For integers $a_1, \ldots a_n$, define the generalized continued fraction expression \begin{align} [a_n; a_{n-1}, \ldots, a_1] = a_n - \frac{1}{[a_{n-1}; a_{n-2}, \ldots, a_1]} \quad \text{and} \quad [a_1; \emptyset] = a_1 \ . \end{align} This means nothing else but \begin{align} [a_n; a_{n-1}, \ldots, a_1] = a_n - \cfrac{1}{a_{n-1} - \cfrac{1}{\cdots - \frac{1}{a_1}}} \end{align}

I have a conjecture about these:

Set $b_i = a_{n + 1 - i}$, such that $b_1, b_2, \ldots, b_n = a_n, a_{n-1}, \ldots, a_1$. Express the continued fraction $[a_n; a_{n-1}, \ldots, a_1]$ as the rational number $\tfrac{p}{q}$ for $p, q$ coprime integers. Then \begin{align} \prod_{i=1}^n [a_i; a_{i-1}, \ldots, a_1] = p = \prod_{i=1}^n [b_i; b_{i-1}, \ldots, b_1] \end{align}

This claim is of course supported by examples, e.g. \begin{align} [2;3,4] \cdot [3;4] \cdot 4 = \frac{18}{11} \cdot \frac{11}{4} \cdot 4 = 18 = \frac{18}{5} \cdot \frac{5}{2} \cdot 2 = [4;3,2] \cdot [3;2] \cdot 2 \ , \end{align} and I also managed to show the first equality in the special case where $a_i$ is the ceiling of $[a_i; a_{i-1}, \ldots, a_1]$ - but of course not every continued fraction is of that form.

But I somehow fail to see it in this general form. If it's true, then it should be known I guess, so my questions would be: Is it true? If yes, how do I see it/what's a reference? If no, what needs to be changed about the assumptions such that it is true?

EDIT 1: A bit of progress. Using induction, one may show that $\prod_{i=1}^n [a_i; a_{i-1}, \ldots, a_1]$ is the determinant of the matrix \begin{align} M(a_1, \ldots, a_n) = \begin{pmatrix} a_1 & 1 & 0 & \ldots & 0 \\ 1 & a_2 & 1 & \ldots & 0 \\ \vdots & \vdots & \ddots & \ldots & \vdots \\ 0 & 0 & \ldots & 1 & a_n \\ \end{pmatrix} \end{align} From this, we get at least \begin{align} \prod_{i=1}^n [a_i; a_{i-1}, \ldots, a_1] &= \det M(a_1, \ldots, a_n) \\ &= \det M(a_n, \ldots, a_1) = \prod_{i=1}^n [b_i; b_{i-1}, \ldots, b_1] \end{align}

EDIT 2: Ok, it's not quite true: choose $2n-1 = [n; 1, 1, n]$. Then $[1; 1, n] = -\tfrac{1}{n-1}$, and the product will be $-(2n - 1)$, rather than $2n-1$. I suspect that the sign will be something like the number of negative eigenvalues of $M(a)$, since $M(n, 1, 1, n)$ has one negative eigenvalue, but I don't see it right now.


1 Answer 1


Here's my attempt to make a proof:

If $[c_1; c_2, \ldots, c_n]$ is any continued fraction, defined in the usual way: $$ [c_1; c_2, \ldots, c_n] = c_1 + \frac{1}{c_2+\frac{1}{\ldots + \frac{1}{c_n}}}$$ and $\frac{p_m}{q_m}$ is its $m^{th}$ convergent, then, for all $m$ between $1$ and $n$,

$$\frac{p_m}{p_{m-1}} = [c_m;c_{m-1}, \ldots, c_1].$$

This can be proved by induction, using the well-known relation $$p_m = c_m p_{m-1} + p_{m-2} ~~~(1 \leq m \leq n, ~p_0 = 1, ~p_{-1} = 0).$$ The sequential product of the quotients $\frac{p_m}{p_{m-1}}$ is clearly a telescoping product,

$$ p_n = \frac{p_n}{p_{n-1}}\frac{p_{n-1}}{p_{n-2}} \ldots \frac{p_1}{p_0} = \prod_{i=1}^n [c_i;c_{i-1}, \ldots,c_1],$$

where $p_n$ is the numerator of the irreducible fraction equal to $[c_1;c_2,\ldots,c_n]$. Substituting $c_i$ for $a_i$ and $b_i$, we have two products, each equal to the numerator of the respective $n^{th}$ convergent, say $p_n^{(a)}$ and $p_n^{(b)}$. Now, as you already pointed out in your definition of the $b_i$, we have $$ [b_1;b_2, \ldots, b_n] = [a_n,a_{n-1},\ldots,a_1] = \frac{p_n^{(a)}}{p_{n-1}^{(a)}},$$ which implies $p_n^{(b)} |~ p_n^{(a)}$. However, it's also clear that $$ [a_1,a_2, \ldots, a_n] = [b_n,b_{n-1},\ldots,b_1] = \frac{p_n^{(b)}}{p_{n-1}^{(b)}},$$ which implies $p_n^{(a)} |~ p_n^{(b)}$. Therefore, $|p_n^{(a)}| = |p_n^{(b)}|=p$, and a more rigorous result follows: $$ p = \left| \prod_{i=1}^n [a_i;a_{i-1}, \ldots,a_1]\right| = \left| \prod_{i=1}^n [b_i;b_{i-1}, \ldots,b_1]\right|.$$ This is also true for the continued fractions of the kind: $$ [c_1';c_2',\ldots,c_n'] = c_1 - \frac{1}{c_2-\frac{1}{\ldots - \frac{1}{c_n}}},$$ since $c_i' = (-1)^{i+1}c_i$. The result will be basically the same, only changing the coefficients ($a_i$ for $a_i'$; $b_i$ for $b_i'$) and the values of the convergents ($p_n$ for $p_n'$).

If I didn't get sloppy anywhere, this shows you were pretty right. I hope this could be helpful, since I can't tell much about this particular result.

  • $\begingroup$ Looks good, thank you :) As per my edits, I managed to prove it up to a pesky sign, but I now changed my assumptions a bit so that I can disregard the sign. $\endgroup$
    – Jo Mo
    May 9, 2022 at 13:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .