S O L V E D
After using the Euclidean algorithm to find the greatest common divisor between $ a = r_{-1} $ and $ b = r_0 $ (see figure) I'm trying to express in a general way the solution (x and y) of the correlated Bezout's identity.
In figure I have indicated the solutions where the Euclidean algorithm has 1, 2, 3, 4, 5 and 6 steps.
I would like a solution for the case where the Euclidean algorithm has n steps.
For the $n^{th}$-case, using "continuants", we have: $y= (-1)^{n-1} K_{n-1}(q_1 , q_2 , … q_{n-1})$ where $K_0=1 ; K_1=q_1$.
$x= (-1)^{n} K_{n-1}(q_1 , q_2 , … q_{n-1})$ where $K_0=0 ; K_1=1$.
Best Answer
What you are looking for is continuants. The Wikipedia article Continued fraction mentions continuants in context. The Wikipedia article Continuants states
The recursion explains why the number of terms are Fibonacci numbers.