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Basic continued fractions arise from recurrence relations such as:

$$ n = a + \frac{b}{n}, $$

This gives rise to the continued fraction:

$$ n = a + \frac{b}{a+\frac{b}{a+\frac{b}{a+...}}}. $$

What about relations such as:

$$ n = a + \frac{n}{b}? $$

Do these give rise to things like:

$$ n = a + \frac{a+\frac{a+\frac{a+\frac{a+\frac{a+...}{b}}{b}}{b}}{b}}{b}? $$

Of course, one could just solve the original equation as:

\begin{align} n &= a + \frac{n}{b} \\ n(1-b^{-1}) &= a \\ n &= \frac{a}{1-b^{-1}}. \end{align}

But couldn't the above "reverse continued fraction" be generalized like the usual one to create interesting structures and definitions of known constants? For instance, one can obtain this strange result for $n=a + \frac{n}{b²}$ by setting $a=1$ and $b=\sqrt{2}^{-1}$:

$$ -1 = 1 + \frac{1+\frac{1+\frac{1+\frac{1+\frac{1+...}{\sqrt{2}^{-1}}}{\sqrt{2}^{-1}}}{\sqrt{2}^{-1}}}{\sqrt{2}^{-1}}}{\sqrt{2}^{-1}}? $$

Is this known/of any particular interest?

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    $\begingroup$ Hm, interesting. One might also wonder about alternating versions of these: $$a+\frac{b}{a+\frac{a+\frac{b}{a+\frac{a+...}{b}}}{b}}$$ $\endgroup$ – Franklin Pezzuti Dyer Dec 21 '17 at 14:43
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    $\begingroup$ $n=a+\frac nb$ is a rather less interesting equation than $n=a+\frac bn$. $\endgroup$ – Lord Shark the Unknown Dec 21 '17 at 14:47
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Note that $$n = a + \frac{a+\frac{a+\frac{a+\frac{a+\frac{a+...}{b}}{b}}{b}}{b}}{b}=a+\frac ab + \frac a{b^2}+ \frac a{b^3} +\ldots$$ gives you a geometric series. If the constants vary it is still just another way of writing an infinite sum, so doesn't give anything new.

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  • $\begingroup$ I would like to add that ascending continued fractions are intimately related to the Engel expansion of real numbers. $\endgroup$ – Klangen May 14 '18 at 7:45
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I see what Shark means: with an infinite periodic conventional c.f. you get a quadratic for the limiting value, but with a reverse c.f. you only get a linear equation.

Playing with your idea: Let $\pi = 3 + (1 + x_1)/n_1$ where $x_1 > 0$ and $n_1$ is an integer. The least $n_1$ that will do is $8$, so use that. Then express $x_1 = (1 + x_2)/n_2$, taking the least $n_2$, which happens to be 8 again. Proceed in this way and get

$$\pi = 3 + {1 + {1 + {1 +{1 + {{1 + \ldots}\over 300}\over 19}\over 17}\over 8}\over 8}$$

The sequence $3; 8, 8, 17, 19, 300, \ldots$ is non-decreasing, and this always happens after the first term (why?). For $\sqrt2$ we get $1; 3, 5, 5, 16, 18, 78, 102, 120, 144, \ldots$, hard to see a pattern there. For $e$ we get (ah!) $2; 2, 3, 4, 5, 6, 7, 8, 9, 10, \ldots$. But how to prove it?

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    $\begingroup$ For sqrt(2) see OEIS A028253 The Engel Expansion of sqrt(2). I assume you may be able to find something for exp(1) also. $\endgroup$ – O. S. Dawg Dec 21 '17 at 22:27
  • $\begingroup$ Thanks for the link. Seems these results are already known as Engel expansions. mathworld.wolfram.com/EngelExpansion.html $\endgroup$ – Michael Behrend Dec 22 '17 at 13:38
  • $\begingroup$ Supposing 0<x_i<1 and choosing n_i to be the largest possible integer yields another interesting expansion. Not sure it has a name, however. $\endgroup$ – O. S. Dawg Dec 23 '17 at 1:38

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