Finding the ith Item in a Continued Fraction Expansion

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The discussion centers on finding a binary operation f(i, x) that retrieves the ith term of a continued fraction expansion for a real number x. It is noted that both rational and irrational numbers have unique continued fraction expansions, prompting the inquiry into a formula that can directly yield the ith term. The initial suggestion involves using the Euclidean algorithm, but the user seeks alternative methods to compute the terms without relying on this approach. The conversation highlights the complexity of extracting specific terms from continued fractions and the need for a more efficient computation method. Overall, the quest for a formula to simplify the extraction of terms in continued fractions remains a key focus.
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Suppose we can write a real number x as a continued fraction like this

x=a0+1/(a1+1/(a2+1/(a2+...=[a0; a1, a2, a3, a4, ... an...].

Is there a binary operation f(i,x) so that f(i,x)=ai? I was wondering if there was a formula which gives the ith item in the sequence of integers which is connected to every x in the context of this expansion.

Every rational number has a unique continued fraction expansion so I think this is a valid question. Moreover, every irrational number has a unique, infinite continued fraction expansion.

My first guess was to combine the inputs a, b in the Euclidean algorithm from which the continued fraction expansion arises but I don't know how to extract the ith item in the sequence of quotients. Any thoughts?
 
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Thanks a lot for answering my question. It's a big step in the right direction.

My question was how to compute the continued fraction expansion of a fraction. For example, using the Euclidean algorithm, we have
<br /> \frac{7}{10}=0+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{3}}} <br />
Therefore, $$a_0=0, a_1=1, a_2=3, a_3=3, a_4=0, a_5=0, a_6=0,\dots$$

Is there a way to compute $$a_n$$ without using the euclidean algorithm?
 
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