TITLE: A Pascal's-triangle -like random process
QUESTION [8 upvotes]: I was exploring Pascal's triangle on a cylinder when I encountered this puzzle-like problem.
It is surely elementary, but perhaps weekend-entertaining.
Start with a permutation of $(1,2,3, \ldots, n)$, call it $(x_1,x_2,x_3,\ldots,x_n)$.
Perform two operations in sequence: Sum, and Shuffle.
Sum computes a new sequence $x'_i = (x_i + x_{i+1})$, with cylindrical index wraparound.
(This is the Pascal's-triangle aspect.)
Shuffle permutes the resulting numbers (below I used a random permutation).
Then the Sum/Shuffle-transformations are repeated.
Here is an example, for $n=4$:

      

I hope the process is clear without further formal definitions.
The numbers in the sequence grow by powers of 2, but approach the same normalized ratio,
and so in that sense "approach" equality.
One could ask many questions (I'll resist, although I welcome speculation!), but this is perhaps the primary question:

For which $n$ could the described process lead to a sequence of identical numbers?

For $n=2$, $(1,2)$ is the only option, which ends in $(3,3)$ after just one step.
For $n=3$, all starting permutations lead to a permutation of $(3,4,5)$, which leads to
a permutation of $(7,8,9)$, which leads to a permutation of $(15,16,17)$, and
then to $(31,32,33)$, and so on.  This is all I know: $n=2$ must end in a constant
sequence, $n=3$ cannot, and $n=4$ might.
Addendum. After Douglas Zare's (nice!) analysis answered the question
for all but $n = 0 \pmod 4$,
I concentrated on $n=8$ and found these two examples:
$$
\left(
\begin{array}{cccccccc}
 7 & 2 & 8 & 1 & 5 & 4 & 6 & 3
   \\
 10 & 9 & 10 & 9 & 6 & 9 & 10
   & 9 \\
 9 & 6 & 9 & 10 & 9 & 10 & 9 &
   10 \\
 15 & 15 & 19 & 19 & 19 & 19 &
   19 & 19 \\
 19 & 19 & 15 & 19 & 19 & 19 &
   19 & 15 \\
 34 & 38 & 34 & 34 & 38 & 38 &
   38 & 34 \\
 34 & 38 & 34 & 38 & 34 & 38 &
   34 & 38 \\
 72 & 72 & 72 & 72 & 72 & 72 &
   72 & 72 
\end{array}
\right)
$$
$$
\left(
\begin{array}{cccccccc}
 4 & 3 & 6 & 1 & 8 & 7 & 2 & 5
   \\
 9 & 7 & 9 & 7 & 9 & 15 & 9 &
   7 \\
 9 & 7 & 9 & 15 & 9 & 7 & 9 &
   7 \\
 16 & 16 & 24 & 24 & 16 & 16 &
   16 & 16 \\
 24 & 16 & 16 & 16 & 24 & 16 &
   16 & 16 \\
 40 & 40 & 32 & 32 & 40 & 40 &
   32 & 32 \\
 40 & 32 & 40 & 32 & 40 & 32 &
   40 & 32 \\
 72 & 72 & 72 & 72 & 72 & 72 &
   72 & 72
\end{array}
\right)
$$
The rows alternate between permutation and sum, i.e.,
each odd row is a permutation of the even row above.

REPLY [3 votes]: It is possible if and only if 4 divides n. Douglas Zare showed that if 4 does not divide n, then it is impossible (btw, his argument might be simpler to state mod 2 - if 4 does not divide n, then it is already impossible to get all numbers divisible by 4). Now, if n=4k, do the following:
1, 12, 2, 10, 4, 8, 6, 7, 5, 9, 3, 11.
This will give 2k-1 n's, 2k-1 n+2's and two n+1's. Next, do
12, 14, 12, 14, 12, 14, 12, 14, 12, 14, 13, 13.
This gives 4k-2 2n+2's, one 2n+3 and one 2n+1.
Now, for simplicity, subtract 2n+2 from each number. We have all zeros but one +1 and one -1. Make the permutation so that we double the number of +1's and -1's in each step, until they become >n/4, then "mix them a bit"* so that in the next step we can exactly n/2 of both. Then an alternating sequence will give all equal numbers.
$^*Edit$: As Douglas pointed out this mixing is not that clear how to do. So instead, I claim that we for any i from 1 to n/2 it is possible to get exactly i +1's and i -1's. The proof is by induction - we either double or double -1, by putting a +1 and a -1 next to each other. Eventually we can get n/2 of each and we are done.
However, this my method only works for the numbers 1, 2, .. , n. What if we start from another sequence? Mod 2 the problem is always solvable, maybe that helps, but over Z already 1, 0, .. , 0 is not clear how to solve.
Edit: As Doublas pointed out this is possible if and only if n is a power of 2. So to summarize, I think if 4 does not divide n>1, then you can not get all equal numbers starting from 1, 2, .. , n, if 4 divides n you can, while if is n>2 is a power of 2, then it seems you can start from any sequence. (Though we have not seen a full proof of this last part yet.)