I claim that $a_n = n$ for all $n$. Note that $[\frac{n+2^{k-1}}{2^k}]$ counts how many numbers are $2^{k-1}$ in mod $2^k$ and not greater than $n$. Summing this for all $n$, we cover all numbers from $1$ to $n$. We are done with our claim, and that proves that $a_{2015} = 2015$ $\blacksquare$