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A curious question recently crossed my mind. And that is can we construct decimal numbers of the form $$\text{"a.bcdefghij..."}$$ where each letter represents a digit $0-9$ (the number may or may not be rational) so that it equals a continued fraction of the form $$a+\cfrac{1}{b+\cfrac{1}{c+\cfrac{1}{d+\cfrac{1}{e+\ddots}}}}$$

For example I found that

$$0.32062241134...=0+\cfrac{1}{3+\cfrac{1}{2+\cfrac{1}{0+\cfrac{1}{6+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{4+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{3+\cfrac{1}{3+\cfrac{1}{5}}}}}}}}}}}}$$

Experimenting a bit, we see we can certainly construct infinitely many numbers that are close solutions, but how can we make them the most efficient? To clarify, we prefer not to go too deeply nested into the continued fraction to make the previous digits work, because then we have to make those new digits work. So some questions are:

~What algorithm will construct such numbers?
~Does that algorithm construct numbers that are “efficient” as described above? Is there one number that is the "most efficient"?
~Does an exact non-repeating rational example exist?

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    $\begingroup$ Duplicate unanswered question: math.stackexchange.com/q/2044747/527345. $\endgroup$ – noedne Apr 13 '18 at 3:43
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    $\begingroup$ zeros do usually not occur in a continued fraction, except in the first entry. Nevertheless, a good example. $\endgroup$ – Peter Apr 13 '18 at 13:28
  • $\begingroup$ Very interesting! $(+1)$ $\endgroup$ – Mr Pie Apr 13 '18 at 13:36
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    $\begingroup$ Trott Constant $\endgroup$ – Ed Pegg Jul 17 '18 at 15:39
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    $\begingroup$ @EdPegg That link is broken now. Use Trott Constants instead $\endgroup$ – Varun Vejalla Sep 18 '20 at 13:15
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A "correct" continued fraction does not contain any zeroes after the first digit. For example, [8;6,7,5,3,0,9] is not a valid continued fraction; the number that computes to will actually have a continued fraction of [8;6,7,5,12]. (This is because 3+1/{0+1/9}=3+9=12). The numbers before and after the zero are added together. Similarly, the continued fraction of the number 0.32062241134... actually begins [0;3,8,2,2,4,1,1,3...]

So, you might be able to generate a sequence of [a;b,c,d,e,...] that gives you a+(b/10)+(c/100)+(d/1000)+(e/10000)+... , but you will not be able to successfully reverse it. You can calculate the decimal digits from the continued fraction, but not the other way around, unless there is a solution that contains no zeroes.

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