#  Irrationality exponents of semi-regular continued fractions

Daniel Duverney  110, rue du chevalier français, 59000 Lille, France
<EMAIL_ADDRESS>and  Iekata Shiokawa  13-43, Fujizuka-cho, Hodogaya-
ku, Yokohama 240-0031, Japan<EMAIL_ADDRESS>
(Date: January 30, 2022)

###### Abstract.

We prove that the formula giving the exact value of the irrationality exponent
of regular continued fractions remains valid for semi-regular continued
fractions satisfying certain conditions.

###### Key words and phrases:

Irrationality exponent, semi-regular continued fraction, negative continued
fraction

###### 1991 Mathematics Subject Classification:

11A55, 11J70, 11J82

##  1\. Introduction

An infinite continued fraction

(1.1)
$\alpha:=b_{0}+\frac{a_{1}}{b_{1}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{2}}{b_{2}}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{a_{n}}{b_{n}}\genfrac{}{}{0.0pt}{}{{}}{+\cdots}$

is called semi-regular if $b_{0}\in\mathbb{Z}$ and

(1.2) $a_{n}\in\left\\{-1,1\right\\},\qquad b_{n}\in\mathbb{Z}_{>0},\qquad
b_{n}+a_{n+1}\geq 1\qquad\left(n\geq 1\right),$

with the additional condition

(1.3) $b_{n}+a_{n+1}\geq 2\qquad\text{infinitely often.}$

It is known that semi-regular continued fractions (SRCF) are well defined and
convergent by Tietze Theorem [24]. Examples of SRCF (among others) are:

(i) The regular continued fractions (RCF), where $a_{n}=1$ for every $n\geq
1.$

(ii) The negative continued fractions (NCF), also known as backward continued
fractions, where $a_{n}=-1$ for every $n\geq 1$ [4].

(iii) The nearest integer continued fractions (NICF), where $b_{n}+a_{n+1}\geq
2$ and $b_{n}\geq 2$ for every $n\geq 1$ [17].

(iv) The singular continued fractions (SCF), where $b_{n}+a_{n}\geq 2$ and
$b_{n}\geq 2$ for every $n\geq 1$ [17].

(v) The Lehner continued fractions (LCF), where $(b_{n},a_{n+1})=(1,1)$ or
$(2,-1)$ for every $n\geq 0$ ([5], [15]).

It should be noted that the condition (1.3) is always realized if there exist
infinitely many $n$ such that $a_{n}=1.$ So (1.3) is always realized, except
if there exists $n$ such that

(1.4)
$x_{n}:=\frac{a_{n+1}}{b_{n+1}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{n+2}}{b_{n+2}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{n+3}}{b_{n+3}}\genfrac{}{}{0.0pt}{}{{}}{+\cdots}\qquad\left(n\geq
0\right)$

is a NCF and $b_{k}=2$ for all large $k$ (see Remark 1 below). Hence in the
definition of SRCF the condition (1.3) can be replaced by

If $a_{n}=-1$ for all large $n,$ then $b_{n}\geq 3$ infinitely often.

This shows that NCF are of special importance among SRCF.

The story of SRCF seems to begin with Lagrange, who mentioned in [14, Sections
6 and 7] the possibility of introducing signs in the partial numerators of
regular continued fractions and gave the way for transforming such continued
fractions into regular ones. Tietze [24] proved the convergence of SRCF and
the irrationality of their values. Chapter 5 of Perron’s classical book [17]
is entirely devoted to SRCF. Holzbaur and Riederle [10] characterized SRCF
which represent quadratic irrationals. Pinner [18] computed certain
inhomogeneous approximation constants by using NCF expansions. Sarma and
Kushwaha [21] examined precisely the algorithm to convert a SRCF into regular,
even or odd continued fractions.

SRCF have also been widely discussed from the metric theoretical point of view
(see [5] and [13]). In particular, the map attached to NCF defined by Rényi
[19] has been studied by many authors (see [2] and the recent papers [11],
[23]).

Let $\alpha$ be the SRCF defined by (1.1). As usual, we define for $n\geq 1$

(1.5)
$\left\\{\begin{array}[c]{lll}p_{-1}=1,&p_{0}=b_{0},&p_{n}=b_{n}p_{n-1}+a_{n}p_{n-2},\\\
q_{-1}=0,&q_{0}=1,&q_{n}=b_{n}q_{n-1}+a_{n}q_{n-2}.\end{array}\right.$

It is well known ([7], [12], [17]) that

(1.6)
$b_{0}+\frac{a_{1}}{b_{1}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{2}}{b_{2}}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{a_{n}}{b_{n}}=\frac{p_{n}}{q_{n}}\qquad\left(n\geq
1\right),$

and an easy induction using (1.5) shows that

(1.7) $p_{n}q_{n-1}-p_{n-1}q_{n}=\left(-1\right)^{n-1}a_{1}a_{2}\cdots
a_{n}\qquad\left(n\geq 1\right),$

which yields immediately

(1.8)
$\frac{p_{n}}{q_{n}}=b_{0}+\sum_{k=1}^{n}\frac{\left(-1\right)^{k-1}a_{1}a_{2}\cdots
a_{k}}{q_{k-1}q_{k}}.$

By (1.7), we observe that $p_{n}$ and $q_{n}$ are coprime for all $n\geq
1.\vskip 6.0pt plus 2.0pt minus 2.0pt$

For a real number $\alpha,$ the irrationality exponent
$\mu\left(\alpha\right)$ is defined by the infimum of the set of numbers $\mu$
for which the inequality

(1.9) $\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{\mu}}$

has only finitely many rational solutions $p/q$, or equivalently the supremum
of the set of numbers $\mu$ for which the inequality (1.9) has infinitely many
solutions. If $\alpha$ is irrational, then $\mu\left(\alpha\right)\geq 2$. If
$\alpha$ is a real algebraic irrational number, then
$\mu\left(\alpha\right)=2$ by Roth’s theorem [20]. If
$\mu\left(\alpha\right)=\infty,$ then $\alpha$ is called a Liouville number.

When $\alpha$ defined by (1.1) is a regular continued fraction, that is when
$a_{n}=1$ for all $n\geq 1,$ it is known that

(1.10) $\mu\left(\alpha\right)=1+\limsup_{n\rightarrow\infty}\frac{\log
q_{n+1}}{\log q_{n}}=2+\limsup_{n\rightarrow\infty}\frac{\log b_{n+1}}{\log
q_{n}}$

(see for example [22, Theorem 1]). The main purpose of this paper is to
examine to what extent this formula applies to SRCF. Our main result is

######  Theorem 1.

Let $\alpha$ be the SRCF defined by (1.1) with (1.2) and (1.3). Assume that
one of the following conditions holds:

(A) The number of consecutive $n$ such that $a_{n}=-1$ is bounded.

(B) $b_{n}\geq 2$ for all large $n$ and the number of consecutive $n$ such
that $b_{n}=2$ is bounded.

(C) $b_{n}+a_{n}\geq 2$ for all large $n.$

(D) $b_{n}\geq 2$ and $b_{n}+a_{n+1}\geq 2$ for all large $n.$

Then (1.10) holds.

By $(C)$ and $(D)$, we see that (1.10) holds for all SCF and NICF
respectively, but it is not valid in general for LCF. Indeed, (1.10) implies
that $\mu\left(\alpha\right)=2$ as soon as $b_{n}$ is bounded since
$q_{n}\rightarrow\infty$ by Tietze Theorem. As any irrational number
$\alpha\in\left]1,2\right[$ can be expanded uniquely in a LCF [15], (1.10) is
false for LCF if $\mu\left(\alpha\right)>2.$ On the other hand, $(A)$ implies
that $\mu\left(\alpha\right)=2$ for all LCF for which the number of
consecutive $n$ such that $\left(b_{n},a_{n+1}\right)=\left(2,-1\right)$ is
bounded.

Formula (1.10) is not valid either for all NCF (see Section 3 below). However,
$(B)$ implies that (1.10) holds for all NCF for which the number of
consecutive $n$ such that $b_{n}=2$ is bounded.

By adding a condition, we can deduce from Theorem 1 an expression of the
irrationality exponent of $\alpha$ in terms of the $b_{n}$’s.

######  Corollary 1.

Let $\alpha$ be as in Theorem 1. Assume that

$\lim_{n\rightarrow\infty}\frac{n}{\log\left(b_{1}b_{2}\cdots
b_{n}\right)}=0.$

Then

(1.11) $\mu\left(\alpha\right)=2+\limsup_{n\rightarrow\infty}\frac{\log
b_{n+1}}{\log(b_{1}b_{2}\cdots b_{n})}.$

Note that (1.11) also results from [8, Corollary 4] when the series $\sum$
$\left(b_{n}b_{n+1}\right)^{-1}$ is convergent and, in the case of regular
continued fractions, from [22, Corollary 2].

In order to prove Theorem 1, we will need to present in some details the bases
of the theory of SRCF. This theory is exposed in Perron’s book [17] and, to
our knowledge, it has not been updated since then. Besides, some proofs can be
simplified, as we are interested here only in infinite SRCF. Thus, we will
study first some basic properties of infinite SRCF in Section 2. In Sections 3
and 4 respectively we will prove formulas for the irrationality exponents of
NCF and LCF by transforming them into RCF. In particular, we will give in
Theorem 2 examples of NCF for which (1.10) is not valid. In Section 5, we will
state and prove Theorem 4, which gives a rather general formula allowing to
compute irrationality exponents. In Section 6 we will prove Theorem 1 and
Corollary 1. Finally, we will give some examples of application of Corollary 1
in Section 7.

##  2\. Convergence and irrationality of infinite SRCF

In this section we recall some basic results on infinite SRCF.

######  Proposition 1.

Let $q_{n}$ be defined by (1.5). We have for $n\geq 0$

$\left\\{\begin{array}[c]{l}q_{n}\geq 1\\\ q_{n}+a_{n+1}q_{n-1}\geq 1\\\
q_{n+1}>q_{n}\quad\text{if}\quad b_{n+1}\geq 2\text{ or }a_{n+1}=1\\\
q_{n+2}>q_{n}>q_{n+1}\quad\text{if}\quad b_{n+1}=1\text{ and
}a_{n+1}=-1.\end{array}\right.$

###### Proof.

We prove first by induction that

(2.1) $q_{n}\geq 1\quad\text{and}\quad q_{n}+a_{n+1}q_{n-1}\geq 1$

by following Offer [16]. When $n=0,$ $q_{0}+a_{1}q_{-1}=q_{0}=1$ by (1.5).
Assume that (2.1) holds for some $n\geq 0.$ Then by (1.1), (1.2) and (1.5)

$q_{n+1}=b_{n+1}q_{n}+a_{n+1}q_{n-1}\geq q_{n}+a_{n+1}q_{n-1}\geq 1$

and $q_{n+1}+a_{n+2}q_{n}=\left(b_{n+1}+a_{n+2}\right)q_{n}+a_{n+1}q_{n-1}\geq
q_{n}+a_{n+1}q_{n-1}\geq 1,$ which proves (2.1). If $b_{n+1}\geq 2,$ then by
(2.1)

$q_{n+1}\geq 2q_{n}+a_{n+1}q_{n-1}\geq q_{n}+1,$

and the same inequation holds evidently if $a_{n+1}=1.$ Finally, if
$b_{n+1}=1$ and $a_{n+1}=-1,$ then $a_{n+2}=1$ by (1.2), and so
$q_{n+2}=b_{n+2}q_{n+1}+q_{n}>q_{n}$ by (2.1), which completes the proof of
Proposition 1. ∎

Proposition 1 shows that the the convergents of the infinite SRCF defined by
(1.1) and (1.2) are well defined since $q_{n}\geq 1$ for all $n\geq 0.$ It
also shows that the sequence $q_{n}$ is not always increasing, for example in
the case of the Lehner continued fraction

$1+\frac{\sqrt{2}}{2}=1+\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{-}\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{-}\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{-}\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+\cdots}.$

However, the sequence $q_{n}$ is increasing in the cases of regular, negative,
nearest integer and singular SRCF.

######  Proposition 2.

(Tietze Theorem) The infinite SRCF defined by (1.1) and (1.2) is convergent
and $\lim_{n\rightarrow\infty}q_{n}=+\infty.$

The proof of Tietze Theorem in the general case uses continuants ([17], [16]).
It is rather long and we don’t reproduce it here. However, when the sequence
$q_{n}$ is increasing, the convergence of $p_{n}/q_{n}$ follows immediately
from (1.8).

######  Proposition 3.

For $n\geq 0,$ let $x_{n}$ be defined by (1.4). Then for $n\geq 0$

$0<x_{n}\leq 1\text{ if }a_{n+1}=1,\qquad-1\leq x_{n}<0\text{ if }a_{n+1}=-1.$

###### Proof.

Define for $n\geq 1$ and $k\geq 0$

(2.2)
$x_{n,k}:=\frac{a_{n+1}}{b_{n+1}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{n+2}}{b_{n+2}}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{a_{n+k}}{b_{n+k}}.$

As $x_{n}=\lim_{k\rightarrow\infty}x_{n,k}$ by Proposition 2, we have to prove
that for $n\geq 0$ and $k\geq 1,$

(2.3) $0<x_{n,k}\leq 1\text{ if }a_{n+1}=1,\qquad-1\leq x_{n,k}<0\text{ if
}a_{n+1}=-1.$

The proof is by induction on $k.$ If $k=1,$ then $x_{n,1}=a_{n+1}/b_{n+1}$ and
so (1.2) implies (2.3) for $k=1$ and any $n\geq 0.$ Assume that (2.3) holds
for some $k\geq 1$ and any $n\geq 0.$ Then by (2.2) we have

(2.4) $x_{n,k+1}=\frac{a_{n+1}}{b_{n+1}+x_{n+1,k}},$

where for any $n\geq 0$

(2.5) $0<x_{n+1,k}\leq 1\text{ if }a_{n+2}=1,\qquad-1\leq x_{n+1,k}<0\text{ if
}a_{n+2}=-1.$

If $b_{n+1}\geq 2,$ then by (2.5)

(2.6) $b_{n+1}+x_{n+1,k}\geq 1.$

Otherwise, we have $b_{n+1}=1$ and therefore $a_{n+2}=1$ by (1.2), and so we
find again (2.6) by the first inequality in (2.5). Hence (2.4) and (2.6) imply

$0<x_{n,k+1}\leq 1\text{ if }a_{n+1}=1,\qquad-1\leq x_{n,k+1}<0\text{ if
}a_{n+1}=-1,$

which proves (2.3) by induction, and Proposition 3 follows. ∎

######  Proposition 4.

The value of the infinite SRCF $\alpha$ defined by (1.1) with (1.2) and (1.3)
is irrational.

###### Proof.

Let $x_{n}$ be defined in (1.4). It is well known, and easy to check by
induction, that for all $n\geq-1$

$\alpha=\frac{p_{n+1}+x_{n+1}p_{n}}{q_{n+1}+x_{n+1}q_{n}}.$

This yields immediately

(2.7)
$\left|\alpha-\frac{p_{n}}{q_{n}}\right|=\frac{1}{q_{n}\left|\xi_{n+1}\right|},$

where $\xi_{n+1}:=q_{n+1}+x_{n+1}q_{n}$ $\left(n\geq 0\right).$ Assume first
that there exist infinitely many $n$ such that $a_{n}=1.$ Then by Proposition
3 there are infinitely many $n$ such that $x_{n+1}>0$ and therefore by (2.7)

(2.8)
$0<\left|\alpha-\frac{p_{n}}{q_{n}}\right|\leq\frac{1}{q_{n}q_{n+1}}\leq\frac{2}{q_{n}q_{n+1}}$

for infinitely many $n.$ Now if $a_{n}=-1$ for all large $n,$ there is no loss
of generality in assuming that $a_{n}=-1$ for all $n\geq 1.$ In this case
$x_{n}$ is a negative continued fraction and there exist infinitely many
$n\geq N$ such that $b_{n+1}\geq 3,$ in which case

$q_{n+1}=b_{n+1}q_{n}-q_{n-1}\geq 2q_{n}+q_{n}-q_{n-1}>2q_{n},$

since $q_{n}>q_{n-1}$ by Proposition 1. As $-1<x_{n+1}\leq 0$ by Proposition
3, we get

$\xi_{n+1}=q_{n+1}+x_{n+1}q_{n}>q_{n+1}-q_{n}\geq\frac{1}{2}q_{n+1}$

and therefore (2.8) holds again for infinitely many $n.$ Now, assume that
$\alpha$ is a rational number, $\alpha=c/d$ with $d>0.$ Then by (2.8) there
exist infinitely many $n$ such that $1\leq\left|cq_{n}-dp_{n}\right|\leq
2d/q_{n+1}$ since $cq_{n}-dp_{n}$ is a non-zero integer. This yields a
contradiction for $n$ large because
$\lim_{n\rightarrow\infty}q_{n+1}=+\infty.$ ∎

######  Remark 1.

The continued fraction

(2.9)
$\omega:=\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{-}\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{-\cdots-}\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{-\cdots}$

is convergent by Proposition 2, and clearly $\omega=1/(2-\omega),$ whence
$\omega=1.$ This explains the condition (1.3) in the definition of SRCF.

##  3\. Negative continued fractions

In this section, we study in more detail negative continued fractions (NCF),
in order to show that formula (1.10) may not hold in this case. We will use
the following notation:

$b_{0}-\frac{1}{b_{1}}\genfrac{}{}{0.0pt}{}{{}}{-}\frac{1}{b_{2}}\genfrac{}{}{0.0pt}{}{{}}{-\cdots-}\frac{1}{b_{n}}=\left[b_{0};b_{1},b_{2},\ldots,b_{n}\right]^{-},$

which is similar to that for regular continued fractions (RCF):

$b_{0}+\frac{1}{b_{1}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{b_{2}}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{1}{b_{n}}=\left[b_{0};b_{1},b_{2},\ldots,b_{n}\right].$

As already observed, NCF are of special importance among SRCF. They present
two characteristics:

(a) $b_{n}\geq 3$ infinitely often, in order for NCF to be irrational (Remark
1).

(b) For all $n\geq 0,$ the convergent $p_{n}/q_{n}$ is an approximation by
excess of the infinite NCF $\alpha$ defined in (1.1). This results immediately
from (1.8).

A constructive process, which goes back to Lagrange [14], allows to transform
any NCF into a RCF by using the following two elementary formulas, which are
easy to prove. The first one is

(3.1)
$x-\frac{1}{y}=x-1+\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{y-1}.$

The second one is

(3.2) $x+\frac{1}{0}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{y}=x+y.$

Assume first that $b_{0},b_{1},\ldots,b_{n}$ are indeterminates, and let

$T_{n}:=\left[b_{0};b_{1},b_{2},\ldots,b_{n}\right]^{-}.$

By using (3.1), we get easily

(3.3)
$T_{n}=\left[b_{0}-1;\underline{1,b_{1}-2},\underline{1,b_{2}-2},\ldots,\underline{1,b_{n}-2},1\right].$

Now assume that $b_{0}$ is a rational integer and that
$b_{1},b_{2},\ldots,b_{n},\ldots$ are rational integers with $b_{n}\geq 2$ for
all $n\geq 1$ and $b_{n}\geq 3$ for infinitely many $n\geq 1.$ Let $n_{k}$ be
defined by $n_{0}=0$ and for all $k\geq 1$

(3.4) $b_{n_{k}}\geq 3\quad\text{and}\quad b_{n}=2\text{ if }n_{k-1}<n<n_{k}.$

Then (3.3) yields by using (3.2)

(3.5)
$T_{n_{k}}=\left[b_{0}-1;\underline{n_{1}-n_{0},b_{n_{1}}-2},\underline{n_{2}-n_{1},b_{n_{2}}-2},\ldots,\underline{n_{k}-n_{k-1},b_{n_{k}}-2},1\right].$

Consider the infinite RCF

$\displaystyle\beta$
$\displaystyle=\left[b_{0}-1;\underline{n_{1}-n_{0},b_{n_{1}}-2},\underline{n_{2}-n_{1},b_{n_{2}}-2},\ldots,\underline{n_{k}-n_{k-1},b_{n_{k}}-2},\ldots\right]$
$\displaystyle=\left[c_{0};c_{1},c_{2},c_{3},c_{4},\ldots,c_{2k-1},c_{2k},\ldots\right],$

where $c_{0}=b_{0}-1$ and for $k\geq 1$

(3.6) $c_{2k-1}=n_{k}-n_{k-1},\quad c_{2k}=b_{n_{k}}-2$

with $n_{k}$ defined in (3.4). We observe by (3.5) that

$T_{n_{k}}=\left[c_{0};c_{1},c_{2},c_{3},c_{4},\ldots,c_{2k-1},c_{2k},1\right].$

Let $P_{n}/Q_{n}$ be the n-th convergent of $\beta.$ Then

$\left|T_{n_{k}}-\frac{P_{2k}}{Q_{2k}}\right|=\left|\frac{P_{2k}+P_{2k-1}}{Q_{2k}+Q_{2k-1}}-\frac{P_{2k}}{Q_{2k}}\right|=\frac{1}{Q_{2k}\left(Q_{2k}+Q_{2k-1}\right)}.$

Hence $\lim_{k\rightarrow\infty}\left(T_{n_{k}}-P_{2k}/Q_{2k}\right)=0,$ which
proves that

(3.7)
$\left[b_{0};b_{1},b_{2},\ldots,b_{n},\ldots\right]^{-}=\left[c_{0};c_{1},c_{2},\ldots,c_{n},\ldots\right],$

where the $c_{n}$ are defined by $c_{0}=b_{0}-1$ and (3.6). This formula
allows to transform any infinite NCF into an infinite RCF.

######  Remark 2.

Formula (3.7) is equivalent to Formula (6) in [11] (see also [5, Proposition
2]), which allows to transform any RCF into a NCF:

$\left[0;a_{1},a_{2},\ldots\right]=[1;\underset{\left(a_{1}-1\right)\text{
{times}}}{\underline{2,\ldots,2}},a_{2}+2,\underset{\left(a_{3}-1\right)\text{
{times}}}{\underline{2,\ldots,2}},a_{4}+2,\ldots]^{-}.$

######  Remark 3.

More generally, any SRCF can be transformed into a regular one by using (3.1)
and (3.2), as shown in [17, p. 159]. The converse is also true [5].

Now we can show that (1.10) doesn’t hold in general for NCF. For this, we
consider an arbitrary increasing sequence $n_{k}$ with $n_{0}=0$ and define
$b_{n}$ by

(3.8) $b_{n_{k}}=3,\qquad b_{n}=2\text{\quad if\quad}n_{k-1}<n<n_{k}.$

Let $\alpha$ be the infinite NCF defined by (3.8). Assume that (1.10) is true
for every NCF. Then $\mu\left(\alpha\right)=2$ since $b_{n}$ is bounded.
However we can show that, for every $s\geq 2,$ it is possible to construct a
sequence $n_{k}$ such that $\mu\left(\alpha\right)=s.$ Indeed, we know that
the expansion in RCF of $\alpha=\lim P_{n}/Q_{n}$ is given by (3.7), and
therefore by (1.10) we see that

(3.9) $\mu\left(\alpha\right)=2+\limsup_{n\rightarrow\infty}\frac{\log
c_{n+1}}{\log Q_{n}}=2+\limsup_{n\rightarrow\infty}\frac{\log c_{2k+1}}{\log
Q_{2k}},$

since $c_{2k}=1.$ But $Q_{2k}$ depends only on $c_{0},c_{1},\ldots,c_{2k}$ and
therefore only on $n_{1},n_{2},\ldots,n_{k}$ by (3.6). Hence it is possible to
construct by induction the sequence $(n_{k})$ in such a way that

(3.10) $c_{2k+1}=n_{k+1}-n_{k}=\left\lfloor Q_{2k}^{s-2}\right\rfloor.$

As $c_{2k+1}\leq Q_{2k}^{s-2}<c_{2k+1}+1,$ by (3.9) we get
$\mu\left(\alpha\right)=s.$ This contradiction proves that (1.10) is not valid
in general for NCF. As a consequence, we have

######  Theorem 2.

For any $s\in\left\\{1\right\\}\cup\left[2,+\infty\right[,$ there exists a NCF

$\alpha=-\frac{1}{b_{1}}\genfrac{}{}{0.0pt}{}{{}}{-}\frac{1}{b_{2}}\genfrac{}{}{0.0pt}{}{{}}{-\cdots-}\frac{1}{b_{n}}\genfrac{}{}{0.0pt}{}{{}}{+\cdots}$

with $b_{n}\in\left\\{2,3\right\\}$ for all $n\geq 1$ such that
$\mu\left(\alpha\right)=s.$

Indeed, the case $s=1$ is given by (2.9). This result, as well as Theorem 3 in
the next section and Example 4 in Section 7, has to be compared with [9,
Theorem 3.4].

##  4\. Lehner continued fractions

In this section we will prove a formula similar to (3.9) by transforming LCF
into RCF. Let $\alpha$ defined by (1.1) be any LCF. Then

$\displaystyle\left(a_{n},b_{n}\right)$
$\displaystyle=\left(1,1\right)\quad\Rightarrow\quad\left(a_{n+1},b_{n+1}\right)=\left(1,1\right)\text{
or }\left(1,2\right),$ $\displaystyle\left(a_{n},b_{n}\right)$
$\displaystyle=\left(1,2\right)\quad\Rightarrow\quad\left(a_{n+1},b_{n+1}\right)=\left(-1,1\right)\text{
or }\left(-1,2\right),$ $\displaystyle\left(a_{n},b_{n}\right)$
$\displaystyle=\left(-1,1\right)\quad\Rightarrow\quad\left(a_{n+1},b_{n+1}\right)=\left(1,1\right)\text{
or }\left(1,2\right),$ $\displaystyle\left(a_{n},b_{n}\right)$
$\displaystyle=\left(-1,2\right)\quad\Rightarrow\quad\left(a_{n+1},b_{n+1}\right)=\left(-1,1\right)\text{
or }\left(-1,2\right).$

Moreover by (1.3) there exist infinitely many $n$ such that $b_{n}=1.$ So
there exist sequences of integers $l_{n}\geq 0$ and $m_{n}\geq 0$ such that
$\alpha=1+\beta$ or $\alpha=2-\beta,$ where

$\beta:=\underset{l_{0}\text{
{times}}}{\underline{\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{1}{1}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{+}\underset{m_{1}\text{
{times}}}{\underline{\frac{-1}{2}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{-1}{2}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{-1}{1}\genfrac{}{}{0.0pt}{}{{}}{+}\underset{l_{1}\text{
{times}}}{\underline{\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{1}{1}}}\\\
\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{+}\underset{m_{2}\text{
{times}}}{\underline{\frac{-1}{2}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{-1}{2}}\genfrac{}{}{0.0pt}{}{{}}{+}}\frac{-1}{1}\genfrac{}{}{0.0pt}{}{{}}{+}\underset{l_{2}\text{
{times}}}{\underline{\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{1}{1}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{+}\underset{m_{3}\text{
{times}}}{\underline{\frac{-1}{2}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{-1}{2}}\genfrac{}{}{0.0pt}{}{{}}{+\cdots}}$

It is easily seen that

$\frac{1}{y}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{-1}{1}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{x}=\frac{1}{y-1}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{1+x}.$

Using this repeatedly we obtain by induction for all $m\geq 0$

$\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{+}\underset{m\text{
{times}}}{\underline{\frac{-1}{2}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{-1}{2}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{-1}{1}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{x}=\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{m+1+x},$

and therefore for all $l,m\geq 0$

(4.1) $\underset{l\text{
{times}}}{\underline{\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{1}{1}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{+}\underset{m\text{
{times}}}{\underline{\frac{-1}{2}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{-1}{2}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{-1}{1}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{x}=\underset{l+1\text{
{times}}}{\underline{\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{1}{1}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{m+1+x}.$

Let us consider the convergent $T_{n_{k}}$ of $\beta$ defined by

$T_{n_{k}}:=\underset{l_{0}\text{
{times}}}{\underline{\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{1}{1}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{+}\underset{m_{1}\text{
{times}}}{\underline{\frac{-1}{2}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{-1}{2}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{-1}{1}\genfrac{}{}{0.0pt}{}{{}}{+}\underset{l_{1}\text{
{times}}}{\underline{\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{1}{1}}}\genfrac{}{}{0.0pt}{}{{}}{+\cdots}\\\
\genfrac{}{}{0.0pt}{}{{}}{\cdots+}\frac{-1}{1}\underset{l_{k-1}\text{
{times}}}{\genfrac{}{}{0.0pt}{}{{}}{+}\underline{\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{1}{1}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{+}\underset{m_{k}\text{
{times}}}{\underline{\frac{-1}{2}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{-1}{2}}\genfrac{}{}{0.0pt}{}{{}}{+}}\frac{-1}{1}.$

Applying (4.1) with $x=\infty$ to the end of $T_{n_{k}},$ we see that

$T_{n_{k}}=\underset{l_{0}\text{
{times}}}{\underline{\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{1}{1}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{+}\underset{m_{1}\text{
{times}}}{\underline{\frac{-1}{2}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{-1}{2}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{-1}{1}\genfrac{}{}{0.0pt}{}{{}}{+}\underset{l_{1}\text{
{times}}}{\underline{\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{1}{1}}}\genfrac{}{}{0.0pt}{}{{}}{+\cdots}\\\
\genfrac{}{}{0.0pt}{}{{}}{+\cdots}\genfrac{}{}{0.0pt}{}{{}}{+}\underset{l_{k-2}\text{
{times}}}{\underline{\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{1}{1}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{+}\underset{m_{k-1}\text{
{times}}}{\underline{\frac{-1}{2}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{-1}{2}}\genfrac{}{}{0.0pt}{}{{}}{+}}\frac{-1}{1}\genfrac{}{}{0.0pt}{}{{}}{+}\underset{l_{k-1}+1\text{
{times}}}{\underline{\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{1}{1}}}.$

Applying again (4.1), we get

$T_{n_{k}}=\underset{l_{0}\text{
{times}}}{\underline{\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{1}{1}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+}\underset{m_{1}\text{
{times}}}{\underline{\frac{1}{m_{1}+1}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{-1}{2}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{-1}{1}\genfrac{}{}{0.0pt}{}{{}}{+}\underset{l_{1}\text{
{times}}}{\underline{\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{1}{1}}}\genfrac{}{}{0.0pt}{}{{}}{+\cdots}\\\
\genfrac{}{}{0.0pt}{}{{}}{+\cdots}\genfrac{}{}{0.0pt}{}{{}}{+}\underset{l_{k-2}+1\text{
{times}}}{\underline{\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{1}{1}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{m_{k-1}+2}\underset{l_{k-1}\text{
{times}}}{\genfrac{}{}{0.0pt}{}{{}}{+}\underline{\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{1}{1}}},$

and so finally

$T_{n_{k}}=[0,\underset{l_{0}+1\text{
{times}}}{\underline{1,\ldots,1}},m_{1}+2,\underset{l_{1}\text{
{times}}}{\underline{1,\ldots,1}},m_{2}+2,,\ldots,\underset{l_{k-2}\text{
{times}}}{\underline{1,\ldots,1}},m_{k-1}+2,\underset{l_{k-1}\text{
{times}}}{\underline{1,\ldots,1}}],$

which yields immediately the expansion in RCF

$\beta=[0,\underset{l_{0}+1\text{
{times}}}{\underline{1,\ldots,1}},m_{1}+2,\underset{l_{1}\text{
{times}}}{\underline{1,\ldots,1}},m_{2}+2,\ldots,\underset{l_{k-1}\text{
{times}}}{\underline{1,\ldots,1}},m_{k}+2,\ldots].$

Define $t_{k}=l_{0}+l_{1}+\cdots+l_{k-1}+k.$ Applying (1.10) we get

(4.2)
$\mu\left(\alpha\right)=\mu\left(\beta\right)=2+\limsup_{k\rightarrow\infty}\frac{\log\left(m_{k}+2\right)}{\log
Q_{t_{k}}}.$

For example, let $l_{k}=0$ for all $k\geq 0$ and

(4.3) $\alpha:=1+\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{+}\underset{m_{1}\text{
{times}}}{\underline{\frac{-1}{2}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{-1}{2}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{-1}{1}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{+}\underset{m_{2}\text{
{times}}}{\underline{\frac{-1}{2}\genfrac{}{}{0.0pt}{}{{}}{+\cdots+}\frac{-1}{2}}\genfrac{}{}{0.0pt}{}{{}}{+}}\frac{-1}{1}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{+\cdots}$

Then $\alpha=\left[1,1,m_{1}+2,m_{2}+2,m_{3}+2,\ldots,m_{k}+2,\ldots\right].$
By (4.2) we have

(4.4)
$\mu\left(\alpha\right)=2+\limsup_{k\rightarrow\infty}\frac{\log\left(m_{k}+2\right)}{\log
Q_{k}}.$

Let $s>2.$ As in (3.10), for every $k$ we can choose $m_{k}$ such that

$m_{k}+2=\left\lfloor Q_{k}^{s-2}\right\rfloor,$

and so by (4.4) we get $\mu\left(\alpha\right)=s>2.$ We have proved

######  Theorem 3.

Let $s>2$ and let $\alpha$ be defined by (4.3). Then there exists a sequence
$m_{k}$ such that $\mu\left(\alpha\right)=s.$

##  5\. A general formula for the irrationality exponent

In this section, we prove the following theorem, which allows to compute the
irrationality exponent of a real number $\alpha$ by knowing suitable rational
approximations.

######  Theorem 4.

Let $\alpha\in\mathbb{R}.$ Let $p_{n}\in\mathbb{Z}$ and
$q_{n}\in\mathbb{Z}_{>0}$ be coprime for all $n\geq 0,$ with
$\lim_{n\rightarrow\infty}q_{n}=+\infty.$ Assume that there exist positive
constants $\rho,$ $\sigma$ and $\tau$ such that $q_{n+1}\geq\rho q_{n}$ and

(5.1)
$\frac{\sigma}{q_{n}q_{n+1}}\leq\left|\alpha-\frac{p_{n}}{q_{n}}\right|\leq\frac{\tau}{q_{n}q_{n+1}}$

for all large $n$. Then $\alpha$ is irrational and its irrationality exponent
is

(5.2) $\mu\left(\alpha\right)=1+\limsup_{n\rightarrow\infty}\frac{\log
q_{n+1}}{\log q_{n}}.$

###### Proof.

There is no loss of generality in assuming that $q_{n+1}\geq\rho q_{n}$ and
(5.1) hold for every $n\geq 0.$ Let $\lambda_{n}:=\log q_{n+1}/\log q_{n}$ and
$\lambda:=\limsup_{n\rightarrow\infty}\lambda_{n}.$ Let $\varepsilon>0.$ By
(5.1), we have

$\left|\alpha-\frac{p_{n}}{q_{n}}\right|\leq\frac{\tau}{q_{n}^{\lambda_{n}+1}}\leq\frac{1}{q_{n}^{\lambda_{n}+1-\varepsilon/2}}$

for all large $n$ since $q_{n}\rightarrow+\infty.$ If $\lambda=\infty,$ then
$\mu\left(\alpha\right)=\infty$ and (5.2) holds. So we may assume that
$\lambda<\infty.$ As $\lambda_{n}\geq\lambda-\varepsilon/2$ for infinitely
many $n,$ the inequality

(5.3)
$\left|\alpha-\frac{p_{n}}{q_{n}}\right|\leq\frac{1}{q_{n}^{\lambda+1-\varepsilon}}$

holds for infinitely many $n.$ Hence the inequality

(5.4) $\left|\alpha-\frac{p}{q}\right|\leq\frac{1}{q^{\lambda+1-\varepsilon}}$

has infinitely many solutions $p/q$ because $p_{n}$ and $q_{n}$ are coprime
and $q_{n}\rightarrow+\infty.$ This proves that
$\mu\left(\alpha\right)\geq\lambda+1.$ We prove that
$\lambda+1\geq\mu\left(\alpha\right).$ Let again $\varepsilon>0.$ We consider
the inequality

(5.5)
$\left|\alpha-\frac{p}{q}\right|\leq\frac{1}{q^{\lambda+1+\varepsilon}}.$

Our aim is to prove that this inequality has only finitely many solutions
$p/q.$ Without loss of generality, we can assume that $p$ and $q>0$ are
coprime. We distinguish two cases.

First case : there exists $n$ such that $p/q=p_{n}/q_{n}.$ Then $p_{n}=p$ and
$q_{n}=q$ since $p/q$ and $p_{n}/q_{n}$ are both irreducible. By (5.1), we
have

$\left|\alpha-\frac{p}{q}\right|=\left|\alpha-\frac{p_{n}}{q_{n}}\right|\geq\frac{\sigma}{q_{n}^{\lambda_{n}+1}}=\frac{\sigma}{q^{\lambda_{n}+1}}=\frac{\sigma
q^{\varepsilon/2}}{q^{\lambda_{n}+1+\varepsilon/2}}.$

For large $q,$ we have $\sigma q^{\varepsilon/2}\geq 1$ and
$\lambda_{n}+1+\varepsilon/2\leq\lambda+1+\varepsilon$ since
$\lim_{n\rightarrow\infty}q_{n}=+\infty.$ Hence

(5.6)
$\left|\alpha-\frac{p}{q}\right|\geq\frac{1}{q^{\lambda+1+\varepsilon}}.$

Second case : for all $n,$ we have $pq_{n}-qp_{n}\neq 0.$ Fix a positive
integer $Q$ such that $2\tau Q>\rho q_{1}$ and assume that $q\geq Q.$ Let
$n=n(q)\geq 1$ be the least integer such that

(5.7) $\frac{\tau}{\rho q_{k}}q<\frac{1}{2}$

for all $k\geq n+1,$ which exists since
$\lim_{k\rightarrow\infty}q_{k}=+\infty.$ Then

(5.8) $q_{n}\leq\frac{2\tau}{\rho}q.$

As $pq_{n+1}-qp_{n+1}\neq 0,$ we can write by (5.1) and (5.7):

$\displaystyle 1$ $\displaystyle\leq\left|pq_{n+1}-qp_{n+1}\right|\leq
q\left|q_{n+1}\alpha-p_{n+1}\right|+q_{n+1}\left|q\alpha-p\right|$
$\displaystyle\leq\frac{\tau}{q_{n+2}}q+q_{n+1}\left|q\alpha-p\right|\leq\frac{\tau}{\rho
q_{n+1}}q+q_{n+1}\left|q\alpha-p\right|<\frac{1}{2}+qq_{n+1}\left|\alpha-\frac{p}{q}\right|.$

By using (5.8), this implies

$\left|\alpha-\frac{p}{q}\right|>\frac{1}{2qq_{n+1}}=\frac{1}{2qq_{n}^{\lambda_{n}}}\geq\frac{\rho^{\lambda_{n}}}{2^{\lambda_{n}+1}\tau^{\lambda_{n}}q^{\lambda_{n}+1}}=\frac{\rho^{\lambda_{n}}q^{\varepsilon/2}}{2^{\lambda_{n}+1}\tau^{\lambda_{n}}q^{\lambda_{n}+1+\varepsilon/2}}.$

However there exist two constants $C$ and $C^{\prime}$ such that
$C\leq\lambda_{n}\leq C^{\prime},$ because $\lambda$ is finite,
$q_{n+1}\geq\rho q_{n}$ and $q_{n}\rightarrow+\infty.$ So there exists
$C^{\prime\prime}>0$ such that

$\left|\alpha-\frac{p}{q}\right|>\frac{C^{\prime\prime}q^{\varepsilon/2}}{q^{\lambda_{n}+1+\varepsilon/2}}.$

Now $n=n\left(q\right)$ tends to infinity with $q,$ and therefore for $q\geq
Q^{\prime}>Q$ we have $C^{\prime\prime}q^{\varepsilon/2}\geq 1$ and
$\lambda_{n}\leq\lambda+\varepsilon/2.$ This yields for large $q$

(5.9) $\left|\alpha-\frac{p}{q}\right|>\frac{1}{q^{\lambda+1+\varepsilon}}.$

By (5.6) and (5.9) we conclude that the inequality (5.5) has only finitely
many solutions for every $\varepsilon>0.$ The proof of Theorem 4 is complete.
∎

##  6\. Proof of Theorem 1 and Corollary 1

First we prepare two lemmas.

######  Lemma 1.

Let $m$ and $h$ be positive integers. Assume that $a_{m}=1$ and $a_{m+k}=-1$
for $1\leq k\leq h.$ Then

(6.1) $\displaystyle q_{m+k}$
$\displaystyle\geq\frac{k+2}{k+1}q_{m+k-1}\qquad\left(0\leq k\leq h-1\right),$
(6.2) $\displaystyle q_{m+h}$ $\displaystyle\geq\frac{b_{m+h}}{h+1}q_{m+h-1}.$

###### Proof.

First we prove (6.1). It is true for $h=1.$ Indeed, in this case we have
$a_{m}=1,$ $a_{m+1}=-1$ and $b_{m}\geq 2,$ whence
$q_{m}=b_{m}q_{m-1}+q_{m-2}\geq 2q_{m-1}.$ Assume that (6.1) is satisfied at
the order $h\geq 1,$ and prove it at the order $h+1.$ By (6.1),

$q_{m+h-2}\leq\frac{h}{h+1}q_{m+h-1}.$

Since $a_{m+h}=a_{m+h+1}=-1,$ we have $b_{m+h}\geq 2$ and

$q_{m+h}\geq 2q_{m+h-1}-\frac{h}{h+1}q_{m+h-1}=\frac{h+2}{h+1}q_{m+h-1},$

which proves (6.1) by induction. By using (6.1) we get

$q_{m+h}\geq b_{m+h}\left(q_{m+h-1}-q_{m+h-2}\right)\geq
b_{m+h}q_{m+h-1}\left(1-\frac{h}{h+1}\right),$

which proves (6.2). ∎

######  Lemma 2.

Let $m$ and $h$ be positive integers such that $b_{m}\geq 3$ and $b_{m+k}=2$
for $1\leq k\leq h$. Assume that $b_{m-1}\geq 2$ if $m\geq 2.$ Then

(6.3) $q_{m+k}\geq\frac{k+2}{k+1}q_{m+k-1}\qquad\left(0\leq k\leq h\right).$

###### Proof.

By Proposition 1, we see that $q_{m-2}<q_{m-1}$ for $m\geq 2,$ and this
remains true for $m=1.$ Therefore for $m\geq 1$ we have

$q_{m}\geq b_{m}q_{m-1}-q_{m-2}\geq(b_{m}-1)q_{m-1}\geq 2q_{m-1}.$

Hence (6.3) is true for $k=0.$ Assume that (6.3) is true for some $k\leq h-1.$
Then

$q_{m+k+1}\geq
2q_{m+k}-q_{m+k-1}\geq\left(2-\frac{k+1}{k+2}\right)q_{m+k}=\frac{k+3}{k+2}q_{m+k},$

which proves (6.3) by induction. ∎

Proof of Theorem 1. We observe that

$\displaystyle\limsup_{n\rightarrow\infty}\frac{\log q_{n+1}}{\log q_{n}}$
$\displaystyle=1+\limsup_{n\rightarrow\infty}\frac{\log b_{n+1}+\log
q_{n}+\log\left(1+a_{n+1}b_{n+1}^{-1}q_{n-1}q_{n}^{-1}\right)}{\log q_{n}}$
$\displaystyle=2+\limsup_{n\rightarrow\infty}\frac{\log b_{n+1}}{\log q_{n}}$

if $q_{n-1}q_{n}^{-1}$ is bounded. Hence we only have to check that Theorem 4
applies when $(A),$ $(B),$ $(C)$ or $(D)$ are satisfied. Besides, it results
from (2.7) that (5.1) holds if there exist two positive constants $\gamma$ and
$\delta$ such that $\gamma q_{n+1}\leq\xi_{n+1}\leq\delta q_{n+1}$ for all
large $n.\vskip 3.0pt plus 1.0pt minus 1.0pt$

Assume that $(A)$ is satisfied. Let $n$ large. As the number of consecutive
$k$ such that $a_{k}=-1$ is bounded, there exists $p\leq n$ such that
$a_{p}=1.$ We distinguish four cases.

Case 1. $a_{n+1}=a_{n+2}=1.$ Then $0<x_{n+1}\leq 1$ by Proposition 3, so that

(6.4) $q_{n+1}\leq\xi_{n+1}\leq q_{n+1}+q_{n}.$

As $q_{n+1}\geq q_{n}$ by Proposition 1, we obtain $q_{n+1}\leq\xi_{n+1}\leq
2q_{n+1}.$

Case 2. $a_{n+1}=-a_{n+2}=1.$ Then $-1\leq x_{n+1}<0$ by Proposition 3, whence

(6.5) $q_{n+1}-q_{n}\leq\xi_{n+1}\leq q_{n+1}.$

By (6.1) with $m=n+1,$ $h=1$ and $k=0,$ we see that $q_{n+1}\geq 2q_{n},$ and
so

$\frac{1}{2}q_{n+1}\leq\xi_{n+1}\leq q_{n+1}.$

Case 3. $a_{n+1}=-a_{n+2}=-1.$ Then (6.4) holds. Let $m\leq n$ be the greatest
integer such that $a_{m}=1.$ By (6.2) with $h=n+1-m,$ we have

$q_{n+1}\geq\frac{1}{h+1}q_{n}\geq\frac{1}{L}q_{n}$

as $h<L$ for some $L\geq 2$ since the number of consecutive $k$ such that
$a_{m+k}=-1$ is bounded. So we obtain
$q_{n+1}\leq\xi_{n+1}\leq\left(L+1\right)q_{n+1}.$

Case 4. $a_{n+1}=a_{n+2}=-1.$ Then (6.5) holds. Let $m\leq n$ be the greatest
integer such that $a_{m}=1.$ By (6.1) with $h=n+2-m$ and $k=h-1,$ we see that

$q_{n+1}\geq\left(1+\frac{1}{h}\right)q_{n}\geq\frac{L+1}{L}q_{n}.$

By (6.5), we obtain

$\frac{1}{L+1}q_{n+1}\leq\xi_{n+1}\leq q_{n+1}.$

Theorem 4 applies with $\rho=1/L,$ $\sigma=1/(L+1)$ and $\tau=L+1.\vskip 3.0pt
plus 1.0pt minus 1.0pt$

Assume that $(B)$ is satisfied. Let $n$ large. As $b_{n+1}\geq 2,$ we have
$q_{n+1}>q_{n}$ by Proposition 1. Therefore by Proposition 3

(6.6) $\xi_{n+1}\leq q_{n+1}+q_{n}\leq 2q_{n+1}.$

For the lower bound of $\xi_{n+1},$ we distinguish two cases.

Case 1. If $b_{n+1}\geq 3,$ then $q_{n+1}\geq b_{n+1}q_{n}-q_{n-1}\geq
2q_{n}.$

Case 2. If $b_{n+1}=2,$ we define $m\leq n$ to be the greatest integer such
that $b_{m}\geq 3.$ By (6.3) with $k=h=n+1-m,$ we have

(6.7) $q_{n+1}\geq\frac{h+2}{h+1}q_{n}\geq\frac{M+1}{M}q_{n}$

as $h<M$ for some $M\geq 2$ since the number of consecutive $k$ such that
$b_{m+k}=2$ is bounded.

So in both cases (6.7) holds and

(6.8) $\xi_{n+1}\geq q_{n+1}-q_{n}\geq
q_{n+1}-\frac{M}{M+1}q_{n+1}\geq\frac{1}{M+1}q_{n+1}.$

Theorem 4 applies with $\rho=1,$ $\sigma=1/2$ and $\tau=M+1.\vskip 3.0pt plus
1.0pt minus 1.0pt$

Assume that $(C)$ is satisfied. Then $q_{n+1}>q_{n}$ for $n$ large by
Proposition 1. Therefore (6.6) holds. For the lower bound of $\xi_{n+1},$ we
distinguish three cases.

Case 1. If $a_{n+2}=1,$ then $x_{n+1}\geq 0$ by Proposition 3, whence
$\xi_{n+1}\geq q_{n+1}.$

Case 2. If $a_{n+2}=-a_{n+1}=-1,$ then $b_{n+1}\geq 2$ and $q_{n+1}\geq
2q_{n}.$ Therefore

(6.9) $\xi_{n+1}\geq q_{n+1}-q_{n}\geq\frac{1}{2}q_{n+1}.$

Case 3. If $a_{n+2}=$ $a_{n+1}=-1,$ then (6.9) holds again, since

$q_{n+1}=\left(b_{n+1}+a_{n+1}\right)q_{n}-a_{n+1}\left(q_{n}-q_{n-1}\right)\geq
2q_{n}.$

Theorem 4 applies with $\rho=1,$ $\sigma=1/2$ and $\tau=2.\vskip 3.0pt plus
1.0pt minus 1.0pt\vskip 3.0pt plus 1.0pt minus 1.0pt$

Assume that $(D)$ is satisfied. Again, $q_{n+1}>q_{n}$ for $n$ large by
Proposition 1 and (6.6) holds again. For the lower bound of $\xi_{n+1},$ we
distinguish two cases.

Case 1. If $a_{n+2}=1,$ then $\xi_{n+1}\geq q_{n+1}.$

Case 2. If $a_{n+2}=-1$, then $b_{n+1}\geq 3$ and $q_{n+1}\geq
b_{n+1}q_{n}-q_{n-1}\geq 2q_{n}$, so that (6.9) holds.

Theorem 4 applies with $\rho=1,$ $\sigma=1/2$ and $\tau=2.\vskip 3.0pt plus
1.0pt minus 1.0pt\vskip 6.0pt plus 2.0pt minus 2.0pt$

Proof of Corollary 1

Assume that $(A),$ $(B)$, $(C)$ or $(D)$ is satisfied. We will prove that
there exists a positive constant $K$ such that

(6.10) $K^{n}\prod_{k=1}^{n}b_{k}\leq q_{n}\leq
F_{n+1}\prod_{k=1}^{n}b_{k}\quad\left(n\geq 0\right),$

where $F_{n}$ is the n-th Fibonacci number. It is clear that Corollary 1
results immediately from Theorem 1 and (6.10). The second inequality in (6.10)
comes from the fact that $q_{0}=1,$ $q_{1}=b_{1}$ and

$q_{n+1}=b_{n+1}q_{n}+a_{n}q_{n-1}\leq q_{n}+q_{n-1}\quad\left(n\geq
1\right).$

Now we prove the first inequality in (6.10) in each of the cases $(A),$ $(B)$,
$(C)$ or $(D).$ For this, it is sufficient to prove the existence of a
positive number $\kappa$ such that $q_{n+1}\geq\kappa b_{n+1}q_{n}$ for all
large $n.\vskip 3.0pt plus 1.0pt minus 1.0pt$

Assume that $(A)$ is satisfied, which means that the number of consecutive $n$
such that $a_{n}=-1$ is less than $L$ $\left(\geq 2\right).$

If $a_{n+1}=1,$ then $q_{n+1}\geq b_{n+1}q_{n},$ and so

(6.11) $q_{n+1}\geq\frac{b_{n+1}}{L}q_{n}.$

If $a_{n+1}=-1,$ let $m<n$ be the greatest integer such that $a_{m}=1.$ Then
(6.2) with $h=n-m+1$ shows that (6.11) also holds in this case.

Assume that $(B)$ is satisfied, which means that the number of consecutive $n$
such that $b_{n}=2$ is less than $M$ $\left(\geq 2\right).$ Then by (6.7) we
have

$q_{n+1}\geq b_{n+1}\left(q_{n}-q_{n-1}\right)\geq
b_{n+1}\left(q_{n}-\frac{M}{M+1}q_{n}\right)=\frac{b_{n+1}}{M+1}q_{n}\qquad\left(n\geq
1\right).$

Assume that $(C)$ is satisfied. If $a_{n+1}=1,$ then $q_{n+1}\geq
b_{n+1}q_{n}.$

If $a_{n+1}=-1,$ then $b_{n+1}\geq 3$. Since $q_{n}>q_{n-1},$ we get

$q_{n+1}\geq\frac{1}{2}b_{n+1}q_{n}+\frac{3}{2}q_{n}-q_{n-1}\geq\frac{1}{2}b_{n+1}q_{n}.$

Assume that $(D)$ is satisfied. If $a_{n+1}=1,$ then again $q_{n+1}\geq
b_{n+1}q_{n}.$

If $a_{n+1}=-1,$ then $b_{n+1}\geq 2$ and $q_{n}>q_{n-1},$ and so

$q_{n+1}\geq\frac{1}{2}b_{n+1}q_{n}+q_{n}-q_{n-1}\geq\frac{1}{2}b_{n+1}q_{n}.$

##  7\. Applications

We give here examples of semi-regular continued fractions and compute their
irrationality exponent by using Corollary 1.

Example 1. For $c\neq 0,-1,-2,\ldots,$ the modified Bessel function is defined
by

${}_{0}F_{1}\left(c;z\right):=\sum_{n=0}^{\infty}\frac{z^{n}}{\left(c\right)_{n}n!},$

where $(c)_{0}=1$ and $(c)_{n}=c\left(c+1\right)\cdots\left(c+n-1\right)$ for
$n\geq 1.$ It is known that

$\frac{{}_{0}F_{1}\left(1;z\right)}{{}_{0}F_{1}\left(2;z\right)}=1+\frac{z}{2}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{z}{3}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{z}{4}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{z}{5}\genfrac{}{}{0.0pt}{}{{}}{+\cdots}$

(cf. [12, Formula (6.1.51)]). Corollary 1 $(B),$ $(C)$ or $(D)$ applies and we
get

$\mu\left(1+\frac{a_{1}}{2}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{2}}{3}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{3}}{4}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{4}}{5}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{5}}{6}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{6}}{7}\genfrac{}{}{0.0pt}{}{{}}{+\cdots}\right)=2$

for any sequence $(a_{n})_{n\geq 1}$ with $a_{n}=\pm 1.\vskip 6.0pt plus 2.0pt
minus 2.0pt$

Example 2. For any integer $b\geq 1,$ we have

$e^{\frac{1}{b}}=1+\frac{1}{b}\genfrac{}{}{0.0pt}{}{{}}{-}\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{3b}\genfrac{}{}{0.0pt}{}{{}}{-}\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{5b}\genfrac{}{}{0.0pt}{}{{}}{-}\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{7b}\genfrac{}{}{0.0pt}{}{{}}{-}\frac{1}{2}\genfrac{}{}{0.0pt}{}{{}}{+\cdots}$

(cf [3, (A.61)]). Then Corollary 1 $(A)$ yields $\mu\left(e^{1/b}\right)=2.$
More generally, Corollary 1 $(B)$ applies and we get

$\mu\left(1+\frac{a_{1}}{b}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{2}}{2}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{3}}{3b}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{4}}{2}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{5}}{5b}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{6}}{2}\genfrac{}{}{0.0pt}{}{{}}{+\cdots}\right)=2$

for any sequence $(a_{n})_{n\geq 1}$ with $a_{n}=\pm 1.\vskip 6.0pt plus 2.0pt
minus 2.0pt$

Example 3. Let $\alpha>0$ be any irrational number with the regular continued
fraction expansion $\alpha^{-1}=\left[c_{0};c_{1},c_{2},c_{3},\ldots\right]$
and let $b$ be any integer with $\left|b\right|>1$. Let $p_{n}/q_{n}$ be the
n-th convergent of the RCF expansion of $\alpha^{-1}$. Adams and Davison ([1],
see also [3, Th. 7.18]) proved that

(7.1)
$S_{b}(\alpha):=\left(b-1\right)\sum_{k=1}^{\infty}\frac{1}{b^{\left\lfloor
k\alpha\right\rfloor}}=b_{0}+\frac{1}{b_{1}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{b_{2}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{b_{3}}\genfrac{}{}{0.0pt}{}{{}}{+\cdots},$

where $b_{0}=c_{0}b$ and
$b_{n}=\left(b^{q_{n}}-b^{q_{n-2}}\right)/\left(b^{q_{n-1}}-1\right)$ for
$n\geq 1$. Note that $b_{n}$ is an integer (positive or negative), since
$q_{n-1}$ divides $q_{n}-q_{n-2}.$ It is clear that (7.1) can be written as a
SRCF. In particular [6], if $\alpha=(1+\sqrt{5})/2,$ then $q_{n}=F_{n+1}$ for
$n\geq-1,$ where $F_{n}$ is the Fibonacci sequence defined by $F_{0}=0,$
$F_{1}=1$ and $F_{n+2}=F_{n+1}+F_{n}$ for all $n\geq 0.$ Applying (7.1), we
get

$\displaystyle S_{2}(\alpha)$
$\displaystyle=\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{2^{F_{1}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{2^{F_{2}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{2^{F_{3}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{2^{F_{4}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{1}{2^{F_{5}}}\genfrac{}{}{0.0pt}{}{{}}{+\cdots},$
$\displaystyle S_{-2}(\alpha)$
$\displaystyle=\frac{1}{1}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{\left(-1\right)^{F_{2}}}{2^{F_{1}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{\left(-1\right)^{F_{3}}}{2^{F_{2}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{\left(-1\right)^{F_{4}}}{2^{F_{3}}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{\left(-1\right)^{F_{5}}}{2^{F_{4}}}\genfrac{}{}{0.0pt}{}{{}}{+\cdots}.$

Returning to the general case, it is easily seen that

$\limsup_{n\rightarrow\infty}\frac{\log\left|b_{n+1}\right|}{\log\left|b_{1}b_{2}\cdots
b_{n}\right|}=\limsup_{n\rightarrow\infty}\left(\frac{q_{n+1}}{q_{n}}\right)-1.$

Corollary 1 $(B)$, $(C)$ or $(D)$ applies and we get

(7.2)
$\mu\left(b_{0}+\frac{a_{1}}{b_{1}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{2}}{b_{2}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{3}}{b_{3}}\genfrac{}{}{0.0pt}{}{{}}{+\cdots}\right)=1+\limsup_{n\rightarrow\infty}\left(\frac{q_{n+1}}{q_{n}}\right)$

for any sequence $(a_{n})_{n\geq 1}$ with $a_{n}=\pm 1.$ For
$\alpha=(1+\sqrt{5})/2,$ we have

(7.3)
$\mu\left(S_{2}(\alpha)\right)=\mu\left(S_{-2}(\alpha)\right)=\frac{3+\sqrt{5}}{2}=2.61803\ldots.$

Returning to the general case (7.2), it is clear that $S_{b}(\alpha)$ is a
Liouville number if and only if $c_{n}$ is unbounded since
$c_{n+1}<q_{n+1}/q_{n}<c_{n+1}+1$ for all $n\geq 0$ by (1.5). For example,
$S_{b}(\alpha)$ is a Liouville number if $\mu\left(\alpha\right)>2.$ Similarly
$S_{b}(e)$ is a Liouville number.

Moreover, let $L=\limsup_{n\rightarrow\infty}\left(q_{n+1}/q_{n}\right).$ Then
$L^{2}\geq L+1$ by (1.5). Therefore $L\geq\left(1+\sqrt{5}\right)/2,$ the
golden number. Consequently
$\mu\left(S_{b}(\alpha)\right)\geq\left(3+\sqrt{5}\right)/2.$

Finally, assume that $\alpha$ is reduced quadratic. Then its RCF expansion is
purely periodic, so that $c_{0}=0$ and $c_{n+H}=c_{n}$ for some $H\geq 1$ and
all $n\geq 1.$ Hence

$\frac{q_{kH}}{q_{kH-1}}=\left[c_{kH},c_{kH-1},\ldots,c_{1}\right]=\left[\underline{c_{H},c_{H-1},\ldots,c_{1}},\ldots,\underline{c_{H},c_{H-1},\ldots,c_{1}}\right]$

for all $k\geq 1.$ Let $\alpha^{\ast}$ and $N(\alpha)$ be the conjugate and
the norm of $\alpha,$ respectively. By applying Galois Theorem ([17, § 23],
[7, Exercise 4.9]), we get

(7.4)
$\lim_{k\rightarrow\infty}\frac{q_{kH}}{q_{kH-1}}=-\frac{1}{\alpha^{\ast}}=-\frac{\alpha}{N(\alpha)}:=\beta_{1}.$

Let $\beta_{n}$ be defined by $\beta_{n}:=c_{n-1}+1/\beta_{n-1}$ for
$n=2,\ldots,H.$ Then by (7.2) and (7.4) we obtain

$\mu\left(b_{0}+\frac{a_{1}}{b_{1}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{2}}{b_{2}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{3}}{b_{3}}\genfrac{}{}{0.0pt}{}{{}}{+\cdots}\right)=1+\max\left(\beta_{1},\ldots,\beta_{H}\right)$

for any sequence $(a_{n})_{n\geq 1}$ with $a_{n}=\pm 1.$ For example, let
$c,d\in\mathbb{Z}_{>0}$ and

$\alpha=\left[\underline{c,d},\underline{c,d},\underline{c,d},\underline{c,d},\ldots\right].$

Then $d\alpha^{2}-cd\alpha-c=0$ and
$\mu\left(S_{b}(\alpha)\right)=1+\max\left(\alpha,\alpha d/c\right).$ This
generalizes (7.3).

In Example 3, we could also have used [8, Corollary 4], and the same is true
for Example 1 (but not for Example 2). We give a last example, where the
irrationality exponent is greater than 2 and [8, Corollary 4] is not
applicable.

Example 4. Let $\sigma>1$. For $k\geq 1,$ define $b_{3k-2}=b_{3k-1}=2,$
$b_{3k}=\left\lfloor 2^{\sigma^{k}}\right\rfloor.$ Then

$\limsup_{n\rightarrow\infty}\frac{\log b_{n+1}}{\log\left(b_{1}b_{2}\cdots
b_{n}\right)}=\limsup_{k\rightarrow\infty}\frac{\log
b_{3k}}{\log\left(b_{1}b_{2}\cdots b_{3k-1}\right)}=\sigma-1.$

Corollary 1 $(B)$ applies, and we get

$\mu\left(b_{0}+\frac{a_{1}}{b_{1}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{2}}{b_{2}}\genfrac{}{}{0.0pt}{}{{}}{+}\frac{a_{3}}{b_{3}}\genfrac{}{}{0.0pt}{}{{}}{+\cdots}\right)=\sigma+1>2$

for any sequence $(a_{n})_{n\geq 1}$ with $a_{n}=\pm 1.$

## References

  * [1] W. W. Adams and J. L. Davison, A remarkable class of continued fractions, Proc. Amer. Math. Soc. 65 (1977), 194-198.
  * [2] R. L. Adler and L. Flatto, The backward continued fraction map and geodesic flow, Ergodic Theory and Dynamical Systems 4 (1984), 487-492.
  * [3] J. Borwein, A. van der Poorten, J. Shallit and W. Zudilin, Neverending Fractions, An Introduction to Continued Fractions, Cambridge, 2014.
  * [4] E. Cahen, Théorie des nombres, Tome Second, Hermann, 1924.
  * [5] K. Dajani and C. Kraaikamp, The mother of all continued fractions, Colloquium Math. 84/85 (2000), 109-123.
  * [6] J. L. Davison, A series and its associated continued fraction, Proc. Amer. Math. Soc. 63 (1977), 29-32.
  * [7] D. Duverney, Number theory, World Scientific, 2010.
  * [8] D. Duverney and I. Shiokawa, Irrationality exponents of numbers related with Cahen’s constant, Monatsh. Math. 191-1 (2020), 53-76.
  * [9] J. Hančl, K. Leppälä, T. Matala-aho and T. Törmä, On irrationality exponents of generalized continued fractions, J. Number Th. 151 (2015), 18-35.
  * [10] U. Holzbaur and M. Riederle, A characterization of the semi-regular continued fractions which represent quadratic irrationals, Arch. Math. 4 (1983), 319-323.
  * [11] H. Ito, Diophantine approximation by negative continued fractions, to appear in Tokyo J. Math. 45(1) (2022).
  * [12] W. B. Jones and W. J. Thron, Continued fractions : analytic theory and applications, Encyclopedia of Math. and its applications 11, Addison-Wesley, 1980.
  * [13] C. Kraaikamp, A new class of continued fraction expansions, Acta Arith. LVII (1991), 1-39.
  * [14] J.-L. Lagrange, Addition aux éléments d’algèbre d’Euler, Paris 1773. English translation: London 1822.
  * [15] J. Lehner, Semi-regular continued fractions whose partial denominators are 1 or 2, Contemp. Math. 169 (1994), AMS, 407-410
  * [16] A. Offer, Continuants and semi-regular continued fractions (2008), unpublished.
  * [17] O. Perron, Die Lehre von den Kettenbrüchen, Teubner, 1910.
  * [18] C. G. Pinner, More on inhomogeneous diophantine approximation, J. Theor. Nombres Bordeaux, 13 (2001), 539-557.
  * [19] A. Rényi, Algorithms for the generation of real numbers (Hungarian), Magyar Tud. Akad., Mat. Fiz. Oszt. Közl. 7 (1957), 265-293 .
  * [20] K. F. Roth, Rational approximations to algebraic numbers, Mathematika 2 (1955), 1-20.
  * [21] H. Sarma and S. Kushwaha, On finite semi-regular continued fractions, Integers 16 (2016), 1-11.
  * [22] J. Sondow, Irrationality measures, irrationality bases and a theorem of Jarník, arXiv 2004.06300 (2004).
  * [23] H. Takahasi, Large deviation principle for the backward continued fraction expansion, to appear in Stochastic processes and their applications.
  * [24] H. Tietze, Über Kriterien für Konvergenz und Irrationalität unendlichen Kettenbrüche, Math. Ann. 70 (1911), 236-265.


# Searching for Possible Spin Configurations of Ferrum Chain via Quantum
Approximate Optimization Algorithm

Saba Arife Bozpolat  Physics Department, Marmara University, 34722 Ziverbey,
Istanbul, Turkey<EMAIL_ADDRESS>
OrcID: 0000-0002-2166-6464

###### Abstract

Calculating the expected spin configuration of the chain consisting of Ferrum
atoms interacting with each other through exchange interaction is
fundamentally a configuration optimization problem. Quantum Approximate
Optimization Algorithm is a suitable candidate to configure such systems on a
quantum device. In this work we have considered Ferrum chains of three
different lengths and calculated their most-probable spin configurations using
Quantum Approximate Optimization Algorithm. We employed a Quantum Feed Forward
Neural Network as the optimizer of Quantum Approximate Optimization Algorithm.
We have successfully obtained the expected spin configuration for the longest
Ferrum Chain.

##  1 Introduction

In Physics, the Ising model is a well established theoretical framework to
describe spin orientations of magnetic members of a magnetic material, among
some other things, considering their interactions with one another and some
other internal or external physical entities [1]. This model is also used
interdisciplinarily to build Quadratic Unconstrained Binary Optimization
(QUBO) models [2, 3, 4] to solve several Combinatorial Optimization Problems
(COP) [5], such as traveling salesperson problem (TSP) [6], workflow
scheduling (WS) [7], the intersection traffic signal control problem (ITSCP)
[8], etc. QUBO helps with the construction of an enunciative square matrix for
considered problem. And this matrix is then subjected to an optimization model
to fine-tune the problem-specific parameters. Quantum Approximate Optimization
Algorithm is a quantum algorithm to do such calibrations in polynomial time
whereas its classical counterparts cannot provide any guarantee on the time
scale [9, 10, 11, 12].

Much as coupling QAOA with classical optimizers to apply this method on some
COPs is a relatively novel and beneficial idea, it is possible to take one
step further and integrate it with Deep Neural Networks (DNN) to capitalize
swiftness of pioneering Machine Learning (ML) algorithms.

In this work, we obtained the optimized configuration of a ferromagnetic (FM)
material structured in 1 dimension, $i.e.$ a Ferrum chain, without an external
magnetic field, using QAOA with Quantum Feed Forward Neural Network (QFFNN).
We have considered a Ferrum chain of three different lengths and expressed
this sistem using Heisenberg Hamiltonian with nearest-neighbor Ising Model. We
have converted this Hamiltonian to a suitable unitary gate representation to
build a suitable quantum circuit. We have implemented this circuit on a
quantum computer simulator using Cirq [13], TensorFlow [14] and TensorFlow
Quantum [15]. By using QAOA with QFFNN as the optimizer to configure a Ferrum
chain under no influence of external field of any kind, We have observed that
the most likely configurations are the most organized ones consistent with
energy minimization.

##  2 Material and Method

Magnetic Moment$0$$1$$2$$\dots$$n-1$$n$

Figure 1: [colored online] Illustration of a chain of $n$ magnetic moments. In
this work we consider three different Ferrum chains with a periodic boundary
condition. Initially, Ferrum atoms have random spin orientations denoted by
thick-black arrows here. Length of these arrows has no physical meaning. See
the text for details.

The simplest way to organize some number of magnetic moments is to place them
as so that they shape a line with equidistant between each successive magnetic
moments. This geometric arrangement is called as a chain of magnetic moments
and is depicted in Fig. (1). The Hamiltonian of a magnetic moment chain can be
described by Ising Model with a Hamiltonian such as;

$H=-\sum_{i=0}^{n}\sum_{j=i\pm 1}J_{ij}S_{i}S_{j},$ (1)

where $n+1$ is total number of magnetic moments in the chain,
$S_{i}\in\\{-1,1\\}$ is the spin momentum quantum number of the magnetic
moment placed on the vertex $i$, and $j=i\pm 1$ is the index to denote nearest
neighbors of $i$. Here $J_{ij}$ is the exchange coupling constant between the
magnetic moments on vertices $i$ and $j$. Since the spin momentum quantum
numbers fit the eigenvalues of Pauli-$Z$ matrix, we may represent spin-up and
spin-down states of the magnetic moments with statevectors $|0\rangle$ and
$|1\rangle$, respectively. Then, we may represent the chain’s Hamiltonian
using Pauli-$Z$ matrix such as,

$\hat{H}=-\sum_{i=0}^{n}\sum_{j=i\pm 1}J_{ij}Z_{i}\otimes Z_{j}.$ (2)

In order to write the unitary operator corresponding to this Hamiltonian, it
is necessary to raise it to exponential. Therefore we get the unitary operator
as,

$\displaystyle U(\gamma)=\prod_{i=0}^{n}\prod_{j=i\pm 1}U_{ij}(\gamma),$ (3)

where $U_{ij}(\gamma_{ij})\equiv\exp(-i\gamma_{ij}J_{ij}Z_{i}\otimes Z_{j})$
is the representative part of the unitary operator for the magnetic moment on
the vertex-$i$. This unitary operator is a parametrized one and it needs to be
optimized to yield the most probable combinations of considered system.
Innately, QAOA can be used to optimize these types of operators, and to do so
a suitable quantum circuit needs to be configured. Note that this operator
contains only second order $Z$ matrices and these matrices can be realized in
a quantum circuit by using an $R_{z}(\gamma)$ gate sandwiched by two
Controlled-X gates [16]. This way, the circuit representation for the $i$-th
magnetic moment in the chain interacting with only its nearest neighbors is
pictured in Fig. (2).

$q_{i-1}$$q_{i}$$q_{i+1}$RZ($\gamma_{i,i-1}$)RZ($\gamma_{i,i+1}$)

Figure 2: Quantum circuit part of exchange interaction for atom on $i$-th
index on the lattice. Only nearest neighbors are considered. Figure 3: Counts
of most-encountered spin configurations for ferrum chain of length (a) 12, (b)
16 and (c) 20 atoms.

We examined three different Ferrum chains of different lengths in our
simulations. Since we map the qubit states $|0\rangle$ and $|1\rangle$ to up
and down spin orientations of Ferrum atoms, respectively, n+1 qubits should be
employed in the quantum circuit in order to represent the entire spin
configuration space of the Ferrum chain with n+1 Ferrum atoms. The qubits were
initially put into an equi-superposed state. Then, we applied QAOA to this
system by applying a phase ten times in which the two operators are applied
alternately; first one is the problem unitary operator given in Eqn. 3 and the
second one is a mixing unitary operator selected as $X$ gate [17] applied on
each qubit on the quantum circuit. Therefore we introduced 20 input parameters
into the circuit and initial values of these parameters are randomly assigned
between 0 and $2\pi$ radians. All of the qubits in the circuit was measured at
the end of the circuit execution and this measurement yielded an (n+1)-bit
string which represents the spin configuration of the Ferrum chain. The energy
eigenvalue corresponding to the obtained spin configuration was used to
optimize the circuit parameters via QFFNN. Adam optimizer [18] was used for
optimization with 125 epoches. After training of QFFNN, we sampled 50 million
spin configurations and counted the number of different spin configurations
occurred during the sampling. To imitate a physically infinite chain in the
calculations, we have applied the periodic boundary conditions [19]. In
literature, $J_{ij}$ is taken as $J_{soft}$, or $J_{s}$ for short, for Ferrum
atoms and its value is 1.5 peV/m for a Ferrum chain [20].

The optimized spin configuration of such systems is either all-spin-up
(“$111\dots 11$”) or all-spin-down (“$000\dots 00$”) both of which has the
same energy value. This orientation situation is independent of the chain
length and only due to the nature of ferromagnetic materials. On the other
hand, dimension of the spin configuration space increases exponentially with
the number of atoms in the chain. This actuality yields a rapid decrease on
the probability of getting one of the optimized spin configurations in a
search by brute force. However, our solution finds the possible configuration
with increasing reliability for increasing chain length.

##  3 Results

After analysing the data, we have counted the most-encountered spin
configurations obtained from sampling via optimized parameters. Since there is
no external magnetic field is introduced to the system, all-spin-up and all-
spin-down configurations are essentially the same. Therefore we have counted a
spin configuration along with its upside-down version. First a few most
captured spin configurations for Ferrum chains of length 12, 16 and 20 atoms
are showed in Fig. (3).

Dimension of the configuration space increases exponentially with the number
of atoms in the chain. Notwithstanding, there are only two expected
configurations. This situation is independent of the chain length and only due
to the nature of ferromagnetic materials. This characteristic yields a rapid
decrease on the probability of getting the right solution, if the ordered seek
algorithms would be employed. The probability of getting one of two all-
aligned spin configurations is 1 in 2048 for the chain of 12 atoms, 1 in 32768
for the chain of 16 atoms and 1 in 524288 for for the chain of 20 atoms.
However, our solution finds the theoretically expected spin configuration with
increasing reliability for increasing chain length.

##  4 Conclusion

In conclusion, we have investigated the spin configuration of a ferromagnetic
chain using cutting-edge quantum computation and quantum machine learning
techniques. In our calculations we have applied QAOA to the Ferrum chain of 3
different lengths with a quantum feed-forward neural-network as optimizer. We
have expressed this system via Ising Hamiltonian consists of only exchange
term. We haven’t considered an external magnetic field or any other
interactions between the magnetic moments of any sort. In case of the Ferrum
chain with 12 and 16 atoms, our algorithm does not yield the anticipated spin
configuration. But when the Ferrum chain was made long enough, our algorithm
was able to calculate the expected spin configuration. This can be explained
by the size effect.

##  5 Acknowledgement

This study was carried out within the scope of the QC570-1 course given in
partnership of QWorld and DF@LU. Author sincerely thanks Abuzer Yakaryilmaz
and course attendants for their valuable input to this work about quantum
computation. Author also appreciates fruitful discussions with her colleague,
Caner Deger about magnetism and magnetic materials.

##  6 Code and Data Availability

Data, and the code from which is used to create this data, will be provided by
the author upon reasonable request.

## References

  * [1] Ernst Ising.  Contribution to the Theory of Ferromagnetism.  PhD thesis, Zeitschrift für Physik, 1925.
  * [2] Gary Kochenberger, Jin-Kao Hao, Fred Glover, Mark Lewis, Zhipeng Lü, Haibo Wang, and Yang Wang.  The unconstrained binary quadratic programming problem: A survey.  J. Comb. Optim., 28(1):58–81, jul 2014.
  * [3] Fred W. Glover and Gary A. Kochenberger.  A tutorial on formulating QUBO models.  CoRR, abs/1811.11538, 2018.
  * [4] Andrew Lucas.  Ising formulations of many np problems.  Frontiers in Physics, 2, 2014.
  * [5] Bernhard Korte and Jens Vygen.  Combinatorial Optimization, Theory and Algorithms.  Springer, Berlin, Heidelberg, 2006.
  * [6] David L. Applegate, Robert E. Bixby, Vašek Chvátal, and William J. Cook.  The Traveling Salesman Problem: A Computational Study.  Princeton University Press, Princeton, New Jersey, 2006.
  * [7] G. Kousalya, P. Balakrishnan, and C. Pethuru Raj.  Workflow Scheduling Algorithms and Approaches, pages 65–83.  Springer International Publishing, Cham, 2017.
  * [8] Myungeun Eom and Byung-In Kim.  The traffic signal control problem for intersections: a review.  European Transport Research Review, 12(1):50, Sep 2020.
  * [9] Edward Farhi, Jeffrey Goldstone, and Sam Gutmann.  A quantum approximate optimization algorithm, 2014.
  * [10] Edward Farhi, Jeffrey Goldstone, and Sam Gutmann.  A quantum approximate optimization algorithm applied to a bounded occurrence constraint problem, 2014.
  * [11] Boaz Barak, Ankur Moitra, Ryan O’Donnell, Prasad Raghavendra, Oded Regev, David Steurer, Luca Trevisan, Aravindan Vijayaraghavan, David Witmer, and John Wright.  Beating the random assignment on constraint satisfaction problems of bounded degree, 2015.
  * [12] Solving combinatorial optimization problems using qaoa.  Online.  Access Date: 23 Feb 2022.
  * [13] Cirq | google quantum ai.  Online.  Access Date: 05 Mar 2022.
  * [14] Tensorflow.  Online.  Access Date: 05 Mar 2022.
  * [15] Tensorflow quantum.  Online.  Access Date: 05 Mar 2022.
  * [16] Justyna Zawalska.  Assessment of tensorflow quantum.  Master’s thesis, AGH University of Science and Technology, al. Mickiewicza 30, 30-059 Krakow, Poland, 2021.  Advisor: Dr. Katarzyna Rycerz.
  * [17] Stuart Hadfield, Zhihui Wang, Bryan O’Gorman, Eleanor Rieffel, Davide Venturelli, and Rupak Biswas.  From the quantum approximate optimization algorithm to a quantum alternating operator ansatz.  Algorithms, 12(2):34, Feb 2019.
  * [18] Diederik P. Kingma and Jimmy Ba.  Adam: A method for stochastic optimization, 2014.
  * [19] D. C. Rapaport.  The Art of Molecular Dynamics Simulation (2nd ed.).  Cambridge University Press, 2004.
  * [20] C Değer, ME Aköz, M Erkovan, and F Yıldız.  Investigation on magnetic properties of soft/hard magnetic bilayers at different temperatures: a monte-carlo study.  2016\.