First notice the sequence is uniquely determined because $b_i>a_i$ and thus you can always find the next $a_i$ if you've determined all the previous $a_i$ and $b_i$ since it has to be the first unoccupied positive integer. Also notice no two $b_i$ can be consecutive, and that the gaps between consecutive $a_i$ are thus always either 0 or 1, and the gaps between consecutive $b_i$ are always either 1 or 2. $\textbf{Lemma 1}:$ $b_n=a_{a_n}+1$. $\textbf{Proof:} $ Let $b_n-1=a_k$. Then $a_k-k$ is the number of elements of $b$ before $a_k$, so we get $a_k-k=n-1$. This means $b_n=k+n$ therefore $k=a_n$. $\textbf{Lemma 2}:$ $b_{a_n}+1=a_{b_n}$. $\textbf{Proof:} $ Calculate $b$ from Lemma 1 to get that we need to prove $a_{a_{a_n}}+2=a_{(a_{a_n}+1)}$. This is equivalent to showing that there is an element of $b$ right after $a_{a_{a_n}}$, and this is obvious from Lemma 1, since that is precisely $b_{a_n}$. Since $a_{b_n}-b_n$ is the number of elements of $b$ before $a_{b_n}$, which is precisely $a_n$ by Lemma 2, we get the desired conclusion.

$a_n = [n\phi]$, where $\phi = \frac{\sqrt 5+1}{2}$

Beatty's theorem, especially the well-known special case with the pair $(\phi,\phi^2)$