Looking more closely at this sequence, it turns out that the sequence $h_n:=a_n-4n$ is not bounded, both from above and from below. In fact, along suitable integer sequences $n_k$ and $m_k$ (of order $O(9^n)$ in fact, satisfying linear recurrences) there holds $$h_{n_k}=k\ ,\qquad h_{m_k}=-k\ .$$
I'll recall some algebraic language, that may be of use to study further properties of this sequence or other similar ones. The convenient reference is Richard Stanley's Enumerative Combinatorics, vol I, especially the chapter on Rational Generating Functions.
We start with the observation that $\Delta b_n:=b_{n+1}-b_n\ge1$, because $b_n$ is strictly increasing; so for the finite difference $\Delta a_n:=a_{n+1}-a_n$ we have $$\Delta a_n=\Delta b_{n-1}+2\Delta b_{n-2}\ge3\quad(\text{for } n\ge2).$$ Therefore, apart from the pair $(a_0,a_1)$, between any two consecutive elements of $A$ we find at least two consecutive integers, which are elements of $B$. That is, no hole of $B$ is larger than $1$. Thus $\Delta b_n\in\{0,1\}$ for all $n$, and for the same recurrence above, the values of the sequence $\Delta a_n$ for $n\ge2$ are taken among the numbers $3,4,$ and $5$ only. To see better the distribution of these three admissible values, we need to look at the sets $A$ and $B$, how they are interlaced locally at $b_n$, or in other words, where is located the maximum element $a_m$ of $A$ smaller than $b_n$ w.r.to the elements of $B$ nearby.
So if $\Delta a_m=3$, which means $$b_{n-2},\ b_{n-1},\ a_m,\ b_n,\ b_{n+1},\ a_{m+1}$$ are consecutive integers, we get $$\Delta a_n=\Delta b_{n-1}+2\Delta b_{n-2}=4\ ,\qquad \Delta a_{n+1}=\Delta b_{n} +2\Delta b_{n-1} =5$$ while $a_{m+1}$ is the maximum element of $A$ smaller than $b_{n+2}$, so that $\Delta a_{n+3}$ depends on $\Delta a_{m+1}$ analogously. Similarly, if $\Delta a_m=4$, that is $$b_{n-2},\ b_{n-1},\ a_m,\ b_n,\ b_{n+1},\ b_{n+2},\ a_{m+1}$$ are consecutive integers, then $$\Delta a_{n}=4\ ,\qquad\Delta a_{n+1}=5\ ,\qquad\Delta a_{n+2}=3,$$ while $\Delta a_{n+3}$ depends on $\Delta a_{m+1}\ .$ Finally, if $\Delta a_m=5$ then $$\Delta a_{n}=4\ ,\qquad\Delta a_{n+1}=5\ ,\qquad\Delta a_{n+2}=3,\qquad\Delta a_{n+3}=3\ ,$$ while $\Delta a_{n+4}$ depends on $\Delta a_{m+1}$.
For the moment we don't want to bother to make the correspondence $n\mapsto m$ explicit; we may rather use a convenient algebraic formalism and see concisely the sequence $(\Delta a_n)_{n\ge2}$ as an infinite string, that we build concatenating recursively new terms induced by the preceding ones. To describe it more precisely, let $\mathcal{A}$ be the set of symbols $\{\rm a,b,c\}$ and $\mathcal{A}^*$ the free monoid on $\mathcal{A}$. Let $\tau:\mathcal{A}^*\to\mathcal{A}^*$ the monoid homomorphism defined on generators by $$\tau({\rm a}):={\rm bc}\ ,\qquad\tau({\rm b}):={\rm bca}\ ,\qquad\tau({ \rm c}):={ \rm bcaa}\ .$$ Note that $\tau$ extends to a map, still denoted $\tau$, on the set of infinite strings, $\tau:\mathcal{A}^\mathbb{N}\to\mathcal{A}^\mathbb{N}$, via the left-action of $\mathcal{A}^*$, that is just $\tau({\bf x})=\tau({\rm x_0)\tau(x_1)\tau(x_2)\tau(x_3})\dots$, for any ${\bf x}={\rm x_0x_1x_2x_3}\dots\in\mathcal{A}^\mathbb{N}.$ Let ${\bf u}\in\mathcal{A}^\mathbb{N}$ be the unique fixed point of the map ${\bf x}\mapsto {\rm a}^5\tau({\bf x})$, that is $${\bf u}={\rm a}^5\tau({\bf u})={\rm a}^5\tau({\rm a}^5)\tau^2({\rm a}^5)\tau^3({\rm a}^5)\tau^4({\rm a}^5)\dots=$$ $$={\rm a}^5({\rm bc})^5 ({\rm bcabcaa })^5 ({\rm bcabcaabcbcabcaabcbc})^5 \dots=$$ $$={\rm aaaaa bcbcbcbcbc bcabcaa bcabcaa bcabcaa bcabcaa bcabcaa bcabcaabcbcabcaabcbc}\dots$$ (Warning: $\tau^3$ e.g. here means the third compositional iterate of $\tau$ that is $\tau\circ\tau\circ\tau$, while ${\rm a}^5$ refers to concatenation: ${\rm aaaaa}$; of course $\tau^3({\rm a}^5)=(\tau^3({\rm a}))^5$.) By the above definition of $\tau$, the string ${\bf u}$, putting $ {\rm a}=3, {\rm b}=4, {\rm c}=5$, produces exactly the sequence of differences $(\Delta a_n)_{n\ge2}$. (Double-check: this way, while editing, by means of paste, copy, find, replace, I quickly got the last value reported in the OEIS link, $a_{56}=221$, and e.g. $a_{100}=398$).
Since we are interested in the finite differences of the sequence $h_n:=a_n-4n$, we may consider the additive weight function $w:\mathcal{A}^*\to(\mathbb{Z},+)$ defined on the generators by $w({\rm a})=-1$, $w({\rm b})=0$, $w({\rm c})=1$. So for any ${\rm f}\in \mathcal{A}^*$ one has $w(\tau({\rm f}))=-w({\rm f})$. Moreover, for any left factor ${\rm f}\in\mathcal{A}^*$ of ${\bf u}$ (written ${\rm f}\dashv{\bf u}$, meaning that ${\bf u}={\rm f }{\bf u'}$ for some ${\bf u'}\in\mathcal{A}^\mathbb{N}$), of length $n$, one has $w({\rm f })=h_{n+2}-h_{2}$.
I'll define inductively a sequence ${\rm f}_k\in\mathcal{A}^*$ with ${\rm f}_k{\rm c}\dashv{\bf u}$, and with $w({\rm f}_k)=-k$, which proves that $\Delta h_n$ is unbounded from below. Indeed we can take ${\rm f}_0={\rm a}^5({\rm bc})^5{\rm b}$; given ${\rm f}_k$ such that $w({\rm f}_k)=-k$ and ${\rm f}_k{\rm c}\dashv{\bf u}$, we have by the fixed point equation,
$${\rm a}^5({\rm bc})^5\tau^2({\rm f}_k){\rm bcaabc}\dashv{\bf u}$$ whence ${\rm f}_{k+1}:={\rm a}^5({\rm bc})^5\tau^2({\rm f}_k){\rm bcaab}$ satisfies both ${\rm f}_{k+1}{\rm c}\dashv{\bf u}$ and $$w({\rm f}_{k+1})=w\big({\rm a}^5({\rm bc})^5\tau^2({\rm f}_k){\rm bcaab}\big)=w({\rm f}_k)+w({\rm bcaab})=-k-1\ .$$ Also, the sequence of left factors ${\rm g}_k:={\rm a}^5\tau({\rm f_k})\dashv{\bf u}$ satisfies $$w({\rm g}_{k})=w({\rm a}^5 \tau ({\rm f}_k)) =-5+k\ $$ so that we conclude that $h_n$ is also unbounded from above.