Problem: The positive integers $a_0$, $a_1$, $a_2$, \dots, $a_{3030}$ satisfy
\[ 2a_{n + 2} = a_{n + 1} + 4a_n \text{ for } n = 0, 1, 2, \dots, 3028.  \]
Prove that at least one of the numbers
$a_0$, $a_1$, $a_2$, \dots, $a_{3030}$ is divisible by $2^{2020}$.

Solution: The idea is this:
\begin{itemize}
  \ii All terms $a_0$, \dots, $a_{3030}$ are integers
  (divisible by $2^0=1$);
  \ii Hence all terms $a_1$, \dots, $a_{3029}$ are divisible by $2$,
  \ii Hence all terms $a_1$, \dots, $a_{3028}$ are divisible by $4$,
  \ii Hence all terms $a_2$, \dots, $a_{3027}$ are divisible by $8$,
  \ii \dots and so on.
\end{itemize}
The $2021$st item in this list reads as $2^{2020} \mid a_{1010}$.

One can also phrase this with induction.
Replace $1010$ by $N$ in the obvious way and proceed by induction on $N \ge 0$
with the base case $N = 0$ being vacuous.
Notice that for any index $k \neq 0, 3N-1, 3N$ we have
\[ a_k = 2a_{k+1} - 4a_{k-1} = 2(2a_{k+2}-4a_k)-4a_{k-1}
  = 4(a_{k+2}-2a_k-a_{k-1}) \]
so it follows that $(a_1, a_2, \dots, a_{3N-2})$
are all divisible by $4$ and satisfy the same relations.
But then $(a_1/4, a_2/4, \dots, a_{3N-2}/4)$ has length $3(N-1)+1$
and so one of them is divisible by $4^{N-1}$;
hence some term of our original sequence is divisible by $4^N$.