Let $a_0 = 1$ and define the sequence $\{a_n\}$ by \[a_{n+1} = \frac{\sqrt{3}a_n - 1}{a_n + \sqrt{3}}.\] If $a_{2017}$ can be expressed in the form $a+b\sqrt{c}$ in simplest radical form, compute $a+b+c$.

[i]2016 CCA Math Bonanza Lightning #3.2[/i]
1. **Define the sequence and initial condition:**
   Given \( a_0 = 1 \) and the sequence \( \{a_n\} \) defined by:
   \[
   a_{n+1} = \frac{\sqrt{3}a_n - 1}{a_n + \sqrt{3}}
   \]

2. **Relate the sequence to the tangent function:**
   Recall the tangent sum formula:
   \[
   \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}
   \]
   Notice the similarity to our sequence definition. Let \( a_n = \tan \theta_n \). We observe that \( \tan \frac{\pi}{3} = \sqrt{3} \). Then:
   \[
   a_{n+1} = \frac{\sqrt{3} \tan \theta_n - 1}{\tan \theta_n + \sqrt{3}}
   \]

3. **Simplify using the tangent subtraction formula:**
   We can rewrite the sequence definition using the tangent subtraction formula:
   \[
   a_{n+1} = \frac{\sqrt{3} \tan \theta_n - 1}{\tan \theta_n + \sqrt{3}} = -\frac{1}{\tan \left( \theta_n + \frac{\pi}{3} \right)}
   \]
   This can be further simplified using the identity \( \tan \left( \frac{\pi}{2} - x \right) = \cot x \):
   \[
   a_{n+1} = -\tan \left( \frac{\pi}{2} - \left( \theta_n + \frac{\pi}{3} \right) \right) = \tan \left( \theta_n + \frac{\pi}{3} - \frac{\pi}{2} \right) = \tan \left( \theta_n - \frac{\pi}{6} \right)
   \]

4. **Determine the general form of \( a_n \):**
   Since \( a_n = \tan \theta_n \) and \( \theta_{n+1} = \theta_n - \frac{\pi}{6} \), starting from \( \theta_0 = \frac{\pi}{4} \) (because \( a_0 = 1 = \tan \frac{\pi}{4} \)):
   \[
   \theta_n = \frac{\pi}{4} - \frac{n \pi}{6}
   \]
   Therefore:
   \[
   a_n = \tan \left( \frac{\pi}{4} - \frac{n \pi}{6} \right)
   \]

5. **Calculate \( a_{2017} \):**
   \[
   a_{2017} = \tan \left( \frac{\pi}{4} - \frac{2017 \pi}{6} \right)
   \]
   Simplify the angle:
   \[
   \frac{\pi}{4} - \frac{2017 \pi}{6} = \frac{3\pi}{12} - \frac{2017 \pi}{6} = \frac{3\pi - 4034 \pi}{12} = \frac{-4031 \pi}{12}
   \]
   Since \( \tan \) is periodic with period \( \pi \):
   \[
   \tan \left( \frac{-4031 \pi}{12} \right) = \tan \left( \frac{-4031 \pi}{12} + 336 \pi \right) = \tan \left( \frac{-4031 + 4032}{12} \pi \right) = \tan \left( \frac{\pi}{12} \right)
   \]
   Therefore:
   \[
   a_{2017} = \tan \left( \frac{\pi}{12} \right)
   \]

6. **Express \( \tan \left( \frac{\pi}{12} \right) \) in simplest radical form:**
   Using the identity \( \tan \left( \frac{\pi}{12} \right) = 2 - \sqrt{3} \):
   \[
   a_{2017} = 2 - \sqrt{3}
   \]

7. **Sum the components:**
   The expression \( 2 - \sqrt{3} \) is in the form \( a + b \sqrt{c} \) where \( a = 2 \), \( b = -1 \), and \( c = 3 \). Therefore:
   \[
   a + b + c = 2 - 1 + 3 = 4
   \]

The final answer is \(\boxed{4}\)