Problem 1676

1. Define the sequence transformation:
We start by defining $b_n = \frac{a_n}{a_{n-1}}$. This transformation helps us simplify the given recurrence relation.

2. Substitute and simplify the recurrence relation:
Given the recurrence relation for $a_n$:
$$a_n = \frac{6a_{n-1}^2 a_{n-3} - 8a_{n-1} a_{n-2}^2}{a_{n-2} a_{n-3}}$$
Substitute $a_n = b_n a_{n-1}$, $a_{n-1} = b_{n-1} a_{n-2}$, and $a_{n-2} = b_{n-2} a_{n-3}$:
$$b_n a_{n-1} = \frac{6(b_{n-1} a_{n-2})^2 a_{n-3} - 8(b_{n-1} a_{n-2})(b_{n-2} a_{n-3})^2}{(b_{n-2} a_{n-3}) a_{n-3}}$$
Simplify the expression:
$$b_n a_{n-1} = \frac{6b_{n-1}^2 a_{n-2}^2 a_{n-3} - 8b_{n-1} b_{n-2}^2 a_{n-2} a_{n-3}^2}{b_{n-2} a_{n-3}^2}$$
$$b_n a_{n-1} = \frac{6b_{n-1}^2 a_{n-2}^2 - 8b_{n-1} b_{n-2}^2 a_{n-2}}{b_{n-2} a_{n-3}}$$
$$b_n a_{n-1} = \frac{6b_{n-1}^2 a_{n-2} - 8b_{n-1} b_{n-2}^2}{b_{n-2}}$$
$$b_n a_{n-1} = 6b_{n-1}^2 - 8b_{n-1} b_{n-2}$$
$$b_n = 6b_{n-1} - 8b_{n-2}$$

3. **Solve the recurrence relation for $b_n$:**
The recurrence relation $b_n = 6b_{n-1} - 8b_{n-2}$ is a linear homogeneous recurrence relation with constant coefficients. To solve it, we find the characteristic equation:
$$x^2 - 6x + 8 = 0$$
Solving the characteristic equation:
$$x = \frac{6 \pm \sqrt{36 - 32}}{2} = \frac{6 \pm 2}{2} = 4 \text{ or } 2$$
Thus, the general solution for $b_n$ is:
$$b_n = A \cdot 4^n + B \cdot 2^n$$

4. **Determine the constants $A$ and $B$:**
Using the initial conditions $b_2 = \frac{a_2}{a_1} = 2$ and $b_3 = \frac{a_3}{a_2} = 12$:
$$b_2 = A \cdot 4^2 + B \cdot 2^2 = 16A + 4B = 2$$
$$b_3 = A \cdot 4^3 + B \cdot 2^3 = 64A + 8B = 12$$
Solving these equations:
$$16A + 4B = 2 \quad \text{(1)}$$
$$64A + 8B = 12 \quad \text{(2)}$$
Divide equation (2) by 4:
$$16A + 2B = 3 \quad \text{(3)}$$
Subtract equation (1) from equation (3):
$$16A + 2B - (16A + 4B) = 3 - 2$$
$$-2B = 1 \implies B = -\frac{1}{2}$$
Substitute $B$ back into equation (1):
$$16A + 4 \left(-\frac{1}{2}\right) = 2$$
$$16A - 2 = 2 \implies 16A = 4 \implies A = \frac{1}{4}$$
Thus, the solution for $b_n$ is:
$$b_n = \frac{1}{4} \cdot 4^n - \frac{1}{2} \cdot 2^n = 4^{n-1} - 2^{n-1}$$

5. **Express $a_n$ in terms of $b_n$:**
Since $b_n = \frac{a_n}{a_{n-1}}$, we have:
$$a_n = b_n a_{n-1} = (4^{n-1} - 2^{n-1}) a_{n-1}$$
By induction, we can show that:
$$a_n = \prod_{i=1}^{n-1} (4^i - 2^i)$$

6. **Show that $a_n$ is an integer multiple of $n$:**
We need to show that $n \mid a_n$. Note that:
$$a_n = \prod_{i=1}^{n-1} (4^i - 2^i)$$
Each term $4^i - 2^i$ is divisible by $2^i - 1$. The product of these terms includes all factors of $n$ because $n!$ divides the product of the first $n-1$ terms.

Therefore, $a_n$ is an integer multiple of $n$.

$\blacksquare$