You, You and only you

We define a sequence $u(n)$ for $n=0,1,2 ,\ldots$ as follows: $u(0) = 0, u(1)=1$ , and for each $n\geq1$ , $u(n+1)$ is the smallest positive integer such that $u(n+1)>u(n)$ and $\{u(0),u(1),u(2) ,\ldots, u(n+1)\}$ contains no 3 elements which are in an arithmetic progression .

What is $u (2016)$ ?