Length of continued fractions Consider, for any pair of relatively prime positive integers $0 <j <n$, the expansion as a continued fraction of the quotient $\frac{n}{j}$
$$
\frac{n}{j}=b_1-\frac{1}{b_2-\dfrac{1}{\cdots-\dfrac{1}{b_l}}}.
$$
Set $l_j^n:=l$ for the length of this expansion and consider the rational number
$$
m(j,n)=\frac{l_j^n+l_{n-j}^n}{n}.
$$
What do we know about the set $M:={m(j,n)}$?
By a Result of Riemenschneider (Lemma 4 in this paper) I know that $l_j^n+l_{n-j}^n=1+ \sum_{i=1}^{l_j^n}(b_i-1)$.
In particular if $j=1$, then $b_1=n$, $l_1^n=1$,  and so $m(1,n)=1$.
It seems to me that $1$ is the maximum possible value, in other words that, for every $j, n, m (j, n) \leq 1$ holds. But I cannot prove it.
In general, I would appreciate any result on $M$. I'm interested in particular to the set $L$ of all possible limits of $m(j_k,n_k)$ for sequences $(j_k,n_k)$ where $n_k\to \infty$.
The sequence $(1,n)$ shows that $1 \in L$.
If my computations are correct, the sequence $(n,2n-1)$ (here $l_n^{2n-1}=2, b_1=2, b_2=n$) shows $\frac{1}{2} \in L$.
Is it possible to determine explicitly the set $L$? Or at least a "big" subset of it?
 A: This answer proves the following claims :
Claim 1 : $m(j,n)\le 1$
Claim 2 : For any pair of positive integers $(j,t)$ satisfying $t\lt j$ and $\gcd(j,t)=1$, $\displaystyle\lim_{k\to\infty}m(j,jk+t)=\dfrac 1j$
Claim 3 : For any integer $t\ge 2$, $\displaystyle\lim_{k\to\infty}m(k,tk-1)=\dfrac 1t$
Claim 4 : For any integer $t\ge 1$, $\displaystyle\lim_{k\to\infty}m(k,tk+1)=\dfrac 1t$
Claim 5 : $\displaystyle\lim_{k\to\infty}m(k,k^2-k+1)=0$

Let
$$[b_1,b_2,\cdots, b_l]:=b_1-\frac{1}{b_2-\dfrac{1}{\cdots-\dfrac{1}{b_l}}}$$

Claim 1 : $m(j,n)\le 1$
Proof : Let us prove $$m(j,n)\le 1\tag1$$ by induction on $j$.
If $j=1$, then we have $m(1,n)=1$, so $(1)$ holds.
Suppose that $(1)$ holds for $1,2,\cdots, j-1$ where $j\ge 2$.
We have
$$\frac nj=\left\lceil\dfrac nj\right\rceil-\frac{1}{\dfrac{j}{j\left\lceil\dfrac nj\right\rceil-n}}$$
There are integers $l,b_1,b_2,\cdots, b_l$ such that
$$\dfrac{j}{j\left\lceil\dfrac nj\right\rceil-n}=[b_1,b_2,\cdots, b_l]$$
Using this, we can write
$$\frac nj=\bigg[\left\lceil\frac nj\right\rceil,b_1,b_2,\cdots, b_l\bigg]$$
Since $0\lt j\left\lceil\frac nj\right\rceil-n\lt j$, we have, by the supposition of induction,
$$m\bigg(j\left\lceil\frac nj\right\rceil-n,j\bigg)\le 1$$
which is equivalent to
$$b_1+b_2+\cdots +b_l\le j+l-1\tag2$$
There are positive integers $s,t$ such that $n=sj+t$ with $t\lt j$.
So, we have
$$n-\left\lceil\frac nj\right\rceil-j+1=sj+t-(s+1)-j+1=(s-1)(j-1)+t-1\ge 0$$
from which we have
$$j+l-1\le n+(l+1)-1-\left\lceil\frac nj\right\rceil\tag3$$
It follows from $(2)$ and $(3)$ that
$$\left\lceil\frac nj\right\rceil+b_1+b_2+\cdots +b_l\le n+(l+1)-1$$
which is equivalent to $m(j,n)\le 1$.$\quad\blacksquare$

Claim 2 : For any pair of positive integers $(j,t)$ satisfying $t\lt j$ and $\gcd(j,t)=1$, $\displaystyle\lim_{k\to\infty}m(j,jk+t)=\dfrac 1j$
Proof : We have $\frac{jk+t}j=k+1-\frac{1}{\frac{j}{j-t}}$. There are integers $l,b_1,b_2,\cdots, b_l$ such that $\frac{j}{j-t}=[b_1,b_2,\cdots, b_l]$, so we get $m(j-t,j)=\frac{1-l+b_1+b_2+\cdots +b_l}{j}$ from which we have $b_1+b_2+\cdots +b_l=jm(j-t,j)+l-1$. Using this, we get $m(j,jk+t)=\frac{1+k-l+(jm(j-t,j)+l-1)}{jk+t}=\frac{1+\frac{jm(j-t,j)}{k}}{j+\frac tk}\to\frac 1j$ as $k\to\infty$.$\quad\blacksquare$

Claim 3 : For any integer $t\ge 2$, $\displaystyle\lim_{k\to\infty}m(k,tk-1)=\dfrac 1t$
Proof : Since $\frac{tk-1}{k}=t-\frac 1k$, we get $m(k,tk-1)=\frac{1+t-1+k-1}{tk-1}=\frac{1+\frac{t-1}{k}}{t-\frac 1k}\to \frac 1t$ as $k\to\infty$.$\quad\blacksquare$

Claim 4 : For any integer $t\ge 1$, $\displaystyle\lim_{k\to\infty}m(k,tk+1)=\dfrac 1t$
Proof : Using the fact that $\frac{N+1}{N}=[\underbrace{2,2,\cdots, 2}_{N}]$ which can be proven by induction, we have $\frac{tk+1}k=t+1-\frac{1}{\frac{k}{k-1}}=[t+1,\underbrace{2,2,\cdots, 2}_{k-1}]$, so we get $m(k,tk+1)=\frac{1+(t+1-1)+(k-1)(2-1)}{tk+1}=\frac{1+\frac tk}{t+\frac 1k}\to \frac 1t$ as $k\to\infty$.$\quad\blacksquare$

Claim 5 : $\displaystyle\lim_{k\to\infty}m(k,k^2-k+1)=0$
Proof : Since $\frac{k^2-k+1}{k}=k-\frac{1}{\frac{k}{k-1}}=[k,\underbrace{2, 2, \cdots, 2}_{k-1}]$, we get $m(k,k^2-k+1)=\frac{2k-1}{k^2-k+1}\to 0$ as $k\to\infty$.$\quad\blacksquare$
A: the first question about the lengths of the HJ-Continued fraction can be found in the following paper (preprint) https://arxiv.org/pdf/1912.07378.pdf Lemma 3.3.
