# 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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