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](https://en.wikipedia.org/wiki/Zeckendorf%27s_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?



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


