TITLE: Asymptotics of a recurrence relation
QUESTION [16 upvotes]: The sequence $(a_n)_{n \ge 0}$ satisfies, $a_0 = a_1 = 1$ and the recursion relation:
$$a_n = \sum\limits_{k=0}^{[n/2]} \frac{a_k}{(n-2k)!}$$
where, $[x]$ is the nearest integer to $x$ not exceeding it.
Alternatively define $a_n$'s as:
$$\sum\limits_{n=1}^{\infty} a_nx^n = \exp\left(\sum\limits_{n=0}^{\infty} x^{2^n}\right)$$
We need to show that: $$\liminf_{n \to \infty} \frac{\log a_n}{\log n} \le \frac{1}{\ln 2} - 1 \le \limsup_{n \to \infty} \frac{\log a_n}{\log n}$$
How do we investigate the asymptotics of this type of recursion relation?

REPLY [3 votes]: Here is a sketch of the standard approach to a problem of this type; its expansion will probably of Lucia's solution.
If in Jeff Shallit's response you introduce the related monotone increasing sequence $A_n=\sum_{k=0}^na_n$ then the difference relations can be used to show that $$C^kA_{\max\{0,[n/2^k]-5\}}\le A_n\le C^kA_{[n/2^k]+1}$$ for any $k$ with the choice $C=1+1/2+1/24+1+1/6+1/120=163/60$. An inductive argument then shows that $C_1n^\xi\le A_n\le C_2n^\xi$ with $\xi=1/\log(2)$ and some positive $C_1,C_2$ and the final passage from $A_n$ to $a_n$ can be performed with the help of the Stolz–Cesàro theorem.