Harald Niederreiter: Dyadic fractions with small partial quotients, Monatshefte f"ur Mathematik 101,4 (December 1986) 309-315 actually showed by induction that for every power of 2, say 2^w, there existed an odd a such that the rational number a/2^w was represented by a continued fraction with only small partial quotients, specifically {1,2,3} only. Essentially, his proof went as follows. First, he found a set of suitable base cases to handle all w<12. Second, if w>11 is even he used his solution for b/2^(w/2) to construct a via a = b*2^(w/2) +- 1. Third, if w>11 is odd, he used his solution for c/2^((w-1)/2) to construct a via a = c*2^((w+1)/2) +- 1. Niederreiter also notes the maximum partial quotient "3" is least possible. That is since there exist w (examples: w=4,5,9,10,11,13,17,18,23,24, which by the way could also be another OEIS sequence) such that every a/w with a odd contains at least one partial quotient >=3 in its continued fraction. And Niederreiter's sequence of a's could well be in the OEIS, but also isn't. Now there are two troubles with Niederreiter's proof. First, the a's it constructs are useless for most applications, because their binary forms end with long strings of 11111111111 or with 0000000001. It would be more useful for almost all applications to find a's that seem more like they are made of "random bits." E.g. call an odd a which does not end with FFF or 001 in hexadecimal, "non-Niederian." QUESTION: For each w>=0, do there exist non-Niederian odd a, such that a/2^w has only {1,2,3} partial quotients in its continued fraction? I conjecture "yes" and have confirmed the conjecture with plentiful examples for each w=0,1,2,3,...,64. HARDER QUESTION: For all w>=0 except for a finite set of exceptions, do there exist non-Niederian odd a, such that a/2^w has only {1,2} partial quotients in its continued fraction? I conjecture "yes" and have found such a for each w=45,46,...,63. (Computer still working on w=64, but it is plausible that w=44, or some smaller w, is the last exceptional case where "3" is needed.) STILL HARDER QUESTION: What is the least K>1 such that: For all b>0 except for a finite set of exceptions (or, different problem version: "without exceptions") there exist a with GCD(a,b)=1, such that a/b has only {1,2,3,...,K} partial quotients in its continued fraction? S.K.Zaremba in 1972 conjectured that K=5 worked for the "without exceptions" problem version. I conjecture that K=2 works for the "finite number of exceptional b" problem version. The reason for my K=2 conjecture is twofold. First, the computer seems to find more a that work, for larger b. Second, if we consider the rational numbers representable as continued fractions [0; q1, q2, q3, ..., qh] with all the qj = 1 or 2, there are 2^h such rational numbers, and hence <=2^h different denominators. Their mean denominator actually grows asymptotic to 2^h. But the median denominator grows with h like C^h where the constant C apparently obeys 1.7<C<1.9. If so, that means a typical number b ought to be a denominator of such a "good" rational a/b, for MANY a, namely at least b^0.05 different a. This growth is fast enough so that, if the distribution "acts random" then the number of exceptional b (for which no good a exists) ought to be finite. This is a reasonably convincing heuristic argument, but I do not know any fast algorithm to find the good a for any given b.