Show that there exists a B so that, for every n > 0, there exists a k relatively prime to n whose continued fraction has

Definitions and mathematical statement

For an integer $n\ge2$, choose $1\le k<n$ with $\gcd(k,n)=1$ and write its regular continued fraction in the canonical form \[ \frac{k}{n}=[0;a_1,\ldots,a_r], \qquad a_i\in\mathbb Z_{>0},\quad a_r\ge2. \] Here the notation means the finite nested fraction $1/(a_1+1/(a_2+\cdots+1/a_r))$. Say that its partial quotients are bounded in prefix average by $B$ if \[ \sum_{i=1}^{j}a_i\le Bj \qquad(1\le j\le r). \] Does an absolute constant $B$ exist such that every denominator $n\ge2$ admits a coprime numerator satisfying all these inequalities? In particular, is $B=3$ sufficient?

The numerator may depend on $n$, but the constant may not. The condition concerns every initial segment, not only the average over the entire continued fraction; large early quotients cannot be compensated by later small ones. It is also weaker than requiring each quotient separately to be at most $B$. The source's reference to a continued fraction of a multiplier $k$ is interpreted as the rational number $k/n$. The trivial denominator $n=1$ is omitted, and the terminal-quotient convention removes the two-expansion ambiguity for rational numbers.

Short English statement

Does every denominator admit a coprime numerator whose continued-fraction partial quotients have a uniformly bounded average on every prefix, perhaps with bound3?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 3100053 (AMR-030-0053), AMR Open Problem Lists. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Cooper - Combinatorial Problems I Like (2020 snapshot), Combinatorial Number Theory, source-order bullet 53. Original source indexed by AMR-030-0053. https://people.math.sc.edu/cooper/combprob.html.