Let $L \subseteq \Sigma^{\ast}$ with $\Sigma = \{0,1\}$ be the language such that two times the number of $1$'s in a word in $L$ plus the number of $0$'s is divisible by $3$, i.e. if we denote by $|w|_1$ the number the symbol $1$ occurs in $w$ and by $|w|_0$ the number the symbols $0$ occurs, then $$ L = \{ w \in \Sigma^{\ast} \mid 2|w|_1 + |w|_0 \equiv 0 \pmod{3} \}. $$ I know this language to be regular, as it is quite easy to give an automaton for it. I also got a (quite complicated) regular expression for it using an algorithm going from the automaton. But is there any regular expression that is easy in the sense that it gives some insight into why it is a regular expression for the language?
Not that for example for the language $\{ w \mid |w|_0 \equiv 0 \pmod{3} \}$ it is quite easy to give a regular expression, it would be $1^{\ast}(01^{\ast}01^{\ast}01^{\ast})^{\ast}$, or for $\{ w \mid |w| \equiv 0 \pmod{3} \}$ as it would be $(\Sigma\Sigma\Sigma)^{\ast}$. But the above language somehow combines these conditions.