In 1973, Freiman found an interesting number in the complement of the Lagrange spectrum in the Markov spectrum .

By carefully analyzing Freiman’s argument, Cusick and Flahive constructed in 1989 a sequence converging to as such that for all , and, as it turns out, was the largest *known* element of .

In our recent preprint, Gugu and I described the structure of the complement of the Lagrange spectrum in the Markov spectrum near , and this led us to wonder if our description could be used to find new numbers in which are larger than .

As it turns out, Gugu and I succeeded in finding such numbers and we are currently working on the combinatorial arguments needed to extract the largest number given by our methods. (Of course, we plan to include a section on this matter in a forthcoming revised version of our preprint.)

In order to give a flavour on our construction of new numbers in , we will prove in this post that a certain number

belongs to .

**1. Preliminaries**

Let

be the usual continued fraction expansion.

We abbreviate periodic continued fractions by putting a bar over the period: for instance, . Moreover, we use subscripts to indicate the multiplicity of a digit in a sequence: for example, .

Given a bi-infinite sequence and , let

In this context, recall that the classical Lagrange and Markov spectra and are the sets

where

As we already mentioned, Freiman proved that

and Cusick and Flahive extended Freiman’s argument to show that the sequence

accumulating on has the property that for all . In particular,

was the largest known number in .

**2. A new number in **

In what follows, we will show that

Remark 1is a “good” variant of in the sense that it falls in a certain interval which can be proved to avoid the Lagrange spectrum: see Proposition 5 below.

Remark 2Note that , i.e., if we center our discussion at , then is almost 25 times bigger than .

Similarly to the arguments of Freiman and Cusick-Flahive, the proof of Theorem 1 starts by locating an appropriate interval centered at such that does not intersect the Lagrange spectrum.

In this direction, one needs the following three lemmas:

Lemma 2If contains any of the subsequences

- (a)
- (b)
- (c)
- (d)
- (e)
- (f)

then where indicates the position in asterisk.

*Proof:* See Lemma 2 in Chapter 3 of Cusick-Flahive’s book.

Lemma 3If contains any of the subsequences:

- (i)
- (ii)
- (iii)
- (iv)
- (v)
- (vi)
- (vii)
- (viii)
- (ix)
- (x)
- (xi)

then where indicates the position in asterisk.

*Proof:* See Lemma 1 in Chapter 3 of Cusick-Flahive’s book and also Lemma 3.2 of our preprint with Gugu.

Lemma 4If contains the subsequence:

- (xii)

then where indicates the position in asterisk.

*Proof:* In this situation,

thanks to the standard fact that if

and

with , then if and only if .

As it is explained in Chapter 3 of Cusick-Flahive’s book and also in the proof of Proposition 3.7 of our preprint with Gugu, Lemmas 2, 3 and 4 allow to show that:

does not intersect the Lagrange spectrum .

Of course, this proposition gives a natural strategy to exhibit new numbers in : it suffices to build elements of as close as possible to the right endpoint of .

Remark 3As the reader can guess from the statement of the previous proposition, the right endpoint of is intimately related to Lemma 4. In other terms, the natural limit of this method for producing the largest known numbers in is given by how far we can push to the right the boundary of .Here, Gugu and I are currently trying to optimize the choice of by exploiting the simple observation that the proof of Lemma 4 is certainly not sharp in our situation: indeed, we used the sequence

to bound , but we could do better by noticing that this sequence provides a pessimistic bound because contains a copy of the word (and, thus, , i.e., can’t coincide with on a large chunk when ).

Anyhow, Proposition 5 ensures that

does not belong to the Lagrange spectrum .

At this point, it remains only to check that belongs to the Markov spectrum.

For this sake, let us verify that

By items (a), (b), (c) and (e) of Lemma 2,

except possibly for with . Since

we have

for all . This proves that belongs to the Markov spectrum.

## Recent Comments