Consider fractions with denominator q such that $\frac{p}{q}$ < 1 and p & q are coprime, e.g. q = 10: $\frac{1}{10}$, $\frac{3}{10}$, $\frac{7}{10}$, $\frac{9}{10}$. Now take the continued fractions of all such fractions, for any given denominator.
Conjecture (or maybe it's well known, or maybe false): for all q (except 6), there is at least one numerator p such that the terms of the continued fraction of $\frac{p}{q}$ are all less than 4.
Questions: is that conjecture true? If so, why is q = 6 the only exception?
I have looked at examples and aside from q = 6 have not found a counter-example. The question arose from looking at the pattern of fractions with a given denominator as they appear in the Stern-Brocot tree. For a given denominator, continued fractions with a long sequence of smaller numbers appear higher up the tree than shorter sequences with larger terms. It is fairly easy to see intuitively why this is so. The Stern-Brocot row is determined by the sum of the terms of the continued fraction, whilst the value of the original fraction's denominator is determined by successive multiplication and then addition of continued fraction terms from the last to the first. But I cannot see why (if it is so) there is always some continued fraction with terms all less than 4. I am tagging this 'combinatorics' because I think the answer may be connected with partitioning.