THis can be shown by a bit less concrete estimates that in @Deld's answer.
We show by induction on $n$ that $$ 4n-2\leq a_n\leq 4n; \qquad(*) $$ while proving that, we show the required equality as well. The base cases $k=1,2$ are trivial.
Assuming $(*)$ for $n=1,2,\dots,k$, we get $$ \frac{4t+1}3\geq t+\left\lceil\frac{t-1}3\right\rceil\geq b_t\geq t+\left\lfloor\frac{t-1}3\right\rfloor\geq \frac{4t}3-1 \qquad(**) $$ for $t\leq 3k$. Indeed, if $s=\left\lceil\frac{t-1}3\right\rceil\leq k-1$, then $a_{s+1}\geq 4s+2$, so there are at least $3s+1$ values of $b$ in $[1,4s+1]$. This yields that there are at most $s$ values of $a$ in $[1,b_{3s+1}]$; so, since $t\leq 3s+1$, we get $b_t\leq t+s$, as desired.
Similarly, setting $p=\left\lfloor\frac{t-1}3\right\rfloor\leq k-1$, we have $a_p\leq 4p$, so $[1,4p+1]$ contains at most $3p+1$ values of $b$. Hence $[1,b_{3p+1}]$ contains at least $p$ values of $a$; so, as $t\geq 3p+1$, we get $b_t\geq t+p$. Thus $(**)$ is proved. Moreover, we have shown that $b_{3p+1}=(3p+1)+p$ for all $p\leq k-1$, as the estimates in $(**)$ agree for $t=3p+1$. That is, we have showed the initially requested equality.
It remains to finish the step of induction, proving $(*)$ for $n=k+1$. Indeed, by $(**)$ we have $$ 4(k+1)+\frac23=\frac{4(k+1)+1}3+\frac{4(2k+2)+1}3\geq b_{k+1}+b_{2k+2}=a_{k+1}\geq \frac{4(k+1)}3+\frac{4(2k+2)}3-2=4(k+1)-2, $$ which yields the required result, as $a_{k+1}$ is integer. (Here we used that $2k+2\leq 3k$, i.e., that $k\geq 2$.)