We claim that $ a<2$ is necessary and sufficient. First, we show that $ a<2$ does indeed give the desired sets $ A_1,A_2,...,A_n$: If $ a\le1$ then we may simply have $ A_1=\mathbb{Z}$. If $ 1<a<2$, then let $ A_1=\{...,-3,-1,1,3,...\}$. For each $ i$ such that $ a^i>2^{i-1}$, let $ A_i$ be the set of integers of the form $ 2^{i-1}(2k+1)$ for integer $ k$. Note that this is valid since each integer is seperated by $ 2^i>a^i$ and since all integers in $ A_j$ such that $ j<i$ have greatest factor of $ 2$ strictly less than $ 2^i$. Also notice that at each iteration, the integers left to be taken are of the form $ 2^ik$ for integer $ k$. For some $ n$, $ a^n\le2^{n-1}$. For this $ A_n$, we have $ A_n$ be the set of integers of the form $ 2^{n-1}k$ for integer $ k$, and we have all of our sets. Suppose, for the sake of contradiction, that there exists $ a\ge2$ such that there exists a valid series of sets $ A_1,A_2,...,A_n$. Let $ l=lcm(\lceil a\rceil,\lceil a^2\rceil,...,\lceil a^n\rceil)$. Consider the set $ S=\{1,2,...,l\}$. The set $ A_i$ can have at most $ \frac{l}{\lceil a^i\rceil}$ elements that are also in $ S$. Thus, the total number of elements of $ \bigcup A_i$ that are also in $ S$ is $ \displaystyle\sum_{i=1}^n\frac{l}{\lceil a^i\rceil}\le\displaystyle\sum_{i=1}^n\frac{l}{a^i}=\frac{l(1/a-1/a^{n+1})}{1-1/a}=\frac{l(a^n-1)}{a^{n+1}-a^n}$. The fraction $ \frac{a^n-1}{a^{n+1}-a^n}$ is less than $ 1$ since $ a\ge2\Rightarrow 2a^n<a^{n+1}+1\Rightarrow a^n-1<a^{n+1}-a^n$. Thus, the members of $ A_1,A_2,...,A_n$ cannot possibly contain all elements of $ S$. Contradiction!