## A Continued Fractions Lemma

I was bored, recently, in a seminar talk, and didn’t have much in front of me besides a piece of borrowed paper. I figured I’d be too distracted to do real work, so started messing around with continued fractions. They’ve always appealed to me, and I’ve only recently started playing with them. So I’ve still got lots to learn. I sat down and just started writing out the continued fraction expansion of several rational values. And I started looking for patterns. I seem to have found one. It’s not exactly brilliant, but I did it myself, so I like it.

Let $q_k(n)=[0;1,1,\ldots,1,n]$, where there are $k$ copies of 1. So, for example,

$q_0(n)=[0;n]=\cfrac{1}{n}$

$q_1(n)=[0;1,n]=\cfrac{1}{1+\cfrac{1}{n}}$

$q_2(n)=[0;1,1,n]=\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{n}}}.$

I wrote down several more, but that should give you a good place to start (and recall some notation, if you haven’t seen these recently).

It’s pretty easy to tell that $q_{k+1}(n)=\cfrac{1}{1+q_k(n)}$, which gives a handy way to prove things about these things. After writing down several terms, I started seeing Fibonacci numbers pop up, and came up with the following.

Define the Fibonacci numbers with $F_{-1}=0$ and $F_0=1$, and then the usual recurrence, $F_n=F_{n-1}+F_{n-2}$, for $n\geq 1$.

Lemma: For $k\geq 2$, $q_k(n)=\dfrac{F_{k-1} n-F_{k-3}}{F_k n-F_{k-2}}$.

This is pretty easy to show by induction, so I guess I’ll leave it as an exercise for the reader (hoping I didn’t make any mistakes). I had thought maybe defining appropriate $F_{-2}$ and maybe even $F_{-3}$ might let this lemma work for $k=0,1$, but I think that’s not the case.