Products of generalized continued fractions of rational numbers

continued-fractionselementary-number-theory

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.

Best Answer

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.