A combinatorial proof of a result for permutation pairs

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.

Full text

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
797

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-binomialcoefficientn
kqisgivenby
n
kq=

nq!
kq!(n−k)q!if 06k6n;
0otherwise.
Notethatthew(d)
n(q)in(1)satisfytherecurrence
n
X
k=0(−1)kn
kd
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=0n
kqq(k2)xk,
andthenequatingthecoefficientsofxr.
Lemma 2.1.
Ifn>r>1,then X
{a1,a2,...,ar}⊆[n]qa1+a2+···+ar=q(r+1
2)n
rq.
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
r2qan−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), isn
r2qan−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,itsufficestoshowthatn
rqaccountsforallpossiblechoicesregardingthefirstr
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
rq,
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+1n
k2qan−k(q), n>1.
ByLemma2.4andtelescoping,wehave
n
X
k=1(−1)k+1n
k2qan−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
rd
qa(d)
n−r(q), n>r>1.
ReasoningasintheproofofTheorem2.5thenshowsthat
a(d)
n(q)= n
X
k=1(−1)k+1n
kd
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
rd
qb(d)
n−r(q), n>r>1.
Notethatthereareanadditional1+2+···+(r−1)=r2inversionsproducedbyeachcomponentofσ∈C(r)
n,d∪C(r+1)
n,d
foratotal ofdr2inversionsin 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)=r2additionalinversionsinthesecondcomponent.
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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)kn
k2qan−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−1Fkn
kdn−k, n>1.(6)
Todoso,fixk,16k6n,andformpermutationsof[n]byfirstselectingak-subsetSof[n],writingitselementsin
increasingorder,andthenwritingtheremainingn−kelementsof[n]aftertheelementsofSaccordingtoapermutation
ρhavingnoevenruns. Firstnotethatifkisodd,thenn
kdn−kcountsallthemembersofDnstartingwithak-runas
wellasallmembersofSn−Dnwhichstartwithanevenrunoflengthatleastk+1andcontainnootherevenruns.
Toseethis,notethattheformercaseoccurswhenthefirstletterofρissmallerthanthelargest(=last)elementofS,
whenceweobtainaninitialrunoflengthk,andthelattercaseoccurswhenthefirstletterofρexceedsthelargest
elementofS,whenceweobtainaninitialevenrunwithlengthexceedingk. Bysimilarreasoning,ifkiseven,then
n
kdn−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−1n
2i−1dn−2i+1 termforeach
i∈[m]andF2mtimes inanegativewaybytheF2mn
2mdn−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
1dn−1−n
`dn−`+n
`+1dn−`−1−n
2`dn−2`+n
2`+1dn−2`−1−··· (7)
Todoso,firstdefinethesetsA(r)
nfor16r6nby
A(r)
n=π∈Sn:πstartswithani-runforsomei,r6i6r+`−1,withallotherrunsoflengthatmost`−1.
ThenthecardinalityofA(r)
nisn
rdn−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
inanegativewaybythen
mdn−mtermontherightsideof(7)aboveandonceinapositivewaybythen
m+1dn−m−1
804

T. Mansour, M. Shattuck
term(andsothesecontributionscancel). If`dividesi,thenπiscountedinapositivewaybythen
i−`+1dn−i+`−1term
andinanegativewaybythenidn−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)=12e−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
rqdn−r(q),(8)
whereA(r)
nisasintheproofofTheorem4.2.Toshowthis,firstnotethatn
rqaccountsforboththechoiceofthesubsetS
comprisingthefirstrlettersofπ=π1π2···πn∈A(r)
naswellastheinversionsthattheycreatewithinπwiththeletters
in[n]−S.Toseethis,letw=w1w2···wnbethebinarywordwherewi=0ifi∈Sand1otherwise.Thenthenumber
ofinversionscausedbytheelementsofSwithinπisthesameasthenumberofinversionsofw,withn
rqthegenerating
functionforthenumberofinversionsinbinarywordsoflengthncontainingexactlyrzeros,see[7,Proposition1.3.17].
Sincedn−r(q)accountsfortheorderingofthefinaln−rlettersofπoncetheelementsofShavebeenchosenaswell
astheinversionswithintheseletters,equation(8)follows.
Theproofof(7)nowappliesandshowsfurtherthat
dn(q)=n
1qdn−1(q)−n
`qdn−`(q)+n
`+1qdn−`−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!
arepolynomialsinqwithpositiveintegercoefficients.Furthermore,theyarethedistributionpolynomialsforinvonthe
setofpermutationsof[n]havingnoevenruns.
Proof. Ifdn(q)denotesthedistributionpolynomialfortheinvstatisticonthesetofpermutationsof[n]havingno
evenruns,thenwemustshowthatdn(q)satisfiestherecurrence
dn(q)= n
X
k=1(−1)k−1Fkn
kqdn−k(q), n>1.(9)
Todo so,we generalizetheproofof (6). Observeinitially thatthe firstparagraphintheproof of(6)carriesover
withn
kdn−kreplacedbyn
kqdn−k(q);notethatn
kqarisesforthesamereasonthatitdidintheproofof(8)above.
Equation(9)nowfollowsfromthesameargumentusedinthesecondparagraphoftheproofof(6).
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