Skiponacci: $p | a_p$ Alternate Solution For the Skiponacci sequence:
$a_0=3, a_1=0, a_2=2,$ and $a_{n+1}=a_{n-1}-a_{n-2}$ for $n>2$,
prove that any prime $p$ divides $a_p$.
Is there any alternate solution other than using characteristic functions and/or frobenius endomorphism? Such as an elementary solution (no advanced topics please)?
 A: I will provide a proof based on this one.
Let $S_n$ denote the set of all subsets of $\mathbb{N}_n = \{1,2,\dots,n\}$.
Define a transformation $T$ on $S_n$ by shifting any integer $s \in \mathbb{N}_n$ by $1$:
$$\left\{\begin{array}{l}
    T(s) = s + 1, \quad s<n, \\
    \tau(n) = 1 .
\end{array}\right.$$
Note that $T(S) = S$ is impossible for $S \in S_n$ because, for $s \in S$ that would imply $s + 1 \in S$. For $n = p$, a prime, we may claim that for
$$S \in S_p, T^k(S) = S,\; 1 \leqslant k \leqslant p,$$
only when $k = p$. That is, shifting $S$ repeatedly returns to $S$ only after $p$ shifts. Indeed, let $1 \leqslant k < p$ and $T^k(S) = S$.
Because $p$ is prime, $\gcd(k,p) = 1$ and $p$ so by Bezout's Identity there exist integers $x$ and $y$, with $$xk + yp = 1.$$
Then we know that $k$ shifts, and subsequently $xk$ shifts leave $S$ invariant, as any multiple of $p$ shifts, so that $T(S) = T^{xk + yp}(S) = S$, which is a contradiction.
Therefore, for any $S \in S_p$, the sets
$$S, T(S), T^2(S), \ldots, T^{p-1}(S)$$
are not only distinct but also pairwise disjoint: $T^k(S) \cap T^m(S) = \emptyset$, for $1 \leqslant k, m < p, k \neq m$. In other words, the orbit under $T$ of any $S \in S_p$, consists of $p$ sets so that the whole of $S_p$ splits into subsets of cardinality $p$ each. Therefore, $p \mid \left|S_n\right|$.
