Definitions
Consider all positive rational numbers with "complexity" at most $n$: $$\mathbb{Q}_n=\{\tfrac{a}{b}\in\mathbb{Q}^+: \gcd(a,b)=1, ab\leq n\}\quad .$$
The ordered elements of the $n^\text{th}$ set could be noted like this $\{q_{m,n}=\frac{a_{m,n}}{b_{m,n}}\}_{m=1, 2, ...}$ with $q_{m,n}<q_{m+1,n}$.
Then let's try to define for a given set indexed by $n$ the relative width of a rational from its mediants $\frac{a}{b}\oplus \frac{a'}{b'}=\frac{a+a'}{b+b'}$ with next neighbours:
$$\delta_n \log q_m = \log \frac{q_{m,n}^+}{q_{m,n}^-}=\log \frac{q_{m,n}\oplus q_{m+1,n}}{q_{m-1,n}\oplus q_{m,n}} \quad ,$$
defining the extreme cases $i=1$ and "last" by introducing $q_{0,n}=\frac{0}{1}$ and $q_{\text{last}+1,n}=\frac{1}{0}$. The reciprocal $u_{m,n}=(\delta_n \log q_m)^{-1}$ of this small number can be thought of as a quality-factors, for an applied field that belongs both to here and there.
Note that a mediant $q_{m,n}^+=q_{m,n}\oplus q_{m+1,n}=q_{m+1,n}^-$ is not included in $\mathbb{Q}_n$.
Problem
Prove the following bounds: $$\tfrac{1}{2}\sqrt{a_{m,n} b_{m,n}}\,n < u_{m,n} < \sqrt{a_{m,n} b_{m,n}}\,n \quad ,\quad \forall m,n\quad.$$
Hints
So far, the width of $q_{2,n}=\frac{1}{n-1}$ (and it's inverse) can be observed always closest to the upper bound for $n>3$, so there should be some way to prove it by induction. Then we have (for $n\geq 2$): $$ u_{2,n} = \Big(\log \frac{n^2-\tfrac{1}{2}}{n^2-\tfrac{3}{2}}\Big)^{-1} <n^2-1< n\sqrt{n^2-1} \quad. $$
For the lower bound, we could think that $q_{m,n}=\frac{1}{1}$ is always closest, but this is not the case. I guess that we might prove by contradiction that a number exceeding the upper bound would yield a mediant that belongs to $\mathbb{Q}_n$.
It may also be easier to prove the lower bound using $$\delta_n' \log q_m = \begin{bmatrix}\frac{q^+-q^-}{\sqrt{q^+q^-}}\end{bmatrix}_{m,n} \quad ,$$ and then use the inequality $x\leq 2\sinh \frac{x}{2} \quad \Rightarrow \quad \delta_n \log q_m \leq \delta_n' \log q_m$.
Remark
The bounds seem to hold for $\delta_n \log q_m$, the variant $\delta_n' \log q_m$ and the two other variants: \begin{equation*} \delta_n'' \log q_m = \begin{bmatrix}\frac{q^+}{q}-\frac{q^-}{q}\end{bmatrix}_{m,n} \qquad \delta_n''' \log q_m = \begin{bmatrix}\frac{q}{q^-}-\frac{q}{q^+}\end{bmatrix}_{m,n} \quad. \end{equation*}
Also, the "last" index (the cardinality of the set) is $\sum_{k=1}^n 2^{\omega(k)}$, where $\omega(k)$ is the number of distinct prime factors of $k$.
To go further
Is the average $Q$-factor $\zeta(2)n$?
$\zeta(1+i \theta)$ looks absolutely beautiful (but does not converge absolutely).
$\zeta(\tfrac{1}{2}+i\tau u)$ sounds sharp, flat and saturated.
$a$ and $b$ are pitches, proportional to frequencies in SI unit $s^{-1}$, and $\frac{a}{b}$ irreducible is a just musical interval.
