It should not be hard to prove upper and lower bounds of the form $n^c$ for some constant $c< 1$ for $a_n$. (Probably different $c$ for upper and lower bound.)

The basic idea is that, for all practical purposes, the recurrence relations are $a_{2n} = a_n + a_{n-1}/2 + a_{n-2}/24$ and $a_{2n+1} = a_n + a_{n-1}/6 + a_{n-2}/120$ (where I've omitted lower order terms). This means that (roughly speaking) $a_n$ is going to be greater than $a_{n/2}(1+ 1/6)$, so by iterating this we get that $a_n > (7/6)^{\log_2 n}$, which gives $a_n > n^c$. For the upper bound, bound $a_{n-k}$ by $a_n$ and use the fact that the sum of inverse factorials is bounded by $e$.

It should not be hard to prove upper and lower bounds of the form $n^c$ for some constant $c< 1$ for $a_n$. (Probably different $c$ for upper and lower bound.)

The basic idea is that, for all practical purposes, the recurrence relations are $a_{2n} = a_n + a_{n-1}/2 + a_{n-2}/24$ and $a_{2n+1} = a_n + a_{n-1}/6 + a_{n-2}/120$ (where I've omitted lower order terms). This means that (roughly speaking) $a_n$ is going to be greater than $a_{n/2}(1+ 1/6)$, so by iterating this we get that $a_n > (7/6)^{\log_2 n}$, which gives $a_n > n^c$. For the upper bound, bound $a_{n-k}$ by $a_n$ and use the fact that the sum of inverse factorials is bounded by $e$.

Local minima should be near $n = 2^k - 2$, and local maxima near $n = \lfloor 2^k/3 \rfloor$.