20

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?

Community
  • 1
Yuriy S
  • 31,474
  • 1
    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

0 Answers0