Let $\{a_j\}_{-1\le j\le 2^n-2}$ and $\{b_j\}_{-1\le j\le 2^n-2}$ be two admissible sequences of length $2^n$. Then it is easy to see that $\{a_j\}_{-1\le j\le 2^n-2}$ can be prolonged to an admissible sequence of length $2^{n+1}$ defining, for $j=0,\dots,2^n-1$ $$a_{2^n-1+j}:=x^{2^n}b_j+a_{2^n-1-j} \ ,$$ and conversely, any admissible sequence of length $2^{n+1}$ is obtained this way. This proves that any sequence can be continued indefinitely .
The reason one can do so is essentially due to the fact that one has $a_k=x^k$ for $k=2^n -1$ as observed by Oleksandr Kulkov in მამუკა ჯიბლაძე's answer (see my comment there for a proof, which can also be done directly by induction, I think). As a consequence, for a sequence of length $2^{n+1}$, $\{a_{2^n-1-j} \mod( x^{2^n})\}_{-1\le j\le 2^n-2}$ is an admissible sequence of length $2^m$.