\[\large a_{n+1} = \sqrt{2+a_n} \]

Consider the recurrence relation \(a_n\) above for \(n = 1,2, 3, \ldots \) and \(a_1 = \sqrt3\).

Evaluate \( \log_2 \left [ \cos^{-1} \left ( \dfrac{a_{2016}}2 \right) \right ] \).

If this number can be expressed as \( \log_2 \pi - \log_2 a - b\), where \(a\) and \(b\) are positive integers with \(a\) odd, submit your answer as \(a+b\).

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