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I've been interested in the numbers of this form because it can be proved that for integer $a \geq 2$ all of them are irrational: $$x_a=\sum_{n=0}^\infty \frac{1}{a^{2^n}}$$

They satisfy the conditions listed in this paper: The Approximation of Numbers as Sums of Reciprocals. This is related to my other question.

Now I decided to look at simple continued fractions of such numbers and noticed a surprising thing. For each $a$ I checked the CF entries consist of only three numbers (except the first entry, which is why I show $1/x$ instead of $x$):

$$x_2=\sum_{n=0}^\infty \frac{1}{2^{2^n}}=0.8164215090218931437080797375305252217$$

$$\frac{1}{x_2}=1+\cfrac{1}{4+\cfrac{1}{2+\cfrac{1}{4+\cfrac{1}{4+\cfrac{1}{6+\dots}}}}}$$

Writing the CF in the more convenient form we obtain for $200$ digits:

$1/x_2=$[1; 4, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 6, 4, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 6, 4, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 6, 4, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 4,...]

Clearly, all of the entries are $2,4$ or $6$.

The same goes for other examples:

$1/x_3=$[2; 5, 3, 3, 1, 3, 5, 3, 1, 5, 3, 1, 3, 3, 5, 3, 1, 5, 3, 3, 1, 3, 5, 1, 3, 5, 3, 1, 3, 3, 5, 3, 1, 5, 3, 3, 1, 3, 5, 3, 1, 5, 3, 1, 3, 3, 5, 1, 3, 5, 3, 3, 1, 3, 5, 1, 3, 5, 3, 1, 3, 3, 5, 3, 1, 5, 3, 3, 1, 3, 5, 3, 1, 5, 3, 1, 3, 3, 5, 3, 1, 5, 3, 3, 1, 3, 5, 1, 3, 5, 3, 1, 3, 3, 5, 1, 3, 5, 3, 3, 1, 3, 5, 3, 1, 5, 3, 1, 3, 3, 5, 1, 3, 5, 3, 3, 1, 3, 5, 1, 3, 5, 3, 1, 3, 3, 5, 3, 1, 5, 3, 3, 1, 3, 5, 3, 1, 5, 3, 1, 3, 3, 5, 3, 1, 5, 3, 3, 1, 3, 5, 1, 3, 5, 3, 1, 3, 3, 5, ,...]

$1/x_5=$[4; 7, 5, 5, 3, 5, 7, 5, 3, 7, 5, 3, 5, 5, 7, 5, 3, 7, 5, 5, 3, 5, 7, 3, 5, 7, 5, 3, 5, 5, 7, 5, 3, 7, 5, 5, 3, 5, 7, 5, 3, 7, 5, 3, 5, 5, 7, 3, 5, 7, 5, 5, 3, 5, 7, 3, 5, 7, 5, 3, 5, 5, 7, 5, 3, 7, 5, 5, 3, 5, 7, 5, 3, 7, 5, 3, 5, 5, 7, 5, 3, 7, 5, 5, 3, 5, 7, 3, 5, 7, 5, 3, 5, 5, 7, 3, 5, 7, 5, 5, 3, 5, 7, 5, 3, 7, 5, 3, 5, 5, 7, 3, 5, 7, 5, 5, 3, 5, 7, 3, 5, 7, 5, 3, 5, 5, 7, 5, 3, 7, 5, 5, 3, 5, 7, 5, 3, 7, 5,...]

How can this phenomenon be explained?

The further questions are:

Can we prove that all the CF entries for these numbers belong to a fixed set of three integers?

If so, can we make any conclusions about transcendentality of these numbers?

The implications are interesting. It is conjectured that algebraic numbers of degree $>2$ should have arbitrarily large CF entries at some point. See this paper. Meanwhile we know, that for degree $2$ the CF is (eventually) periodic.


Notice also the same 'pattern' which goes for the examples with odd $a$ here. We have a list of CF entries going the same way.

If we subtract the list of entries for $a=3$ from the list of entries for $a=5$, we obtain:

$$L_5-L_3=[0;2,2,2,2,2,2,2,2,2,...]$$

$$L_7-L_3=[0;4,4,4,4,4,4,4,4,4,...]$$

$$L_{113}-L_3=[0;110,110,110,110,110,110,...]$$

For $6$ and $2$ it goes the same way, but not for $4$ and $2$, there is some 'scrambling' there.

How can we prove/explain this facts? The same pattern for different $a$ seems very strange to me, especially if the numbers are transcendental.

Basically, if this turns out to be true, then from the CF for $x_3$ we will immediately obtain all the CFs for every $x_{2n+1}$


Important update. See http://oeis.org/A004200 for the case $a=3$, it seems like these continued fractions have pattern. And the following paper is linked: Simple continued fractions for some irrational numbers

The pattern is the same for every $a$ except $2$, so not only for the odd $a$.

Morevoer, look at the continued fractions for:

$$y_{ap}=a^p x_a$$

For integer $p$ you will notice a very apparent pattern.

My questions are largely answered by the linked paper. I will try to write up a short summary and post it as an answer, but anyone is three to do it before me.


Turns out a very close question was asked before Continued fraction for $c= \sum_{k=0}^\infty \frac 1{2^{2^k}} $ - is there a systematic expression?

Yuriy S
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    There are uncountably many simple continued fractions with coefficients drawn from any given finite set of $2$ or more positive integers. So I don't think the conjecture you have in mind in your last two paragraphs can be correct without some further qualification. – Rob Arthan Aug 28 '16 at 23:12
  • @RobArthan, how do you prove there are uncountably many such CFs? – Yuriy S Aug 28 '16 at 23:19
  • @YuriyS: There is a bijection between them and the interval $[0, 1]$ expressed in base $b$, where $b$ is the number of integers in the set, right? – Brian Tung Aug 28 '16 at 23:24
  • Two simple CFs are equal iff their sequences of coefficients are equal. If the set of allowed coefficients includes two numbers $c_0$ and $c_1$, then you get a distinct CF for every real number $\alpha \in (0, 1)$, by mapping the binary representation of $\alpha$, a sequence of $0$s and $1$s, to the corresponding sequence of $c_0$s and $c_1$s. – Rob Arthan Aug 28 '16 at 23:25
  • @YuriyS: I am intrigued by the conjecture. Can you give a pointer to the paper you found about this conjecture, please. – Rob Arthan Aug 28 '16 at 23:30
  • What **is** conjectured is that the continued fraction of any real algebraic number of degree $> 2$ has arbitrarily large entries. – Robert Israel Aug 28 '16 at 23:31
  • @YuriyS: thanks for the link. I think you misunderstood the conjecture (which is far from trivial). the conjecture was that the simple continued fractions for **algebraic numbers** is either periodic or contains arbitrarily large coefficients. Read the abstract of the paper in your link (and then the rest of the paper) for clarification. – Rob Arthan Aug 28 '16 at 23:35
  • @RobArthan, yes, thank you. I get it now – Yuriy S Aug 28 '16 at 23:36
  • @RobertIsrael, I have corrected my question, thank you – Yuriy S Aug 28 '16 at 23:38

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