Chain of sequences, such that $a_{k+1}(n)$ completes $a_k(n)$

Question

We say that the sequence $a_{k+1}(n)$ is a complete sequence of $a_k(n)$ if:
(1) Every term of $a_k(n)$ can be written as a sum of distinct terms of $a_{k+1}(n)$.
(2) $\lim_{n\to\infty} \frac{a_k(n)}{a_{k+1}(n)}=0$.
(Possibly there are more than one choices for $a_{k+1}(n)$ if $a_k(n)$ is given).

We can see that letting $a_1(n)=n$, then $a_{2}(n)=F_n$ (the Fibonacci numbers) is an easy (but possibly not unique) choice since by Zeckendorf's theorem every positive integer can be represented as the sum of one or more distinct Fibonacci numbers.
We can see also that an easy choice is $a_3(n)=2^n$ since every natural number is a sum of distinct powers of $2$ (and so are the Fibonacci numbers) and obviously, $\lim_{n\to\infty} \frac{F_n}{2^n}=0$.

Question 1: Can we find $a_4(n)$ given the previous $a_3(n)$.?
Question 2: Is there a general method to compute $a_{k+1}(n)$ if $a_k(n)$ is given?

Answer 1

No, such $a_4$ does not exist.

Surely, we may assume $a_4$ to be increasing. Now consider a representation of some $a_3(n)$. If it contains only elements of $a_4$ which appear in the representations of $a_3(k)$ with $k<n$, then $a_3(n)\leq a_3(1)+\dots+a_3(n-1)$ which is wrong.

Therefore, the representations of $a_3(k)$ with $k\leq n$ contain at least $n$ distinct terms of $a_4$, so $a_4(n)\leq a_3(n)$. This contradicts the limit condition