jmc

Note that $f(2)=5$ implies $f^{-1}(5)=2$. Applying $f^{-1}(x+4)=2f^{-1}(x)+1$ repeatedly, we have: $$\begin{array}{rl}{f}^{-1}(5)& =2\\ \Rightarrow f^{-1}(9)& =2f^{-1}(5)+1=5\\ \Rightarrow f^{-1}(13)& =2f^{-1}(9)+1=11\\ \Rightarrow f^{-1}(17)& =2f^{-1}(13)+1=23.\end{array}$$So $f^{-1}(17)=\boxed{23}$.