In 2018 this question was posted on AoPS:
Prove that$$\sum_{k=0}^\infty\frac{\tan\left( \frac{2k+1}{2}\pi\color{red}{\sqrt{15}}\right)}{(2k+1)^3}=-\frac{\pi^3}{32\color{red}{\sqrt{15}}} $$
Before attempting to prove it, I noticed that $\sqrt{15}$ appears on both sides of the equation. This lead almost immediately to the question:
For what natural numbers $n$ the following holds?$$\sum_{k=0}^\infty\frac{\tan\left( \frac{2k+1}{2}\pi\color{red}{\sqrt{n}}\right)}{(2k+1)^3}=-\frac{\pi^3}{32\color{red}{\sqrt{n}}}$$
I tabled the values for $n$ ranging from $1$ to $200$ and found that the only values satisfying our equation in this range are $$n=3,15,35,63,99,143,195$$ A quick check on the OEIS revealed that these numbers are the sequence $a_m=4m^2-1$, or in other words the numbers that can be written as the product of two consecutive odd integers.
User @pisco linked a previous answer where it is shown that $\sum_{k=0}^\infty\frac{\tan\left( \frac{2k+1}{2}\pi\color{red}{\sqrt{15}}\right)}{(2k+1)^3}$ is of the form $\frac{\pi^3}{\sqrt{n}}\cdot q(n)$ where $q(n)\in\mathbb{Q}$ for every positive integer $n$. So the big question becomes:
What is the relation between $n$ and $q(n)$?
Is it true that $q(4m^2-1)=-\frac{1}{32}$?
Further investigation suggested another conjecture:
Is it true that $q(4m^2+1)=\frac{1}{32}$?
In the link provided, there is an algorithm to calculate $q(n)$ for each $n$, but I couldn't elaborate on it to understand what is going on, so I ask your advice.