2016th term of the sequence

\[a_{m+n} + a_{m-n} = \frac{1}{2}(a_{2m}+a_{2n})\]

The sequence \(a_0,a_1,a_2,...\) satisfies the above relation for all non-negative integers \(m\) and \(n\) with \(m \geq n\).

If \(a_1 = 1\), determine \(a_{2016}\) .

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