From: Allan Wechsler <acwacw@gmail.com> Is not B=6 unsolvable? The only reduced rationals with denominator 6 are 1/6 and 5/6. 1/6 has continued fraction 0;6. 5/6 has 0;1,5. Both have terms that exceed 2. Have I misunderstood the question? If I have understood correctly, then 9 and 14 are also unachievable. I have an algorithm for listing these, but I haven't implemented it. I'll go into it in more detail if Warren confirms that I have understood the problem correctly. --WDS: what AW says sounds good to me. Indeed see http://oeis.org/A228804 and http://oeis.org/A228803 for the first 1000 terms. It appears from that the problem is solvable for approximately half the numbers B and unsolvable for the other half. Assuming "approximately half" tends to some limiting constant C with 0<C<1 it would be of some interest to know what C is. It also is interesting to ask about useful subsequences of the integers. For example when B is restricted to the subsequence "powers of 2" we find we have solubility when B=1, 2, 4, 8, 64, 128, 256, but no others up to and including 2048. The other half of the problem, which AW has not yet addressed, is given a number B, is there some fast way to find A's such that A/B has only 1s and 2s in continued fraction, and e.g. what is the answer when B=2^32. (I do have a handwavy untested solution to this problem which in theory a lot faster than the most-naive brute force approach, but there might easily be an idea much better than mine.)