Suppose I take a finite string of positive reals 1 4 19 3 In the first step, I find the absolute difference between consecutive numbers, the above string becomes
(4-1) (19-4) (19-3) (3-1) ⟹ 3 15 16 2
3 15 16 2
12 1 14 1
11 13 13 11
2 0 2 0
2 2 2 2
0 0 0 0
Taking any arbitrary string of n numbers, is it possible to determine whether or not this ends in 0s?
By trial and error, I found out that strings of length 2,4 always end at 0, however strings of length 3,5,8 can sometimes loop in binary.
Write your starting string as a column vector $$ \bf v = \pmatrix{v_1\cr v_2\cr \ldots\cr v_n}$$ If the initial values are integers, all subsequent ones will also be integers. Modulo $2$, your differencing operation corresponds to multiplication by an $n \times n$ matrix
$$ A = \pmatrix{1 & 1 & 0 & \ldots& 0\cr 0 & 1 & 1 & \ldots &0 \cr \ldots & \ldots & \ldots & \ldots & \ldots\cr 1 & 0 & 0 & \ldots & 1\cr} = I + S $$ where $S$ corresponds to a "shift" with $S^n = I$. If $n$ is not a power of $2$, the characteristic polynomial $p(z) = (z+1)^n + 1$ of $A$ over the binary field $GF(2)$ is not $z^n$, and if $\bf v \ne \bf 0$ is in the null space of $q(A)$ where $q(z) | p(z)$ with $q(0) \ne 0$ it can't be in the null space of a power of $A$. The conclusion, then, is that if $n$ is not a power of $2$, there is some initial string of $0$'s and $1$'s such that all the iterates will always have at least one $1$.
EDIT: For example, for $n = 6$, $z^2+z+1$ is a factor of $(z+1)^6 + 1 = z^6 + z^4 + z^2$, and $$A^2 + A + I \equiv \pmatrix{1 & 1 & 1 & 0 & 0 & 0\cr 0 & 1 & 1 & 1 & 0 & 0\cr 0 & 0 & 1 & 1 & 1 & 0\cr 0 & 0 & 0 & 1 & 1 & 1\cr 1 & 0 & 0 & 0 & 1 & 1\cr 1 & 1 & 0 & 0 & 0 & 1\cr} \mod 2$$ The vector $\bf v = \pmatrix{1 \cr 1 \cr 0 \cr 1 \cr 1 \cr 0\cr}$ is in the null space of this mod $2$, and the iterations go $$ \pmatrix{1 \cr 1 \cr 0 \cr 1 \cr 1 \cr 0\cr} \to \pmatrix{0 \cr 1 \cr 1 \cr 0 \cr 1 \cr 1 \cr } \to \pmatrix{1 \cr 0 \cr 1 \cr 1 \cr 0 \cr 1 \cr } \to \pmatrix{1 \cr 1 \cr 0 \cr 1 \cr 1 \cr 0\cr} \to \ldots$$
EDIT: On the other hand, if $n$ is a power of $2$, $p(z) = z^n$ and $A$ is nilpotent mod $2$. Thus if you start out with a vector of integers, after $n$ iterations you have even integers. After $kn$ iterations, you have multiples of $2^k$. But after the first iteration, which produces a vector of nonnegative entries, the maximum of the entries never increases; thus eventually the only multiple of $2^k$ that is small enough is $0$. So in this case you do get all $0$'s.
The same applies, by scaling, if you start with a vector of rational numbers.