Define three sequences $a_n,b_n,c_n$ as $a_0=b_0=c_0=1$ and
\begin{align*}
    a_{n+1}&=a_n+3b_n+3c_n \\
    b_{n+1}&=a_n+b_n+3c_n \\
    c_{n+1}&=a_n+b_n+c_n
\end{align*}
for $n\geq0$. Let $A,B,C$ be the remainders when $a_{13^4},b_{13^4},c_{13^4}$ are divided by $13$. Find the ordered triple $\left(A,B,C\right)$.

[i]2019 CCA Math Bonanza Team Round #10[/i]
1. Define the sequences \(a_n, b_n, c_n\) with initial conditions \(a_0 = b_0 = c_0 = 1\) and recurrence relations:
   \[
   \begin{aligned}
   a_{n+1} &= a_n + 3b_n + 3c_n, \\
   b_{n+1} &= a_n + b_n + 3c_n, \\
   c_{n+1} &= a_n + b_n + c_n.
   \end{aligned}
   \]

2. Observe that the sequences can be expressed in terms of roots of unity. Let \(\omega = \sqrt[3]{3}\) and \(\omega^2 = \sqrt[3]{9}\). Then, we can write:
   \[
   a_n + b_n \omega + c_n \omega^2 = (1 + \omega + \omega^2)^{n+1}.
   \]

3. Note that \(1 + \omega + \omega^2 = 0\) in the field of complex numbers. However, we need to consider this in the finite field \(\mathbb{F}_{13}\).

4. Compute \((1 + \omega + \omega^2)^{13^4 + 1}\) in \(\mathbb{F}_{13}\). Using Fermat's Little Theorem, we know that \(a^{13} \equiv a \pmod{13}\) for any integer \(a\). Therefore:
   \[
   (1 + \omega + \omega^2)^{13} = 1^{13} + \omega^{13} + (\omega^2)^{13} = 1 + \omega + \omega^2.
   \]

5. Since \((1 + \omega + \omega^2)^{13} = 1 + \omega + \omega^2\), it follows that:
   \[
   (1 + \omega + \omega^2)^{13^2} = (1 + \omega + \omega^2)^{13} = 1 + \omega + \omega^2,
   \]
   and similarly,
   \[
   (1 + \omega + \omega^2)^{13^3} = 1 + \omega + \omega^2.
   \]

6. Therefore, we have:
   \[
   (1 + \omega + \omega^2)^{13^4} = 1 + \omega + \omega^2.
   \]

7. Finally, we need to compute:
   \[
   (1 + \omega + \omega^2)^{13^4 + 1} = (1 + \omega + \omega^2) \cdot (1 + \omega + \omega^2) = 1 + 3\omega + 9\omega^2.
   \]

8. To find the remainders when \(a_{13^4}, b_{13^4}, c_{13^4}\) are divided by 13, we observe the coefficients of \(1, \omega, \omega^2\) in the expression \(1 + 3\omega + 9\omega^2\):
   \[
   A = 1, \quad B = 3, \quad C = 9.
   \]

The final answer is \(\boxed{(1, 3, 9)}\).