# Palindromic continued fraction

Here's what I hope is a final question outside of my area that I need to understand a problem about stable vector bundles on $\mathbb{P}^2$. Thank you everybody for your help so far!

Suppose I have a rational number $0<\alpha<1$ with a palindromic continued fraction expansion, i.e.

$$\alpha = [0; a_1,\ldots,a_k],$$ where $a_i = a_{k+1-i}$, so that the sequence $a_1,\ldots,a_k$ is a palindrome. I believe from working with several examples that the last two convergents $p_k/q_k$, $p_{k-1}/q_{k-1}$ of this continued fraction satisfy $p_k = q_{k-1}$. That is, the denominator of the penultimate convergent is the numerator of $\alpha$.

I assume this is well known, and if so a reference would be great! Thanks!

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Yes, I believe this is well-known. See, for example, the proof of Theorem 1 (p.8) in math.ru.nl/~bosma/onderwijs/FrContPal.pdf (in which this fact is stated). –  Benjamin Dickman Sep 3 '12 at 22:33
See also page 10 of Rockett and Szüsz or page 32 of Die Lehre von den Kettenbrüchen by Perron. –  Henry Cohn Sep 4 '12 at 4:27

It's straightforward to prove by induction on $k$ that $[a_k; a_{k-1},\ldots,a_1] = q_k/q_{k-1}$. (Let the left-hand side be $r_k$, observe that $r_{k+1} = a_{k+1} + 1/r_k$, and then note that the sequence $q_k/q_{k-1}$ satisfies the same recurrence.) Then $[0; a_k,\ldots,a_1] = q_{k-1}/q_k$; deduce for your palindromic $\alpha$ that $\alpha = q_{k-1}/q_k$. But of course $\alpha = p_k/q_k$ as well, and so we conclude.