<s>[INST] Expanding product of binomials $(z^k + z^{-k})$ Suppose $z$ is a complex number, and consider the product
$$f_m(z)=\prod_{k=1}^m \left(z^k + \frac 1 {z^k} \right),$$
for $m = 1,2,\dots$ . Of course, one should be able to expand this into a sum of terms that are either powers of $z$ or of $1/z$. It is easy to see that the highest and lowest order terms will be $z^{1 + 2 + \cdots + m} = z^{\frac 1 2m(m+1)}$ and its reciprocal, respectively.
Here's the situation for low values of $m$:
$$\begin{split}
f_1(z) = z &+ \frac 1 z \\
f_2(z) = z^3 + z &+ \frac 1 z + \frac 1 {z^3} \\
f_3(z) = z^6+z^4+z^2+\ &\color{red}2 + \frac 1 {z^2}+ \frac 1 {z^4} + \frac 1 {z^6} \\
f_4(z) = z^{10} + z^8 + z^6 + \color{red}2z^4 + \color{red}2z^2 +\  &\color{red}2 + \frac {\color{red}{2}} {z^2} + \frac {\color{red}2} {z^4} + \frac 1 {z^6} + \frac 1 {z^8} + \frac{1}{z^{10}} \\
f_5(z)=z^{15} + z^{13}+ z^{11} + \color{red}2z^{9} + \color{red}2z^7 + \color{orange}3z^5 + \color{orange}3z^3 + \color{orange}3z &+ \frac{\color{orange}3} z + \frac {\color{orange}3}{z^3} + \frac {\color{orange}3} {z^5} + \frac {\color{red}2} {z^7} + \frac {\color{red}2} {z^9} + \frac 1 {z^{11}} + \frac 1 {z^{13}} + \frac 1 {z^{15}}
\end{split}$$
It seems that


*

*In each $f_m$ the exponent of $z$ falls from $\tau_m = \frac 1 2 m(m+1)$ down to $-\tau_m$, skipping every other value;

*For all $m$ the coefficient $a_{m,n}$ of the $n$-th term in the analytic part is equal to the coefficient $a_{m,-n}$ of the $n$-th term in the singular part (which makes sense intuitively).



So:
  
  
*
  
*Is there a closed-form expression for $a_{m,n}$?
  
*How can I prove fact 1. above?
  

 [/INST] This is just a start, not a solution, but I can't fit it in a comment.  Maybe it's enough for you to answer the question yourself.  We can think of the problem this way.  Partition $[m] = \{1,2,\dots\}$ into two sets $A$ and $B$ and let $\sum A$ and $\sum B$ be the sum of the elements of $A$ and $B$ respectively.  Then $a_m,k= [z^k]f_m(z)$ is the number of such partitions with $\sum A-\sum B = k$.  We have $$
a_{m,k}=\cases{
1,& $m=1,\ |k|=1$\\
0,& $m=1,\  |k|\neq 1$\\
a_{m-1,k-m}+a_{m-1,k+m},&otherwise
}
$$   
This is because the partitions of $[m]$ arise from adding the element $m$ to one of the sets in a partition of $[m-1].$  Notice that I am assuming that $a_{m,k}$ is defined for all positive integers $m$ and all integers $k.$  Just make $a_{m,k}=0$ when $z^k$ does not appear in $f_m.$ 
I have no idea whether this can be solved for $a_{m,k}$ in closed form, but at least it makes it easy to calculate.  
</s>