The sum of digits functions of the Zeckendorf and the base phi expansions

The sum of digits functions of the Zeckendorf and the base phi expansions

Michel Dekking, F.

DOI

10.1016/j.tcs.2021.01.011

Publication date

2021

Document Version

Final published version

Published in

Theoretical Computer Science

Citation (APA)

Michel Dekking, F. (2021). The sum of digits functions of the Zeckendorf and the base phi expansions.

Theoretical Computer Science, 859, 70-79. https://doi.org/10.1016/j.tcs.2021.01.011

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Theoretical Computer Science 859 (2021) 70–79

Contents lists available atScienceDirect

Theoretical

Computer

Science

www.elsevier.com/locate/tcs

The

sum

of

digits

functions

of

the

Zeckendorf

and

the

base

phi

expansions

F. Michel Dekking

DIAM,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:

Received6August2020

Receivedinrevisedform23November2020 Accepted5January2021

Availableonline8January2021 CommunicatedbyM.Sciortino Keywords: Zeckendorfexpansion Basephi Wythoffsequence Fibonacciword

GeneralizedBeattysequence

We consider the sum of digits functions for both base phi, and for the Zeckendorf expansionofthenaturalnumbers.Forbothsumofdigitsfunctionswepresentmorphisms oninfinitealphabetssuchthatthesefunctionsviewedasinfinitewordsareletter-to-letter projectionsoffixedpointsofthesemorphisms.Wecharacterizethefirstdifferencesofboth functionsa)withgeneralizedBeattysequences,orunionsofgeneralizedBeattysequences, andb)withmorphicsequences.

©2021TheAuthor(s).PublishedbyElsevierB.V.Thisisanopenaccessarticleunderthe CCBYlicense(http://creativecommons.org/licenses/by/4.0/).

1. Introduction

PerhapsthemostfamouswordinlanguagetheoryistheThue-Morseword s2

:=

0110100110010110

. . . ,

fixed point ofthe morphism

μ

:

0

→

01

,

1

→

10.Here a morphismis a map frominfinitewords toinfinite wordsthat preservesthe concatenationoperationonthe setofwords.The remarkablepropertyof s2 is thatit canalsobe obtained frombinaryexpansions:s2

(

N

)

givestheparityofthesumofdigitsinthebinary expansionofthenaturalnumberN.The sumofdigitsinbase2writtenasaninfinitewordequals

sTM

:=

0112122312232334

, . . . ,

andsTMisfixedpointofthemorphism j

→

j

,

j

+

1 ontheinfinitealphabet

{

0

,

1

,

2

,

. . .

}

.Thereasonforthisissimple:the number2N hasthesamenumberofdigitsasN,and2N

+

1 hasonemoredigitthanN.

Thequestionarises:dosimilarconnectionstolanguagetheoryholdforexpansionsinotherbases? Forexpansionsinintegerbasesb,b anaturalnumber,itisnothardtoestablishthattheanswerispositive.

Inthispaperweconsiderthecasewherethepowersof2arereplacedbytheFibonaccinumbers(theZeckendorfexpansion), respectivelythepowersofthegoldenmean

ϕ

= (

1

+

√

5

)/

2 (thebasephiexpansion).

Forbothexpansionswegive inTheorem3,respectivelyTheorem11a morphismonan infinitealphabet,suchthat the sumofdigits functionsof theseexpansionsconsideredasinfinitewords are letter-to-letterprojections offixedpoints of

E-mailaddress:F.M.Dekking@TUDelft.nl.

https://doi.org/10.1016/j.tcs.2021.01.011

0304-3975/©2021TheAuthor(s).PublishedbyElsevierB.V.ThisisanopenaccessarticleundertheCCBYlicense (http://creativecommons.org/licenses/by/4.0/).

these morphisms. This means that the sumof digits functionsare morphic words, defined in generalas letter-to-letter projectionsoffixedpointsofmorphisms.

We then willshow how theseresults permit to give precise informationon thefirst differencesofthe sumof digits functions.Thefirstdifferences ofafunction f

: N

0

→ N

0aregivenbythefunction



f definedby



f

(

N

)

=

f

(

N

+

1

)

−

f

(

N

),

for N

=

0

,

1

,

2

. . . .

Weshallfocusonthesignsof



f .AnumberN iscalledapointofincrease ofafunction f

: N

0

→ N

0 if



f

(

N

)

>

0.Itis calledapointofconstancy if



f

(

N

)

=

0,andapointofdecrease if



f

(

N

)

<

0.

For basetwo it is simple to seethat the points ofincrease of the sumof digits function sTM are given by the even numbers,andthatthepointsofconstancyanddecreasearegivenbythenumbers1 mod 4,respectively3 mod 4.

Wewillprove(Theorem4andTheorem12)forboththeZeckendorfrepresentationandthebasephiexpansionthatthe pointsofincrease,constancyanddecreaseareallgivenbyunionsofgeneralizedBeattysequences,asstudiedin[1].These aresequences V ofthetype

V

(

n

)

=

p



n

α

 +

q n

+

r

,

n

≥

1

,

where

α

isarealnumber,andp

,

q,andr areintegers.Herewedenotedthefloorfunctionby

·

.

Wewillalsoprovethatthefirstdifferencesofthesequencesofpointsofincrease,constancyanddecreaseareallmorphic sequences.SeeTheorem5fortheZeckendorfrepresentation,andTheorem13forthebasephiexpansion.

Aprominentrole inthispaper, bothforbasephiandthe Zeckendorfexpansion, isplayed by

(



n

ϕ

)

,the well known lowerWythoffsequence.

A standard result(see, e.g., [14]) isthat the sequence

(



n

ϕ

)

is equalto the Fibonacciword x1,2

=

1211212112

. . .

onthealphabet

{

1

,

2

}

,i.e., theuniquefixedpointofthemorphism1

→

12

,

2

→

1.Moregenerally,wehavethefollowing simplelemma.

Lemma 1. ([1]) Let V

= (

V

(

n

))

n≥1bethegeneralizedBeattysequencedefinedbyV

(

n

)

=

p



n

ϕ



+

qn

+

r,andlet



V bethesequence

ofitsfirstdifferences.Then



V istheFibonacciwordonthealphabet

{

2p

+

q

,

p

+

q

}

.Conversely,ifxa,bistheFibonacciwordonthe

alphabet

{

a

,

b

}

,thenanyV with



V

=

xa,bisageneralizedBeattysequenceV

= ((

a

−

b

)



n

ϕ

)

+ (

2b

−

a

)

n

+

r

)

forsomeintegerr.

Let A

(

n

)

= 

n

ϕ



,andB

(

n

)

= 

n

ϕ

2

_

_{.Itiswellknownthat A and B formapairofBeattysequences,i.e.,theyaredisjoint} withunion

N

.Inthenextlemma,V A isthecompositiongivenbyV A

(

n

)

=

V

(

A

(

n

))

.

Lemma 2. ([1]) Let V beageneralizedBeattysequencegivenbyV

(

n

)

=

p



n

ϕ



+

qn

+

r,n

≥

1.ThenV A andV B aregeneralized Beattysequenceswithparameters

(

pV A

,

qV A

,

rV A

)

= (

p

+

q

,

p

,

r

−

p

)

and

(

pV B

,

qV B

,

rV B

)

= (

2p

+

q

,

p

+

q

,

r

)

.

2. The Zeckendorf sum of digits function

Let F0

=

0

,

F1

=

1

,

F2

=

1

,

. . .

betheFibonaccinumbers.Ignoringleadingandtrailingzeros,anynaturalnumberN can bewrittenuniquelywithdigitsdi

=

0 or1,as

N

=



i≥0 diFi+2

,

wheredidi+1

=

11 isnotallowed.WedenotetheZeckendorfexpansionofN asZ

(

N

)

,withdigitsdi

(

N

)

.

LetsZbethesumofdigitsofsuchanexpansion:forN

≥

0 sZ

(

N

)

=



i≥0 di

(

N

).

Wehave

(

sZ

(

N

))

= (

0

,

1

,

1

,

1

,

2

,

1

,

2

,

2

,

1

,

2

,

2

,

2

,

3

,

1

,

2

,

2

,

2

,

3

,

2

,

3

,

3

,

1

,

2

,

2

,

2

, . . . )

Ourfirst resultis that sZ isa morphic sequence.Thealphabet willconsist ofsymbols



j 0



and



j 1



. Notethatin the theorem



j 0

 

j+1 1



isawordoflength2overthisalphabet.

Theorem 3. ThefunctionsZ,asasequence,isamorphicsequenceonaninfinitealphabet,i.e.,

(

sZ

(

N

))

isalettertoletterprojectionof

F. Michel Dekking Theoretical Computer Science 859 (2021) 70–79

τ

(



j 0



)

=



j 0

 

j+1 1



,

τ

(



j 1



)

=



j 0



.

Theletter-to-lettermapisgivenbytheprojectiononthefirstcoordinate:



j i



→

j fori

=

0

,

1.Thefixedpointxτ of

τ

withinitial symbol



0

0



projectedonthefirstcoordinateequals

(

sZ

(

N

))

.

Proof. See theCommentsofsequenceA007895in[15] foraproofofthis.



LetIZ

,

CZandDZ bethefunctionslistingthepointsofincrease,constancy,anddecreaseofthefunctionsZ.Wehave1 IZ

= (

0

,

3

,

5

,

8

,

11

,

13

,

16

, . . . ),

CZ

= (

1

,

2

,

6

,

9

,

10

,

14

, . . . ),

DZ

= (

4

,

7

,

12

,

17

,

20

,

25

, . . . ).

TostateourresultsitisactuallyconvenienttodefineDZ

= (−

1

,

4

,

7

,

12

,

17

,

20

,

25

,

. . . )

.

When

(

an

)

and

(

bn

)

are two increasing sequences, indexed by

N

, then we mean by the union of

(

an

)

and

(

bn

)

the

increasingsequencewhosetermsgothroughtheset

{

an

,

bn

:

n

∈ N}

.

Theorem 4. ThefunctionIZ,thepointsofincreaseofthefunctionsZ,isgivenforn

=

1

,

2

,

. . .

by IZ

(

n

)

= 

n

ϕ

 +

n

−

2

.

ThefunctionCZ,thepointsofconstancyofthefunctionsZ,isgivenforn

=

1

,

2

,

. . .

bytheunionofthetwogeneralizedBeattysequences

withterms

2



n

ϕ

 +

n

−

2 and 3



n

ϕ

 +

2n

−

3

.

ThefunctionDZ,thepointsofdecreaseofthefunctionsZ,isgivenforn

=

1

,

2

. . .

by DZ

(

n

)

=

2



n

ϕ

 +

n

−

4

.

Proof. Let IZ bethesequenceofthepointsofincreaseofthefunctionsZ.

Projectiononthesecondcoordinateof

τ

yieldstheFibonaccimorphism

σ

Fgivenby

σ

F

(

0

)

=

01

,

σ

F

(

1

)

=

0

.

Thus thesecond coordinatesof thefixedpoint of

τ

equaltheinfiniteFibonacciword x0,1 =0100101001001....Obviously, theincrease points ofsZ occurifandonlyiftheword

(

j

,

0

)

(

j

+

1

,

1

)

occursinthe fixedpoint xτ of

τ

ifandonlyifthe word 01occursin x0,1. Since11 doesnot occur inx0,1,this meansthat we haveto shiftthe positionsof 1’sin x0,1 by 1.Itis wellknownthat thepositionsof1are givenbytheupperWythoff sequence

(

n

ϕ

2

_)

_{= (}

_n

_ϕ

_

₊

_n

₎

_{.Sincethe first} coordinateofthefixedpointof

τ

startsfromindex0,andthesecondfromindex1,wehavetoreplacen byn

+

1,andthis yieldsthefirstresultofTheorem4.

Thepointsofconstancyaremoredifficulttocharacterizewiththefixedpointxτ thanthepointsofincrease.We there-foretakeanotherapproach.Write Z

(

N

)

= . . .

w,wherew isawordoflength4.Then w canbeanywordofthe0-1-words oflength4containingno11.Obviously,thethreewords w

=

0000

,

w

=

0100 and w

=

1000 givepointsofincrease.

FurthermorethenumbersN with Z

(

N

)

endinginw

=

0001

,

1001 andw

=

0010 give

Z

(

N

)

= . . .

001

⇒

Z

(

N

+

1

)

= . . .

.

002

= . . .

.

010

,

Z

(

N

)

= . . .

0010

⇒

Z

(

N

+

1

)

= . . .

.

0011

= . . .

.

0100

.

Weseethatthesegivepointsofconstancy.

Finally,we show thatthe N with Z

(

N

)

having suffix w

=

0101 or w

=

1010 givepoints ofdecrease.Inthe following twocomputationsthe

=

.

-signindicatesthatweusealsonon-admissibleZeckendorfrepresentations.

Z

(

N

)

= . . .

0101

⇒

Z

(

N

+

1

)

= . . .

.

0102

= . . .

.

0110

= . . .

.

1000

,

Z

(

N

)

= . . .

01010

⇒

Z

(

N

+

1

)

= . . .

.

01011

= . . .

.

01100

= . . .

.

10000

.

Inbothcasesatleastonedigit1islost,sotheseN arethepointsofdecrease.

Withthisknowledge we canapply Theorem2.3 andProposition 2.8inthe paper[9],obtaining that one partof IZ is givenbythegeneralizedBeattysequence

(

2



n

ϕ



+

n

−

2

)

andtheotherpartisgivenby

(

3



n

ϕ



+

2n

−

3

)

.

1 _I

ZisthesequenceA026274in[15].

Again from Theorem 2.3 and Proposition 2.8 in the paper [9], we obtain that

(

DZ

(

n

+

1

))

is the union of the two generalizedBeattysequences

(

3



n

ϕ



+

2n

−

1

)

and

(

5



n

ϕ



+

3n

−

1

)

.

ItisnotasimplemattertoseethatthisunionisgivenbythesinglegeneralizedBeattysequence

(

2



n

ϕ



+

n

−

4

)

,where theindexstartsatn

=

2.

Let us write V

(

p

,

q

,

r

)

= (

p



n

ϕ



+

qn

+

r

)

n≥1. We have proved so far that IZ

=

V

(

1

,

1

,

−

2

)

, and CZ is the union of

V

(

2

,

1

,

−

2

)

andV

(

3

,

2

,

−

3

)

.Ifweadd2toalltermsofthesesequences,weobtainthethreesequencesV

(

1

,

1

,

0

)

,V

(

2

,

1

,

0

)

, andV

(

3

,

2

,

−

1

)

.

Thetripleofsequences

{

V

(

1

,

1

,

0

),

V

(

2

,

1

,

0

),

V

(

1

,

1

,

−

1

)

}

isknownasthe‘firstclassicalcomplementarytriple’,i.e.,thesearethreedisjointsequenceswithunion

N

.Seepage334in [1].Thethirdsequenceofthistriple,V

(

1

,

1

,

−

1

)

,canbewrittenasadisjointunionofthetwosequences V

(

3

,

2

,

−

1

)

and

V

(

2

,

1

,

−

2

)

,byLemma2.Thus

{

V

(

1

,

1

,

0

),

V

(

2

,

1

,

0

),

V

(

3

,

2

,

−

1

),

V

(

2

,

1

,

−

2

)

}

formsacomplementaryquadruple.Ifwesubtract2fromalltermsofthesefoursequences,thefirstgivesIZ,thesecondand thethirdtogether,CZ.Since

{

IZ

,

CZ

,

DZ

}

isacomplementarytriple,withunion

{−

1

,

0

,

1

,

2

,

. . .

}

thisimpliesthat

(

DZ

(

n

+

1

))

hastobeequalto V

(

2

,

1

,

−

4

)

.



Next,wegiveacharacterizationofIZ

,

CZandDZ intermsofmorphisms.

Theorem 5. ThepointsofincreaseofthefunctionsZaregivenbythesequenceIZ,whichhasIZ

(

1

)

=

0,and



IZisthefixedpointof

theFibonaccimorphism3

→

32

,

2

→

3.

ThepointsofconstancyofthefunctionsZaregivenbythesequenceCZ,whichhasCZ

(

1

)

=

1,and



CZisthefixedpointofthe

2-blockFibonaccimorphismonthealphabet

{

1

,

4

,

3

}

givenby1

→

14

,

3

→

14

,

4

→

3.

ThepointsofdecreaseofthefunctionsZaregivenbythesequenceDZ,whichhasDZ

(

1

)

= −

1,and



DZisthefixedpointofthe

Fibonaccimorphism5

→

53

,

3

→

5.

FortheproofofTheorem5wehavetomakesomepreparations.Let



3

:= {

2

},



3

:= {

0

,

1

}

=: [

0

,

1

]

,anddefineforn

≥

4 theintervalsofintegers



n and



n by



n

:= [

Fn

,

Fn+1

−

1

], 

n

:= [

0

,

Fn

−

1

].

The

(

n

)

formapartitionof

N

0

\ {

0

,

1

}

,andthe

(

n

)

satisfy



n+1

= 

n

∪ 

n

.

(1)

ForanintervalI,letCZ

(

I

)

denotethepoints ofincreaselyingintheinterval I.Also,let



CZ

(

I

)

denotethefirstdifferences ofthepoints ofincrease lyinginthe interval I,consideredasa wordon thealphabet

{

1

,

2

,

3

,

4

}

.Atfirstsight,the latter definitionisproblematic,asonehastoknowthefirstpointofincreaseafterthelast elementofCZ

(

I

)

.However,we shall onlyconsiderintervalsI

= 

n andI

= 

n+1,whichbotharefollowedby



n+1,andoneverifieseasilythatthefirstpoint ofincreasein



n+1 isalwaysthesecondpoint.Actually,thisfollowsdirectlyfromthefollowinglemma.

Lemma 6. Foralln

≥

3 onehasCZ

(

n+1

)

=

CZ

(

n

)

+

Fn+1.

Proof. We used the notation A

+

y

= {

x

+

y

:

x

∈

A

}

for a set A, anda number y. The lemma follows from the basic Zeckendorfrecursion:thenumbersN in



n+1 allhaveadigit1addedtotheexpansionofthenumberN

−

Fn+1.



Leth bethemorphismonthealphabet

{

1

,

3

,

4

}

givenby h

(

1

)

=

14

,

h

(

3

)

=

14

,

h

(

4

)

=

3

.

Proposition 7. Foralln

≥

5 onehas (i)



CZ

(

n

)

=

hn−4

(

3

)

(

ii

)



CZ

(

n

)

=

hn−5

(

3

)

.

Proof. The proof isby induction. Forn

=

5,we have



5

= [

0

,

4

]

,which hastwo points of constancy: N

=

1 and N

=

2. ThereforeCZ

(

5

)

=

14

=

h

(

3

)

.Herethedifference4iscomingfromN

=

6,thesecondpointoftheinterval



5.Wefurther have



5

= [

5

,

7

]

,whichhasonepointofconstancyN

=

6.ThereforeCZ

(

5

)

=

3.

F. Michel Dekking Theoretical Computer Science 859 (2021) 70–79

(i)Byequation(1),



CZ

(

n+1

)

= 

CZ

(

n

)

CZ

(

n

)

=

hn−4

(

3

)

hn−5

(

3

)

=

hn−5

(

h

(

3

)

3

)

=

hn−5

(

143

)

=

hn−5

(

h2

(

3

))

=

hn−3

(

3

).

(ii)DirectlyfromLemma6:



CZ

(

n+1

)

= 

CZ

(

n

)

=

hn−4

(

3

)

.



Proof of Theorem5. The statementsonIZandDZfollowimmediatelyfromLemma1.

ThestatementonCZ followsfromProposition7,part(i),sincehn

(

3

)

=

hn

(

1

)

foralln

>

0.



3. The base phi expansion

AnaturalnumberN iswritteninbasephi([2])ifN hastheform N

=

∞



i=−∞

di

ϕ

i

,

withdigits di

=

0 or1,andwheredidi+1

=

11 isnotallowed. Ignoring leadingandtrailing0’s,thesumisactually finite, andthebasephirepresentationofanumberN isunique([2]).

Wewritetheseexpansionsas

β(

N

)

=

dLdL−1

. . .

d1d0

·

d−1d−2

. . .

dR+1dR

.

LetforN

≥

0 sβ

(

N

)

:=

k=R



k=L dk

(

N

)

bethesumofdigitsfunctionofthebasephiexpansions.Wehave

(

sβ

(

N

))

= (

0

,

1

,

2

,

2

,

3

,

3

,

3

,

2

,

3

,

4

,

4

,

5

,

4

,

4

,

4

,

5

,

4

,

4

,

2

,

3

,

4

,

4

,

5

,

5

,

5

,

4

,

5

,

6

,

6

,

7

,

5

,

5

,

5

,

6

, . . . ).

Thecaseofbasephiisconsiderablymorecomplicatedthan theZeckendorfcase. Weneedseveralpreparations,before wecanproveTheorem11inSection3.2,Theorem12inSection3.3andTheorem13inSection3.4.

3.1. Therecursivestructuretheorem

Theresultofthissectionwasanticipatedin[10],[11],and[16],andprovedin[8]. TheLucasnumbers

(

Ln

)

= (

2

,

1

,

3

,

4

,

7

,

11

,

18

,

29

,

47

,

76

,

123

,

. . . )

aredefinedby

L0

=

2

,

L1

=

1

,

Ln

=

Ln−1

+

Ln−2 for n

≥

2

.

Forn

≥

2 weareinterestedinthreeconsecutiveintervalsgivenby In

:= [

L2n+1

+

1

,

L2n+1

+

L2n−2

−

1

],

Jn

:= [

L2n+1

+

L2n−2

,

L2n+1

+

L2n−1

],

Kn

:= [

L2n+1

+

L2n−1

+

1

,

L2n+2

−

1

].

To formulate the next theorem,it is notationally convenient to extendthe semigroup of words to the free group of words.Forexample,onehas110−101−100

=

100.

Theorem 8. [Recursive Structure Theorem] I For all n

≥

1 and k

=

1

,

. . . ,

L2n−1 one has

β(

L2n

+

k

)

= β(

L2n

)

+ β(

k

)

=

10

. . .

0

β(

k

)

0

. . .

01.

IIForalln

≥

2 andk

=

1

,

. . . ,

L2n−2

−

1

In

: β(

L2n+1

+

k

)

=

1000

(

10

)

−1

β(

L2n−1

+

k

)(

01

)

−11001

,

Kn

: β(

L2n+1

+

L2n−1

+

k

)

=

1010

(

10

)

−1

β(

L2n−1

+

k

)(

01

)

−10001

.

Moreover,foralln

≥

2 andk

=

0

,

. . . ,

L2n−3

Jn

: β(

L2n+1

+

L2n−2

+

k

)

=

10010

(

10

)

−1

β(

L2n−2

+

k

)(

01

)

−1001001

.

74

Itis crucialtoouranalysisto partitionthenaturalnumbersinwhatwe calltheLucas intervals,givenby



0

:= [

0

,

1

]

, andforn

=

1

,

2

. . .

by



2n

:= [

L2n

,

L2n+1

], 

2n+1

:= [

L2n+1

+

1

,

L2n+2

−

1

].

IfI

= [

k

,

]

and J

= [

+

1

,

m

]

aretwoadjacentintervalsofintegers,thenwewriteI J

= [

k

,

m

]

.

WecodetheLucasintervalswithfoursymbols



0,



1,



2 and



3 (forextrareadabilitythesesymbolsareincolorinthe webversionofthisarticle),byacode



inthefollowingway:

(

0

)

=



0

,(

1

)

=



1

,(

2

)

=



2

,(

3

)

=



3

.

We thencode

(

4

)

= (

0

)(

1

)(

2

)

=



0



1



2,

(

5

)

= (

3

)(

2

)(

3

)

=



3



2



3,andingeneralby induc-tion,suggestedbyTheorem8:

(

2n+2

)

= (

0

)(

1

)(

2

) . . .

(

2n

),

(

2n+1

)

= (

2n−1

)(

2n−2

)(

2n−1

).

Let

σ

bethemorphismonthealphabet

{



0,



1,



2,



3

}

definedby

σ

(



0

)

=



0



1

,

σ

(



1

)

=



2



3

,

σ

(



2

)

=



0



1



2

,

σ

(



3

)

=



3



2



3

.

Lemma 9. Foreachn

≥

0 wehave

(

2n+2

)

=

σ

n

(



2

)

,

(

2n+3

)

=

σ

n

(



3

)

.

Proof. By induction.Forn

=

0:

(

2

)

=



2,

(

3

)

=



3.Theinductionstep:



(



2n+5)

= 

(



2n+3)



(



2n+2)



(



2n+3)

=

σ

n(



3)

σ

n₍



₂₎

_σ

n₍



₃₎

₌

_σ

n+1₍



_3). Also,usingthesimpleidentity

σ

(



2)



3

σ

(



2)

=

σ

2₍



_{2)inthelaststep:}

(

2n+4

)

= (

0

)(

1

)(

2

)

. . .(

2n

)(

2n+1

)(

2n+2

)

= (

2n+2

)(

2n+1

)(

2n+2

)

=

σ

n₍



₂₎

_σ

n−1₍



₃₎

_σ

n₍



₂₎

₌

_σ

n+1₍



₂₎

_

Wewillnowshowthatthefixedpointxσ ofthemorphism

σ

isquasi-Sturmian,anddetermineitscomplexityfunction

pσ ,i.e., pσ

(

n

)

isthenumberofwordsoflengthn thatoccursinxσ .Letga,b themorphismonthealphabet

{

a

,

b

}

givenby

ga,b

(

a

)

=

baa

,

ga,b

(

b

)

=

ba

.

(2)

Themorphismga,biswell-known,andcloselyrelatedtotheFibonaccimorphism.Infact,xg

=

bxa,b,ifxg isthefixedpoint

of ga,b,andxa,bisthefixedpointoftheFibonaccimorphisma

→

ab

,

b

→

a (see[3]).

Proposition 10. Thefixedpointxσ of

σ

isequaltothedecoration

δ(

xg

)

ofthefixedpointxgofg

=

ga,b.Thedecorationmorphism

δ

isgivenby

δ(

a

)

=



2



3,

δ(

b

)

=



0



1.Foralln

≥

1 onehaspσ

(

n

)

=

n

+

3. Proof. For thetwowords



0



1 and



2



3 occurringinxσ wefind

σ

(



0



1

)

=



0



1



2



3

,

σ

(



2



3

)

=



0



1



2



3



2



3

.

Inotherwords,

σ

(δ(

a

))

= δ(

baa

)

= δ(

g

(

a

)),

σ

(δ(

b

))

= δ(

ba

)

= δ(

g

(

b

)).

Thus

σ

δ

= δ

g,whichimplies

σ

n

_δ

_{= δ}

_gn _{foralln.Since}

xσ hasprefix



0



1

= δ(

b

)

,withb theprefixofxg,thisimpliesthe

firstpartoftheproposition.

For the second part, Proposition 8 in [4] is not conclusive, as we do not know a priori the constant n0. But there is a direct computation possible. The complexity function of the Sturmian word xg is given by p

(

n

)

=

n

+

1. We have,

distinguishing betweenwords ofeven andodd length, andthen splitting accordingto words occurring ateven orodd positionsinxσ ,

pσ

(

2n

)

=

p

(

n

)

+

p

(

n

+

1

)

=

n

+

1

+

n

+

2

=

2n

+

3

,

pσ

(

2n

+

1

)

=

p

(

n

+

1

)

+

p

(

n

+

1

)

=

2n

+

4

.



Proposition 10 incombination withthe main resultof the paper[13], explains whythe factors of xσ have a simple returnwordstructure.ThisliesatthebasisofTheorem12inSection3.3.

F. Michel Dekking Theoretical Computer Science 859 (2021) 70–79

3.2. Amorphicsequencerepresentationofsβ

Theimageunderamorphism

δ

ofthefixedpointx ofamorphism,willbecalledadecoration ofx.Itiswellknownthat sucha

δ(

x

)

isamorphicsequence,i.e.,thelettertoletterprojectionofthefixedpointofamorphism.Thisisthewaywe formulatethemorphicsequenceresultinthenexttheorem.

Theorem 11. Thefunctionsβ,asasequence,isadecorationofamorphicsequenceonaninfinitealphabet,i.e.,

(

sβ

(

N

))

isanimage

underamorphism

δ

ofafixedpointofamorphism

γ

.Thealphabetis

{

0

,

1

,

...,

j

,

...}

× {



0

,



1

,



2,



3

}

,and

γ

isthemorphismgiven forj

≥

0 by

γ

(



_j



0



)

=



_j



0

 

_j



1



,

γ

(



_j



1



)

=



_j



2

 

_j



3



,

γ

(



_j



2



)

=



_j+2



0

 

_j+2



1

 

_j+2



2



,

γ

(



_j



3



)

=



_j+1



3

 

_j+2



2

 

_j+1



3



.

Thedecorationmapisgivenbythemorphism

δ

:

δ(



_j



0



)

=

0

+

j

,

1

+

j

,

δ(



_j



1



)

=

2

+

j

,

δ(



_j



2



)

=

2

+

j

,

3

+

j

,

δ(



_j



3



)

=

3

+

j

,

3

+

j

.

Theimage

δ(

xγ

)

ofthefixedpointxγ of

γ

withinitialsymbol



₀



0



equals

(

sβ

(

N

))

.

Proof. One combinesTheorem8withLemma9.Weseefrompart I ofTheorem8,thatthenumberof1’sintheexpansion of N from



2n+2 is2more thanthe numberof1’s inthecorresponding N in



0



1

. . .

2n. Thisgivesthethree upper indices j

+

2 in

γ

(



_j



2



)

.Similarly,part II givesthatthenumberof1’sinthethree intervals



2n−1,



2n−2,and



2n−1 is increasedby1,by2,andrespectively1forthecorresponding Nintheinterval



2n+1.Thisgivesthethreeupperindices in

γ

(



_j



3



)

.The lower indices aregivenby the morphism

σ

.Thisall happensatthe levelof theshifted versionsofthe fourintervals



0

,



1

,



2and



3.Here



0

= [

0

,

1

]

withsβ

(

0

)

=

0andsβ

(

1

)

=

1;



1

= {

2

}

withsβ

(

2

)

=

2;



2

= [

3

,

4

]

with

sβ

(

3

)

=

2andsβ

(

4

)

=

3;



3

= [

5

,

6

]

withsβ

(

5

)

=

3andsβ

(

6

)

=

3.Thisyields thedecorations

δ

,takingintoaccountthe

correspondingincrementsofthesumofdigits.



WeillustrateTheorem11withthefollowingtable.

N 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

sβ

(

N

)

0 1 2 2 3 3 3 2 3 4 4 5 4 4 4 5 4 4

Lucas interval



0



1



2



3



4



5

shifted Lucas intervals



0



1



2



3



0



1



2



3



2



3



0,



1,



2,



3-coding



0



1



2



3



0



1



2



3



2



3

Remark. Inthepaper[7] the basephianalogueoftheThue-Morsesequence,i.e., thesequence

(

sβ

(

N

)

mod 2

)

,isshown

tobeamorphicsequence.ThisresultfollowsalsofromTheorem11,bymapping2 j to0,and2 j

+

1 to1.Themorphisms foundinthiswayareonalargeralphabetthanthemorphismin[7].

3.3. GeneralizedBeattysequencesforsβ

LetIβ bethesequencelistingthepointsofincreaseofsβ

(

N

)

.Weseethatthefirstsixpointsofincreaseare Iβ

(

1

)

=

0,

Iβ

(

2

)

=

1, Iβ

(

3

)

=

3,Iβ

(

4

)

=

7, Iβ

(

5

)

=

8,Iβ

(

6

)

=

10.SimilarlywedefineCβ andDβ.

Theorem 12. ThesequenceIβ,thepointsofincreaseofthefunctionsβ,isgivenbytheunionofthetwogeneralizedBeattysequences

(



n

ϕ

 +

2n

)

n≥0

,

and

(

4



n

ϕ

 +

3n

+

1

)

n≥0

.

ThesequenceCβ,thepointsofconstancyofthefunctionsβ,isgivenbytheunionofthefourgeneralizedBeattysequences

(

3



n

ϕ

 +

n

+

1

)

n≥1

, (

4



n

ϕ

 +

3n

+

2

)

n≥0

, (

7



n

ϕ

 +

4n

+

2

)

n≥0

,

and

(

11



n

ϕ

 +

7n

+

4

)

n≥1

.

ThesequenceDβ,thepointsofdecreaseofthefunctionsβ,isgivenbytheunionofthethreegeneralizedBeattysequences

(

4



n

ϕ

 +

3n

−

1

)

n≥1

, (

7



n

ϕ

 +

4n

)

n≥1

,

and

(

7



n

ϕ

 +

4n

+

4

)

n≥1

.

Proof. I: Points of increase

Anyoccurrenceofa



0 givestwopointsofincrease,namelythepair0

+

j

,

1

+

j,andthepair1

+

j

,

2

+

j.Hereweuse that



0 isalwaysfollowedby



1.Similarly,anyoccurrenceofa



2 givesapointofincrease2

+

j

,

3

+

j.

AsaconsequenceweobtainthenumbersN whicharepointofincreasebythesequencesofoccurrencesof



0,andthose of



2.Howdoweobtainthesesequences?Wehavetostudythereturnwordsto



0,and



2.Thesetsofthesereturnwords arerespectively

{



0



1



2



3,



0



1



2



3



2



3

}

,and

{



2



3,



2



3



0



1

}

.

Both



0, and



2 induce the descendant morphism ga,b (the descendant morphism is a generalization of the derived

morphism,see[12]).Herewecodedb

:=



0



1



2



3,a

:=



0



1



2



3



2



3,respectivelyb

:=



2



3,a

:=



2



3



0



1.

Theoccurrencesof



0 inthefixedpointof

σ

occuratdistancesgivenbythelengthsof

δ(



0



1



2



3)and

δ(



0



1



2



3



2



3). Theseare

|δ(



0



1



2



3)

| =

7,and

|δ(



0



1



2



3



2



3)

| =

11.It thenfollowsfromLemma1that theincrease pointsare given bytheunionofthetwogeneralizedBeattysequences V

(

4

,

3

,

0

)

andV

(

4

,

3

,

1

)

,wheretheindicatesthatthesestartfrom index0.Similarly,theoccurrencesof



2 havefirstdifferences7and4,givingthegeneralizedBeattysequence V

(

3

,

1

,

−

1

)

.

This isnot yetthe first resultinTheorem 12,butby Lemma2 thesequence V

(

1

,

2

,

0

)

splits into the twosequences

V

(

3

,

1

,

−

1

)

andV

(

4

,

3

,

0

)

.AddingN

=

0 toV

(

1

,

2

,

0

)

andtoV

(

4

,

3

,

0

)

thenyieldstheresulton Iβ inTheorem12.

II: Points of constancy

Anyoccurrenceofa



1 givesapointofconstancy, namelythepair2

+

j

,

2

+

j. Hereweusethat



1 isalways followed by



2.Similarly,anyoccurrenceofa



3 givesapointofconstancy3

+

j

,

3

+

j.

But there are more points of constancy. At the inner boundary of



2



3 in the quadruple



0



1



2



3 occurs 3

,

3. However,thisisnotthecaseattheinnerboundaryoftheinterval



2



3inthetriple



3



2



3 in



5.Since

(

0



2



3



4

)

=



0



1

σ

(



1

)

,and

(

3



2



3

)

=

σ

(



3

)

thesepointsofconstancyoccurifandonlyif

σ

(



1

)

occursinthefixedpointof

σ

. This still doesnot yet exhaust all possibilities:there is thepoint N

=

14 with sβ

(

N

)

=

sβ

(

N

+

1

)

=

4 in



5, not yet covered bythe previous sequences.Thisinduces points ofconstancy occurringatallshifted



5,whichoccur ifandonly if

σ

(



3

)

occursinthefixedpoint of

σ

.Sinceany



k fork

>

5 canbe writtenasaunionofshiftedversionsofthethree

intervals



0



1



2



3,



4,and



5,wehavecoveredallpossibilities.

As a consequence we obtain the numbers N which are point ofincrease by the sequences of occurrences of



1,



3,

σ

(



1),and

σ

(



3

)

.As before,all fourhavea setoftwo returnwords,anda descendantmorphism thatis equalto g.For



1 the

δ

-imageshavelengths11and7,for



3 the

δ

-imageshavelengths7and4,for

σ

(



1

)

,the

δ

-imageshavelengths29 and18,andfor

σ

(



3

)

the

δ

-imageshavelengths18and11.ApplicationofLemma1thengivesthefourgeneralizedBeatty sequencesofCβ inTheorem12.

III: Points of decrease

The first point of decrease is N

=

6, which occursat the endof



3, so N

+

1

=

7 occurs atthe beginning of



4

=



0



1



2.Thisgivesoccurrencesofpointsofdecreaseateveryoccurrenceof



3



0.Thiswordhastworeturnwords:b

:=



3



0



1



2,anda

:=



3



0



1



2



3



2.These induceasdescendant morphismthe morphism g,oncemore. As

|δ(

a

)

|

=

11,and

|δ(

b

)|

=

7,thisleadstothesequenceV

(

4

,

3

,

−

1

)

.

The next point of decrease is at N

=

11, occurringat the inner boundary ofthe adjacent



4



5. The third point of decreaseisatN

=

15,whichliesinside



5.Thecodingof



5 is

(

5

)

=



3



2



3 =

σ

(



3

)

.Asintheprevioussection,this givesthesequence V

(

7

,

4

,

0

)

fortheoccurrencesofthedecreasepoints N

=

11,andlatershifts.Then V

(

7

,

4

,

4

)

givesthe occurrencesofthedecreasepointsN

=

15

=

11

+

4,andlatershifts.Again,sinceany



kfork

>

5 canbewrittenasaunion

ofintervals



0



1



2



3,



4,and



5,wehavecoveredallpossibilities.Thisfinishesthe Dβ partofTheorem12.



3.4. Morphismsforthefirstdifferences

AsfortheZeckendorfexpansion,wehaveseenintheprevioussectionthatthepointsofconstancyhaveamore compli-catedstructurethanthepointsofincreaseorthepointsofdecrease.Thisphenomenonexpressesitselfalsointhe‘morphic versions’ofthecharacterization.

Theorem 13. ThepointsofincreaseofthefunctionsβaregivenbythesequenceIβ,whichhasIβ

(

1

)

=

0,and



Iβisthefixedpointof

themorphismonthealphabet

{

1

,

2

,

4

}

givenby

1

→

12

,

2

→

4

,

4

→

1244

.

ThepointsofconstancyofthefunctionsβaregivenbythesequenceCβ,whichhasCβ

(

1

)

=

2,and



Cβisamorphicsequence,givenby

theletter-to-letterprojection1

→

1

,

2

→

2

,

3

→

3

,

3

→

3

,

4

→

4 ofthefixedpointofthemorphismonthealphabet

{

1

,

2

,

3

,

3

,

4

}

givenby

F. Michel Dekking Theoretical Computer Science 859 (2021) 70–79

ThepointsofdecreaseofthefunctionsβaregivenbythesequenceDβ,whichhasDβ

(

1

)

=

6,and



Dβistheshiftbyoneofthefixed

pointsof themorphismonthealphabet

{

2

,

4

,

5

,

7

}

givenby

2

→

542

,

4

→

542

,

5

→

7

,

7

→

7542

.

Proof. We use in all three cases the return words to



0 which are b

:=



0



1



2



3 anda

:=



0



1



2



3



2



3 to follow the occurrencesofthepointsofincrease,constancyanddecrease.Theimportantpropertyofthesereturnwordsisthatthefirst

occurrenceofthepointsofincreaseisatthesamepositioninthedecorateda andb,andthesameholdsforthepointsof constancyanddecrease.

Proof. I: Points of increase

Wetakeintoaccounttheincreaseinthedifferencesoftheoccurrencesoftheincreasepointsinthedecorations

δ(



_j



0

 

_j



1

 

_j



2

 

_j



3



)

=

0

+

j

,

1

+

j

,

2

+

j

,

2

+

j

,

3

+

j

,

3

+

j

,

3

+

j,

δ(



_j



0

 

_j



1

 

_j



2

 

_j



3

 

_j



2

 

_j



3



)

=

0

+

j

,

1

+

j

,

2

+

j

,

2

+

j

,

3

+

j

,

3

+

j

,

3

+

j

,

2

+

j

,

3

+

j

,

3

+

j

,

3

+

j,

oftheextendedreturnwordsa andb.Fora thesedifferencesare1

,

2

,

4 and4.Forb thedifferencesbetweentheoccurrences oftheincrease points are1

,

2,and4.Recall here,that thelast4 comesfromthefirstincrease pointofthenext word.It followsthat wecanobtain



Iβ bydecoratingthefixed pointofthemorphism g givenbya

→

baa

,

b

→

ba withthetwo

words124and1244.Toturnthisdecoratedfixedpointintoafixedpoint,weapplythenaturalalgorithm(cf.theproofof Corollary9in[5]).Inthiscasethisgivesthefollowingblockmaponthealphabet

{

a1

,

a2

,

a3

,

a4

,

b1

,

b2

,

b3

}

:

a1a2a3a4

→

b1b2b3a1a2a3a4a1a2a3a4 b1b2b3

→

b1b2b3a1a2a3a4

.

Themostefficientwaytoturnthisintoamorphism: a1

→

b1b2

,

a2

→

b3

,

a3

→

a1a2a3a4

,

a4

→

a1a2a3a4 b1

→

b1b2

,

b2

→

b3

,

b3

→

a1a2a3a4

.

The associated letter-to-letter map

λ

is givenby

λ(

a1a2a3a4

)

=

1244,

λ(

b1b2b3

)

=

124.We see that we can consistently mergea1 andb1totheletter1,a2 andb2 totheletter2,anda3 andb3totheletter4.Renaminga4 by4,thisthenyields themorphism1

→

12

,

2

→

4

,

4

→

1244 asgeneratingmorphismfor



Iβ.

II: Points of constancy

We follow thesame strategy asin part I. The differencesof the occurrencesof points of constancy inthe decorated versionsofa andb arenow 2

,

1

,

4 and3

,

1

,

3

,

4.Decoratingthefixed pointofthemorphism g on

{

a

,

b

}

bya

→

214,and

b

→

3134 thistimeleadstoamorphismonthealphabet

{

1

,

2

,

3

,

3

,

4

}

givenby

1

→

43

,

2

→

21

,

3

→

21

,

3

→

1343

,

4

→

134

.

The letter-to-letterprojection 1

→

1

,

2

→

2

,

3

→

3

,

3

→

3

,

4

→

4 of thefixedpoint ofthismorphismonthealphabet

{

1

,

2

,

3

,

3

,

4

}

yieldsthesequence



Cβ (whereCβ

(

1

)

=

2).

III: Points of decrease

Thedifferencesoftheoccurrencesofpointsofdecreaseinthedecoratedversionsofa andb are7and5

,

4

,

2.Decorating the fixed point ofthemorphism ga,b by a

→

7,andb

→

542 thistime leads to a morphismon the alphabet

{

2

,

4

,

5

,

7

}

givenby

2

→

542

,

4

→

542

,

5

→

7

,

7

→

7542

.

The uniquefixed pointofthismorphism onthealphabet

{

2

,

4

,

5

,

7

}

yields thesequence



Dβ,whenwe put Dβ

(

1

)

=

−

1.



4. Alternative proofs of Theorem12and 13

TheproofsofTheorem12and13havebeenbasedentirelyonthepropertiesoftheinfinitemorphism

γ

ofTheorem11. The question rises whether thereis also amore local approachbased on thedigit blocks ofthe expansion aswas used for thepoints ofconstancy, andthe points of decrease ofthe Zeckendorf sumof digits function. Here we give a sketch of howthismight be achievedforthe points ofincrease ofthe basephiexpansion. We saya number N is oftype B if

d1d0d−1

(

N

)

=

000,andoftypeE ifd2d1d0

(

N

)

=

001.Onecanthenprovethefollowing. 78

Proposition 14. AnumberN isapointofincreaseof

(

sβ

(

N

))

ifandonlyifN isoftypeB oroftypeE.

Next,Theorem 5.1fromthepaper[6] gives that type B occursalong thegeneralizedBeatty sequence

(

n

ϕ



+

2n

)

n≥0, andonecandeducefromRemark6.3inthesamepaperthattypeE occursalongthegeneralizedBeattysequence

(

4



n

ϕ



+

3n

+

1

)

n≥0.ThisgivesthealternativeproofoftheIB-partofTheorem12,basedonProposition14.

Wenextgiveaproofofthe



IB partofTheorem13,directlyfromTheorem12byapurelycombinatorialargument. Alternative proof of Theorem13. Let

IB

:= (

n

ϕ

 +

2n

)

n≥0

,

IE

:= (

4



n

ϕ

 +

3n

+

1

)

n≥0

.

By Lemma 1, the difference sequence of the sequence

(

n

ϕ



+

2n

,

n

≥

1

)

is equal to the Fibonacci word x4,3

=

4344344344

. . .

on the alphabet

{

4

,

3

}

, and the difference sequence of the sequence

(

4



n

ϕ



+

3n

+

1

,

n

≥

1

)

is the Fi-bonacciwordx11,7

=

11

,

7

,

11

,

11

,

7

,

. . .

.However,inTheorem12thesequences startatn

=

0,yieldingthetwodifference sequences



IB

=

3x4,3

=

34344344344

. . . ,



IE

=

7x11,7

=

7

,

11

,

7

,

11

,

11

,

7

, . . . .

Recallthatthesequencesbxa,barefixedpointsofthemorphisms ga,bfromEquation(2) givenbyga,b

(

a

)

=

baa,ga,b

(

b

)

=

ba.

Thereturnwordsof3in



IB are34and344.Wecodethesewordsby thedifferencesthattheyyieldbetweensuccessive occurrencesof3’s,i.e.,bytheletters7and11.Then,since

g4,3

(

34

)

=

34 344

,

g4,3

(

344

)

=

34 344 344

,

thereturnwordsinduceaderivedmorphism

7

→

7

,

11

,

11

→

7

,

11

,

11

.

Thisderivedmorphismhappenstobeequaltog11,7,themorphismgivingthesequence



IE.Thisimpliesthattomergethe twosequences IB andIE toobtain I,onehastoreplacethe3’sin



IB by1

,

2.Thisdecorationof



IB,inducesamorphism

μ

onthealphabet

{

1

,

2

,

4

}

intheusualway,givenby

μ

(

1

)

=

12

,

μ

(

2

)

=

4

,

μ

(

4

)

=

1244

.

Thisprovesthetheorem.



Declaration of competing interest

The authors declare that they haveno known competingfinancial interests or personal relationships that could have appearedtoinfluencetheworkreportedinthispaper.

References

[1] J.-P.Allouche,F.M.Dekking,GeneralizedBeattysequencesandcomplementarytriples,Mosc.J.Comb.NumberTheory8(2019)325–342,https:// doi.org/10.2140/moscow.2019.8.325.

[2]G.Bergman,Anumbersystemwithanirrationalbase,Math.Mag.31(1957)98–110.

[3]J.Berstel,P.Séébold,AremarkonmorphicSturmianwords,RAIROTheor.Inform.Appl.28(1994)255–263.

[4]J.Cassaigne,Sequenceswithgroupedfactors,in:DLT’97,DevelopmentsinLanguageTheoryIII,Thessaloniki,AristotleUniversityofThessaloniki,1998, pp. 211–222.

[5] F.M.Dekking,Morphicwords,BeattysequencesandintegerimagesoftheFibonaccilanguage,Theor.Comput.Sci.809(2019)407–417,https://doi.org/ 10.1016/j.tcs.2019.12.036.

[6]F.M.Dekking,Basephirepresentationsandgoldenmeanbeta-expansions,FibonacciQ.58(2020)38–48.

[7]M.Dekking,Thesumofdigitsfunctionofthebasephiexpansionofthenaturalnumbers,Integers20 (#A45)(2020)1–6.

[8]F.MichelDekking,Howtoaddtwonaturalnumbersinbasephi,FibonacciQ.(2021),inpress.

[9]F.MichelDekking,ThestructureofZeckendorfexpansions,arXiv:2006.06970v1,2020,toappearinIntegers(2021).

[10]E.Hart,Onusingpatternsinthebeta-expansionstostudyFibonacci-Lucasproducts,FibonacciQ.36(1998)396–406.

[11]E.Hart,L.Sanchis,OntheoccurrenceofFnintheZeckendorfdecompositionofnFn,FibonacciQ.37(1999)21–33.

[12]C.Holton,L.Q.Zamboni,Descendantsofprimitivesubstitutions,TheoryComput.Syst.32(1999)133–157.

[13]Y.Huang,Z.Y.Wen,ThesequenceofreturnwordsoftheFibonaccisequence,Theor.Comput.Sci.593(2015)106–116.

[14]M.Lothaire,AlgebraicCombinatoricsonWords,CambridgeUniversityPress,2002.

[15] On-lineencyclopediaofintegersequences,foundedbyN.J.A.Sloane,electronicallyavailableathttp://oeis.org.