<s>[INST] Shortest universal bit string: One string to contain all others Let $s$ be a string of bits. Treat it as a cycle, with the first bit following
the last.
Say that $s$ is universal for $n$ if all the $2^n$ strings
of $n$ bits can be found in $s$ as consecutive, left-to-right bits
(with wrap-around).
Define $u(n)$ as the length of the shortest string universal for $n$.
Examples:
$u(1)=2$:
$$
\begin{matrix}
\color{red}{0} & 1 \\
0 & \color{red}{1}
\end{matrix}
$$
$u(2)=4$:
$$
\begin{matrix}
\color{red}{0} & \color{red}{0} & 1 & 1 \\
0 & \color{red}{0} & \color{red}{1} & 1 \\
\color{red}{0} & 0 & 1 & \color{red}{1} \\
0 & 0 & \color{red}{1} & \color{red}{1} 
\end{matrix}
$$
$u(3)=8$:
$$
\begin{matrix}
\color{red}{0} & \color{red}{0} & \color{red}{0} & 1 & 1 & 1 & 0 & 1\\
0 & \color{red}{0}  & \color{red}{0}  & \color{red}{1} & 1 & 1 & 0 & 1\\
\color{red}{0} & 0 & 0 & 1 & 1 & 1 & \color{red}{0} & \color{red}{1}\\
0 & 0 & \color{red}{0} & \color{red}{1} & \color{red}{1} & 1 & 0 & 1\\
\color{red}{0} & \color{red}{0} & 0 & 1 & 1 & 1 & 0 & \color{red}{1}\\
0 & 0 & 0 & 1 & 1 & \color{red}{1} & \color{red}{0}  & \color{red}{1}\\
0 & 0 & 0 & 1 & \color{red}{1} & \color{red}{1} & \color{red}{0}  & 1\\
0 & 0 & 0 & \color{red}{1} & \color{red}{1} & \color{red}{1} & 0 & 1
\end{matrix}
$$

Q1. What is $u(n)$?

This may be simple, but the wrap-around seems to make it not so straightforward.

Q2. What is the generalization to strings of $k$ symbols?
  Let $u(k,n)$ be the length of the shortest string on $k$ symbols
  that contains all strings of length $n$.

Likely this is all known...
 [/INST] What you're looking for is related to De Bruijn sequences. The formula seems to be very simple: $u(k,n) = k^n$ (and the special case $u(n) = 2^n$).

(source: Wikipedia, by Eviatar Bach)
</s>