Subject: Infinite periodic continued fractions & their equivalent fractions of the format A/(B+√C).

The equations in question:

continuedfraction{0;{a}} = 2/(a+√(4+a2))

continuedfraction{0;{a,b}} = 2b/(ab+√(ab(4+ab)))

continuedfraction{0;{a,b,c}} = 2(1+bc))/(-b+c+a+(abc)+√((b−c−a−(abc))²–(4(1+(bc))(−1−ba))))

continuedfraction{0;{a,b,c,d}} = (2(d+bcd+b))/((cd)+(ad)+(bcad)-(bc)+(ab)+√((−(cd)−(ad)—(bcad)+(bc)−(ba))²−4(−c−a−(bac))(d+(bcd)+b)))

continuedfraction{0;{a,b,c,d,f}} = 2(df+bdcf+bf+1+bc)/(-d-bcd+dcf+adf+bcadf+f+baf-b+c+a+bac+√((d+bcd-dcf-adf-bcadf-f-baf+b-c-a-bca)²-4(-cd—ad—bcad—1—ba)(df+bdcf+bf+1+bc)))

https://www.desmos.com/calculator/ezxy7sfowu

What I'm doing: I am looking for an understanding of infinite continued fractions with repeating periodic blocks such as x=[0;a,b,a,b,a,b,...] and x=[0;a,b,c,a,b,c,a,b,c,...].

Why I am doing it: It may not be immediately obvious, but the equations above are of the format A/(B+√C), and that's what I was going for. If you search 0.26556443707 on WolframAlpha, it will tell you this number is equal to a truncation of ∞CF[0;{3,1,3}], and it will tell you this CF is equivalent to a calculation of 4/(7 + sqrt(65)) or 1/4 (sqrt(65) - 7). But what is that? Where did that 4 7 & 65 come from? What links this equivalence to ∞CF[0;{4,5,6}] being equal to a calculation of 62/(125 + sqrt(18229))? I wanted a look at the underlying mechanism.

How I started: Getting to that first equation was a slog through tables of curves, described here: https://twitter.com/hezooss/status/1556361128313344025 .
This was a horrible process, but it gave me the first two equations.

While attempting to start something similar with 3 variables, I came up with a shortcut: I couldn't ask Wolfram Alpha continuedfraction{0;{a,b}}, but I could ask it continuedfraction{0;{pi,e}}, and then swap pi and e out for a and b in the resulting equation. For more variables, I just had to use more transcendental constants and switch them out. That's how I got the rest of the equations.

So, is it useful? Well, I think so. I was having a look at that nice old problem “divide 10 into two parts, the product of which is 40”. While I haven't yet managed that problem as I was planning (without complex numbers), I came across a number, −3.26556443707. As mentioned above, 0.23844795843 is a truncation of ∞CF[0;{3,1,3}], so I made use of the relevant equation to get this number into Desmos: https://www.desmos.com/calculator/kzuo5c5tvf

I then used these equations to make a crude calculator.
https://www.desmos.com/calculator/txhbonn6kc

I say it's crude because the simplification process used is imperfect, as you can see if you change list a to [2,3], and note that the simplified fraction is not the same as when a=[2,3,2,3]. The simplification was never the point, but I'd be happy to be shown where I went wrong. I suspect the problem might be in the (mis)use of the gcd function. It is also probably crude that I made this file before having the rule.

I am looking for help in finding the rule linking the equations so that I can evaluate ∞CFs with more and more repeated variables. I could use Wolfram to work out the equation to use for 6 repeating variables, but I wasn't sure it would help, and I'd prefer to see and then use the rule. I did note that these equations appear to be related to the quadratic formula moved up a level as an extra variable is added.

Comments/assistance/collaboration most welcome.