Let the sides be $a, b, c$. From the sine rule, we have
$$\begin{aligned} \frac{a}{b} &= \frac{\sin 3B}{\sin B} = 4\cos^2 B - 1 \\ \frac{c}{b} &= \frac{\sin C}{\sin B} = \frac{\sin 4B}{\sin B} = 8\cos^3 B - 4\cos B \end{aligned}$$
Thus
$$2 \cos B = \frac{a^2 + c^2 - b^2}{ac} \in \mathbb{Q}.$$

Hence there exist coprime positive integers $p, q$ such that $2 \cos B = \frac{2}{q}$. Hence
$$\frac{a}{b} = \frac{p^2}{q^2} - 1 \quad \Leftrightarrow \quad \frac{a}{p^2 - q^2} = \frac{b}{q^2};$$
$$\frac{c}{b} = \frac{p^3}{q^3} - \frac{2p}{q} \quad \Leftrightarrow \quad \frac{c}{p^3 - 2pq^2} = \frac{b}{q^3}.$$
Thus
$$\frac{a}{(p^2 - q^2)q} = \frac{b}{q^3} = \frac{c}{p^3 - 2pq^2} = \frac{e}{f}, \quad \text{gcd}(e, f) = 1.$$
Since perimeter is minimum, $\text{gcd}(a, b, c) = 1$. From $\text{gcd}(e, f) = 1$, we have $f \mid q^3$ and $f \mid p^3 - 2pq^2$. We'll prove that $f = 1$.
If $f > 1$, then it has a prime divisor $f' > 1$ such that $f' \mid q^3$ and $f' \mid p^3 - 2pq^2$. Thus $f' \mid q$ and $f' \mid p$, contradicting $\text{gcd}(p, q) = 1$. Thus $f = 1$. From $\text{gcd}(a, b, c) = 1$, we conclude that $e = 1$. Thus
$$a = (p^2 - q^2)q, \quad b = q^3, \quad c = p^3 - 2pq^2.$$
From $0^\circ < \angle A + \angle B = 4\angle B < 180^\circ$, we get $0^\circ < \angle B < 45^\circ$ and hence $\sqrt{2} < 2 \cos B < 2$ implying that $\sqrt{q} < p < 2q$. The smallest positive integers satisfying this inequality is $p = 3, q = 2$. Since $a + b + c = p^2q + p(p^2 - 2q^2)$ and $p^2 - 2q^2 = 1$, we see that the minimum perimeter is achieved when $p = 3, q = 2$ and the value is $21$.