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.

How do we investigate the asymptotics of this type of recursion relation?

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?

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.

How do we investigate the asymptotics of this type of recursion relations?

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.

How do we investigate the asymptotics of this type of recursion relation?