TITLE: Polylogarithm ladders for the tribonacci and n-nacci constants
QUESTION [17 upvotes]: While reading about polylogarithms, I came across the nice polylogarithm ladder,
$$6\operatorname{Li}_2(x^{-1})-3\operatorname{Li}_2(x^{-2})-4\operatorname{Li}_2(x^{-3})+\operatorname{Li}_2(x^{-6}) = \frac{7\pi^2}{30}\tag{1}$$
where $x = \phi = \frac{1+\sqrt{5}}{2}$, the golden ratio, or the root $1<x<2$ of, 
$$x^n(2-x) = 1\tag{2}$$
for the case $n=2$.  I wondered if there was anything for $n=3$ (the tribonacci constant $T$, or the real root of $x^3-x^2-x-1 = 0$).  After a little expt with Mathematica's integer relations command, I found,
$$4\operatorname{Li}_2(x^{-1})-4\operatorname{Li}_2(x^{-3})-3\operatorname{Li}_2(x^{-4})+\operatorname{Li}_2(x^{-8}) = \frac{\pi^2}{6}\tag{3}$$
However, for general $n$ (including the tetranacci constant and higher), it seems they obey,
$$4\operatorname{Li}_2(x^{-1})-\operatorname{Li}_2(x^{-2})+\operatorname{Li}_2(x^{-n+1})-2\operatorname{Li}_2(x^{-n}) = 2\operatorname{Li}_2(-x)+\frac{\pi^2}{2}\tag{4}$$
where $x$ is the root $1<x<2$ of $(2)$.
Q: Anybody knows how to prove if $(4)$ is indeed true for all integer $n\geq2$?

REPLY [6 votes]: I'll change $x \rightarrow 1+x$ to have
\begin{equation}
\left(1+x\right)^n\left(1-x\right) = 1,\quad 0 < x < 1 \label{eq1} \tag{1}
\end{equation}
Consider
\begin{align}
f_n &= 4\operatorname{Li}_2\left(\dfrac{1}{1+x}\right) - \operatorname{Li}_2\left(\dfrac{1}{\left(1+x\right)^2}\right) + \operatorname{Li}_2\left(\dfrac{1+x}{\left(1+x\right)^n}\right) - 2\operatorname{Li}_2\left(\dfrac{1}{\left(1+x\right)^n}\right) - \phantom{a} \\
&- 2\operatorname{Li}_2\left(-1-x\right)
\end{align}
where $x$ is the solution to $(1)$ with parameter $n$.
I restate here main functional equations for the dilogarithm (most of them can be found here)
\begin{align}
\operatorname{Li}_2(x)+\operatorname{Li}_2(-x) &= \dfrac{1}{2}\operatorname{Li}_2(x^2) \label{dl2} \tag{2} \\ 
\operatorname{Li}_2\left(1-x\right)+\operatorname{Li}_2\left(1-\dfrac{1}{x}\right) &= -\dfrac{1}{2}\ln^2 x \label{dl3} \tag{3} \\
\operatorname{Li}_2(x)+\operatorname{Li}_2\left(1-x\right) &= \dfrac{1}{6}\pi^2 - \ln\left(1-x\right)\ln x \label{dl4} \tag{4} \\
\operatorname{Li}_2(-x)-\operatorname{Li}_2\left(1-x\right)+\dfrac{1}{2}\operatorname{Li}_2\left(1-x^2\right) &= -\dfrac{1}{12}\pi^2 - \ln\left(1+x\right)\ln x \label{dl5} \tag{5} \\
\operatorname{Li}_2(-x)+\operatorname{Li}_2\left(-\dfrac{1}{x}\right) &= -\dfrac{1}{6}\pi^2 - \dfrac{1}{2}\ln^2 x \label{dl6} \tag{6}
\end{align}
Now use $\left(1+x\right)^n\left(1-x\right) = 1$ to get rid of $n$-th power of $(1+x)$ in $f_n$ and start to use dilogarithm identities
\begin{align*}
f_n &\overset{(1)}{=}  4\operatorname{Li}_2\left(\dfrac{1}{1+x}\right) - \color{red}{\operatorname{Li}_2\left(\dfrac{1}{\left(1+x\right)^2}\right)} + \operatorname{Li}_2\left(1-x^2\right) - 2\operatorname{Li}_2\left(1-x\right) -
2\operatorname{Li}_2\left(-1-x\right) \\
&\overset{(2)}{=} \color{red}{2}\operatorname{Li}_2\left(\dfrac{1}{1+x}\right) - \color{red}{2\operatorname{Li}_2\left(-\dfrac{1}{1+x}\right)} + \color{blue}{\operatorname{Li}_2\left(1-x^2\right)} - \color{blue}{2\operatorname{Li}_2\left(1-x\right)} - 2\operatorname{Li}_2\left(-1-x\right) \\
&\overset{(5)}{=} 2\operatorname{Li}_2\left(\dfrac{1}{1+x}\right) - \color{magenta}{2\operatorname{Li}_2\left(-\dfrac{1}{1+x}\right)} - \color{blue}{\dfrac{1}{6}\pi^2} - \color{blue}{2\ln\left(1+x\right)\ln x} - \color{blue}{2\operatorname{Li}_2(-x)}- \color{magenta}{2\operatorname{Li}_2\left(-1-x\right)} \\
&\overset{(6)}{=} 2\operatorname{Li}_2\left(\dfrac{1}{1+x}\right) + \color{magenta}{\dfrac{1}{3}\pi^2} + \color{magenta}{\ln^2\left(1+x\right)} - \dfrac{1}{6}\pi^2 - 2\ln\left(1+x\right)\ln x - \color{green}{2\operatorname{Li}_2(-x)} \\
&\overset{(3)}{=} 2\operatorname{Li}_2\left(\dfrac{1}{1+x}\right) + \dfrac{1}{3}\pi^2 + \ln^2\left(1+x\right) - \dfrac{1}{6}\pi^2 - 2\ln\left(1+x\right)\ln x + \color{green}{\ln^2\left(1+x\right)} + \phantom{a} \\
&+ \color{green}{2\operatorname{Li}_2\left(\dfrac{x}{1+x}\right)} \\
&= \color{red}{2\operatorname{Li}_2\left(\dfrac{1}{1+x}\right)} + \color{red}{2\operatorname{Li}_2\left(\dfrac{x}{1+x}\right)} + \dfrac{1}{6}\pi^2 + 2\ln^2\left(1+x\right)- 2\ln\left(1+x\right)\ln x \\
&\overset{(4)}{=} \color{red}{\dfrac{1}{3}\pi^2} - \color{red}{2\ln\left(\dfrac{1}{1+x}\right)\ln\left(\dfrac{x}{1+x}\right)} + \dfrac{1}{6}\pi^2 + 2\ln^2\left(1+x\right)- 2\ln\left(1+x\right)\ln x \\
&= \dfrac{\pi^2}{2}
\end{align*}