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!

  • 1
    $\begingroup$ 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). $\endgroup$ Sep 3, 2012 at 22:33
  • $\begingroup$ See also page 10 of Rockett and Szüsz or page 32 of Die Lehre von den Kettenbrüchen by Perron. $\endgroup$
    – Henry Cohn
    Sep 4, 2012 at 4:27

2 Answers 2


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.


See Edward B Burger, A tail of two palindromes, American Mathematical Monthly 112, April 2005, pages 311 to 321, but especially page 317, Lemma 1 and discussion thereof.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.