Consider following polynomial sequence.

$$\begin{cases}a_{-1}=0,~a_0=1, \\a_{n+1}=x \cdot a_n \pm a_{n-1}\end{cases}$$

Here $+$ or $-$ is taken in such a way that all coefficients in $a_n$ do not exceed $1$ by absolute value (i.e. $\forall k \hookrightarrow |[x^k]a_n(x)|\leq 1$). The sequence seems to be infinite, but what is the strict proof of it and what are the properties of such sequence? Maybe it has unique name?

Also it seems that if we will write $0$ each time we use $-$ and $1$ each time we use $+$ in sequence $s_i$, there will be $2^{\lfloor\log_2 n\rfloor+1}$ sequences $s_i$ which generate correct $a_i$. Here is all $16$ possible patterns of first $15$ terms of $s_i$:

001011001011010 110100110100101
001011010011010 110100101100101
001101001010110 110010110101001
001101010010110 110010101101001
010010101101001 101101010010110
010010110101001 101101001010110
010100101100101 101011010011010
010100110100101 101011001011010

Can you see any pattern in here? Note that in the same row it is the sequence and same sequence reversed.

UPD: It seems that $s_0$ does not matter (since it produces $a_1=x$ in any way) and if we consider sequence $g_i$ which generates $\{s_i\}_{i=1}^{2^{n}-2}$, its prefixes can be generated as

$$\begin{cases}g_1=\varepsilon,\\g_{n+1}=g_n + (01|10) + g_n^r\end{cases}$$

where $g_n^r$ stands for reversed $g_n$.