, 2023

Given a good sequence, consider the following for all $n$ for fixed $i$, $j$, $k$:
$$n \mid a_{i+k} - a_i \Leftrightarrow f(n) \mid k \Leftrightarrow n \mid a_{j+k} - a_j.$$
Thus one has
$$|a_{i+k} - a_i| = |a_{j+k} - a_j|$$
for all $i$, $j$, $k$. Now consider two cases separately:

a. If $a_1 = a_3$, then $a_{i+2} = a_i$ for all $i$, which means
$$a_i = \begin{cases} c_0, & i \equiv 0 \pmod 2, \\ c_1, & i \equiv 1 \pmod 2 \end{cases}$$
for some $c_0$, $c_1$ which are not necessarily distinct. Indeed, then the function
$$f(n) = \begin{cases} 1, & n \mid c_0 - c_1, \\ 2, & n \nmid c_0 - c_1 \end{cases}$$
satisfies the condition.

b. If $a_1 \neq a_3$, then $a_{i+2} \neq a_i$ for all $i$, thus $(a_{i+2} - a_{i+1}) \neq -(a_{i+1} - a_i)$, hence $(a_{i+2} - a_{i+1}) = (a_{i+1} - a_i)$, which means
$$a_i = ki + c$$
for some $k$ and $c$. Indeed, then the function
$$f(n) = \frac{n}{\gcd(n, k)}$$
satisfies the condition.