 votes]: [Just as I was finishing this I saw user8268's answer. I suspect the explanations are related.]
I wasn't intending to answer my own question in this case, but I've now found an explanation. The character theory analogy turned out to be more useful than I expected. Thinking of the values $\tan\frac{kl\pi}{2n+1}$ for fixed $k$ as vectors composed of integer multiples of mutually orthogonal vectors made me wonder what these mutually orthogonal vectors might be. A natural choice was a Fourier-style set of sines or cosines, and indeed it turns out that
$$
\sum_{l=1}^{2n}\sin\frac{2jl\pi}{2n+1}\tan\frac{kl\pi}{2n+1}=
\begin{cases}
\pm(2n+1)&\gcd(k,2n+1)\mid j\;,\\
0&\text{otherwise.}
\end{cases}
$$
I found this surprising at first, but it's actually not too difficult to explain. We have
$$
\def\ex#1{\mathrm e^{#1}}\def\exi#1{\ex{\mathrm i#1}}\def\exm#1{\ex{-\mathrm i#1}}
\ex{2n\mathrm i\phi}-\ex{-2n\mathrm i\phi}=\left(\exi{\phi}+\exm{\phi}\right)\left(\ex{(2n-1)\mathrm i\phi}-\ex{(2n-3)\mathrm i\phi}+\dotso+\ex{-(2n-3)\mathrm i\phi}-\ex{-(2n-1)\mathrm i\phi}\right)\;,
$$
so
$$
\sin(2j\phi)=2\cos\phi\left(\sin((2j-1)\phi)-\sin((2j-3)\phi)+\dotso+(-1)^{j+1}\sin\phi\right)\;.
$$
Thus, for $k=1$ the cosine in the denominator of the tangent is cancelled, and the remaining sine picks out the last term in the sum of alternating sines with odd frequencies, which yields
$$
\sum_{l=1}^{2n}\sin\frac{2jl\pi}{2n+1}\tan\frac{l\pi}{2n+1}=(-1)^{j+1}(2n+1)\;.
$$
But since the integers $jl$ and $kl$ in the arguments of both factors only matter $\bmod2n+1$, if $k$ is coprime to $2n+1$, we can sum over $kl$ instead of $l$ and will get the result for $k^{-1}j\bmod(2n+1)$, so for $k$ coprime to $2n+1$
$$
\sum_{l=1}^{2n}\sin\frac{2jl\pi}{2n+1}\tan\frac{kl\pi}{2n+1}=(-1)^{\sigma_k(j)+1}(2n+1)\;,
$$
where $\sigma_k$ is the permutation effected by multiplication with $k^{-1}\bmod(2n+1)$. If $1\lt\gcd(k,2n+1)\mid j$, the sum reduces to $\gcd(k,2n+1)$ identical copies, whereas if $\gcd(k,2n+1)\nmid j$, cancellation lets the sum vanish.
Thus, the $m_{jkn}$ are integers because the vectors $\left\{\tan\frac{kl\pi}{2n+1}\right\}_l$ are integer linear combinations of vectors $\left\{\sin\frac{2jl\pi}{2n+1}\right\}_l$ whose dot products are all either $0$ or $\pm1$, and these values can be obtained using only elementary number theory, namely permutations induced by multiplicative inverses.