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. On Wed, Sep 11, 2013 at 9:14 PM, Warren D Smith <warren.wds@gmail.com>wrote:
Consider a rational number (in reduced form) X=A/B.
Suppose I tell you B, and I ask you to "find A such that X has only 1s and 2s in its partial fraction expansion." (Even better, "...and with the fewest 2s.")
For example if B=2^32. You could by brute force just try all 2^32 possible A (actually only 2^31 if we insist A=odd) computing the continued fraction of each A/B,
Is there a faster method? How fast can you get? For which (if any) B's is the problem unsolvable?
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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