Find a way from 2011 to 2 in four steps using a special movement USAMTS 6/2/22 states:

The roving rational robot rolls along the rational number line. On each turn, if the
  robot is at $\frac{p}{q}$, he selects a positive integer $n$ and rolls to $\frac{p+nq}{q+np}$. The robot begins at the rational number 2011. Can the roving rational robot ever reach the rational number 2?

Now, of course, I know it is true (I proved it and got the answer correct). However, I'm interested to see whether or not the robot can move from 2011 to 2 in four steps. I know it must move an even number of steps and cannot move in two steps, but that's as far as I was able to prove. I was able to find a set of six steps, so I know six is possible ($\displaystyle 2011 \rightarrow \frac{1}{671} \rightarrow 111 \rightarrow \frac{1}{23} \rightarrow 7 \rightarrow \frac{1}{3} \rightarrow 2$ works).
Is a set of four steps even possible? I figured out that for all $n > 2$, it is possible to move from $\frac{n-2}{2n-1}$ to 2 (with the robot using that same $n$). However, I have not been able to extend all the way from 2011 to that, even with a lot of time brute forcing in Mathematica.

Not an answer, but maybe some help.  It turns out your steps commute.  That is, instead of using $(1007,133,29,10,5,5)$ you use $(5,5,10,29,133,1007)$ (or any other order) you get to $2$ as well.  If the integers you use are $(n_1,n_2,n_3,n_4),$ after the four iterations you are at $\frac{2011(1+e_2+e_4)+(e_1+e_3)}{(1+e_2+e_4)+(e_1+e_3)2011}$, where $e_i$ is the $i^{\text{th}}$ degree symmetric polynomial.  For example, $e_2=n_1n_2+n_1n_3+n_1n_4+n_2n_3+n_2n_4+n_3n_4$  Setting this equal to $2$ gives $2009(1+e_2+e_4)=4021(e_1+e_3)$.  This will allow you to set bounds on the search-if the $n$'s are too large, $e_4$ will be too much larger than $e_3$
