Morphic words, Beatty sequences and integer images of the Fibonacci language

Delft University of Technology

Morphic words, Beatty sequences and integer images of the Fibonacci language

Dekking, Michel

DOI

10.1016/j.tcs.2019.12.036

Publication date

2020

Document Version

Final published version

Published in

Theoretical Computer Science

Citation (APA)

Dekking, M. (2020). Morphic words, Beatty sequences and integer images of the Fibonacci language.

Theoretical Computer Science, 809, 407-417. https://doi.org/10.1016/j.tcs.2019.12.036

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Theoretical

Computer

Science

www.elsevier.com/locate/tcs

Morphic

words,

Beatty

sequences

and

integer

images

of

the

Fibonacci

language

Michel Dekking

DelftUniversityofTechnology,FacultyEEMCS,P.O.Box5031,2600GADelft,theNetherlands

a

r

t

i

c

l

e

i

n

f

o

a

b

s

t

r

a

c

t

Articlehistory:

Received29September2019

Receivedinrevisedform18December2019 Accepted29December2019

Availableonline7January2020 CommunicatedbyR.Giancarlo Keywords:

Morphicword HD0L-system

IteratedBeattysequence Frobeniusproblem Goldenmeanlanguage

Morphic words are letter-to-letter images of fixed points x of morphisms on finite alphabets. Thereare situationswhere theseletter-to-lettermapsdo not occur naturally, but have to be replaced by amorphism. We call this adecoration ofx. Theoretically, decorationsofmorphicwordsareagainmorphicwords,butinseveralproblemstheidea ofdecoratingthefixedpointofamorphismisuseful.Wepresenttwoofsuchproblems. The first considersthe so called A A sequences, where

α

isa quadraticirrational, A is the Beatty sequence defined by A(n)= 

α

n,and A A is the sequence (A(A(n))). The secondexample considershomomorphic embeddingsofthe Fibonaccilanguageintothe integers,whichturnsouttoleadtogeneralizedBeattysequenceswithtermsoftheform V(n)=p

α

n+qn+r,wherep,q andr areintegers.

©2020ElsevierB.V.Allrightsreserved.

1. Introduction

Thegoalofthispaperistoshowthat ifonesuspectsaninfinitewordona finitealphabettobea morphicword,i.e., theletter-to-letterimage ofafixedpoint ofamorphism,thenthewaytoachieve thisisnottotrytodothisdirectly,but indirectly.Bythelatterwe meanthatone replacesthe searchforafixed pointandaletter-to-lettermap bya searchfor afixedpointandamoregeneralobject:amorphism.Toemphasizethisprinciple,wecallthismorphismadecoration,and theinfinitewordwillthenbeadecoration ofafixedpoint.Itiswell-knownthattheclassofdecorationsoffixedpointsof morphisms isequaltothe classofmorphic words,see,e.g.,Corollary 7.7.5inthemonographby AlloucheandShallit[2]. Theirproof, although algorithmic, issomewhat indirect.We willbe using a‘natural’ algorithm togo fromthe decorated fixedpointtoamorphicword,given,e.g.,in[16].WedescribethisalgorithmintheproofofCorollary9.

Weillustratetheusefulnessofthis‘decorationprinciple’bygivingtwoexamples:iteratedBeattysequencesinSection3

andintegerimagesoftheFibonaccilanguage inSection 4.Inthat sectionwe solvethe Frobeniusproblemfor homomor-phicembeddingsofthe Fibonaccilanguage intheset ofintegers,which meansthat we givea precisedescription ofthe complementofthisembedding.

Althoughthe two examples are seemingly unrelated,they are connected by theappearance ofgeneralized Beatty se-quences,whichwedefineinSection2.

Inthe appendixwe give adifferentproof thatthe differencesequence oftheiteratedBeatty sequence A A definedby A A

(

n

)

= 

n



n

√

2



isa morphic word.Thisleads toa morphic wordonan alphabetofsize4.We conjecturethat thisis thesmallestsizepossible,whichisequivalenttotheconjecturethatthedifferencesequenceofA A isnotafixedpointofa morphism.

E-mailaddress:F.M.Dekking@math.tudelft.nl. https://doi.org/10.1016/j.tcs.2019.12.036

Forsomegeneralresultsforaspecialclassofdecorationsoffixedpointsofmorphismssee[15].In[15] thedecorations are socalledmarkedmorphisms,whichinsomesense aretheopposite ofthedecorationsthatone willencounterinthe presentpaper.Wementionalsothatdecorations ofmorphisms arecloselyconnectedtoHD0L-systems.See[19] forsome recentresultsontheseinthecontextofBeattysequences,whichinsomesensearealsooppositetoourresults.

2. GeneralizedBeattysequences

Let

α

beanirrationalnumberlargerthan1,then A definedby A

(

n

)

= 

n

α



forn

≥

1 isknownastheBeattysequenceof

α

.Here,

·

denotesthefloorfunction.Following[3] wecallanysequence V oftheform V

(

n

)

=

p A

(

n

)

+

qn

+

r for n

≥

1

where p

,

q

,

r areintegers,ageneralizedBeattysequence,forshortaGBS.

If S isasequence,wedenoteitssequenceoffirstorderdifferencesas



S,i.e.,



S isdefinedby



S

(

n

)

=

S

(

n

+

1

)

−

S

(

n

),

for n

=

1

,

2

. . .

HowdoesonerecognizeGBS’s?Ingeneralthisisnoteasy,butthereisausefulcharacterizationforquadraticirrational numbers

α

,whichhavethepropertythat

α

∈ (

0

,

1

)

andtheiralgebraicconjugate

α

∈ (

/

0

,

1

)

.TheseareknownastheSturm numbers.Ingeneral,thesequencesoffirstdifferences

cα

:=



(

n

+

1

)

α

 − 

n

α





=





A

(

n

)



are calledSturmiansequences. The characterization is derived from the followingkey result, which isalso proved in the monographs[2] and[17].

Proposition1.([10],[1]) Let

α

beaSturmnumber.Thenthereexistsamorphism

σ

α onthealphabet

{

0

,

1

}

,suchthat

σ

α

(

cα

)

=

cα. Inthefollowingwe willconsiderthevariantsof

σ

α onvariousotheralphabetsthan

{

0

,

1

}

,butwillnotindicatethisin thenotation.Asnotedin[3],thefollowinglemmafollowsdirectlyfromProposition1byrealisingthat

V

=

p A

+

q Id

+

r

⇒

V

(

n

+

1

)

−

V

(

n

)

=

p

(

A

(

n

+

1

)

−

A

(

n

))

+

q

=

p cα

(

n

)

+

q

.

Lemma2. (AlloucheandDekking[3]) Let

α

beaSturmnumber.LetV

= (

V

(

n

))

n≥1bethegeneralizedBeattysequencedefined

byV

(

n

)

=

p

(



n

α

)

+

qn

+

r,andlet



V bethesequenceofitsfirstdifferences.Then



V isthefixedpointof

σ

α onthealphabet

{

q

,

p

+

q

}

.

3. IteratedBeattysequences

RecallthataBeattysequenceisasequenceA

= (

A

(

n

))

n≥1,withA

(

n

)

= 

n

α



forn

≥

1,where

α

isapositiverealnumber.

WhatBeattyobservedisthatwhenB

= (

B

(

n

))

n≥1isthesequencedefinedby B

(

n

)

= 

n

β



,with

α

and

β

satisfying

1

α

+

1

β

=

1

,

(1)

then A andB arecomplementary sequences,thatis,thesets

{

A

(

n

)

:

n

≥

1

}

and

{

B

(

n

)

:

n

≥

1

}

aredisjointandtheirunionis thesetofpositiveintegers.Inparticularif

α

=

ϕ

=

1+₂√5 isthegoldenmean,thisgivesthatthesequences

(

n

ϕ

)

n≥1 and

(



n

ϕ

2

_)

n≥1arecomplementary.

In thispaperwe lookatsequences asfunctionsfrom

N

to

N

.Inthiswaycompositions Z

=

X Y oftwosequences X andY aredefinedasthesequencegivenby Z

(

n

)

=

X

(

Y

(

n

))

forn

∈ N

.

AwellknownresultonthecompositionofBeattysequencesinthegoldenmeancaseisthefollowing.

Theorem 3. (Carlitz-Scoville-Hoggatt [7]) Let U

= (

U

(

n

))

n≥1 be a composition of the sequences A

= (

n

ϕ

)

n≥1 and B

=

(



n

ϕ

2

_)

n≥1,containingi occurrencesofA andj occurrencesofB,thenforalln

≥

1

U

(

n

)

=

Fi+2 jA

(

n

)

+

Fi+2 j−1n

− λ

U

,

whereFkaretheFibonaccinumbers(F0

=

0,F1

=

1,Fn+2

=

Fn+1

+

Fn)and

λ

Uisaconstant.

Thismeansthatanycompositionof A and B canbewrittenasanintegerlinearcombinationp A

+

qId

+

r, whereId is definedbyId

(

n

)

=

n.Thelength2compositionsinTheorem3give

A A

=

B

−

1

=

A

+

Id

−

1

,

A B

=

A

+

B

=

2 A

+

Id

,

B A

=

A

+

B

−

1

=

2 A

+

Id

−

1

,

B B

=

A

+

2B

=

3 A

+

2Id

.

AresultasTheorem3doesnotholdforallquadraticirrationals.Ifwetake,forexample,

α

=

√

2,i.e.,we considerthe Beattysequencegivenby A

(

n

)

= 

n

√

2



,thenthecomplementaryBeatty sequenceB isgivenby B

(

n

)

= 

n

(

2

+

√

2

)



.Itis provedin[8] (seealso[14])thatforn

≥

1

A B

(

n

)

= 

√

2



n

(

2

+

√

2

)

 =

A

(

n

)

+

B

(

n

)

=

2 A

(

n

)

+

2n

.

However,noexpressionforA A isgiven.1 Infact,onecaneasilyprovethattheredonotexistintegers p

,

q andr suchthat A A

=

p A

+

qId

+

r.ThisfollowsfromLemma2inSection2,sincethefirstorderdifferencesequenceof A A takesmorethan 2values.Still,expressionsforA A areknowninvolvingthesequence



√

2

{

n

√

2

}

,seeTheorem1in[13],andsee[5].

Whydoes thegolden mean always yieldGBS’s forthe difference sequences ofthe compositionsof A and B,butthe silvermeandoesnot?OurTheorem5clarifiesthesituation.

From nowon we focuson theiterated Beatty sequence A A givenby A A

(

n

)

= 

n

α



α



.It hasbeenstudied by many authors. See, among others, [7], [8], [13], [4], [5]. The main effort in these papers has been to express A A as a linear combinationof A,Id andtheconstantfunction.

HereisanimportantbasicresultontheiteratesoftheBeattysequence A

(

n

)

= 

n

α



foralgebraic

α

ofdegreed.Inthe following,

{

n

α

}

=

n

α

− 

n

α



isthefractionalpartofn

α

.

Theorem4. (Fraenkel[13]) Letd

≥

1 anda0

,

...,

ad

,

n

,

K

,

L

,

M

∈ Z

.Supposethatadxd

+

ad−1xd−1

+ · · · +

a1x

+

a0

=

0 hasareal

nonzeroroot

α

.LetA

(

n

)

= 

n

α



.Then A



M

+

Ln

+



_id₌−₀2Ai

₍

_{K a}

i+2A

(

n

))



= (

L

−

K a1

)

A

(

n

)

−

K a0n

+

D,whereD isboundedinn,namely,

D

= 

M

α

+ (

L

+

K a0

α

−1

)

{

n

α

}

− θ

α



,where

θ

=



di=−12

(

K ai+2A

(

n

)

α

i

−

Ai

(

K ai+2A

(

n

)))

.

Weareinterestedonlyinthecased

=

2.Let

(

x

−

α

)(

x

−

α

)

betheminimalpolynomialofaquadraticirrational

α

.

Theorem5.Let

α

>

1 beaquadraticirrationalwithminimalpolynomialin

Z

_[

x

]

.LetA

(

n

)

= 

n

α



.ThesequenceA A isageneralized Beattysequenceifandonlyif

|

α

|

<

1.

Proof. Ifonesubstitutes K

=

1,L

=

M

=

0,d

=

2 anda2

=

1 inTheorem4,oneobtains

A A

(

n

)

= −

a1A

(

n

)

−

a0n

+

D

(

n

),

where

(

x

−

α

)(

x

−

α

)

=

x2

₊

_a 1x

+

a0,and D

(

n

)

=



_a 0

α

{

n

α

}



.

Thetheoremnowfollows,since

αα

=

a0,andsincethesequence

(

{

n

α

})

isequidistributedover

[

0

,

1

]

.

Hereweusedthat

θ

=

0 ford

≤

2,asindicatedbyFraenkelintheNotesonpage642of[13].

2

Example6.Let

α

=

1

+

√

2, with corresponding A

(

n

)

= 

n

(

1

+

√

2

)



. As in the proof of Theorem 5 one computes that A

(

A

(

n

))

=

2 A

(

n

)

+

n

−

1.Thenumber

α

−

2

=

√

2

−

1 is aSturmnumber.The correspondingSturmiansequence isfixed point ofthe morphism

σ

α−2 given by 0

→

01

,

1

→

010,asfollowsfroma computation by continuedfractions ([10],[2]).

Since A

(

n

)

= 

n

(

α

−

2

)



+

2n,



A isfixed pointofthemorphism givenby2

→

23

,

1

→

232.Since A A

(

n

+

1

)

−

A A

(

n

)

=

2

(

A

(

n

+

1

)

−

A

(

n

))

+

1,



A A isfixedpointofthemorphismgivenby5

→

57

,

7

→

575.Sointhisparticularcase



A A is puremorphic.

2

Whatisthestructureof A A if

α

>

1 and

|

α

|

>

1?

WedeterminethisfortheFraenkelfamily,alsoknownasthemetallicmeans,whicharethepositivesolutionstox2

_{+ (}

_t

₋

2

)

x

=

t,wherethenaturalnumbert istheparameter.Fort

=

1 oneobtainsthegoldenmean,fort

=

2 thesilvermean

√

2.

Theorem7.Let

α

=



2

−

t

+

√

t2

₊

₄



_/

_2,fort

₌

₂

_,

₃

_,

_{. . .}

_,andletA

₍

_n

₎

_{= }

_n

_α

_

_forn

_≥

_1.Then

_

_{A A isamorphicword.Infact,}

_

_{A A}

isadecoration

δ

ofafixedpointofamorphism

τ

,bothdefinedonthealphabet

{

1

,

2

. . . ,

t

+

1

}

.Fort

=

2 andt

=

3 themorphisms

τ

and

δ

aregivenrespectivelyby

τ

(

1

)

=

12

,

τ

(

2

)

=

131

,

τ

(

3

)

=

121

, δ(

1

)

=

13

, δ(

2

)

=

222

, δ(

3

)

=

132

,

τ

(

1

)

=

123

,

τ

(

2

)

=

124

,

τ

(

3

)

=

1141

,

τ

(

4

)

=

1241

, δ(

1

)

=

113

, δ(

2

)

=

122

, δ(

3

)

=

2122

, δ(

4

)

=

1222

.

Fort

≥

4 themorphism

τ

isgiven2by

τ

(

1

)

=

1

...

[

t

−

1

]

t

,

τ

(

2

)

=

1

...

[

t

−

1

]

[

t

+

1

]

,andforj

=

3

,

...,

t

−

1

τ

(

j

)

=

1

...

[

t

−

j

] [

t

−

j

+

1

] [

t

−

j

+

1

] [

t

−

j

+

2

] . . . [

t

−

2

] [

t

+

1

],

τ

(

t

)

=

112

. . .

[

t

−

2

] [

t

+

1

]

1

,

τ

(

t

+

1

)

=

1223

. . .

[

t

−

2

] [

t

+

1

]

1

.

For t

≥

4 the morphism

δ

is given by

δ(

1

)

=

1t−13

,

δ(

2

)

=

1t−222

,

δ(

j

)

=

1t−j21j−22 for j

=

3

,

...,

t

−

1, and

δ(

t

)

=

21t−2₂₂

_,

_δ(

_t

₊

₁

₎

₌

₁₂₁t−3_22.

IntheproofofthistheoremweneedthecombinatorialLemma8.Weknowthat



A isfixedpointofthemorphism

σ

onthealphabet

{

1

,

2

}

givenby

σ

(

1

)

=

1t−12

,

σ

(

2

)

=

1t−121

,

(2)

ascanbefoundinCrispetal. [10],orAlloucheandShallit[2].Hereoneusesthat

α

hasaverysimplecontinuedfraction expansion:

α

= [

1

;

t

,

t

,

t

,

. . .

]

.

Lemma8.Lett

≥

2 beaninteger.Fort

=

2,definethethreewordsu1

=

121

,

v

=

2112,andw

=

1212.

Fort

≥

3,definethet

−

1 wordsuj

=

1t−j21jforj

=

1

,

...,

t

−

1,andthetwowordsv

=

21t2

,

w

=

121t−12.

Let

σ

bethemorphismin (2),thenfort

=

2,onehas

σ

(

u1

)

=

u1v

,

σ

(

v

)

=

u1wu1

,

σ

(

w

)

=

u1vu1.

Fort

=

3 onehas

σ

(

u1

)

=

u1u2v

,

σ

(

u2

)

=

u1u2w

,

σ

(

v

)

=

u1u1wu1

,

σ

(

w

)

=

u1u2wu1.

Fort

≥

4 onehas

σ

(

u1

)

=

u1

...

ut−1v

,

σ

(

u2

)

=

u1

...

ut−1w,andforj

=

3

,

...,

t

−

1 onehas

σ

(

uj

)

=

u1

...

ut−jut−j+1ut−j+1ut−j+2

. . .

ut−2w

,

σ

(

v

)

=

u1u1u2

. . .

ut−2wu1

,

σ

(

w

)

=

u1u2u2u3

. . .

ut−2wu1

.

Proof. First we take t

=

2.Then

σ

is given by

σ

(

1

)

=

12

,

σ

(

2

)

=

121. One easily verifies the statement ofthe lemma:

σ

(

u1

)

=

1212112

=

u1v

,

σ

(

v

)

=

1211212121

=

u1wu1

,

σ

(

w

)

=

1212112121

=

u1vu1.

Thecaset

=

3 followsfromananalogouscomputation.

Next,thecaset

≥

4.Wefirstmentionfourrelations,directlyimpliedbythedefinitions,whichwillbeusedintheproof: v

=

21

σ

(

1

),

w

=

12

σ

(

1

),

w

=

ut−12

,

u1

=

σ

(

2

).

Wealsouserepeatedly

σ

(

1j

)

=

u1

. . .

uj−11t−j2 for j

=

2

, . . .

t

−

1

,

whichcanbeprovedbyinduction:

σ

(

1j+1

₎

₌

_u

1

. . .

uj−11t−j2

σ

(

1

)

=

u1

. . .

uj−11t−j21t−12

=

u1

. . .

uj1t−j−12.

Wethenhave

σ

(

u1

)

=

σ

(

1t−121

)

=

u1

. . .

ut−212

σ

(

2

)

σ

(

1

)

=

u1

. . .

ut−2121t−121

σ

(

1

)

=

u1

. . .

ut−121

σ

(

1

)

=

u1

...

ut−1v

,

σ

(

u2

)

=

σ

(

1t−2211

)

=

u1

. . .

ut−31121t−1211t−212

σ

(

1

)

=

u1

. . .

ut−112

σ

(

1

)

=

u1

...

ut−1w

.

Now foruj,with3

≤

j

≤

t

−

1:(interpretingu1

. . .

u0 asanemptyprefixinthecase j

=

t

−

1;sointhatcasetheoutcome

is

σ

(

ut−1

)

=

u1u2u2u3

. . .

ut−2w (ift

≥

4).)

σ

(

uj

)

=

σ

(

1t−j

)

σ

(

21j

)

=

u1

. . .

ut−j−11j2 1t−121

σ

(

1j

)

=

u1

. . .

ut−j−1ut−j1j−121 1t−12

σ

(

1j−1

)

=

u1

. . .

ut−j−1ut−jut−j+11j−12

σ

(

1j−1

)

=

u1

. . .

ut−j−1ut−jut−j+11j−12 1t−12

σ

(

1j−2

)

=

u1

. . .

ut−j−1ut−jut−j+1ut−j+11j−22

σ

(

1j−2

)

=

u1

. . .

ut−j−1ut−jut−j+1ut−j+11j−22 1t−12

σ

(

1j−3

)

=

u1

. . .

ut−j−1ut−jut−j+1ut−j+1ut−j+21j−32

σ

(

1j−3

)

= · · ·

=

u1

. . .

ut−j−1ut−jut−j+1ut−j+1

. . .

ut−212

σ

(

1

)

=

u1

. . .

ut−jut−j+1ut−j+1ut−j+2

. . .

ut−2w

.

Forv and w onederives:

σ

(

v

)

=

σ

(

2

)

σ

(

1t−1

)

σ

(

12

)

=

u1u1

. . .

ut−212

σ

(

1

)

σ

(

2

)

=

u1u1u2

. . .

ut−2w u1

σ

(

w

)

=

σ

(

ut−12

)

=

u1u2u2u3

. . .

ut−2w

σ

(

2

)

=

u1u2u2u3

. . .

ut−2w u1

.

2

ProofofTheorem7. Inview of thecomplexity of theproof we first give theproof forthe caset

=

3,i.e., the case

α

=

(

√

13

−

1

)/

2,thebronzemean.

We then have toshow that



A A is a decoration

δ

ofa fixed point ofa morphism

τ

, both defined onthe alphabet

{

1

,

2

,

3

,

4

}

,where

τ

isgivenby

τ

(

1

)

=

123

,

τ

(

2

)

=

124

,

τ

(

3

)

=

1141

,

τ

(

4

)

=

1241

,

andthedecoration

δ

isgivenby

δ(

1

)

=

113

,

δ(

2

)

=

122

,

δ(

3

)

=

2122

,

δ(

4

)

=

1222

.

ThewordsfromLemma8areinthiscase

u1

=

1121

,

u2

=

1211

,

v

=

21112

,

w

=

12112

,

andtheirimagesunder

σ

are

σ

(

u1

)

=

u1u2v

,

σ

(

u2

)

=

u1u2w

,

σ

(

v

)

=

u1u1wu1

σ

(

w

)

=

u1u2wu1

.

Thecodingu1

→

1

,

u2

→

2

,

v

→

3

,

w

→

4 transforms

σ

workingon

{

u1

,

u2

,

v

,

w

}

into

τ

.

LetL bethemapthatassignstoanyworditslength,so,e.g.,L

(

u1

)

=

4

,

L

(

v

)

=

5.

CLAIM:1)Theword



A canbewrittenas



A

=

x1x2

. . .

whereeachxi isanelementfrom

{

u1

,

u2

,

v

,

w

}

.

2)Thewordr

:=

L

(

x1

)

L

(

x2

)

. . .

isfixedpointofthemorphism

σ

4,5 givenby4

→

445

,

5

→

4454.

Proof of part1) of the claim: we know that



A is the unique fixed point of the morphism

σ

=

σ

1,2 givenby 1

→

112

,

2

→

1121.Since1121

=

u1 isaprefixof



A,theword

σ

n

(

u1

)

isalsoaprefix of



A foralln

≥

1.SowithLemma8

thisprovestheCLAIM,part1). Part2)ofthe claimthenfollowsfrom L

(

u1

)

=

L

(

u2

)

=

4

,

L

(

v

)

=

L

(

w

)

=

5,whichinduces

themorphism

σ

4,5fortheinfinitewordr oflengths.

Howdoweobtain



A A from



A?SinceA

(N)

=

A A

(N)

∪

A B

(N)

,adisjointunion,oneobtains A A fromA byremoving theintegers A B

(

n

)

,which,ofcourse,haveindexB

(

n

)

inthesequence A.The differencesequence



B ofthissequenceis the unique fixed point of the morphism

σ

4,5,since

β

=

α

+

3. It follows then from the CLAIMthat the integers A B

(

n

)

occuratpositionswhichcorrespondtothethird letterinthewordxi.Hereit isthethird letter, becausethefirsttermof

thesequence

(

A

(

B

(

n

))

=

5

,

10

,

15

,

22

,

. . .

occursatposition4inthesequence

(

A

(

n

))

=

1

,

2

,

3

,

5

,

. . .

.Removalofthe A B

(

n

)

is then performed by adding the third and the fourth letterin the xi. This operation turns u1

=

1121 into

δ(

1

)

=

113,

u2

=

1211 into

δ(

2

)

=

122, v

=

21112 into

δ(

3

)

=

2122, and w

=

12112 into

δ(

4

)

=

1222. The conclusion is that this

decoration

δ

turnstheuniquefixedpointof

τ

into



A A.Thisendstheproofforthecaset

=

3.

Forgeneralt,thecodingu1

→

1

,

...,

ut−1

→

t

−

1

,

v

→

t

,

w

→

t

+

1 transforms

σ

workingon

{

u1

,

...,

ut−1

,

v

,

w

}

into

τ

.

Ananalogousclaimasforthet

=

3 caseholds,andnowthemap L satisfies L

(

u1

)

=

L

(

u2

)

= ... =

L

(

ut−1

)

=

t

+

1

,

L

(

v

)

=

L

(

w

)

=

t

+

2

,

whichinduces themorphism

σ

t+1,t+2 fortheinfinitewordr oflengths.One continuesinthesameway,usingnow that

β

=

α

+

t. Thistime, the integers A B

(

n

)

occur at positions in A with correspond tothe tth letter inthe words xi from

{

u1

,

...,

ut−1

,

v

,

w

}

.Hereitisthetth letter,becausethefirsttermofthesequence

(

A

(

B

(

n

))

occursatpositionB

(

1

)

=

t

+

1

inthesequence

(

A

(

n

))

.HereB

(

1

)

= β

= 

α

+

t



=

t

+

1,sinceasimplecomputationshowsthat1

<

α

<

2 forallt. Removal of the A B

(

n

)

is then performed by adding the tth _{and the}

₍

_t

₊

₁

₎

th _{letter in the x}

i. This operation turns

u1

=

1t−121 into

δ(

1

)

=

1t−13,u2

=

1t−2211 into

δ(

2

)

=

1t−222 anduj

=

1t−j21j into

δ(

j

)

=

1t−j21j−22,for j

=

3

,

...,

t

−

1.

Moreover,thetwowordsv

=

21t2

,

w

=

121t−12 areturnedinto

δ(

t

)

=

21t−222,respectively

δ(

t

+

1

)

=

121t−322. Theconclusionisthatthisdecoration

δ

mapsthefixedpointof

τ

tothefirstdifferences



A A.

2

Corollary9.Hereisawaytowrite



A A

=

11312221222

. . .

asamorphicwordforthecaset

=

3,i.e.,

α

= (

√

13

−

1

)/

2,thebronze mean.Let

θ

on

{

1

,

. . . ,

6

}

bethemorphismgivenby

θ

:

1

→

123

,

2

→

164

,

3

→

5145

,

4

→

1645

,

5

→

123

,

6

→

164

.

Lettheletter-to-lettermorphism

λ

begivenby

λ

:

1

→

1

,

2

→

1

,

4

→

2

,

5

→

2

,

6

→

2

,

3

→

3

.

Then



A A

= λ(θ

∞

(

1

))

.

Proof. This corollary isderived fromTheorem 7 byusing the‘natural’ algorithm given, forexample,by Honkala in[16], Lemma4.Honkala’srequirementof‘cyclicity’inthatlemmaisnotnecessary.

Tomakethispapermoreself-containedwegiveadescriptionofthis‘natural’algorithm.

Let x be a fixed point of a morphism

τ

on an alphabet A, and

δ

:

A

→

B a decoration. Let d

(

a

)

:= |δ(

a

)

|

, so that

δ(

a

)

= δ

1

(

a

)

. . . δ

d(a)

(

a

)

foreacha

∈

A.The‘natural’ algorithmconsistsofreplacingeach lettera inx by d

(

a

)

copiesofthe

lettera,denotedasC1

(

a

),

. . . ,

Cd(a)

(

a

)

.Thelettertolettermap

λ

onthealphabetAC

:= {

Cj

(

a

)

:

a

∈

A

,

j

=

1

,

. . . ,

d

(

a

)

}

isthen

definedas

λ(

Cj

(

a

))

= δ

j

(

a

)

.Themorphism

τ

inducesalargenumberofmorphisms

θ

onAC,byfirstmappingforeacha the

wordC

(

a

)

:=

C1

(

a

),

. . .

Cd(a)

(

a

)

totheconcatenationofwordsCτ

(

a

)

:=

C

(

τ

1

(

a

))

. . .

C

(

τ

t(a)

(

a

))

,when

τ

(

a

)

=

τ

1

(

a

)

. . .

τ

t(a)

(

a

)

,

andthensplittingCτ

(

a

)

intod

(

a

)

words,defining

θ (

Ci

(

a

))

astheith wordinthissplitting.Thesplittingshouldbedonein

suchawaythataprimitivemorphism

θ

results.

IntheproofofCorollary9thealphabetAC hasapriorid

(

1

)

+

d

(

2

)

+

d

(

3

)

+

d

(

4

)

=

14 letters.Thesituationisspecialhere,

since d

(

a

)

=

t

(

a

)

fora

=

1

,

2

,

3

,

4.This suggeststo define

θ

by splittingthe Cτ

(

a

)

intothewords C

(

τ

1

(

a

))

till C

(

τ

t(a)

(

a

))

.

Afterprojectingletterswiththesame

θ

-imageandthesame

λ

-imageonasingleletter,thenumberoflettersreducesto6, andoneobtainsthemorphism

θ

inthecorollary.

2

Fraenkel’sTheorem4withthe‘defect’function D

=

D

(

n

)

suggeststhat the



A A sequencescantakemanyvalues.This isnotthecase.

Proposition10.Foranyirrational

α

largerthan1 thesequence



A A

=



(

n

+

1

)

α



α



− 

n

α



α





takesvaluesinanalphabetof sizetwo,threeorfour.

Proof. Weillustratetheproofwiththecase1

<

α

<

2.Then s

:= 

A isaSturmianwordtakingvaluesd

=

1 ord

=

2.So



A A

(

n

)

=

A

(

A

(

n

+

1

))

−

A

(

A

(

n

))

=

A

(

A

(

n

)

+

d

)

−

A

(

A

(

n

)),

where d

=

1 or 2

.

We puti

:=

A

(

n

)

.Incased

=

1, A

(

A

(

n

)

+

d

)

−

A

(

A

(

n

))

=

A

(

i

+

1

)

−

A

(

i

)

=

1 or2.Incased

=

2, A

(

A

(

n

)

+

d

)

−

A

(

A

(

n

))

=

A

(

i

+

2

)

−

A

(

i

)

=

A

(

i

+

2

)

−

A

(

i

+

1

)

+

A

(

i

+

1

)

−

A

(

i

)

.SoeitherA

(

A

(

n

)

+

d

)

−

A

(

A

(

n

))

=

2 or3,orA

(

A

(

n

)

+

d

)

−

A

(

A

(

n

))

=

3 or4,respectivelyif11,12and21arethesubwordsoflength2ofs,orif12,21and22arethesubwordsoflength2ofs. Whatwefoundisthat



A A takesvaluesin

{

1

,

2

,

3

}

if1

<

α

<

3

/

2,and



A A takesvaluesin

{

1

,

2

,

3

,

4

}

if3

/

2

<

α

<

2.In somecases



A A maytakeonly2values,forexample,if

α

isthegoldenmean.

Theproofforothervaluesof

α

issimilar,exploitingbalancednessoftheSturmianwords

= 

A.

2

Example.Take

α

=

√

11

/

2

=

1

.

658

. . .

.Then A A

(

n

)

=

1

,

4

,

6

,

9

,

13

,

14

,

18

,

21

,

23

. . .

,so



A A takesthe fourvalues 1

,

2

,

3 and4.

Remark.Oncemore,let

α

=

√

2.Thedifferencesx2,k

:= (

A B

)

k

− (

B A

)

k,wherek

≥

1,arethe‘commutator’functions.They

are extensivelystudied in[9]. Theyare all similarto x2,1,whichis equalto x2,1

=

A B

−

B A

=

2Id

−

A A.One can derive

fromthisthatallcommutatorfunctionsaremorphicwords.

4. EmbeddingsoftheFibonaccilanguageintotheintegers

Let

L

bealanguage,i.e.,asub-semigroupofthefreesemigroupgeneratedbyafinitealphabetundertheconcatenation operation.Ahomomorphismof

L

intothenaturalnumbersisamapS

:

L

→ N

satisfying

S

(

v w

)

=

S

(

v

)

+

S

(

w

),

for all v

,

w

∈

L

.

The classical Frobenius problemasks whetherthe complementof S

(L)

inthe naturalnumbers will be infinite orfinite, andinthelattercasethevalueofthelargestelementinthiscomplement.IntheclassicalFrobeniusproblem

L

isthefull language consistingofallwords overafinitealphabet.We willsolvethisproblemwhen

L

=

L

F i.e.,thesetofall words

occurringinxF,wherexFistheFibonacciword,theinfinitewordfixedbythemorphism0

→

01

,

1

→

0.

Recallthat

ϕ

= (

1

+

√

5

)/

2.ThekeyingredientinthissectionisthelowerWythoffsequence

(



n

ϕ

)

n≥1

=

1

,

3

,

4

,

6

,

8

,

9

,

Theorem11.([12]) LetS

:

L

F

→ N

beahomomorphism.Definea

=

S

(

0

),

b

=

S

(

1

)

.ThenS

(L

F)istheunionofthetwogeneralized

Beattysequences



(

a

−

b

)



n

ϕ



+ (

2b

−

a

)

n



and



(

a

−

b

)



n

ϕ



+ (

2b

−

a

)

n

+

a

−

b



.

WhatremainstobedoneistodeterminethecomplementofthesetS

(L

F)in

N

.Weshallshowthatthecorresponding

infinitewordisalways amorphicword,byrepresentingitasadecorationofafixedpointofamorphism.Itappearsthat thisisamatterofcomplicatedbookkeeping,especiallywhenthetwovaluesS

(

0

)

andS

(

1

)

aresmall.

Thereare threemorphisms f

,

g andh thatplay animportantrole inthissection, whereitisconvenientto lookata andb bothasintegersandasabstractletters.Themorphismsaregivenby

f

:



a

→

ab b

→

a

,

g

:



_a

_→

_baa b

→

ba

,

h

:



_a

_→

_aab b

→

ab

.

Lemma12.LetxFbetheFibonaccisequenceonthealphabet

{

a

,

b

}

,fixedpointof f .ThenthefixedpointxGofg isthesequencebxF,

andthefixedpointxHofh isaxF.

Proof. BerstelandSééboldprovein[6] (Theorem3.1)thatforanymorphicSturmianwordcα onthealphabet

{

a

,

b

}

both acα andbcα areagainmorphicwords.Theirproofisconstructive,andtheygivethemorphisms g andh forcα

=

xFintheir

Example1.

2

HereisaresultthatgivesanideaoftheproofingeneralforthecaseS

(

0

)

>

S

(

1

)

.

Theorem13.LetS

:

L

F

→ N

beahomomorphismdeterminedbya

=

S

(

0

),

b

=

S

(

1

)

.Supposethat

a

+

2

<

2b

+

1

<

2a

−

1

.

Thenthefirstdifferencesofthecomplement

N

\

S

(

L

F)ofS

(

L

F)isthewordobtainedbydecoratingthefixedpointxHofthemorphism

h bythemorphism

δ

givenby

δ(

a

)

=

1b−22 1a−b−22

,

δ(

b

)

=

12b−a−22 1a−b−22

.

Proof. ThesequenceoffirstdifferencesofageneralizedBeattysequence



p



n

ϕ



+

qn

+

r

)

isthefixedpointoftheFibonacci morphism f on the alphabet

{

2p

+

q

,

p

+

q

}

.This followsdirectly formLemma2, see alsoLemma8 in[3]. Sothe two generalizedBeattysequencesG1

:=



(

a

−

b

)



n

ϕ



+(

2b

−

a

)

n



andG2,givenbyG2

(

n

)

=

G1

(

n

)

+

a

−

b inTheorem11havethe

propertythat



G1

= 

G2isthefixedpointxFoftheFibonaccimorphismonthealphabetwithsymbols2

(

a

−

b

)

+

2b

−

a

=

a

anda

−

b

+

2b

−

a

=

b.

Weillustratetheproofbyfirstconsideringthecasea

=

8

,

b

=

5.Inthiscasewehave

G1

=

5

,

13

,

18

,

26

,

34

,

39

,

47

,

52

,

60

, . . . ,

G2

=

G1

+

3

=

8

,

16

,

21

,

29

,

37

,

42

,

50

,

55

,

63

, . . . ...

Partitionthepositiveintegers

N

intoadjacentsetsVi

,

i

=

1

,

2

,

. . .

definedby

Vi

= {

G2

(

i

−

1

)

+

1

, . . . ,

G2

(

i

)

}.

Here we put G2

(

0

)

=

0. As a consequence, Card

(

Vi

)

=

8 if xH

(

i

)

=

a and Card

(

Vi

)

=

5 if xH

(

i

)

=

b, where xH

=

xH

(

1

)

xH

(

2

)

· · · =

aabaab

. . .

is the fixed point of h. The reason that the directive sequence is xH instead of xF is that

thelast elementofeach Vi isequaltoG2

(

i

)

fori

=

1

,

2

,

. . .

.

V1 V2 V3

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

     

Inthetableabove,theintegersinG1

(N)

aremarkedwith



,thoseinG2

(N)

with



,andthoseinthecomplementwitha

.Byconstruction,all theVi withcardinality8havethesamepattern

 

fortheirmembers.Alsoall Vi with

cardinality5havethesamepattern

 

.Notethatthelasttwosymbolsare



,forbothsize5andsize8Vi’s,and

theirfirstsymbolsare

forboth.Thisimpliesthatifwegluethepatternstogether,thentheinfinitesequenceofdifferences ofthepositionsof

intheinfinitepatternyieldsfirstdifferencesofthesequenceofelementsin

N

_{\ (}

G1

(N)

∪

G2

(N))

.For

Vi ofsize8thesedifferences(includingthe‘jumpover’lastvalue2)aregivenby1,1,1,2,1,2,andforVi ofsize5by2,1,2.It

followsthatthefirstdifferencesareobtainedbydecoratingthefixedpointxHbythemorphism

δ

givenby

δ

:

a

→

111212

,

b

→

212

.

Forthe generalcase one considers sets Vi ofconsecutive integers ofsize a or sizeb, wherethe order isagaindictated

positioneda

−

b placesbeforetheend.ItfollowsagainthatovertheVi’sthefirstdifferencesofthecomplementsetendin

2(the‘jumpover’value),areprecededbya

−

b

−

2 1’s,whichisprecededbya2.Thefirstdifferencesstartwithanumber of1’s,whichis

(

a

−

2

)

−

1

− (

a

−

b

−

2

)

−

1

=

b

−

2 forthe Vi’soflengtha,and

(

b

−

2

)

−

1

− (

a

−

b

−

2

)

−

1

=

2b

−

a

−

2

fortheVi’soflengthb.Thisyieldsthedecoration

δ

statedinthetheorem.

2

WenowgiveanexampleofthedifficultiesoneencounterswhenS

(

0

)

orS

(

1

)

are(relatively)small.

Theorem14.LetS

:

L

F

→ N

bethehomomorphismdeterminedbya

=

S

(

0

)

=

3

,

b

=

S

(

1

)

=

1.Thenthesequenceoffirstdifferences

ofthecomplement

N

\

S

(

L

F

)

ofS

(

L

F

)

isthewordobtainedbydecoratingthefixedpointxHofh by

δ

: {

a

,

b

}

→ {

7

,

11

}

givenby

δ(

a

)

=

7

,

11,and

δ(

b

)

=

11.

Proof. AccordingtoTheorem11,S

(L

F

)

istheunionofthetwosets G1

(N)

andG2

(N)

givenby

G1

(

N)

= {

2

n

ϕ

 −

n

,

n

≥

1

} =

1

,

4

,

5

,

8

,

11

,

12

. . . ,

G2

(

N)

= {

2

n

ϕ

 −

n

+

2

,

n

≥

1

} =

3

,

6

,

7

,

10

,

13

,

14

. . . .

Thefirstdifferences



G1

= 

G2aretheFibonacciwordonthealphabet

{

3

,

1

}

.Imitatingtheproofoftheprevioustheorem,

weobtainthefollowingtable,inducedbythemorphismh givenby1

→

331

,

3

→

31.OnehasCard

(

Vi

)

=

a

=

3 ifxH

(

i

)

=

a

andCard

(

Vi

)

=

b

=

1 ifxH

(

i

)

=

b,wherexH

=

xH

(

1

)

xH

(

2

)

· · · =

aabaababaa

. . .

isthefixedpointofh.

V1 V2 V3 V4 V5 V6 V7 V8 V9 V10

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

                    

Thereareatleasttwothingswrongwiththis:

[E1] The Vi’soflength3donotallhavethesamepattern,

[E2] Therearepatternsthatdonotcontaina

.

To counter theseproblems,wego fromtheletters a

=

3

,

b

=

1 tothewordsh

(

3

),

h

(

1

)

, yieldinga partition with Wi’sof

length7and4.Thetableweobtainis

W1 W2 W3 W4

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

                    

Problem[E1]iscausedbythefactthatVi’soflength3havedifferentpatternsdependingonwhethertheyarefollowedby

a Vioflength1oroflength3.Problem[E1]isnowsolvedwiththeWi’s,since33canonlyoccurasaprefixofh

(

1

)

=

331,

and31canonlyoccurasasuffixofeitherh

(

1

)

orh

(

3

)

.

However,[E2]isnotyetsolved,since W3doesnotcontaina

.Thewaytotacklethisistopasstothesquareofh,i.e.,

taketheW_i’soflength18and11correspondingtoh2

(

1

)

=

33133131 andh2

(

3

)

=

33131.

It isobviousfromthecorresponding patterns,that thedifferencesofthecomplement

N

\

S

(L

F)are givenby the

dec-oration W₁

→

7

,

11

,

W₃

→

11 oftheW_i’s.Butsinceh2

₍

_x

H

)

=

xH,thisisthesameasdecoratingthelettersa

→

7

,

11,and

b

→

11 inxH.

2

Remark15.Theorem25in[3] statesthatthethreesequences

(

2



n

ϕ



−

n

,

n

≥

1

)

= (

1

,

4

,

5

,

8

,

11

,

12

. . . ),

(

2



n

ϕ



−

n

+

2n

≥

1

)

,andz

:= (

4



n

ϕ



+

3n

+

2

,

n

≥

0

)

= (

2

,

9

,

20

,

27

,

. . .)

formacomplementarytriple.FromLemma2appliedwiththeSturm number

α

=

ϕ

−

1 onededucesthat



z

=

7x11,7 theFibonaccisequenceonthealphabet

{

11

,

7

}

,precededbytheletter7

(seealsoLemma8in[3],whichstatesthatifV

(

n

)

=

p

(



n

ϕ

)

+

qn

+

r then



V

=

x2p+q,p+q).

Ontheother hand,we haveTheorem14,tellingusthat



z

= δ(

xH

)

,where

δ

isthedecorationa

→

7

,

11,andb

→

11.

Applyingthe‘natural’algorithmto

δ(

xH

)

,weobtainthat

δ(

xH

)

isthemorphicwordobtainedbyapplyingtheletter-to-letter

map

λ(

a

)

=

11

,

λ(

b

)

=

7 tothefixedpointxGofthemorphism g.Thus

δ(

xH

)

= λ(

xG

)

= λ(

bxF

)

= λ(

b

)λ(

xF

)

=

7

λ(

xF

)

=

7x11,7

.

Conclusion:Theorem14isessentiallyequaltoTheorem25in[3],buthasacompletelydifferentproof.

2

WeletC betheincreasingsequenceofintegersinthecomplementofS

(

L

F

)

,soC

(N)

= N \

S

(

L

F

)

.

Theorem16.LetS

:

L

F

→ N

beahomomorphism.Thenthesequence



C offirstdifferencesofthecomplement

N

\

S

(

L

F)ofS

(

L

F)