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This real advanced Sequences & series lesson teaches why a repeated state controls a distant recurrence term—and when a similar recurrence needs a different state definition.
A recurrence is a state machine: once exactly the same required state reappears, determinism guarantees that the entire future repeats. For a second-order rule, the ordered pair—not one value—is the state that must repeat.
Compute only until a full state repeats, record the cycle's first index and length, then reduce the distant index modulo that length.
Let $u_1=10$ and $u_{n+1}=|u_n-3|$. Find $u_{2026}$.
$10\to7\to4\to1\to2\to1$. The value $1$ repeats from $u_4$, so the deterministic first-order rule now cycles through $1,2$. Since $2026$ is even, $u_{2026}=1$.
Let $v_1=v_2=1$ and $v_{n+2}=v_{n+1}+v_n$. For which indices is $v_n$ even?
Reduce modulo $2$ and track ordered pairs: $(1,1)\to(1,0)\to(0,1)\to(1,1)$. The full pair repeats, so $v_n$ is even exactly when $3\mid n$.
This rule is second-order, so repeating one value is not enough; the ordered pair is the state.
Let $b_1=1$, $b_2=2$ and $b_{n+2}=b_{n+1}+b_n$. How many of the first $20$ terms are even?
Modulo $2$, even terms occur when $n\equiv2\pmod3$: $2,5,8,11,14,17,20$. There are $7$.
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