More experimental facts. I think it should not be too hard  prove them by induction, taking into account that we know exactly when the coefficients of $a_n$ are non-zero (see my comment to  მამუკა ჯიბლაძე's answer).

Given a sequence $\epsilon_{2^n}\in\{-1,+1\}$ for $n\ge1$,  let's define inductively the sequence  $(\epsilon_j)_{j\ge1}$  for all positive integers according to the recurrence relations:

$$\epsilon_{2^n+1}=-\epsilon_{2^n}\qquad\text{for }n\ge1 $$

$$\epsilon_{2^n+j}=\epsilon_{2^n-j+1}\qquad\text{for }2\le j< 2^n\ ,$$ 

and also define 

$$
\eta_j:=\begin{cases}
(-1)^{ j\over 2},& \text {for even }\ j \\ \\
(-1)^{  j-1\over 2}\epsilon_{j+2}, &\text {for odd } j \ .\\ 
 \end{cases}
$$
and
$$
\delta_j:=\begin{cases}
-1,& \text {for }\ j=0 \\ \\
1, &\text {for  } j\neq0 \ .\\ 
 \end{cases}
$$


Then (experimentally) the polynomials $a_k= a_k(x,\epsilon_2,\epsilon_4,\dots,\epsilon_{2^n})$ 
are determined inductively by  $a_{-1}:=0, a_0:=1$, and by the recurrence relation, for any $0\le n$ and any $0\le j< 2^n\ $

$$a_{2^n+j}=x^{2^n}b_j+  {\delta_{2^{n-1}-1-j}}\ \eta_j\ \epsilon_{2^n}\ a_{2^n-2-j},$$

where  
$$b_j:=a_j(x,\epsilon_2,\epsilon_4,\dots,\epsilon_{2^{n-2}},-\epsilon_{2^{n-1}})\ .$$

 

For, instance the above gives $$a_{51}={x}^{51}-\epsilon_{{4}}\epsilon_{{16}}{x}^{43}+ \epsilon_{{8}}
\epsilon_{{16}}{x}^{35}+ \epsilon_{{4}}\epsilon_{{32}}{x}^{11}-\epsilon_
{{8}}\epsilon_{{32}}x^3$$