SAUDI ARABIAN IMO Booklet, 2023

We have the general formula of the given sequence as
$$u_n = \frac{(1 + \sqrt{3})^n - (1 - \sqrt{3})^n}{2\sqrt{3}}.$$
Considering the periodicity of the remainder when divided by $3$ of the given sequence, we have $0, 1, 2, 0, 1, 2, \ldots$ So obviously $u_{3k+1}$ and $u_{3k+2}$ are not divisible by $3$, and then $v_3(u_{3k+1}) = v_3(u_{3k+2}) = 0$.

Next, consider $n = 3k$ with $k \in \mathbb{Z}^+$. We have
$$\begin{aligned} u_{3k} &= \frac{(1 + \sqrt{3})^{3k} - (1 - \sqrt{3})^{3k}}{2\sqrt{3}} \\&= \frac{[(1 + \sqrt{3})^k - (1 - \sqrt{3})^k][(4 + 2\sqrt{3})^k + (4 - 2\sqrt{3})^k + (-2)^k]}{2\sqrt{3}} \\&= u_k[(4 + 2\sqrt{3})^k + (4 - 2\sqrt{3})^k + (-2)^k].\end{aligned}$$

Put $a_k = (4 + 2\sqrt{3})^k + (4 - 2\sqrt{3})^k + (-2)^k$ then it is easy to check that $a_0 = 3$, $a_1 = 6$, $a_2 = 60$ and
$$a_{n+3} = 6a_{n+2} + 12a_{n+1} - 8a_n, \forall n \ge 0.$$
Notice that $a_0, a_1, a_2$ are all divisible by $3$ so $3|a_n$ for all $n$. Thus $9|6a_{n+2}+12a_{n+1}$ and followed by $a_{n+3} \equiv -8a_n \equiv a_n \pmod 9$. On the other hand, the first three terms of the sequence are not divisible by $9$, so the same applies to all terms of the sequence. From that we have $v_3(a_n) = 1$ for all $n$. So
$$v_3(u_{3k}) = v_3(u_k) + v_3(a_k) = 1 + v_3(u_k).$$
From here it is easy to see that if we put $n = 3^t m$ with $\gcd(m, 3) = 1$ and $t \in \mathbb{Z}^+$ then
$$v_3(u_n) = v_3(u_{3^{t-1}m}) + 1 = \cdots = v_3(u_m) + t = t.$$