I was playing around with the Euclidean algorithm when I noticed the following. Assume that some number $r_1$ breaks down via the Euclidean algorithm as follows:
$$ r_1 = q_1r_2 + r_3 \\ r_2 = q_2r_3 + r_4 \\ r_3 = q_3r_4 + r_5 \\ \vdots \\ r_{k-1} = q_{k-1}r_{k} + r_{k+1}\\ r_k = q_kr_{k+1}$$
Then, if we write $r_1$ in terms $r_{k+1}$ we will get something that looks like
$$ r_1 = q_1(q_2( \cdots (q_{k-1}(q_kr_{k+1}) + r_k) + r_{k - 1}) \cdots ) + r_4) + r_3$$
Now, actually figuring out what that looks like after expansion with regards to coefficients on $r_{k + 1}$ is rather mind-bending (for me at least), so I'll give a concrete example for $k = 6$:
$$ r_1 = q_1q_2q_3q_4q_5q_6r_7 + q_1q_2q_3q_4r_7 + q_1q_2q_3q_6r_7 + q_1q_2q_5q_6r_7 + q_1q_2r_7 + q_1q_4q_5q_6r_7 + q_1q_4r_7 + q_1q_6r_7 + q_3q_4q_5q_6r_7 + q_3q_4r_7 + q_3q_6r_7 + q_5q_6r_7 + r_7$$
I really hope I did that correctly, because I noticed a neat pattern. The subscripts on the $q$'s form all the ways of deleting two consecutive elements from the sequence $S = 1, 2, 3, 4, 5, 6$ union with all the ways of performing two deletions of two consecutive elements of $S$. The "empty sum" $0r_7$, or the only way to perform three deletions of two consecutive elements from $S$ is included as well, I suppose.
In other words, the family of sets of $q$-subscripts on summands above is:
$$\{\{1, 2, 3, 4, 5, 6\}, \{1, 2, 3, 4\}, \{1, 2, 3, 6\}, \{1, 2, 5, 6\}, \{1, 2\}, \{1, 4, 5, 6\}, \{1, 4,\}, \{1, 6\}, \{3, 4, 5, 6\}, \{3, 4\}, \{3, 6\}, \{5, 6\}, \emptyset\}$$
The Question Does the above mean anything? If so, what in the world did I find? Do I need to go outside more often?
P.S. Seeing as I don't really know what I'm asking for, I may be tagging this incorrectly. Please edit tags if you feel it necessary.