I have been give then continued fraction
$\dfrac{1}{\dfrac{1}{\dfrac{1}{x-1}-1}-1}$
If I let x=5, I get the following pattern...
$\frac{1}{4}, \, -\frac{4}{3}, \,-\frac{3}{7}, \,-\frac{7}{10}, \,-\frac{10}{17}, \,...$
It appears that (excluding the first case) the previous denominator becomes the numerator and the denominator becomes the sum of the previous denominator and numerator. Is there a way to write that pattern as a sequence so I can find the 653th instance where x=56?
Rather than starting with $x=5$, it's more enlightening to just write down the sequence in terms of $x$: $$\frac1{x-1}, -\frac{x-1}{x-2}, -\frac{x-2}{2x-3}, -\frac{2x-3}{3x-5}, -\frac{3x-5}{5x-8}, -\frac{5x-8}{8x-13}, \dots$$ The coefficients in these fractions might look familiar: these are the Fibonacci numbers! More precisely, the $n^{\text{th}}$ fraction in the list is $$-\frac{ F_{n-1}x - F_n}{F_n x - F_{n+1}}$$ where $F_n$ is the $n^{\text{th}}$ Fibonacci number. Since each fraction is obtained by subtracting $1$ from the previous then taking the reciprocal, this pattern is not hard to show by induction.
So the answer to your question should be $$-\frac{56 \cdot F_{652} - F_{653}}{56 \cdot F_{653} - F_{654}}.$$ If you want an explicit answer, you can either substitute in the closed-form expression for Fibonacci numbers, or use the matrix form of the Fibonacci recurrence together with exponentiation by squaring to compute $F_{652}$, $F_{653}$, and $F_{654}$, which have quite a few digits.
