Looking for an "intuitive" regular expression for $\{ w \in \Sigma^{\ast} \mid 2|w|_1 + |w|_0 \equiv 0 \pmod{3} \}$
The automaton for that language is, as you say, extremely simple: it consists of a triangle, in which 0s move clockwise and 1s move counterclockwise:



A classic way to turn this into a regular expression is to eliminate states one at a time, converting transitions through each state into regular expressions. Eliminating state 2 is easy since there are no epsilon transitions or self-loops, so we end up with:

$$\begin{align}s0 \to s0&amp;: 10 \\
s0 \to s1&amp;: 0 + 11 \\
s1 \to s0&amp;: 1 + 00 \\
s1 \to s1&amp;: 01 \\
\end{align}$$

When we then eliminate state 1, we produce a relatively simple regular expression:

$$(10 + (0+11)(01)^*(1+00))^*$$

Is that regex intuitive? I guess that depends on your intuitions. $10$ and $01$ both have no effect on $2|w|_1 + |w|_0 \pmod{3}$; $0+11$ adds 1, and $1+00$ subtracts 1. So the regex definitely shows the path around the three possible values of $2|w|_1 + |w|_0 \pmod{3}$