The growth of a cyclic sequence with $n$ terms under the rule $x_i \mapsto x_{i+1}+x_i$ and $x_{n}\mapsto x_1+x_n$ Say we have $100$ terms connected in a circle, starting with $x_1$ and going clockwise to $x_2$, to $x_3$, etc. and with $x_{100}$ going to $x_1$. Now initially $x_1 = 1$ and $x_2=-1$ and the rest of the terms are $0$. What happens when we a apply a rule where each $x_i$ simultaneously becomes the sum of itself and the term to its left? (for example $x_1\mapsto x_1+x_2$) Will the terms grow without bound? How can I prove it?
I attempted to observe what happens when there are only $4$ terms 
\begin{matrix}
&&a_{1} \\ &\swarrow && \nwarrow \\
a_{4}&&&&a_{2}\\ &\searrow&&\nearrow\\ && a_3 
\end{matrix}
and on every other iteration of the rule I see that there is a pair of terms opposite each other which equal powers of two. Does this tell me anything about the case with $100$ terms? or in general with $n$ terms?
 A: Let $x_{i,j}$ be the $i$-th term after $j$ operations . Let also $x_{i,0}=x_i$
It's not hard to prove using induction that :
$$x_{i,j}=\sum_{k=0}^{j} \binom{j}{k} x_{i+k}$$ where we make the convention that $x_{n+l}=x_l$ for $l \geq 1$ .
This can be seen by looking for patterns :
$$x_{k,1}=x_k+x_{k+1}$$
$$x_{k,2}=x_{k,1}+x_{k+1,1}=x_k+2x_{k+1}+x_{k+2}$$
$$x_{k,3}=x_{k,2}+x_{k+1,2}=x_k+3x_{k+1}+3x_{k+2}+x_{k+3}$$ which looks a lot like Pascal's triangle and then to prove it it's easy .
Also notice that most of the terms in the summation will be $0$ so now all the terms will look like this :
$$x_{s,j}=\binom{j}{n+1-s}-\binom{j}{n+2-s}+\binom{j}{2n+1-s}-\binom{j}{2n+2-s}+\ldots$$ (this is for $s>2$ . )
For $s=1$ there's also $\binom{j}{0}-\binom{j}{1}$ in the sum and for $s=2$ there's a $-\binom{j}{0}$ in the sum.
Now you need to choose that $j$ carefully to get a big  (or small) sum. 
I'll let you handle the remaining details (which seem pretty complicated )
EDIT :
I'll explain the induction :
For $j=1$ it's just : $$x_{i,1}=x_i+x_{i+1}=x_i \binom{1}{0}+x_{i+1} \binom{1}{1}$$
Assume that it's true for some $j$ and then prove it for $j+1$ :
We know that :
$$x_{i,j+1}=x_{i,j}+x_{i+1,j}$$ 
We also know that :
$$x_{i,j}=\sum_{k=0}^{j} \binom{j}{k} x_{i+k}$$
$$x_{i+1,j}=
\sum_{k=0}^{j} \binom{j}{k} x_{i+1+k}=x_{i+j+1}\binom{j}{j}+\sum_{k=1}^{j} \binom{j}{k-1} x_{i+k}$$
Now group the terms and use Pascal's theorem :
$$x_{i,j+1}=x_{i,j}+x_{i+1,j}=\sum_{k=0}^{j} \binom{j}{k} x_{i+k}+x_{i+j+1}\binom{j}{j}+\sum_{k=1}^{j} \binom{j}{k-1} x_{i+k}=x_i+x_{i+j+1}+\sum_{k=1}^{j} x_{i+k} \left ( \binom{j}{k}+\binom{j}{k-1} \right )=x_i \binom{j+1}{0}+x_{i+j+1} \binom{j+1}{j+1} + \sum_{k=1}^{j} x_{i+k} \binom{j+1}{k}=\sum_{k=0}^{j+1} \binom{j+1}{k}x_{i+k}$$ QED .
