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Cent. Eur. J. Math. • 10(2) • 2012 • 797-806

DOI: 10.2478/s11533-012-0001-2

Central European Journal of

Mathematics

A combinatorial proof of a result for permutation pairs

Research Article

Touﬁk Mansour1∗, Mark Shattuck1†

1 Mathematics Department, University of Haifa, Mount Carmel, Haifa 31905, Israel

Received 27 April 2011; accepted 13 December 2011

Abstract:

In this paper, a direct combinatorial proof is given of a result on permutation pairs originally due to Carlitz, Scoville,

and Vaughan and later extended. It concerns showing that the series expansion of the reciprocal of a certain

multiply exponential generating function has positive integer coeﬃcients. The arguments may then be applied to

related problems, one of which concerns the reciprocal of the exponential series for Fibonacci numbers.

MSC:

05A15, 05A05

Keywords:

Exponential generating function • Combinatorial proof • Permutations

©Versita Sp. z o.o.

1. Introduction

Let Nand Pdenote the sets of nonnegative and positive integers, respectively. Given an indeterminate q,

letnq=1+q+···+qn−1ifn∈P,with0q=0.Letnq!=Qn

i=1iqifn∈P,with0q!=1.Put

fq(z)= ∞

X

n=0(−1)nzn

(nq!)dand 1

fq(z)=∞

X

n=0w(d)

n(q)zn

(nq!)d,(1)

whered>2isaﬁxedpositiveinteger.Thegeneratingfunctionfq(z)isbothEulerianandmultiplyexponentialinthe

senseof[7,Section3.15].Uponreplacingzwithz/(1−q)d,weseethat(1)mayalsoberewrittenintheform

gq(z)= ∞

X

n=0(−1)nzn

(q:q)d

nand 1

gq(z)=∞

X

n=0w(d)

n(q)zn

(q:q)d

n,(2)

∗E-mail:tmansour@univ.haifa.ac.il

†E-mail:maarkons@excite.com

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A combinatorial proof of a result for permutation pairs

where(a:q)n=(1−a)(1−qa)···(1−qn−1a)sothat(q:q)n=(1−q)nnq!Inthispaper,weshowthatthecoeﬃcients

w(d)

n(q)in(1),andhence(2),arepolynomialsinqwithpositiveintegercoeﬃcientsbyadirectcombinatorialargument,

seeTheorem2.6. Thecaseq=1andd=2wasﬁrstshownbyCarlitz,Scoville,andVaughan[2]usingalgebraic

methods,andwaslaterextendedbyStanley[6],whoconsideredamoregeneralversionofthisresultrelatedtobinomial

posets.Seealso[3,4]foradditionalrelatedresults.

LetSndenotethesetofpermutationsof[n]={1,2,...,n}andletπ=π1π2···πnandρ=ρ1ρ2···ρnbetwomembers

ofSn.Ariseofπisapairπi,πi+1 withπi<πi+1;afallisapairπi,πi+1 withπi>πi+1.Thus,fori=1,2,...,n−1,

thetwopairsπi,πi+1 andρi,ρi+1 areeitherbothrises(whichwewilldenoteRR),bothfalls(FF),theﬁrstisarise

andthesecondafall(RF),ortheﬁrstisafallandthesecondarise(FR).LetTndenotethesubsetofSn×Snwhose

membersavoidRR(i.e.,anyoccurrenceofRRisforbidden).

In[2],itwasshownusingrecurrencesthatifthewnarethecoeﬃcientsdeterminedby

∞

X

n=0(−1)nzn

n!n!!−1=∞

X

n=0wnzn

n!n!,

thentheyarepositiveintegerswhichgivethecardinalityofTnforalln. Here,weprovideacombinatorialproofin

thesenseof[1]. Indeed,weshow,moregenerally,thatthecoeﬃcientsw(2)

n(q)occurringin(1)ariseasdistribution

polynomialsforthestatisticonTnwhichrecordsthetotalnumberofinversionsoccurringinapermutationpair(π,ρ),

seeTheorem2.5.Takingq=1showsthatwngivesthecardinalityofTnforalln. Thisargumentmaythenbeextended

toprovideacombinatorialproofofthecomparableresultforpermutationd-tuples,seeTheorem2.6.

Theq-binomialcoeﬃcientn

kqisgivenby

n

kq=

nq!

kq!(n−k)q!if 06k6n;

0otherwise.

Notethatthew(d)

n(q)in(1)satisfytherecurrence

n

X

k=0(−1)kn

kd

qw(d)

n−k(q)=0, n>1,(3)

withw(d)

0(q)=1,uponwriting1=fq(z)fq(z)−1andcollectingcoeﬃcients.

2. Main results

Firstassumed=2in(1). Wewillneedthefollowinglemma,whichresultsfromreplacingxwithqxintheq-binomial

theorem(see,e.g.,[7,p.162]), (1+x)(1+qx)···(1+qn−1x)= n

X

k=0n

kqq(k2)xk,

andthenequatingthecoeﬃcientsofxr.

Lemma 2.1.

Ifn>r>1,then X

{a1,a2,...,ar}⊆[n]qa1+a2+···+ar=q(r+1

2)n

rq.

798

T. Mansour, M. Shattuck

Recallthatthenumberofinversionsofamemberπ=π1π2···πn∈Sn,denotedbyinv(π),isthenumberofordered

pairs(i,j)with16i<j6nandπi>πj.

Deﬁnition 2.2.

Ifn∈P,thenlet an(q)= X

(π,ρ)∈Tnqinv(π,ρ),

witha0(q)=1,whereinv(π,ρ)=inv(π)+inv(ρ).

Deﬁnition 2.3.

Ifn>r>1, thenletB(r)

n⊆Sn×Snconsistofpermutationpairs(π, ρ)= (π1π2···πn,ρ1ρ2···ρn)satisfyingthe

followingthreeconditions:

(a)π1<π2<...<πr,

(b)ρ1<ρ2<...<ρr,and

(c)thepartialpermutationpair(πrπr+1···πn,ρrρr+1···ρn)avoidsRR.

Bydeﬁnition,wehaveB(1)

n=TnandB(n+1)

n=∅.

WewillneedthefollowingresultconcerningthedistributionofinvontheB(r)

n.

Lemma 2.4.

Ifn>r>1,then X

(π,ρ)∈B(r)

n∪B(r+1)

nqinv(π,ρ)=n

r2qan−r(q).(4)

Proof. Supposeπ=π1π2···πnandρ=ρ1ρ2···ρnarebothmembersofSn.Then(π,ρ)∈B(r)

n∪B(r+1)

nifandonlyif

(i)π1<π2<...<πr,

(ii)ρ1<ρ2<...<ρr,and

(iii)thepartialpermutationpair(πr+1···πn,ρr+1···πn)hasnooccurrencesofRR.

Weclaimthatthetotalinv-weightofallthemembersofSn×Snsatisfying(i)–(iii), isn

r2qan−r(q). Toseethis,ﬁrst

choosetwor-elementsubsetsof[n]independentlyandwritetheirelementsinincreasingordersoastoobtaintheﬁrst

rpositionsofπandρ. Thecomplementsofthesetwosetsbothcontainn−relementsandcomprisetheﬁnaln−r

positionsofπandρ. Notethatthecomplementstogethermaybewrittenasanypermutationpairsatisfying(iii)and

thatoncethetwor-elementsubsetsareﬁxed,thetotalweightofallthepossiblepairssatisfying(iii)isan−r(q),by

deﬁnition. Tocompletetheproof,itsuﬃcestoshowthatn

rqaccountsforallpossiblechoicesregardingtheﬁrstr

positionsofπ=π1π2···πn. Notethateachletterπi,16i6r,contributesπi−1−(i−1)=πi−iinversionssince

thereareexactlyi−1letterssmallerthanπiandtoitsleft.Thus,thechoiceforthesepositionscontributes

X

{π1,...,πr}⊆[n]qPri=1(πi−i)=q−(r+1

2)X

{π1,...,πr}⊆[n]qPri=1πi=n

rq,

asrequired,byLemma2.1.

Theorem 2.5.

Ifn>0,thenan(q)=w(2)

n(q).

799

A combinatorial proof of a result for permutation pairs

Proof. By(3)whend=2,weneedtoshowthatthean(q)satisfytherecurrence

an(q)= n

X

k=1(−1)k+1n

k2qan−k(q), n>1.

ByLemma2.4andtelescoping,wehave

n

X

k=1(−1)k+1n

k2qan−k(q)= n

X

k=1(−1)k+1 X

(π,ρ)∈B(k)

n∪B(k+1)

nqinv(π,ρ)=X

(π,ρ)∈B(1)

nqinv(π,ρ)+(−1)n+1 X

(π,ρ)∈B(n+1)

nqinv(π,ρ)=an(q),

whichcompletestheproof,sinceB(1)

n=TnandB(n+1)

n=∅.

Letusnowassumed>2. Fortheremainingpartofthissection,wewillletσ=(σ1,σ2,...,σd)denotead-tupleof

permutationsof[n],whereσj=σj,1σj,2···σj,n foreachj∈[d]. Fixi∈[n−1].Letusformawordw=w1w2···wdinthe

alphabet{R,F},wherewjisRorFdependingonwhetherσj,iσj,i+1 isariseorafallinthepermutationσjforeach

j∈[d].Wewillsaythatσcontainsthewordwifthewordwarisesasdescribedforsomeiandthatσavoidswifit

doesnot.

LetT(d)

n⊆Sd

ncomprisethosed-tuplesσ=(σ1,σ2,...,σd)ofpermutationswhichavoidRR···R(dtimes).Letusdenote

thesum,inv(σ1)+inv(σ2)+···+inv(σd),byinv(σ).Leta(d)

n(q)denotethedistributionpolynomialoverT(d)

nforthestatistic

whichrecordsinv(σ),i.e., a(d)

n(q)= X

σ∈T(d)

nqinv(σ).

Theorem 2.6.

Ifn>0andd>2,thena(d)

n(q)=w(d)

n(q).

Proof. WeextendtheproofgivenaboveforTheorem2.5asfollows. Forn>r>1,letB(r)

n,d ⊆Sd

nconsistof

σ=(σ1,σ2,...,σd)satisfyingthefollowingconditions:

(i)σj,1<σj,2<...<σj,r foreachj∈[d],

(ii)thed-tupleofpartialpermutations(σ01,σ02,...,σ0d),whereinσ0j=σj,rσj,r+1···σj,n foreachj,avoidsRR···R.

Bydeﬁnition,wehaveB(1)

n,d=T(d)

nandB(n+1)

n,d =∅.

Notethatσ=(σ1,σ2,...,σd)∈B(r)

n,d∪B(r+1)

n,d ifandonlyifcondition(i)aboveholdswith(ii)replacedbythecondition

thatthed-tuple(σ00

1,σ00

2,...,σ00

d),whereinσ00

j=σj,r+1σj,r+2···σj,n foreachj∈[d],avoidsRR···R.Reasoningasinthe

proofofLemma2.4showsthat X

σ∈B(r)

n,d∪B(r+1)

n,dqinv(σ)=n

rd

qa(d)

n−r(q), n>r>1.

ReasoningasintheproofofTheorem2.5thenshowsthat

a(d)

n(q)= n

X

k=1(−1)k+1n

kd

qa(d)

n−k(q), n>1,

whencew(d)

n(q)=a(d)

n(q)foralln>0,uponcomparisonwith(3).

Remark.

AnalgebraicproofofTheorem2.6whichmakesuseofseveralrecurrencesisgivenin[3].

800

T. Mansour, M. Shattuck

ByconsideringinsteadFF···F(dtimes),RF,orFR,oneobtainsthefollowingadditionalgeneralizationsoftheq=1

caseof(1),whichdonotseemtohavebeenpreviouslynoted.

Theorem 2.7.

Supposed>2andlet f∗

q(z)= ∞

X

n=0(−1)nqd(n

2)zn

(nq!)d.

Thenthecoeﬃcientsu(d)

n(q)intheexpansion 1

f∗

q(z)=∞

X

n=0u(d)

n(q)zn

(nq!)d

arepolynomialsinqwithpositiveintegercoeﬃcients.Furthermore,theyarethedistributionpolynomialsforinvonthe

subsetofd-tuplesofpermutationsof[n]whichavoidFF···F.

Proof. LetU(d)

n⊆Sd

ncomprisethosemembersσ=(σ1,σ2,...,σd)whichavoidFF···F. Deﬁnethepolynomial

b(d)

n(q)by b(d)

n(q)= X

σ∈U(d)

nqinv(σ), n>1,

withb(d)

0(q)=1.Ifn>r>1,thenletC(r)

n,d⊆Sd

nconsistofσsatisfyingthefollowingconditions:

(i)σj,1>σj,2>...>σj,r foreachj∈[d],

(ii)thed-tupleofpartialpermutations(σ01,σ02,...,σ0d),whereinσ0j=σj,rσj,r+1···σj,n foreachj,avoidsFF···F.

Bydeﬁnition,wehaveC(1)

n,d=U(d)

nandC(n+1)

n,d =∅.ReasoningasintheproofofLemma2.4shows

X

σ∈C(r)

n,d∪C(r+1)

n,dqinv(σ)=qd(r2)n

rd

qb(d)

n−r(q), n>r>1.

Notethatthereareanadditional1+2+···+(r−1)=r2inversionsproducedbyeachcomponentofσ∈C(r)

n,d∪C(r+1)

n,d

foratotal ofdr2inversionsin allsincetheﬁrstrlettersofeachcomponentareindescendingorder,whencethe

additionalfactorofqd(r2)intheaboveformula.ReasoningasintheproofofTheorem2.5nowshowsthatb(d)

n(q)satisﬁes

thesamerecurrenceasu(d)

n(q)forn>1andhenceu(d)

n(q)=b(d)

n(q).

Theorem 2.8.

Let efq(z)= ∞

X

n=0(−1)nq(n

2)zn

(nq!)2.

Thenthecoeﬃcientsvn(q)intheexpansion 1

efq(z)=∞

X

n=0vn(q)zn

(nq!)2

arepolynomialsinqwithpositiveintegercoeﬃcients.Furthermore,theyarethedistributionpolynomialsforinvonthe

subsetoforderedpairsofpermutationsof[n]whichavoidRF(orFR).

Proof. ProceedasintheproofofTheorem2.5andthestepsleadinguptoit,makingtheappropriatemodiﬁcations,

themainonebeingtoDeﬁnition2.3. There,condition(b)shouldnowbeρ1>ρ2>...> ρr,with RF replacing

RR incondition(c). NotealsoinLemma2.4thattherewouldbeafactorofq(r2)ontheright-handsideof(4)dueto

1+2+···+(r−1)=r2additionalinversionsinthesecondcomponent.

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A combinatorial proof of a result for permutation pairs

3. A further extension

Theorem2.5maybereﬁnedbyconsideringthenumberofoccurrencesofRRjointlywiththetotalnumberofinversions

withinanarbitrarymemberofSn×Sn.Moreprecisely,letr(π,ρ)denotethetotalnumberofoccurrencesofRRcontained

within(π,ρ)∈Sn×Snanddeﬁnethedistributionpolynomialan(q,x)by

an(q,x)= X

(π,ρ)∈Sn×Snqinv(π,ρ)xr(π,ρ), n>1,

witha0(q,x)=1. Notethatan(q,0)=an(q). ThereasoningusedintheproofofLemma2.4andTheorem2.5above

maybeextendedtoshowthat n

X

k=0(−1)k(1−x)kn

k2qan−k(q,x)=xan(q,x), n>1.(5)

Wewillleavethedetailsoftheproofof(5)asanexercisefortheinterestedreader.Using(5),onethengets

1−x+x∞

X

n=0an(q,x)zn

nq!nq!=fq(z(1−x)) ∞

X

n=0an(q,x)zn

nq!nq!,

where fq(z)= ∞

X

n=0(−1)nzn

nq!nq!,

whichimpliesthefollowingresult.

Theorem 3.1.

Wehave ∞

X

n=0an(q,x)zn

nq!nq!=1−x

fq(z(1−x))−x.

Remark.

LangleyandRemmel[4,Theorem3.4]provideadiﬀerentcombinatorialproofofthisresult,makinguseofasign-changing

involution.

Lettingx=0inTheorem3.1yieldsTheorem2.5.Takingthelimitasx→1inTheorem3.1implies

∞

X

n=0an(q,1) zn

nq!nq!=lim

x→11−x

fq(z(1−x))−x=lim

x→1−1

−zf0q(0)−1=1

1−z,

byL0Hôpital’srule,whichisinaccordwiththefactthatan(q,1)=nq!nq!.Weremarkthattheq=1caseofTheorem3.1

isgivenas(2.15)of[2].

802

T. Mansour, M. Shattuck

4. Some related results

Supposef(z)=cosz=P∞

n=0(−1)nz2n/(2n)!Thenthecoeﬃcientsofthereciprocal

f(z)−1=secz=∞

X

n=0E2nz2n

(2n)!

arepositiveintegers whichenumerate,among otherthings(see, e.g.,[7]), thealternatingpermutationsoflength2n

(i.e.,π=π1π2···π2n∈S2nsuchthatπ1>π2<π3>π4<...>π2n),aresultattributedtoAndré.Inthissection,we

considertwofurtherexamplesinthisdirectioninwhichbothf(z)andf−1(z)areexponentialgeneratingfunctions(egf’s)

withintegercoeﬃcientsandsupplycombinatorialinterpretationsforthecoeﬃcientsofthelatter,giventheformer.These

resultsmaythenbeq-generalizedbyargumentssimilartothoseofthesecondsectionabove.

Inwhatfollows,ifw=w1w2··· denotesaword,thenamaximalsequenceofconsecutiveletterswiwi+1···wi+k−1inw

suchthatwi<wi+1 <...<wi+k−1willbecalledak-run.Runswillbesaidtobeevenorodddependingontheparity

ofk. Forexample,withinthewordw=12434385471∈[8]11,therearetwo1-runs(the5andthelast1),three2-runs

(34,38,and47),andone3-run(124).

LetFn,n∈N,denotetheFibonaccisequencedeﬁnedbytherecurrenceFn=Fn−1+Fn−2ifn>2,withF0=0and

F1=1.OurﬁrstresultrevealsaconnectionbetweenFnandthepermutationsof[n]avoidingrunsofevenlength.

Theorem 4.1.

Let h(z)=1+X

n>1(−1)nFnzn

n!.

Thenthecoeﬃcientspnintheexpansion 1

h(z)=X

n>0pnzn

n!

arepositiveintegerswhichcountthepermutationsof[n]havingnoevenruns.

Proof. Forn>0, letDndenotethesetofpermutationsof[n]havingnoevenrunsandletdn=|Dn|. Toshow

dn=pnforalln,wemustshowthatdnsatisﬁes

dn=n

X

k=1(−1)k−1Fkn

kdn−k, n>1.(6)

Todoso,ﬁxk,16k6n,andformpermutationsof[n]byﬁrstselectingak-subsetSof[n],writingitselementsin

increasingorder,andthenwritingtheremainingn−kelementsof[n]aftertheelementsofSaccordingtoapermutation

ρhavingnoevenruns. Firstnotethatifkisodd,thenn

kdn−kcountsallthemembersofDnstartingwithak-runas

wellasallmembersofSn−Dnwhichstartwithanevenrunoflengthatleastk+1andcontainnootherevenruns.

Toseethis,notethattheformercaseoccurswhentheﬁrstletterofρissmallerthanthelargest(=last)elementofS,

whenceweobtainaninitialrunoflengthk,andthelattercaseoccurswhentheﬁrstletterofρexceedsthelargest

elementofS,whenceweobtainaninitialevenrunwithlengthexceedingk. Bysimilarreasoning,ifkiseven,then

n

kdn−kcountsallmembersofDnstartingwitharunhavinglengthatleastk+1aswellasallmembersofSn−Dn

whichstartwithak-runandcontainnootherevenruns.

Tocompletetheproof,wewilluseacharacteristicfunctionargument. Supposeα∈Sn−Dnstartswitha2m-runfor

somem,16m6bn/2c,andcontainsnootherevenruns.Thentheabovereasoningimpliesthatαiscounted

(F1+F3+···+F2m−1)−F2m=0

803

A combinatorial proof of a result for permutation pairs

timesbytherightsideof(6)sinceitiscountedF2i−1timesinapositivewaybytheF2i−1n

2i−1dn−2i+1 termforeach

i∈[m]andF2mtimes inanegativewaybytheF2mn

2mdn−2mterm. Onthe otherhand, ifα∈Dnstarts witha

(2m−1)-runforsomem,thenαiscounted

F2m−1−(F2+F4+···+F2m−2)=1

timebytherightsideof(6),byasimilarreasoning.Thus,allmembersofDn(andonlythey)arecountedonce,which

implies(6),asdesired.

Remark.

ThesequencepnoccursasA097597in[5],wherenomentionismadetotherelationinTheorem4.1ortotherecurrence(6).

Wenowconsidertheproblemofavoidingrunsofacertainﬁxedlengthormore.Suppose`>2isaﬁxedintegerand

letλn,n∈N,denotethesequencedeﬁnedby

λn=

1ifn≡0(mod`);

−1ifn≡1(mod`);

0otherwise.

Wehavethefollowingresultconcerningtheegfoftheλn.

Theorem 4.2.

Let j(z)=X

n>0λnzn

n!.

Thenthecoeﬃcientstnintheexpansion 1

j(z)=X

n>0tnzn

n!

arepositiveintegerswhichcountthepermutationsof[n]allofwhoserunsareoflengthlessthan`.

Proof. Letdndenotethenumberofpermutationsof[n]allofwhoserunsareoflengthlessthan`. Toshowthat

dn=tnforalln,wemustshowthatthednsatisfythefollowingrecurrencewhenn>1:

dn=n

1dn−1−n

`dn−`+n

`+1dn−`−1−n

2`dn−2`+n

2`+1dn−2`−1−··· (7)

Todoso,ﬁrstdeﬁnethesetsA(r)

nfor16r6nby

A(r)

n=π∈Sn:πstartswithani-runforsomei,r6i6r+`−1,withallotherrunsoflengthatmost`−1.

ThenthecardinalityofA(r)

nisn

rdn−r. Tosee this,note thatmembersofA(r)

nmaybeformedbyﬁrstselectingan

r-elementsubsetSof[n],writingtheelementsofSinincreasingorder,andthenarrangingthen−rmembersof[n]−S

aftertheelementsofSaccordingtoapermutationρcontainingnorunsoflength`ormore. Theresultingmemberof

A(r)

nwillstartwitharunoflengthriftheﬁrstletterofρissmallerthanthelargestelementofSandwillstartwitha

runwhoselengthisstrictlybetweenrandr+`otherwise.

Tocompletetheproof,weagainuseacharacteristicfunctionargument. Supposeπ∈Sn

r=1A(r)

nstartswithani-run,

wherei>`. If`doesnotdivideiandm=`bi/`c,thenπbelongstobothA(m)

nandA(m+1)

nandisthereforecountedonce

inanegativewaybythen

mdn−mtermontherightsideof(7)aboveandonceinapositivewaybythen

m+1dn−m−1

804

T. Mansour, M. Shattuck

term(andsothesecontributionscancel). If`dividesi,thenπiscountedinapositivewaybythen

i−`+1dn−i+`−1term

andinanegativewaybythenidn−iterm.Thus,ifπ∈Sn

r=1A(r)

nstartswithani-runforsomei>`,thenπcontributes

zerotothealternatingsumontherightsideof(7). Ifπstartswithani-runforsomei<`,thenπiscountedonce

andonlybytheﬁrstterm. Thus,therightsideof(7)givesthecardinalityofallthemembersofA(1)

nwhichstartwith

ani-runforsomei<`,whicharesynonymouswiththepermutationsof[n]allofwhoserunsareoflengthlessthan`.

Thisestablishes(7)andcompletestheproof.

Taking`=2inTheorem4.2givesj(z)=e−zandthustheegfenumeratingthepermutationsof[n]avoidingallrunsof

lengthtwoormoreisj−1(z)=ezasanticipatedsinceonlyπ=n(n−1)···1ispermitted.If`=4inTheorem4.2,then

theegfwhichenumeratesallthepermutationsof[n]havingnorunsoflengthfourormoreisj−1(z),where

j(z)=12e−z+cosz−sinz.

Notethatfor`ingeneral,thefunctionj(z)maybeexpressedusing`th rootsofunity. Forexample,letting`=3in

Theorem4.2impliesthattheegfforthenumberofpermutationsof[n]whichavoidallrunsoflengththreeormoreis

j−1(z),where j(z)=(1−ω2)

3eωz+(1−ω)

3eω2z

andwisaprimitivecuberootofunity.

Onecanprovideaq-generalizationofTheorem4.2.

Theorem 4.3.

Let jq(z)=X

n>0λnzn

nq!.

Thenthecoeﬃcientstn(q)intheexpansion 1

jq(z)=X

n>0tn(q)zn

nq!

arepolynomialsinqwithpositiveintegercoeﬃcients.Furthermore,theyarethedistributionpolynomialsforinvonthe

setofpermutationsof[n]allofwhoserunsareoflengthlessthan`.

Proof. Letdn(q)bethedistributionpolynomialfortheinvstatisticonthesetofpermutationsof[n]allofwhoseruns

areoflengthlessthan`.WeﬁrstobservethatX

π∈A(r)

nqinv(π)=n

rqdn−r(q),(8)

whereA(r)

nisasintheproofofTheorem4.2.Toshowthis,ﬁrstnotethatn

rqaccountsforboththechoiceofthesubsetS

comprisingtheﬁrstrlettersofπ=π1π2···πn∈A(r)

naswellastheinversionsthattheycreatewithinπwiththeletters

in[n]−S.Toseethis,letw=w1w2···wnbethebinarywordwherewi=0ifi∈Sand1otherwise.Thenthenumber

ofinversionscausedbytheelementsofSwithinπisthesameasthenumberofinversionsofw,withn

rqthegenerating

functionforthenumberofinversionsinbinarywordsoflengthncontainingexactlyrzeros,see[7,Proposition1.3.17].

Sincedn−r(q)accountsfortheorderingoftheﬁnaln−rlettersofπoncetheelementsofShavebeenchosenaswell

astheinversionswithintheseletters,equation(8)follows.

Theproofof(7)nowappliesandshowsfurtherthat

dn(q)=n

1qdn−1(q)−n

`qdn−`(q)+n

`+1qdn−`−1(q)−···, n>1,

whichimpliestn(q)=dn(q)foralln,uponcomparingtherecurrences.

805

A combinatorial proof of a result for permutation pairs

WealsohavethefollowingextensionofTheorem4.1.

Theorem 4.4.

Let hq(z)=1+X

n>1(−1)nFnzn

nq!.

Thenthecoeﬃcientspn(q)intheexpansion 1

hq(z)=X

n>0pn(q)zn

nq!

setofpermutationsof[n]havingnoevenruns.

Proof. Ifdn(q)denotesthedistributionpolynomialfortheinvstatisticonthesetofpermutationsof[n]havingno

evenruns,thenwemustshowthatdn(q)satisﬁestherecurrence

dn(q)= n

X

k=1(−1)k−1Fkn

kqdn−k(q), n>1.(9)

Todo so,we generalizetheproofof (6). Observeinitially thatthe ﬁrstparagraphintheproof of(6)carriesover

withn

kdn−kreplacedbyn

kqdn−k(q);notethatn

kqarisesforthesamereasonthatitdidintheproofof(8)above.

Equation(9)nowfollowsfromthesameargumentusedinthesecondparagraphoftheproofof(6).

References

[1]BenjaminA.T.,QuinnJ.J.,ProofsthatReallyCount,DolcianiMath.Exp.,27,MathematicalAssociationofAmerica,

Washington,2003

[2]CarlitzL.,ScovilleR.,VaughanT.,Enumerationofpairsofpermutations,DiscreteMath.,1976,14(3),215–239

[3]FedouJ.-M.,RawlingsD.,Statisticsonpairsofpermutations,DiscreteMath.,1995,143(1-3),31–45

[4]LangleyT.M.,RemmelJ.B.,Enumerationofm-tuplesofpermutationsandanewclassofpowerbasesforthespace

ofsymmetricfunctions,Adv.inAppliedMath.,2006,36(1),30–66

[5]SloaneN.J.,TheOn-LineEncyclopediaofIntegerSequences,http://oeis.org

[6]StanleyR.P.,Binomialposets,Möbiusinversion,andpermutationenumeration,J.CombinatorialTheorySer.A,1976,

20(3),336–356

[7]StanleyR.P.,EnumerativeCombinatorics,Vol.1,CambridgeStud.Adv.Math.,49,CambridgeUniversityPress,Cam-

bridge,1997

806

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