I do know how the Euclidean algorithm works $(I=0,1,\cdots)$:
$$r_{I-1}=p_{I+1}r_I+r_{I+1}\quad(0\leq r_{I+1}<r_I)$$
here, if we let $r_{J-1}=0$, we have $r_{J-2}=gcd(r_{-1},r_0)$ according to the Euclidean algorithm.
Under this circumstance, I have found that the recurrence relation for $\alpha_I$,
$$\alpha_{I+1}=\alpha_{I-1}+p_{I+1}\alpha_I\quad(\alpha_{-1}=0,\alpha_0=1),$$
yields the following non-trivial result in every example that I have tried (recall how $J$ is defined above):
$$\alpha_{J-1}=\frac{r_{-1}}{r_{J-2}}=\frac{r_{-1}}{gcd(r_{-1},r_0)}.$$
For now I am quite sure it is true. On top of that, in order to prove the above statement, I figured out it suffices to show that
$$\alpha_I=\left(p_1+\frac{1}{p_2+\frac{1}{\cdots}}\right)\left(p_2+\frac{1}{p_3+\frac{1}{\cdots}}\right)\cdots\left(p_{I-2}+\frac{1}{p_{I-1}+\frac{1}{p_I}}\right)\left(p_{I-1}+\frac{1}{p_I}\right)p_I$$
since this implies
$$\alpha_{J-1}=\frac{r_{-1}}{r_0}\frac{r_0}{r_1}\cdots\frac{r_{J-4}}{r_{J-3}}\frac{r_{J-3}}{r_{J-2}}=\frac{r_{-1}}{r_{J-2}}.$$
And I am quite sure that this claim is true, since the first 5~6 terms perfectly agree. But I couldn't prove it yet... I thought mathematical induction could be useful at first but it turned out that such a method is also not that helpful.
If there is anyone who has seen such a recurrence relation in the context of Euclidean algorithm or has proven similar statement before, please give me some ideas...!