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This is related to another question on this site, but it's not a duplicate, because the actual questions I ask are completely different.

In one of the answers Jeffrey Shallit provided a very useful link on the topic http://www.jstor.org/stable/2305906.

The author proves a very important theorem, which I will present here with changes to suit my question:

Egyptican fraction expansion of a real number in $(0,1)$ by the greedy algorithm is finite if and only if the number is rational.


So the question I ask is this:

What are the known greedy algorithm EF expansions of an irrational number where the denominators form some kind of a pattern.

For example in the same answer Jeffrey Shallit pointed out that the following number has a pattern in its expansion:

$$\frac{3-\sqrt{5}}{2}=2-\phi=\frac{1}{3}+\frac{1}{21}+\frac{1}{981}+\dots$$

The denominators here are $2^n$th Fibonacci numbers. Which is not surprising, since Fibonacci numbers are closely related to the Golden Ratio. But I know of no other example (especially for popular constants such as $\pi$, $e$, etc)

An other question:

Was EF expansion by the greedy algorithm ever used to prove the irrationality of a number, like continued fraction expansions are often used?


According to the paper, every such expansion for a real number has the form:

$$x=\frac{1}{a_1}+\frac{1}{a_2}+\dots$$

$$a_{k+1}=a_k(a_k-1)+b_k,~~~a_1 \geq 2,~~b_k > 1,~~~~a_k,b_k \in \mathbb{N}$$

Conversely, every expansion of this form converges to some real number. So we can make an infinity of such expansions with some kind of pattern. I'm going to try and check some patterns numerically.

As a simple example we have:

$$b_k=a_k,~~~~~a_1=a \geq 2,~~~a \in \mathbb{N}$$

Then we have a simple series:

$$x=\sum_{n=0}^{\infty} \frac{1}{a^{2^n}}$$

Which by the above theorem is irrational.


Update:

It turns out that many quadratic irrationals have a pattern in their expansion defined by a linear recurrence. See this question.

The question still stands.


There is another useful paper on the topic here. It also considers the alternating Engel expansion, it appears that AES has the same property of being finite if and only if the number is rational.

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  • $\begingroup$ Just for .. curiosity, have you also considered Engel Expansion ? $\endgroup$ – G Cab Aug 29 '16 at 0:33
  • $\begingroup$ @GCab, no I haven't. My question is about Greedy algorithm expansions. Engel Expansion can probably work as greedy expansion for some numbers, I'm not sure $\endgroup$ – Yuriy S Aug 29 '16 at 0:39

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