Let $\omega$ denote the set of non-negative integers. For which integers $n>1$ is there a sequence $b_n: \omega\to\omega$ with the following property?
Whenever $v\in\omega^n$ there is a unique $i_v\in\omega$ such that for all $j\in\{0,\ldots,n-1\}$ we have $v(j) = b_n(i_v+j)$.
$\newcommand\N{\mathbb N}$For every $n\ge1$ there is such sequence $b_n$. I replace $\omega$ with $\N$. Endow $\N^n$ with any total order $<$. Let $b_{n,0}$ consist of $n$ zeros. For $k\in\N$ let $v=v_k$ be the minimal element of $\N^n$ that hasn't appeared in $b_{n,k}$. Take $n$ elements $l=l(k)=(l_1,\ldots,l_n)$ of $\N$ that are not in the finite sequence $b_{n,k}\cup v$ and let $b_{n,k+1}$ be concatenation of $b_{n,k}$, $l$ and $v$. Every subsequence of $b_{n,k+1}$ of length $n$ is either in $b_{n,k}$, or overlaps with $l$ (and then, by construction of $l$, can be uniquely determined by length of its $l$-head/$l$-tail), or is $v$. Therefore, for every $k\in\N$, $v\in\N^n$ there is at most one such $i_v$. By construction, for every $v\in\N^n$, for all large enough $k\in\N$, $k>k_v$, there is at least one such $i_v$. Thus, one may take $b_n=\lim_{k\to\infty}b_{n,k}$.
