smc

This is a recursive function, which means the function refers back to itself to calculate subsequent terms. To solve this, we must identify the base case, $f(1)=2$. We also know that when $n$ is odd, $f(n)=f(n-2)+2$. Thus we know that $f(2017)=f(2015)+2$. Thus we know that n will always be odd in the recursion of $f(2017)$, and we add $2$ each recursive cycle, which there are $1008$ of. Thus the answer is $1008\ast 2+2=2018$, which is answer $\boxed{2018}$. Note that when you write out a few numbers, you find that $f(n)=n+1$ for any $n$, so $f(2017)=2018$