Recover number sequence from sliding sum There is unknown sequence of numbers. Also there is known sequence of "sliding sum" of last n numbers. Example for n=3:
unknown sequence:   2 3 2 1 0 3 2 ...
"sliding sum":          7 6 3 4 5 ...

So 7 is 2+3+2, 6 is 3+2+1 etc.
Question: is it possible to recover original sequence based on that "sliding sum" sequence? Could it be done unambiguously? Or it could be done, but few answers exist?
 A: It cannot be done unambiguously. Let the unknown sequence be $\{a_i\}$ and the sliding sums $\{s_i\}$ where $s_i = a_{i-(n-1)} + a_{i-(n-2)} + \cdots + a_i$. I'll use $n=3$, as you did in your example.
Assign any value to $a_1$ and $a_2$. Then we have $s_3 = a_1 + a_2 + a_3$, so $a_3 = s_3 - a_1 - a_2$. Next, $s_4 = a_2 + a_3 + a_4$, so $a_4 = s_4 - a_2 - a_3$. You can see that you can continue this way to determine a unique sequence $\{a_i\}$ that depends on your choice of $a_1$ and $a_2$.
For example, starting with $\{0, 0, \ldots\}$, your sequence would be this:
unknown sequence:   0  0  7 -1 -3  8  0 ...
"sliding sum":            7  6  3  4  5 ...

or starting with $\{4, 3, \ldots\}$, this:
unknown sequence:   4  3  0  3  0  1  4 ...
"sliding sum":            7  6  3  4  5 ...

and so on.
A: For $n=3$, the sequence $a_1,a_2,a_3,\ldots$ and the sequence
$$a_1+a,\,a_2+b,\,a_3-a-b,\,a_4+a,\,a_5+b,\,a_6-a-b,\,\ldots$$
where $a$ and $b$ are arbitrary, will have the same sequence of sliding sums. This generalizes to any $n$:  repeatedly add arbitrary numbers to the first $n-1$ entries and subtract their sum from the $n$th.
