Let CW be an infinite binary tree where each node
has exactly two child nodes designated left and
right. If a node contains the fraction i/j, then
the left child node contains the fraction i/(i+j)
and the right child node contains the fraction
(i+j)/j. The root node contains the fraction 1/1.

If we did a breadth first search of CW, then we 
would encounter the fractions 

   1/1, 1/2, 2/1, 1/3, 3/2, 2/3, 3/1, ...

in that order. If we designate this sequence as
{an}, then prove that it can be 
generated by the following formula:

   ak+1 = 1/(2⌊ak⌋ - ak + 1) 

where a1 = 1/1 and ⌊ak⌋ is the integer part of ak.