# Numbers whose decimal digits are the coefficients of its continued fraction form

A curious question recently crossed my mind: can we construct decimal numbers of the form $$\text{"a.bcdefghij…"}$$ where each letter represents a digit $$0-9$$ (where the number may or may not be rational), so that it is equal to 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

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

Experimenting a bit, we see that 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?

• Duplicate unanswered question: math.stackexchange.com/q/2044747/527345. Apr 13 '18 at 3:43
• zeros do usually not occur in a continued fraction, except in the first entry. Nevertheless, a good example. Apr 13 '18 at 13:28
• Very interesting! $(+1)$ Apr 13 '18 at 13:36
• Trott Constant Jul 17 '18 at 15:39
• @EdPegg That link is broken now. Use Trott Constants instead Sep 18 '20 at 13:15