J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
D. C
RISP
W. M
ORAN
A. P
OLLINGTON
P. S
HIUE
Substitution invariant cutting sequences
Journal de Théorie des Nombres de Bordeaux, tome
5, n
o1 (1993),
p. 123-137
<http://www.numdam.org/item?id=JTNB_1993__5_1_123_0>
© Université Bordeaux 1, 1993, tous droits réservés.
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
Substitution invariant
cutting
sequencesby
D.CRISP,
W.MORAN,
A. POLLINGTON AND P. SHIUE1. Introduction
This work arose from a
problem
discussed in an articleby
TomBrown,
[1].
He considers the sequencesn >
1,
where a is apositive
irrational and[x]
is thegreatest
integer
notexceeding
x . It is clear thatf a
= and so without loss ofgenerality
weassume 0 a
1,
in which case,There is a
large
literature on these sequences(see
for.example
[3]
and[6]).
Each sequence
fa
consists of 0’s and 1’sonly
and we can, withoutambiguity,
omit the commas. A substitution W for such a sequence is a
pair
of mapswhere
Wo
andWI
are finitestrings
of 0’s and l’s. The result ofapplying
this substitution to
fa
is the sequence of 0’s and 1’s obtainedby
replacing
each 0 infa by
Wo
and each 1by
Wi ,
we denote itby
W(f,,,).
Brown showsthat if
1/a
has apurely periodic
continued fractionexpansion,
then thereis a non-trivial
substitution, W ,
leaving
fa invariant,
thatis,
W(fa)
=fa,
and he describes W. He also demonstrates an a
(for
whichI/a
does nothave a
purely periodic
continued fractionexpansion)
with theproperty
thatfa
is fixedonly by
the trivial substitution.We show that the obvious
conjecture
is notentirely
true. We do thisby
giving
acomplete
characterisation of those a for whichfa
is invariantunder some non-trivial substitution. We also
give
a fulldescription
of thesubstitutions concerned.
Ito and
Yasutomi,
[4],
have also considered substitution invariance for the sequencesf a
but with n > 0 instead of n > 1.They
obtain a resultsimilar to Brown’s. Because
they
include the extra termf a (o) =
0,
the substitutions which arise in their work are different to both Brown’s and our’s. Ito and Yasutomipoint
out that their substitutions may be viewedas
automorphisms
of the free groupgenerated by
0 and 1.(The
same istrue of Brown’s and of
our’s.)
In this context we also mention the work ofH.
Cohn,
~2~,
where, automorphisms
of the free group of rank two and the invariance of certain sequences under substitutions are discussed.Our
approach
is to reformulate theproblem
in terms of thecutting
se-quences of lines in the
plane.
Aprecise
definition ofcutting
sequences andsome of their
properties
will begiven
in the nextsection, §2.
Theirconnec-tion with the sequences
f a
is described in the theorem at the end of that section. Our main theorem for substitution invariantcutting
sequences isproved
in section§3.
The translation of that theorem into acorresponding
one for sequences of the formf a together
with a discussion of therelation-ship
between our work and that of Brown and also of Ito and Yasutomi isprovided
in the finalsection,
§4.
2.
Cutting
Sequences
The
cutting
sequence of ageneral
line in theplane
is describedby Series,
[5].
Herehowever,
weonly
consider lines of the form y =¡3x
with~3
apositive
irrational and we restrict our attention to thepositive quadrant.
Thecutting
sequences of such lines are definedby
thefollowing
procedure.
Construct,
in the firstquadrant
of theplane,
the squaregrid consisting
of all vertical and horizontal linesthrough integer
points.
Labelthe
intersectionsof y
=/3x
with thegrid using
0 if thegrid
line crossed is vertical and 1 ifit is horizontal. The sequence of
labels,
read from theorigin
out, is called thecutting
sequenceof y
=3x
and is denoted hereby
Indescribing
these sequences we shall use the standard abbreviation
0~
for a block of1~ consecutive 0’s. The
expression 1 ~
will beinterpreted similarly.
When I~ = 0 theempty
block is meant.We will consider the effect of certain substitutions on the sequences
S{3
and hence establish some of their
properties. First,
however,
we note that#
isuniquely
determinedby
thatis,
if q
is anotherpositive
irrational andSq
=S{3
then q =/3.
To seethis,
observe thatif y
~
#
then there arepoints
of self-intersection of thegrid lying
between thelines y _
-ix andy =
,~x
and henceSq #
Since reflection
in y
= ~interchanges
the(3x
and y = andalso the horizontal and vertical lines of the
grid,
it can be seen thatIf 0
#
1 then1/p
> 1 and it follows that we can, ifdesired,
use R todeduce the
properties
of thecutting
sequences of lines withslope
less than 1 from those of lines withslope
greater
than 1.For k > 0 let
Gk
denote the substitutionWe claim that
In order to prove
this,
we allow the verticalgrid
lines topartition y
=/~x
into
segments.
Specifically,
setPn =
(n, n(3)
and for n > 0 denote thesegment
of y =~~
fromPn
up toincluding Pn+i
butexcluding
P~,,
by
Evidently
the block ofS/3
corresponding
to thesegment L~,
is1Z0
where i =
[(n
+1)/~~ - [no].
Set(~’
_ /~
+ k andsimilarly partition
the liney = into
segments
Ln,
sothat,
the block ofSo,
corresponding
toL’n
is1jO where j
=[(n
+1)/?] - [n,8’].
Since(3’
=#
+ k,
weknow j
= i-t- k
andthus
1~0
=Gk
(li0).
The truth of the claim is now clear. We remark thatthe linear transformation
(x, y)
r-~(x, y + kx)
transformsLn
intoLi
and°
We have
just
seen thatSo
can bepartitioned
into blocks of the formli0
where i =
[(n
+1),8] - [no].
Since theonly possible
values for i are b andb-f-1 where b =
[,8],
theonly possible
blocksoccurring
in such apartitioning
are1b0
and1 (b+1)0.
Because,8
is irrational both must occur.Further,
ifn = 0 then i = b and so
So
starts with1 bOo
For k > 0 let the substitution
Hk
be definedby
Hk
=Gk
oR,
sothat,
The sequence can also be
partitioned
into the blocks 1 and1b0
andtherefore is the result of
applying
~b
to some other sequence of 0’s and l’s.By
using
Hb
=Gb
and theproperties
ofGb
and R describedabove,
itwhere
~3’
=1/(/3 " b).
We can nowrepeat
theargument
starting
withSO,
and so on. The
resulting algorithm
is best understoodby
expanding
(3
as a continuedfraction,
say .where
bo -
b > 0 andbl, b2, b3, ...
arepositive
integers.
If for i > 0 wedefine
then bi
=[pi]
and(3i+l
=l/((3i - bi)
and soS,;
=Hb;
(S¡3i+l).
By
induction we obtain
As i increases the
strings defining
thecomposition
Hbo
o o ... oHb=
lengthen
and ourdescription
ofS¡3
improves.
While it is not necessary forthe purposes of this paper, we
point
out that thisexpansion
leads to anexpression
forSo
as the limit as i increases toinfinity
of thestrings
-- -- -- I- 1
The
following
theorem describes the connection between the sequencesSo
and the sequencesfa.
In itsproof
(and
that of the lemma in section§3)
we will use a cancellativeproperty
of the substitutionsR, Gk
andH.
Specifically,
if S and S’ are any sequences of 0’s and 1’s and W one ofR,
Gk
orHk
then W(S)
= W(S’)
if andonly
if S = S’.THEOREM 1. Let a and
{3
be irrationals with 0 a 1and 0
> 0 thenfa
=S{3
if andonly
if-t
Proof.
Consider theline y
= cxx and itscutting
sequenceSa.
Asbefore,
let the vertical
grid
linespartition y =
ax intosegments.
Thatis,
set(n, na)
and for n > 0 letLn
denote thesegment
fromPn
up toincluding
butexcluding
Pn.
The block inSa
corresponding
toLn
is1 ~ 0
whereThis block is 0 if
fa(n)
= 0 and 10 iffa(n)
= 1.Clearly,
applying
theto
fa yields
thecutting
sequenceSa
of y = axexcept
for the block due tothe
segment Lo .
Themissing
block is 0 and thusapplying
the substitutionyields
the entirecutting
sequence, thatis,
V =S a .
We can nowdeduce
that,
fa
=S{3
if andonly
ifSa
= V(S~).
The substitution V isequal
to l~ oHi
and sowhere q
=1 / ( 1 / ~ -~-
1).
It follows thatf a
=S¡3
if andonly
ifS a
=Sq
butS a
=5,
if andonly
if a = 7 and theproof
iscomplete.
3. Substitution Invariant
Cutting
Sequences
Theorem 1 shows that Brown’s
problem
is solve’d if we can characterisethose #
for which there is a substitutionleaving
S3
invariant. In order todo that we
require
the lemmaimmediately
below.Following
the lemma westate and prove our main theorem. It contains the characterisation
sought.
LEMMA. Let
~3
> 1and y
> 1 beirrationals,
=[bo,
bl,
b2,...
]
and7 =
[co,Ci,C2?...].
If there is a substitution W such that W =5.~,
then, either,
(1)
co >bo,
W =Cco-bo
and~3 - bo
= y - co, or,(2)
there exists a substitution W’ such that W = andfurther,
W = where II =
[Cj
~2,c3, ... ~.
Proof.
Let~31
=[bi,
b2,
b3, ...]
>1,
qi = > 1 and 12 =IC21
C37c4, ... ~
>1,
sothat,
S¡3
=Hbo
(SO,)7
5,
=H Co
(5,1)
and8’1
=HC1
(S12)
and suppose W where W is the substitutionThe two
possible
outcomes in the lemmadepend
on the twopossible
formsof WI.
Case
( 1 ) :
Thestring
WI
does not contain a 0. In this case we will showthat
co >
bo
and W =G Co -bo.
It will then follow thatIf
WI
does not contain a 0 then it is of the form12
for some i > 1(the
case i = 0clearly
does notarise).
By counting
the number of l’s between0’s in
S,~
we will showthat,
infact,
i = 1. First note thatWo
contains azero, so that we can write
Wo =
where j
and k arenon-negative
integers
andXo
is some block of 0’s and l ’s which is eitherempty
or endswith a 0. We know
So
starts with1 boO
and contains the blockthus,
S
starts with and contains Onsubstituting
theexpressions
forWo
andWI
we find thatSq
starts with andcontains Thus
ibo -~ j
= co =which can
only
happen
if i = = 0and i
= co -bo,
sothat,
To see that W =
Gco-bo it
remains to show thatXo
isempty.
We supposenot and obtain a contradiction. Since
Xo
occursimmediately
after a 0 inSq
every 0 in it ispreceded by
1 Co at least. It follows thatXo
=Hco
for some
string
Let V be the substitutionand note that
H Co
o V = W oHbo
and thusIt follows that V
(S{31).=:=
We use this lastequality
to examine the form ofXi.
By counting
the number of l’s between 0’s inS,y1,
we seethat
Xi
contains a 0(by
assumption
Xo
isnon-empty)
and we can writeXo
=I’OYoli
where iand j
arenon-negative
integers and Yo
is some blockwhich is either
empty
or ends with a 0. We knowSol
starts with a 1 andthat 110 occurs:
Therefore,
starts with1Xo
andIXO’LXO’O
occurs. Onsubstituting
Xo-
=1iOYo1j
we find that starts with and that and01jO
both occur. It follows that i + 1 = Cland j
+ 1 -f- iand j
are either el or el + 1. Theonly possibility
is that ci =1, i
= 0 andj - 1.
ThusXo
= OYo 1
andSince ci = 1
and Yo
occursimmediately
after a 0 in there is someand observe that
~I1
o U = ~ oG1
and henceIt follows that U
(5,61-1) ==
S,y2 .
We can now examineY¿.
The sequenceS,-i
contains10’1
and for some i > 0 and therefore8’2
con-tains
O(IYÖl)i1Y¿0
andAgain,
by counting
1’s in theseexpressions,
it is clear thatYo
is notempty
and in fact contains a zero.Further,
sinceYo
is followedby
a 0 in8’2
it must end with 1~2.But YO,
is also followedby
12
implying
that1(12+2)
occurs. This contradicts theproperties
of and ourassumption
aboutXo
was not correct. It follows thatXo
is theempty
string
and W =G Co -bo as
initially
indicated.Case
(2):
Thestring
Wl
contains a 0. In this case we will show thatH~o
(Wó)
for somestring
Wo .
Wealready
know thatWI
=( W1 )
for some
string
W1
(since
W1
is the initialsegment
ofS~
and thus every 0in it is
preceded
by
atleast).
It will followthen,
that W = o W’where W’ is the substitution
and in this event, since
we know
W’
(S{3) ==
and statement(2)
of the lemma is true.We will show that
Wo =
H,,,
(Wó)
by
supposing
otherwise andobtaining
a contradiction. If
Wo
is not of the stated form then it mustbegin
withli0
where i co . We write
Wo
=l i 0Xo
where 0 i co andXo
ispossibly
empty. We are
assuming
WI
contains a 0 and since it is the initialsegment
of
S~
we can writeW1 =
where j
is anon-negative integer
andX,
is eitherempty
or ends with a 0. The blocksW1 Wo
andW1 W1
bothoccur in
S~
and thus and do also. Thisimplies
i =co - l,
j == 1 and
.The blocks and
X,
arepreceded by
0’s inSq
and thus there exist(possibly empty)
Xo
andX’
such thatXo =
Hco
(Xo )
andX, -
so that
H Co
o V = W oG1
and thereforeIt follows that V
(So-,)
= We use this lastequality
to examineX’ 0
and There are two
possibilities
forX1..
S ubcase
( 1 ) :
Thestring
X{,’contains
a 0. Since1X{
is the initialsegment
of andX1
occurs inS’1
preceding
both a 0 and a1,
we canwrite
.X1
= whereY1
is either empty or ends with a 0. Thus the occurrenceof X{ 1XÖ1 in
implies
thatXo
is notempty
and in factbegins
with 0. We writeXo
= oYo
where Yo
ispossibly
empty.
HenceAs
usual,
there exist(possibly empty) Yo’
andYl’
such that~
=H,,
and
Y1
=H,,
(~7)’
The form ofY/
may be elicitedusing
the fact that~7
(S/3-l)
=S’)’2
whereThat this is true follows from the
equalities
H~1
o U = ~ a,nd V(8(3-1)
=S’l
=(S’2).
Since containsOFO
and01 (i+1)0
for some i >0,
the sequence
SY2
contains and * Asbefore,
by counting
l’s between0’s,
it is clear thatY1
contains a 0.Further,
since
Yl
is followedby
a 0 andby
a 1 in it must end with 112.But Yi,
is also followedby
12
implying
thatlO2+2>
occurs. This is a contradictionand our
assumption
aboutX1
was not correct.Subcase
(2):
Thestring
X1
does not contain 0. Asbefore,
we use thefact that V
(8(3-1) =
where V : 0 -1 X11 Xo ,
1 -1 X10
to determinethe form of
X’ 0
andClearly
Xo is
notempty
and in fact contains a0. We write
X’
andX{
= 1’~
where i, j
and k arenon-negative
integers
and Yo is
eitherempty
or ends with a 0. Note thatYo
for some
(possibly
empty)
Assume for the moment that{3 -
1 >1,
that
is, bo >
2. In this case,S’l
starts with1XiO
and both01Xi1X¿
andXo 1X10
occur. Onsubstituting
theexpressions
forXo
andX1
we find thatS’l
starts with1 (l+k)O
and that both01(k+2+i) 0
and occur. Theonly possibility
is that~+l=ci,z==0 and j =
0 or 1. We have shownwhere j
is 0 or 1. We now claim that the result of the substitutionin is
51’2.
Again
this can be verifiedby checking
thatH Cl
o U = V.The
equation
U(S/3-i)
=S,y2
and the form of Usatisfy
thehypothesis
of case
(1)
and hence U = This isclearly impossible
and ourassumption
that~3 -
1 > 1 was wrong. Assume now that;~ -
11,
so
that, bo
= 1 andf31
=1/(~3 - 1).
Since R(S~1 )
= we knowthat V o R =
5,1.
Clearly then,
starts withl.LYi 1X¿
and bothXo’lX’IXO’
and occur. Onsubstituting
theexpressions
forXö
andX1
we find that starts withl~~+2+Z}0
and that both01~~+~+2+i~o
andOl~j+1+~~0
occur. Theonly possibility
is that 1~ + 2 =cl, i
= 0and j
=1,
that
is,
Xo
=OYo1
andX’
=1(cl-2),
sothat,
Let U be the substitution
and note that and thus
It follows that U
(5,~1 _1 )
=8"’2.
However,
we havealready
shown in thefinal
argument
of case(1)
that this situtation leads to a contradiction.Again,
we conclude that ourassumption
aboutX’
was not valid.There are no more
possibilities
and theproof
of the lemma iscomplete.
THEOREM 2.(1)
Let~3
> 1 be irrational. Thecutting
sequenceSo
of theline y
=/3x
is invariant under some non-trivial substitution W if and
only
has a continued fraction
expansion
of the formf3
=[bo,
bl, ... ,
bn]
where
bn > bo >
1.Further,
if that is the case and n is minimalthen W must be a power of
and all such substitutions leave
Sp
invariant.(2)
Let 0,~
1 be irrational. Thecutting
sequenceS~
of theif and
only
has a continued fractionexpansion
of the form(3 =
wherebn
>bo .
Further,
if that is the caseand n is minimal then W must be a power of
- - .. - ..
-and all such substitutions leave
So
invariant.Proof.
Because the substitution Rinterchanges
the sequencesSo
andSl/o,
part
(2)
of the theorem is an easy consequence of part(1).
Wegive
our fullattention to part
(1).
We start with the reverse
implication,
thatis,
given #
_[bo,
bl,
... ,bn]
where
bn >
bo
we will show thatSo
is invariant under the non-trivialsubstitution
By
directcalculation,
W( S ~ ) =
S,~
where q =[be ,
We are
given
that/1
=[bo,
b1,
... , bn)
and therefore/1
=and
W fixes asrequired.
Next we prove the forward
implication,
thatis,
given
thatSo
is invariantunder some non-trivial substitution W we will show that
#
=(bo,
b1, ... ,
bnl
where bn
>bo
> 1. We write#
=~bo, b1, b2, ... ) .
Since W =S~,
thelemma
applies
andthus, either,
W =Go,
or, there exists a substitution W’such that W =
Hbo
o W’ and W’(8ø)
=Sol
where~31
=[61,62~3~ ... , ) .
Because W is
non-trivial,
the latter must be true. A furtherapplication
of the lemma to W’(S~)
=S ,
implies that, either,
If condition
(1)
istrue,
we are finished since we have shown thatbo ,
W =
Hb,,
oGbl-bO
and j3
=[bo,
If condition(2)
istrue,
the lemma canbe
applied again
toW"
(S (3)
=S02
and so on.Repeated
applications
of the lemma musteventually
lead to condition(1)
because eachapplication strictly
reduces the combinedlength
of thestrings
defining
the relevant substitution. Let the total number ofapplications
beAgain,
direct calculation shows W(S,3)
=S.~
whereBy
assumption
W =S~
and so q= ~3 and # =
[bo,
bl, ...,
The first statement in
part
(1)
is now proven and we deal with thesecond.
Suppose
0 =
(bo, bl, ... , b~)
wherebo >
1 and that V is anon-trivial substitution
fixing
Further suppose n is minimal. We willshow that
17 = Wk
for some k > 1 where(By
Wk
we mean thecomposition
copies
ofW).
From theproof
of thefirst statement we know
(bo,
b1, ... ,
wherebm > bo
and thatSince n is minimal there is some l~ > 1 such that m = kn. Thus
But
Hbn
= o.Hbo
and so V =W’,
as claimed. That all suchsubstitutions fix
S{3
is trivial and theproof
iscomplete.
4. S ubst it ut ion invariance for t he sequences
f a
Using
Theorem1,
we can express our results for the sequencesS{3
interms of the sequences
f a .
The mathematics is notcomplicated
and weobtain the
following
theorem. THEOREM 3.~ 1 )
Let1/2
oe 1 be irrationals. The sequencef a
is invariantun-der some non-trivial substitution W if and
only
if a has acontin-ued fraction
expansion
of the form c~ _[0, 1, a2, a3, ... , an~
wherean > c~2 .
Further,
if that is the case and n is minimal then W mustbe a power of
and all such substitutions leave
fa
invariant.(2)
Let 0 a1/2
be irrational. The sequencefa
is invariantun-der some non-trivial substitution W if and
only
if a has aan + 1 ~ 2.
Further,
if that is the case and n is minimal thenW must be a power of
and all such substitutions leave
fa
invariant.Observe that if 0152 =
(0~,
l, a2,
a3, ... , a~~
where an >
a2 then 1 - a =(0, a2
+ 1, a3, ...,
an]
wherean + 1 > a2 + 1 > 2
and vice versa. Thistype
ofrelation is to be
expected
when it is noticed thatapplying R interchanges
fa
andf, - .
Next,
we compare our results with those of Brown. We start with hisTheorem 1. It states
that,
if a =[0,
a1, ... ,am]
thenf a
is invariant underthe substitution
where,
for k >1,
the substitutionhk
is definedby
An easy calculation shows
hk
= and since =Hk
we haveThere are two
possibilities
for
namely,
a1 = 1 and 2. If al = 1then a =
[0, 1,
a2, a3, ...,,
am,1,
a2
andby
part
(1)
of ourtheorem,
fex
isfixed
by
the substitutionSince
Go "is
trivial and ~so is R oHal-1
when a, =1,
wefind,
asexpected,
that V = W . If ai > 2 then a =
[0,
a,, a2, ... , am,all
andby
part
(2)
ofour
theorem,
fex
is ~fixed
by
the
substitutionAgain
T~ =
W,
this time becauseI~1
=G1
0 .~. The theorems areclearly
consistent.
Brown also shows that when cx =
~0, 5, l,1,1, ... ~
there are no non-trivialsubstitutions
fixing
fa.
This is evident from part(2)
of our theorem. Theremainder of Brown’s paper focuses on
general
quadratic
irrationals and substitutions of the form s -C1,
t -~C2 ,
where s and t are finitestrings
of 0’s and l’s and
Cl
andC2
are finitestrings
of s’s and t’s. While suchthings
are not dealt withhere,
we remark that our methods can also beapplied
to them.We conclude with a discussion of the relevant results in the paper
by
Ito and Yasutomi. As mentioned Ito and Yasutomi consider substitution invariance for the sequences where n = 0 is allowed. Since the initial
term is
fa(O)
=0,
we denote these sequences in the natural mannerby
0 f a .
For 0 a 1
they
define a sequencex(a)
of O’s and 1’s and show that if ais irrational with continued fraction
expansion,
say a =[0,
aI, a2, a3,...],
then
They
also introduce the substitutionsand
Observe that 7~ = R o
Hl
but no such relation exists for 71. Their mainresult on substitution invariance
(for
the sequencesOf a)
is contained inTheorem 2.4. While a
slightly
weaker result isactually stated,
it is apparentfrom their work that if
x(a)
isperiodic
with minimalperiod,
saythen
0 fa
is fixedby
the substitutionAgain
we areabbreviating
thecomposition
of icopies
of a substitution 7by
~y~.
The connection between this result and ours may be foundby noting
that if S and S’ are any, two sequences of 0’s and l’s then ,0
(OS)
= OS’ ifand
only
if R o = S’ and 71(OS)
= 0 S’ if andonly
if = S’. Itfollows
that,
if W =0 ... o .y1 ~
then,
W(0/~)
=if and
only
if V =f a
whereGiven that 1 a2 oa3 ... la- 0~‘ is a
period
of1r( a),
there are threepos-sibilities, namely,
It is not hard to
verify
now that Ito and Yasutomi’s work agrees with ours.We stress
however, that,
the index m is(by definition)
even andtherefore,
although
theperiod
of isminimal,
theexpressions
for ajust
listed maynot be. When this occurs, V =
U2
(where
I~ is the minimal substitutionfixing
fa)
and v itself has nocorresponding
W.Ito and Yasutomi have not discussed the
question
of whether there areother solutions to the
equation
W(Ofa)
= It appears that ourre-sults for the sequences
fa
cannot be used to answer thisquestion
directly.
However,
we believe that our methods areapplicable
(with
appropriate
modifications)
and could be used to show that there are no other solutions.Since W =
0 fa
if andonly
if R o W o R(lf1-a)
=1 fl-a,
anequiva-lent
problem
is the solution of theequation
W(1 fa)
=1 fa.
Note that thesequence
1 fa
is anequally justified
extension offa
whenfa
is viewed asthe
cutting
sequence of a line(the
line y
=/3~
where/3
=1/(1/a - 1)
tobe
precise).
In the final section of their paper, Ito and Yasutomi also
provide
someinsight
into thequestion
of substitution invariance forinhomogeneous
se-quences, that
is,
sequences of the formSuch sequences are of course the
cutting
sequences ofarbitrary
lines inthe
plane.
One of the authors(Crisp)
of this paper iscurrently preparing
forpublication
acomplete
solution to this moregeneral problem.
Thesequences
0 fa
and1 fa
will appear asspecial
cases.REFERENCES
[1]
T. C. BROWN, A characterisation of the quadratic irrationals, Canad. Math. Bull.34 (1991), 36-41.
[2]
H. COHN, Some direct limits of primitive homotopy words and of Markoff geodesics,Discontinuous groups and Riemann surfaces, Ann. of Math. Studies No. 79, Princeton Univ. Press, Princeton, N.J., (1974), 81-98.
[3]
A. S. FRAENKEL, Determination of [n03B8] by its sequence of differences, Canad.Math. Bull. 21 (1978), 441-446.
[4]
Sh. ITO and S. YASUTOMI, On continued fractions, substitutions andcharacter-istic sequences [nx + y] - [(n - 1)x + y], Japan J. Math. 16 (1990), 287-306.
[5]
C. SERIES, The geometry of Markoff numbers, Math. Intelligencer 7 (1985), 20-29.[6]
K. B. STOLARSKY, Beatty sequences, continued fractions, and certain shiftoper-ators, Canad. Math. Bull. 19 (1976), 473-482.
D. Crisp, W. Moran Mathematics Department
Flinders University of South Austrlia Bedford Park
SA 5042 AUSTRALIE A. Pollington
Mathematics Department
Talmage Math. / Computer Building
Brigham Young University Provoh, Utah 84602 U. S. A. P. Shiue
Mathematics Department
University of Nevada at Las Vegas
4505 Maryland Pkwy