A game from Exercise in Artin's Algebra (Chapter 2 M.13) I found an interesting problem in Chapter 2 for Artin's Algebra (2nd Ed) in the Miscellaneous section that I haven't been able to figure out. The text of the problem is quoted below. 

M.13 (a game) The starting position is the point $(1,1)$, and a permissible "move" replaces a point $(a,b)$ by one of the points $(a+b, b)$ or $(a, a+b)$. So the position after the first move will be either $(2,1)$ or $(1,2)$. Determine the points that can be reached.

I've written a small python script to generate and plot the point and I saw the pattern in the image  but no idea how to come up with a general description of the points.  
link to code is here.

I appreciate a solution or a link to a solution. 
 A: The game is taking you through all pairs of $(a,b)$ such that $a ,b > 0 $ and $\gcd(a,b)=1$.  
Proof: Take $A$ be the set of all pairs that we get from the game.
We first show that if $(a,b) \in A$ then $\gcd(a,b)=1$: it follows form Euclidean algorithm.  
Now by induction we show that if $a,b > 0$ and $\gcd(a,b)=1$ then $(a,b) \in A$: we know this is true for $a,b<2$, now if this is true for $a,b<n$ then this is true for $a=n$ or $b=n$, because for instance if $2 \leq a=n$ and $1 \leq b<n$ then by induction hypothesis we know $(a-b,b)$ is in $A$ and by the game rule we conclude $(a,b) \in A$.  

Hint for Patterns in the image:
  For every irreducible $\frac{m}{n}$ we have infinite $k$ such that $\frac{mk+1}{nk}$ is irreducible.
Proof: Set $k=an, \quad a\in \mathbb N$

A: Can't add a comment so posting as an answer:
As pointed out by angryavian, the final co-ordinates are obtained by:
$$ E_1E_2 ...E_n\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} m \\ n \end{bmatrix} $$
where each $ E_i$ is one of the two matrices
$\begin{bmatrix} 1 & 1 \\ 1 & 0\end{bmatrix} and \begin{bmatrix} 1 & 0 \\ 1 & 1\end{bmatrix} $
This means
$$ E_n^{-1} ...E_2^{-1}E_1^{-1}\begin{bmatrix} m \\ n \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$
But each $ E_i^{-1} $ encodes a step of Euclid's Algorithm, and the remainder is on the R.H.S. This implies $ gcd(m,n) =1 $
