sequences Hyperquadratic continued fractions and automatic Finite Fields and Their Applications

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Finite Fields and Their Applications

www.elsevier.com/locate/ffa

Hyperquadratic continued fractions and automatic sequences

Alain Lasjaunias^a,∗, Jia-Yan Yao^b

a InstitutdeMathématiquesdeBordeaux,CNRS-UMR5251,Talence33405, France

b DepartmentofMathematics,TsinghuaUniversity,Beijing100084,People’s RepublicofChina

a r t i c l e i n f o a bs t r a c t

Article history:

Received30November2015 Receivedinrevisedform4March 2016

Accepted7March2016 Availableonlinexxxx

CommunicatedbyArneWinterhof

MSC:

primary11J70,11T55 secondary11B85

Keywords:

Finitefields

Powerseriesoverafinitefield Continuedfractions

Finiteautomata Automaticsequences

Weshow thatthree differentfamiliesof hyperquadraticele- ments,studiedintheliterature,havethefollowingproperty:

Fortheseelements,theleadingcoefficientsofthepartialquo- tientsintheircontinuedfractionexpansionform2-automatic sequences. We also show that this isnot true for algebraic elementsinF(q) ingeneral.Indeed,weuseanelementstud- iedbyMillsandRobbinsascounterexample.Thiselementis algebraicinF(3) buttheprincipalcoefficientsof thepartial quotientsdo notformanautomaticsequence.

© 2016ElsevierInc.All rights reserved.

* Correspondingauthor.

E-mailaddresses:Alain.Lasjaunias@math.u-bordeaux1.fr(A. Lasjaunias), jyyao@math.tsinghua.edu.cn(J.-Y. Yao).

http://dx.doi.org/10.1016/j.ffa.2016.03.002 1071-5797/© 2016ElsevierInc.All rights reserved.

1. Introduction

LetFqbethefinitefieldcontainingqelements,withq=p^swherepisaprimenumber ands1 isaninteger.Weconsiderthefieldofpowerseriesin1/T,withcoefficientsin Fq,whereTisaformalindeterminate.WewilldenotethisfieldbyF(q).Henceanon-zero elementofF(q) iswrittenasα=

kk0c_kT^k withk₀∈Z,c_k ∈Fq,andc_k0 = 0.Noting theanalogy ofthis expansionwithadecimalexpansionforareal number,itis natural toregardtheelementsofF(q) as(formal)numbersandindeedtheyareanaloguetoreal numbersinmanyways.

Itiswellknownthatthesequenceofcoefficients(ordigits)(ck)kk0forαisultimately periodicifand onlyif αisrational, thatis αbelongsto Fq(T). However,and this isa singularity of the formal case, this sequence of digits can also be characterized for all elements inF(q) which are algebraic over Fq(T). The origin of the following theorem canbefoundintheworkofChristol[8](seealsothearticleofChristol,Kamae,Mendès France,andRauzy[9]).

Theorem1(Christol).Letαin F(q) with q=p^s.Let(ck)kk0 be thesequenceof digits of αandu(n)=c₋n forallintegers n0. Thenαis algebraicover Fq(T)if andonly ifthefollowingset ofsubsequences of(u(n))_n0

K(u) =

(u(pⁱn+j))_n0|i0,0j < pⁱ

isfinite.

The sequences having the finiteness property stated in this theorem were first in- troduced in the 1960s by computer scientists. Considered in a larger setting (see the beginning of Section3), they are now called automatic sequences, and form a class of deterministicsequenceswhichcanbedefinedinseveraldifferentways.Afullaccounton thistopicandaverycompletelistofreferencesaretobefoundinthebookofAllouche and Shallit [2]. In this note we want to show a different type of connection between automaticsequencesandsomeparticularalgebraicpowerseriesinF(q).

Firstly,letusdescribetheseparticularalgebraicelements.LetαbeirrationalinF(q).

Wesaythatαis hyperquadratic,ifthere exists r=p^twith t0 anintegersuchthat theelementsα^r+1,α^r,α,and1 arelinkedoverFq(T).Thusanhyperquadraticelement isalgebraicoverFq(T) ofdegreer+ 1,andthereadermayconsult[6]whereaprecise definitionwasintroduced.Thesubsetofhyperquadratic elementsinF(q) isdenotedby H(q). Note thatthis subset contains the quadratic power series (take r = 1) and also thecubicpowerseries(taker=p).Originally,thesealgebraicelementswereintroduced in the 1970s by Baum and Sweet (see [4]), in the particular case q = 2, and later consideredinthe1980s byMillsandRobbins[21] andVoloch[25],inallcharacteristic.

ItappearsthatH(q) containselements having anarbitrarylargealgebraic degree.But hyperquadraticpowerseries arerare:analgebraic powerseriesofhigh algebraicdegree

has a small probability to be hyperquadratic. For different reasons, this subset H(q) could beregardedastheanalogueofthesubsetof quadraticrealnumbers.

Besides, it is well known thatany irrational element αin F(q) can be expanded as an infinite continued fraction where the partial quotients an are polynomials inFq[T], all of positive degree, except perhaps for the first one, and we will write it as α = [a₁,a₂,. . . ,a_n,. . .]. The explicit description of continued fractions for algebraic power seriesoverafinitefieldgoesbacktotheworks[4,5]ofBaumandSweet,againwhenthe basefieldisF2.ItwascarriedontenyearslaterbyMillsandRobbinsin[21].Inthereal case, noexplicit continuedfraction expansion, algebraicof degreen>2,isknown.On theotherhandthisexpansionforquadraticrealnumbersiswellknowntobeultimately periodic.Intheformalcase,thesituationismorecomplex.Quadraticpowerserieshave alsoanultimatelyperiodiccontinuedfractionexpansion,butmanyotherhyperquadratic continuedfractionscanalsobeexplicitlydescribed.MostoftheelementsinH(q) havea sequence ofpartial quotientswithunboundeddegrees.IndeedTheorem 4in[21, p. 402]

implies thefollowing: Ifαis hyperquadratic satisfyinguα^r+1+vα^r+wα+z= 0 and wehaver >1+ deg(uz−vw),thenthesequenceofthedegreesofthepartial quotients is unbounded.Butthereare alsoexpansionswithallpartial quotientsofdegree1.This last phenomenonwas discoveredfirstly byMills and Robbinsin[21], andlater studied moredeeplybyLasjauniasand Yaoin[19].Eventhoughthepatternofhyperquadratic expansionscansometimesbeverysophisticated(seeforinstancethework[14]ofFiricel, whereageneralizationofthecubicintroducedbyBaumandSweetispresented),itisyet doubtfulwhetherthisdescription,evenpartial,is possibleforallpowerseriesinH(q).

Power series in H(q) have particular propertiesconcerning Diophantine approxima- tion and this is also why they were first considered. The work [20] of Mahler in this area, is fundamental. There, a first historical example of hyperquadratic power series, on whichwe come backbelowinthis note,wasintroduced. Note thattheirrationality measure (also called approximation exponent,see forinstance [17, p. 214])of a power series canbe computed if the explicitcontinued fraction for this element is known. In this way,formanyelements inH(q), theirrationalitymeasure,oftengreaterthan2 for non-quadratic elements, has been given. Hence, contrary to the real case, many alge- braic power series of degree> 2,most of them hyperquadratic, are knownto have an irrationality measure greater than 2.Actually, for algebraic powerseries whichare not hyperquadratic, concerning their continued fraction expansions and their irrationality measure,notso muchisknown.ThereadermayconsultSchmidt[22],Thakur[24],and Lasjaunias [17],forinstance,formoreinformation and referencesonthis matter.

WitheachinfinitecontinuedfractioninF(q),wecanassociateasequenceinF^∗qinthe following way: ifα= [a1,a2,. . . ,an,. . .], thenfor n≥1, wedefine u(n) asthe leading coefficient of thepolynomial an. For several examples in H(q), we have observed that the sequence (u(n))_n≥1 is automatic. Indeed, a first observationin this area is due to Allouche [1].In thearticle [21] of Millsand Robbins, forall p5, aparticular family of continued fractions inH(p), having a_n =λ_nT for all integersn 1 with λ_n inF^∗p, wasintroduced.Shortlyafterthepublicationof[21],Allouchecouldprovein[1]thatthe

sequence(λn)_n1inF^∗p isautomatic(seealsothelastsectionof[19],wherethisquestion isdiscussedinalargercontext).Inthepresentnote(Section2),weshalldescribeseveral familiesofhyperquadraticcontinuedfractionsandweshowinSection3thatthesequence associatedwiththem,asindicatedabove,isalsoautomatic.

Yetit is an open question to know whetherthis is true for all elements in H(q). If theanswerwere negative,itwouldbeinterestingtobeabletocharacterizetheelements inH(q) whichhavethisproperty. Asmentionedabove,verylittleisknown,concerning continued fractions, for algebraic power series which are not hyperquadratic. However anelement inF(3),algebraic ofdegree4,wasintroduced byRobbinsandMills in[21].

Thiselementisnothyperquadratic.Inthisnote (Section4),weshowthatthesequence inF^∗3, associatedasabovewithitscontinuedfraction expansion,isnotautomatic.

2. Threefamiliesofhyperquadraticcontinuedfractions

Inthissectionweshallusethenotationandresultsfoundin[18].

Let α be an irrational element in F(q) with α = [a₁,. . . ,a_n,. . .] as its continued fraction expansion. We denote by F(q)⁺ the subset of F(q) containing the elements havinganintegralpartofpositivedegree(i.e.withdeg(a1)>0).Forallintegersn1, we putαn = [an,an+1,. . .] (α1 =α), and we introducethecontinuants xn,yn ∈ Fq[T] suchthatx_n/y_n = [a₁,a₂,. . . ,a_n].Asusualweextendthelatternotationton= 0 with x0 = 1 andy0 = 0.Observethatthenotation usedherefor thecontinuantsxn andyn

isdifferentfrom theoneusedin[18],andhopefully simplified.

As above we set r = p^t, where t 0 is an integer. Let P,Q ∈ Fq[T] such that deg(Q) < deg(P) < r. Let 1 be an integer and A = (a₁,a₂,. . . ,a) a vector in (Fq[T]) such thatdeg(a_i) >0 for 1i . Then by Theorem 1in [18], there exists an infinite continued fraction in F(q) defined by α = [a1,a2,. . . ,a,α+1] such that α^r=P α+1+Q.ThiselementαishyperquadraticanditistheuniquerootinF(q)⁺ of thefollowingalgebraicequation:

yX^r+1−xX^r+ (y−1P−yQ)X+xQ−x−1P = 0. (1) Notethatifr= 1,thenαisquadratic.InthiscaseP isanonzeroconstantpolynomial, i.e. P = ε ∈ F^∗q and Q = 0. Given and A, we have α = εα₊₁, and this implies a+m = ε^(−1)mam, for all integers m 1. Hence the continued fraction expansion is purely periodic, and a simple computation shows that 2 (resp. (q−1)) is a period (maybenottheminimumone) ifis odd(resp.even).

Weshalldescribethreefamiliesofcontinuedfractions generatedasabove.

Firstfamily:F1.Thesimplest andfirstcasethatweconsideris(P,Q)= (ε,0) where ε∈F^∗q andconsequentlyα^r=εα_l+1.Due to theFrobeniusisomorphism, weobtain,in thesameway asabovefor r= 1, therelationa_+m=ε^(−1)ma^r_m forallintegersm1.

Hencethecontinuedfraction,dependingonthearbitrarygivenfirstpartialquotients,

is fully explicit. These hyperquadratic continued fractions were studiedindependently bySchmidt[22]andThakur[23](particularlyforε= 1).Letusrecallthattheelements in H(q), called here hyperquadratic, were first named in [16] as algebraic of class I, and then the elements studied by Schmidt and Thakur were called algebraic of class IA.Thepossibility ofchoosingarbitrarilythevectorA hasanimportantconsequence.

Even though this is not truly the matter of this note,we have already mentioned the irrationalitymeasureν(α) ofαinF(q).Inhisfundamentalwork[20],Mahlerestablished (followinganoldresultofLiouvilleintherealcase)thatifα∈F(q) isalgebraicofdegree d2 overFq(T),thenwehaveν(α)∈[2,d]. Besidesν(α) isdirectlydepending onthe sequence ofthedegreesofthepartialquotientsforα(seeforinstance [17,p. 214]).For anelementofF1,thissequenceofdegrees(dn)n1,satisfiesd+m=rdmforallintegers m1,andconsequentlyd_n dependsdirectlyonthefirstdegrees.Hence,Schmidtand Thakur, independently,couldobtain(byasophisticatedcomputation) theirrationality measure forsuch an element,depending onr and the first degrees. In thisway they could establish the following result: for each rational number μ inthe range [2,+∞[, there existsαinF1 suchthatν(α)=μ.

Let us make a last observation on the simplest element in F1. We take = 1 and a1 = T, with (P,Q) = (1,0), then the corresponding continued fraction is Θ₁ = [T,T^r,T^r2,. . . ,T^rn,. . .], and Θ₁ satisfies X = T + 1/X^r. Here the irrational- itymeasure is easyto compute,and indeedwe haveν(Θ₁)=r+ 1. Sincethedegree d of Θ1 satisfies dr+ 1,using Mahler’sargument, weseethatΘ1hasalgebraic degree equaltor+ 1.NotethatforageneralelementαinF1,itsexactalgebraic degreeis an undecidedquestion.

Second andthirdfamilies:F2andF3.Foranelement αinF2,weassume(P,Q)= (ε1T,ε2);andforanelementαinF3,weassume(P,Q)= (ε1T²,ε2T),where (ε1,ε2) is apairin(F^∗q)².Notethatherewehaver >1 forelementsinF2,butr >2 forelements in F3. Our aim is to give theexplicit continued fraction expansion forα. The integer 1,as above,is chosen arbitrarily. Howeverwe need to impose arestriction on the choice of the vector A, in both cases: we assume that T divides ai for all integers i with 1i.Thenwecandescribethesequenceofpartial quotientsinthecontinued fraction expansioninboth cases.Given A, chosenas indicated, forall integersn0, we have:

a_+4n+2=−ε₁ ε2

T, a_+4n+3=−ε²₂a^r_2n+2

ε1T , a_+4n+4= ε₁ ε2

T,

while a+4n+1 = ^a

r 2n+1

ε1T for α∈ F2, and a+4n+1 = ^a

r 2n+1

ε1T² for α ∈ F3. Themethod to obtaintheseformulasisthesameinbothcasesandithasbeenexplainedin[18].Actually theformulasforαinF2 werepublishedin[18,topofp. 334].However,note thatthere is amistake inthefinal statementgiventhere andthepair(k,k+ 1) totheright hand sideofthefirstandthirdformulasmustbereplacedbythepair(2k−1,2k).Concerning

theformulas for αin F3,they weregiven by Firicelin [14] (caseε1 = ε2 =−1).The readeris invited to consultthis last work where themethodis clearly explained.Note thattheparticular choice ofthe vector A, has been madein orderto have aninteger (polynomial)whendividingbyT orT²(howeveralargerchoicecouldhavebeenpossible forelementsinF3,inaparticularcase:forexample,ifisodd,thenitwouldbeenough to assumethatT dividesai only foroddindices i).Note thatifwedo notassumethe particularchoicewemadeforthevector A, thecontinuedfraction forαexists,butan explicitdescriptionisnotgiven.

WemakeacommentconcerningthelastfamilyF3.Ithasbeenintroducedtocovera particularcaseofhistoricalimportance.Inhisfundamentalpaper[20],Mahlerpresented thefollowing powerseries Θ2= 1/T + 1/T^r+· · ·+ 1/T^rn+· · ·. NotethatΘ2 canbe regardedas adualof theelement Θ1 inF1,presentedabove. Howeverwe clearlyhave Θ2= 1/T+Θ^r₂,henceΘ2ishyperquadraticandalgebraicofdegreer.Mahlerobserved thatwehaveν(Θ2)=r,andthereforethealgebraicdegreeofΘ2isequaltor.Nowletus considertheelementofF3 definedasabovebythepair(P,Q)= (−T²,−T) with= 1 anda1=T.Thenαis theuniquerootinF(p)⁺ ofthefollowingalgebraic equation:

X^r+1−T X^r+T X = 0.

Setβ= 1/α,andwegetβ= 1/T+β^r. Since1/β isinF(p)⁺,weobtain β = Θ2= 1/T+ 1/T^r+ 1/T^r2+· · ·

Let us recall that the sequence of partial quotients for Θ₂ has been long known (see forexample[14, p. 633]with thereferencestherein. Seealso [16,p. 224]for adifferent approach).

Aswewroteintheintroduction,weareinterestedintheleadingcoefficientsofpartial quotients. If (an)_n1 is thesequence of partial quotients for α, we denote by u(n) the leadingcoefficient of an. If αis inF1, then u(+m)=ε^(−1)m(u(m))^r for allintegers m1,andthusu(s+m)=ε⁽₁^−1)mu(m) with

ε1=ε^{s−1j=0(−1)(s−j−1)rj} ∈F^∗q,

forwehaver^s=p^st=q^t.Asabove,onecandeduceatoncethatthesequence(u(n))_n1 ispurelyperiodic,and2s(resp.(q−1)s)isaperiod(maybenottheminimumone)if sisodd(resp.even).

NowweturntothecasewhereαbelongstoF2 orF3.Bothrecursivedefinitionsfor thesequence of partial quotients, whether αis inF2 or F3,imply the samerecursive definitionforthecorrespondingsequence(u(n))_n1.Moreprecisely,forallintegers1, givenu(1),u(2),. . . ,u() in F^∗q,wehave,forallintegersn0,

u(+ 4n+ 1) =ε⁻¹₁ (u(2n+ 1))^r, u(+ 4n+ 2) =−ε1ε⁻¹₂ , u(+ 4n+ 3) =−ε⁻₁¹ε²₂(u(2n+ 2))^r, u(+ 4n+ 4) =ε1ε⁻₂¹.

InthenextsectionweshallseeinTheorem 3thatsequencesofthistype,forallintegers 1,are2-automatic sequences.

To conclude this section, we go back to the special element 1/Θ2 in F3, and the associated sequence (u(n))_n1 where= 1 and u(1)= 1.Thelatteris remarkableand hasbeenstudiedextensively(seeforexample[2,Section6.5]).Wedefinerecursivelythe sequence offinitewords(Wn)_n1byW1= 1,andWn+1=Wn,−1,W_n^R,where commas indicatehereconcatenationofwords,and W_n^R isthereverseofthefinite wordWn.Let W be theinfinite word beginningwith Wn for allintegers n1. Thenone cancheck thatthesequence(u(n))_n1coincideswithW,andforp= 2,itisaspecialpaperfolding sequence whichisknownto be2-automatic(seeforexample[2,Theorem6.5.4]).

3. Afamilyofautomaticsequences

Inthissection,webeginwiththedefinitionofautomaticsequences.Formoredetails aboutthis subject,seethebook[2]ofAlloucheand Shallit.

Let Abe afinite nonempty set,called analphabet, ofwhich everyelement iscalled aletter. Fix∅anelementnotinAand call itanemptyletterover A.Letn0 be an integer.Ifn= 0,defineA⁰={∅}.Forn1,denotebyAⁿ thesetofallfinitesequences inAoflengthn.PutA^∗=_+∞

n=0Aⁿ.EveryelementwofA^∗iscalledawordoverAand itslengthisnoted|w|, i.e.|w|=nifw∈Aⁿ.

Takew,v∈A^∗.Theconcatenationbetweenwandv(denotedbyw∗vormoresimply bywv)isthewordoverA whichbegins withwandiscontinued byv.

Now wegivebelowadefinitionoffinite automaton(seeforexample[13]):

A finiteautomatonA= (S,S0,Σ,τ) consistsof

• analphabetS ofstates;onestateS₀ isdistinguishedandcalled initialstate;

• amapping τ : S×Σ→S, called transition function,where Σ is analphabet con- tainingatleasttwo elements.

∀Q∈S,putτ(Q,∅)=Q.Extendτ overS×Σ^∗(notedagainτ)suchthat

∀Q∈S and∀l, m∈Σ^∗, we haveτ(Q, lm) =τ(τ(Q, l), m).

Letk2 beanintegerandΣ_k ={0,1,. . . ,k−1}.Wecallv= (v(n))_n0ak-automatic sequence ifthere exist a finite automaton A = (S,S₀,Σ_k,τ) and a mappingo defined on S suchthatv(0)=o(S0), andforallintegersn1 withstandardk-adicexpansion n=l

j=0njk^j,wehavev(n)=o(τ(S0,nl· · ·n0)).

Werecallthatallultimatelyperiodicsequencesarek-automaticforallk2,addinga prefixtoasequencedoesnotchangeitsautomaticity,andthatasequenceisk-automatic if and only if it is k^m-automatic for all integers m 1. In this work, we consider sequences of theform v = (v(n))n1, andwe saythatv is k-automatic ifthesequence

(v(n))_n0 is k-automatic, with v(0) fixed arbitrarily. We havethe following important characterization:asequencev= (v(n))n1 isk-automaticifandonlyifitsk-kernel

Kk(v) =

(v(kⁱn+j))n1|i0, 0j < kⁱ

is a finite set. The origin of this characterization for automatic sequences is due to Cobham [11, p. 170, Theorem 1](see also Eilenberg [13, p. 107, Proposition 3.3], who wasoneofthefirsttopublishageneraltreatise onthismatter).

Letv= (v(n))_n1 be asequence.Forallintegersn1,wedefine (T₀v)(n) =v(2n) and (T₁v)(n) =v(2n+ 1).

Thenforallintegersn,a1,and0b<2^awith binaryexpansion b=

a−1

j=0

bj2^j (0bj <2), withthehelpoftheoperators T0 andT1,weobtain

v(2^an+b) = (Tb_a−1◦Tb_a−2◦ · · · ◦Tb0v)(n).

Inparticular,weobtainthatvis2-automaticifandonlyifbothT0v,T1vare2-automatic, forwehaveK2(v)={v}∪K2(T0v)∪K2(T1v).

Withthese definitions,we havethefollowing theorem,whichcanbe compared with aresultofAlloucheandShallit(see[3,Theorem 2.2]).

Theorem2.Letm0beaninteger, v= (v(n))_n1 asequenceinan alphabetA,andσ abijectionon A.

(0m) If T0v is2-automatic,and (T1v)(n+m)=σ(v(n)) forallintegers n1,then v is2-automatic;

(1_m) If T₁v is2-automatic,and (T₀v)(n+m)=σ(v(n)) forallintegers n1,then v is2-automatic.

Proof. Firstly we note that A is finite, thus there exists an integer l 1 such that σ^l= idA,theidentitymappingonA.

Secondlyweshowthat(0m)implies(1m).Forthis,putu(n)=v(n+1),forallintegers n1.Then T₀u=T₁v,andforallintegersn1,wehave

(T1u)(n+m) =u(2n+ 2m+ 1) =v(2n+ 2m+ 2)

= (T0v)(n+ 1 +m)

=σ(v(n+ 1)) =σ(u(n)).

Henceuis 2-automaticby virtueof(0m),and sois v, forthelatter isobtainedfrom u byadding aletterbefore.

Inthefollowingweshall show(0m)byinductiononm.

Ifm= 0,thenundertheconditionsof(0₀),wehaveT₁v=σ(v) and

K2(v) ={σ^j(v)|0j < l} ∪

l−1

j=0

σ^j(K2(T0v)).

Thus K2(v) isfinite sinceT0v is2-automatic.

If m= 1,thenT0v is2-automatic,and(T1v)(n+ 1)=σ(v(n)) forallintegersn1.

HenceT₁T₁v=σ(T₀v),andforallintegersn1,wehave

(T0T1v)(n+ 1) = (T1v)(2n+ 2) =σ(v(2n+ 1)) =σ((T1v)(n)).

Thus T1T0T1v=σ(T0T1v),and forallintegersn1,wehave

(T0T0T1v)(n+ 1) = (T0T1v)(2n+ 2) =σ((T1v)(2n+ 1))

=σ²(v(2n)) =σ²((T0v)(n)).

So T₀T₀T₁v is 2-automatic sinceit isobtained from σ²(T₀v) by addinga letterbefore, and thelatteris2-automatic,forT0v is. Setw=T0T1v.Then T0wis 2-automatic,and T1w=σ(w).Thusby(00),weobtainthatT0T1vis2-automatic.ButT1T1v=σ(T0v) is also 2-automatic,consequentlyT1v is2-automatic,and thenv is 2-automatic,forboth T₀v andT₁v are2-automatic.

Now let m1 be aninteger, and assumethat (0_j) hold for0j m. Then (1_j) also hold for0jm,and weshall showthat(0m+1)holds. Forthis, wedistinguish two casesbelow:

Case1:misodd.Thenforallintegersn1,wehave (T₀T₁v)(n+ [m

2] + 1) = (T₁v)(2n+m+ 1) =σ(v(2n)) =σ((T₀v)(n)), (T1T1v)(n+ [m

2] + 1) = (T1v)(2n+m+ 2) =σ(v(2n+ 1)) =σ((T1v)(n)).

HenceT₀T₁v is2-automatic foritis obtainedfrom σ(T₀v) byadding aprefixof length [^m₂]+ 1.Put w=T₁v.ThenT₀wis2-automatic,and forallintegersn1,

(T₁w)(n+ [m

2] + 1) =σ(w(n)).

Note that[^m₂]+ 1 m. Thus wecan apply (0_[^m

2]+1) with w, and we obtainthatw is 2-automatic.Hencevis2-automatic sincebothT0v andT1v are2-automatic.

Case2:m iseven.Thenforallintegersn1,wehave (T₀T₁v)(n+ [m

2] + 1) = (T₁v)(2n+m+ 2) =σ(v(2n+ 1)) =σ((T₁v)(n)), (T₁T₁v)(n+ [m

2]) = (T₁v)(2n+m+ 1) =σ(v(2n)) =σ((T₀v)(n)).

SoT1T1v is2-automaticforitisobtainedfrom σ(T0v) byaddingaprefixoflength[^m₂].

Put w = T1v. Then T1w is 2-automatic, and (T0w)(n+ [^m₂]+ 1) = σ(w(n)), for all integersn1.Byapplying(1_[^m_2]+1)withw,weobtainthatwis2-automatic,andthen v is2-automaticsincebothT0v andT1v are2-automatic.

Finallyweconcludethatboth (0m)and(1m)holdforallintegersm0. 2 We can now prove thatthe sequences, associated with the elements of F2 and F3 anddescribed attheendoftheprevioussectionare2-automatic.Thisfollowsfrom the moregeneralresultstatedbelow,whichisadirectapplicationofTheorem 2.

Theorem 3.Letp be aprime number, s1an integer,and q =p^s. Denote by Fq the finitefieldinqelements.Set r=p^t,witht0aninteger.Fix α,β,γ,δ∈F^∗q,and1 aninteger. Letu= (u(n))_n1 beasequencein F^∗q suchthat forallintegersn0,

u(+ 4n+ 1) =α(u(2n+ 1))^r, u(+ 4n+ 2) =β,

u(+ 4n+ 3) =γ(u(2n+ 2))^r, u(+ 4n+ 4) =δ. (2) Thenthesequenceuis2-automatic.

Proof. For all x,y ∈ F^∗q, we put σ_y(x) = yx^r. Then σ_y is a bijection on F^∗q. For all integersn1, set u₀(n)=u(2n) andu₁(n)=u(2n+ 1),and we needonly showthat bothu₀ andu₁ are2-automatic.Forthis,wedistinguishbelowfourcases.

Case I: = 4m+ 1, with m 0 an integer. Then for all integers n 0, from Formula(2),wededuce

(T₀u₁)(n+m+ 1) =u(4(n+m+ 1) + 1) =u(+ 4n+ 4) =δ, (T₁u₁)(n+m) =u(4n+ 4m+ 3) =u(+ 4n+ 2) =β.

Since both T0u1 and T1u1 areultimately constant, then u1 isultimately periodic,and thus2-automatic.

Similarly,forallintegersn0,we alsohave

(T0u0)(n+m+ 1) =u(4(n+m+ 1)) =u(+ 4n+ 3)

=γ(u(2n+ 2))^r=γ(u0(n+ 1))^r, (T1u0)(n+m) =u(4n+ 4m+ 2) =u(+ 4n+ 1)

=α(u(2n+ 1))^r=α(u1(n))^r.

So T1u0 is ultimately periodic as u1, and (T0u0)(n+m) = σγ(u0(n)) for all integers n1.ThenbyTheorem 2,weobtainthatu0 is2-automatic.

Case II: = 4m+ 2, with m 0 an integer. Then for all integers n 0, from Formula (2),wededuce

(T0u0)(n+m+ 1) =u(4(n+m+ 1)) =u(+ 4n+ 2) =β, (T1u0)(n+m+ 1) =u(4n+ 4m+ 6) =u(+ 4n+ 4) =δ.

Sou₀ isultimatelyperiodic,and thus2-automatic.

Similarly,forallintegersn0,wealsohave

(T₀u₁)(n+m+ 1) =u(+ 4n+ 3) =γ(u(2n+ 2))^r=γ(u₀(n+ 1))^r, (T1u1)(n+m) =u(+ 4n+ 1) =α(u(2n+ 1))^r=α(u1(n))^r.

HenceT0u1 isultimatelyperiodicasu0, and(T1u1)(n+m)=σα(u1(n)) forallintegers n1.ThenbyTheorem 2,weobtainthatu1 is2-automatic.

Case III: = 4m+ 3, with m 0 an integer. Then for all integers n 0, from Formula (2),wededuce

(T0u1)(n+m+ 1) =u(4(n+m+ 1) + 1) =u(+ 4n+ 2) =β, (T1u1)(n+m+ 1) =u(4n+ 4m+ 7) =u(+ 4n+ 4) =δ.

Sou1 isultimatelyperiodic,and thus2-automatic.

Similarly,forallintegersn0,wealsohave

(T₀u₀)(n+m+ 1) =u(4(n+m+ 1)) =u(+ 4n+ 1)

=α(u(2n+ 1))^r=α(u₁(n))^r, (T₁u₀)(n+m+ 1) =u(4n+ 4m+ 6) =u(+ 4n+ 3)

=γ(u(2n+ 2))^r=γ(u0(n+ 1))^r.

Thus T0u0 isultimately periodicasu1,and (T1u0)(n+m)=σγ(u0(n)) for allintegers n1.ThenbyTheorem 2,weobtainthatu0 is2-automatic.

Case IV: = 4m+ 4, with m 0 an integer. Then for all integers n 0, from Formula (2),wededuce

(T0u0)(n+m+ 2) =u(4(n+m+ 2)) =u(+ 4n+ 4) =δ, (T1u0)(n+m+ 1) =u(4n+ 4m+ 6) =u(+ 4n+ 2) =β.

Sou0 isultimatelyperiodic,and thus2-automatic.

Similarly,forallintegersn0,we alsohave

(T0u1)(n+m+ 1) =u(+ 4n+ 1) =α(u(2n+ 1))^r=α(u1(n))^r, (T₁u₁)(n+m+ 1) =u(+ 4n+ 3) =γ(u(2n+ 2))^r=γ(u₀(n+ 1))^r.

Hence T1u1 is ultimately periodic as u0, and (T0u1)(n+m+ 1) = σα(u1(n)) for all integersn1.ThenbyTheorem 2,weobtainthatu₁ is2-automatic. 2

Remark1.AsitispointedoutattheendofSection2,forp= 2,thesequenceassociated with the special element 1/Θ2 in F3, is a paperfolding sequence, thus not ultimately periodic(seeforexample[2,Theorem6.5.3]).Inspiredbythisexample,onecanthenask whetherthesequenceudiscussedinTheorem 3canbeultimatelyperiodicornotforall q3 (inthecasethatq= 2,thesequenceisconstant).Unfortunatelyforthemoment, wedonotknow theanswerofthis problem.

4. Asubstitutive butnotautomaticsequence

In this section we are concerned with the following question: is it also true for an algebraicpowerseries, nothyperquadratic,thatthesequenceoftheleadingcoefficients of thepartial quotients form anautomatic sequence? Tosuch awide question,we will onlygive avery partial answer, byconsideringa last example.As we remarked inthe introduction,thepossibilityofdescribingexplicitlythecontinuedfractionexpansionfor analgebraicpowerseries,whichisnothyperquadratic,appearsto beremote.However, aparticular example, which was introduced by chance in[21], exists. This example is theobjectofthetheorembelow.

First,werecallnotionsonsubstitutivesequences(seeforexample[2]).

LetAbeanalphabetwithA={A1,A2,...,AN}.AsubstitutiononAisamorphismσ: A^∗→A^∗.Withthemorphismσ,thereisassociatedamatrixMσ= (mi,j)1i,jN,where m_i,j is thenumberofoccurrencesofA_i inthewordσ(A_j).SinceM_σ is anon-negative integersquarematrix,bythefamous Frobenius–Perrontheorem(see forexample[15]), Mσhasarealeigenvalueα,calledthedominatingeigenvalueofMσ,whichisanalgebraic integerandgreaterthanorequaltothemodulusofanyothereigenvalue,thusaPerron number. If there exists an integer i (1 i N) such that σ(A_i) = A_ix for some x∈A^∗\ {∅},andlim_n→∞|σⁿ(Ai)|= +∞,thenσissaidtobeprolongableonAi.Since forallintegersn0,σⁿ(Ai) isaprefixofσⁿ⁺¹(Ai),and|σⁿ(Ai)|tendstoinfinitywith n→ ∞,thusthesequence (σⁿ(A_i))_n0 converges, andwedenote itslimitsbyσ^∞(A_i).

Thelatter is theuniqueinfinite fixed point ofσ starting withAi.Let o be amapping definedonA,extendedpointwiselyoverA^∗∪A^N.Weputv=o(σ^∞(Ai)),andcallitan α-substitutivesequence.

We havethe following important characterization for automatic sequences interms ofsubstitutivesequences, duetoCobham[11]:asequencev= (v(n))n1 isk-automatic

if andonlyifv isasubstitutivesequence withσ suchthat|σ(Aj)|=k,for allintegers 1jN.Notethatinthiscasev isk-substitutive.

Now letα, β be two multiplicativelyindependent Perron numbers.By generalizing another classicaltheorem ofCobham[10],Durandhasfinallyshownin[12, Theorem 1, p. 1801] theremarkableresultthatasequenceisbothα-substitutiveandβ-substitutive ifandonlyifitisultimatelyperiodic.

Wecannowstateandprovethefollowing theorem.

Theorem4.The algebraicequation X⁴+X²−T X+ 1= 0 hasauniquerootαinF(3).

Letα= [0,a₁,a₂,. . . ,a_n,. . .]beitscontinuedfraction expansionandu(n)be theleading coefficient of an forallintegers n1.The sequence W = (u(n))n1 isthelimit of the sequence(Wn)n0 offinitewordsoverthealphabet{1,2},definedrecursivelyasfollows:

W₀=∅, W₁= 1, andW_n =W_n−1,2, W_n−2,2, W_n−1, for all integers n2, (3) where commas indicatehere concatenationofwords.Then αisnothyperquadratic, and thesequenceW = (u(n))n1 is (1+√

2)-substitutivebutnot automatic.

Proof. TheexistenceinF(p) of theroot ofthequartic equationstated inthistheorem was observed firstly byMills and Robbins in[21], for all primenumbersp. For p= 3, inthesamework,aconjecture onitscontinued fraction expansion,basedoncomputer observation,wasgiven.BuckandRobbinsestablishedthisconjecturein[7].Shortlyafter anotherproofofthisconjecturewasgivenin[16].Wehaveα= [0,a1,a2,. . . ,an,. . .] and thesequenceofpolynomials(a_n)_n1isobtainedasthelimitofasequenceoffinitewords (Ω_n)_n0 withlettersinF3[T], definedby:

Ω₀=∅, Ω₁=T, and Ω_n= Ω_n−1,2T,Ω⁽³⁾_n−2,2T,Ω_n−1, (4) wherecommasindicateconcatenationofwords,andΩ⁽³⁾_n−2denotes thewordobtainedby cubing each letterof Ωn−2. Sincex³ =xfor allxinF^∗3, we obtainimmediately forW thedesiredformulas(3)from (4).

Thefactthatαisnothyperquadraticwasprovedin[16](seetheremarkafterTheo- rem A,p. 209).Indeedtheknowledgeofthecontinuedfraction allowstoshowthatthe irrationality measure is equalto2.Howeverthe sequenceof partial quotientsisclearly unbounded,anditwasprovedbyVoloch[25]thatifαwerehyperquadraticwithanun- bounded sequence ofpartial quotients,then theirrationality measurewouldbe strictly greater than 2 (the reader may consult [17, pp. 215–216], for a presentation of these generalstatements).

Now weshowthatW is(1+√

2)-substitutive,butnotautomatic.

Put A={a,b,c},and define

σ(a) =abca, σ(b) =ca, σ(c) =c, o(a) = 1, o(b) =o(c) = 2.

Forallintegersn0,setVn=σⁿ(a).Thenforallintegersn2,wehave V_n =σⁿ(a) =σⁿ⁻¹(abca) =V_n−1σⁿ⁻²(ca)cV_n−1=V_n−1cV_n−2cV_n−1.

ButwealsohaveW₁=o(a),andW₂= 1221=o(abca)=o(σ(a))=o(V₁),thuso(σⁿ(a)) satisfiesthesamerelationsasW_n+1,consequentlytheycoincide.Setv= lim_n→∞σⁿ(a).

Thenσ(v)=v, andW =o(v).SoW issubstitutive.Finally

M_σ =

⎛

⎜⎝

2 1 0 1 0 0 1 1 1

⎞

⎟⎠,

whosecharacteristicpolynomialisequalto(λ−1)(λ²−2λ−1),and1+√

2 isthedom- inating eigenvalue.Hence W is (1+√

2)-substitutive. Since 1+√

2 ismultiplicatively independent with all integers k 2, according to Cobham’s characterization and Du- rand’s theorem,we see thatW cannot be k-automatic unless it isultimately periodic.

Toconcludetheproof,weneedonlyprovethatW is notultimatelyperiodic.Todoso, we compute thefrequency of 2 in W. For all integersn 0, putl_n = |W_n|. Then we have

l₀= 0, l₁= 1, andl_n = 2l_n−1+l_n−2+ 2, for all integersn2, fromwhichwededucel_n =−1+²⁺₄^√2(1+√

2)ⁿ+²⁻₄^√2(1−√

2)ⁿ,forallintegersn0.

Forallintegersn0,letm_n bethenumberofoccurrences of2 inW_n.Then m₀=m₁= 0, andm_n= 2m_n−1+m_n−2+ 2, for all integersn2, from whichweobtainmn =−1+¹₂

(1+√

2)ⁿ+ ((1−√ 2)ⁿ

,forallintegersn0.If W wereultimatelyperiodic,thenthefrequencyof2 inW wouldexist,anditwouldbe arationalnumber,incontradictiontolim_n→∞m_n/l_n= 2−√

2. 2 Acknowledgments

Jia-Yan Yaowould like to thankthe National NaturalScienceFoundation of China (Grantsno. 10990012and11371210)andtheMorningsideCenterofMathematics(CAS) for partial financial support. Finally we would like to thank heartily the anonymous refereesforpertinentcommentsand valuablesuggestions.

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