Solving $a_1+...+a_n=2011$ , $a_1\leq |a_2-a_3|, \ldots a_n\leq|a_1-a_2|$ with non-negative integers Let $a_1,...,a_n $ non negative integers s.t. $a_1+...+a_n=2011$, where $n $ is a positive integer,  $n\geq3$ and $a_1\leq |a_2-a_3|$, $a_2\leq|a_3-a_4|$,..., $a_n\leq|a_1-a_2|$.
This problem is similar to What can I say about the minimum and the maximum value of $n $?, only that
here the $a_n$ are allowed to be zero.
I have an answer but I am not sure if it is right.
I have to take two cases: $a_3\leq a_2$ or $a_2\leq a_3$.
If $a_3\leq a_2\Rightarrow |a_2-a_3|=a_2-a_3\Rightarrow a_1\leq a_2-a_3\Rightarrow a_1\leq a_2\Rightarrow |a_1-a_2|=a_2-a_1\Rightarrow a_n\leq a_2\Rightarrow a_{n-1}\leq|a_n-a_1|\leq a_2\Rightarrow...\Rightarrow a_4\leq a_2\Rightarrow |a_3-a_4|\leq a_2.$
But $a_2\leq |a_3-a_4|$. Then $|a_1-a_2|=|a_3-a_4|=...=|a_n-a_1|=a_2$. 
So  $a_1=a_3=a_5=...=0$ and $a_2=a_4=a_6=...$. 
Therefore $k\cdot a_2=2011\Rightarrow a_2=1$ and $k=2011$ $\Rightarrow n=4022$.
The other case is similar.
I don't want to post it but I am just curious!!!
It is right my argument? 
 A: The conclusion 

Then $|a_1-a_2|=|a_3-a_4|=...=|a_n-a_1|=a_2$.

is not correct, as can be seen by considering the sequence
$$
 1, 1, 0, 1, 0, \ldots 1, 0
$$
but I think you were close.
Let
$$
 x = \max(a_1, \ldots, a_n)
$$
The $a_i$ cannot all be equal, so (by rotating the indices if necessary)
we can assume that $a_1 = x$ and $a_n \ne x$.
If $a_2 \le a_3$
then
$$
 x = a_1 \le a_3 - a_2 \le x \Longrightarrow a_2 = 0, a_3 = x \, .
$$
Otherwise
$$
 x = a_1 \le a_2 - a_3 \le x \Longrightarrow a_2 = x, a_3 = 0
$$
which implies
$$
 x = a_2 \le |a_3 - a_4| = a_4 \le x \Longrightarrow a_4 = x \, .
$$
So the sequence starts either with
$$
  x, 0, x, \ldots
$$
or with
$$
 x, x, 0, x, \ldots
$$
Now the same argument can be repeated with $a_3=x$ or $a_4=x$, 
respectively, and one sees that the entire sequence consists of
the “building blocks”
$$
   x, 0 \quad \text{and} \quad x, x, 0 
$$
which can be arbitrarily combined. Finally, since the sum of
all elements is $2011$ (which is a prime number), we necessarily have
$x = 1$.
Therefore the solutions are exactly the sequences 
made of the building blocks
$$
   1, 0 \quad \text{and} \quad 1, 1, 0 
$$
with a total number of $2011$ one's (and rotations thereof).
