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# A combinatorial proof of a result for permutation pairs

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## 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.
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
1. Introduction
Let Nand Pdenote the sets of nonnegative and positive integers, respectively. Given an indeterminate q,
letnq=1+q+···+qn1ifnP,with0q=0.Letnq!=Qn
i=1iqifnP,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/(1q)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=(1a)(1qa)···(1qn1a)sothat(q:q)n=(1q)nnq!Inthispaper,weshowthatthecoeﬃcients
w(d)
seeTheorem2.6. Thecaseq=1andd=2wasﬁrstshownbyCarlitz,Scoville,andVaughan[2]usingalgebraic
methods,andwaslaterextendedbyStanley[6],whoconsideredamoregeneralversionofthisresultrelatedtobinomial
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,...,n1,
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!(nk)q!if 06k6n;
0otherwise.
Notethatthew(d)
n(q)in(1)satisfytherecurrence
n
X
k=0(1)kn
kd
qw(d)
nk(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+qn1x)= 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···πnSn,denotedbyinv(π),isthenumberofordered
pairs(i,j)with16i<j6nandπi>πj.
Deﬁnition 2.2.
IfnP,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)
nSn×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)
nB(r+1)
nqinv(π,ρ)=n
r2qanr(q).(4)
Proof. Supposeπ=π1π2···πnandρ=ρ1ρ2···ρnarebothmembersofSn.Then(π,ρ)B(r)
nB(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
r2qanr(q). Toseethis,ﬁrst
choosetwor-elementsubsetsof[n]independentlyandwritetheirelementsinincreasingordersoastoobtaintheﬁrst
rpositionsofπandρ. Thecomplementsofthesetwosetsbothcontainnrelementsandcomprisetheﬁnalnr
positionsofπandρ. Notethatthecomplementstogethermaybewrittenasanypermutationpairsatisfying(iii)and
deﬁnition. Tocompletetheproof,itsuﬃcestoshowthatn
rqaccountsforallpossiblechoicesregardingtheﬁrstr
positionsofπ=π1π2···πn. Notethateachletterπi,16i6r,contributesπi1(i1)=πiiinversionssince
X
{π1,...,πr}⊆[n]qPri=1(πii)=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).
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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
k2qank(q), n>1.
ByLemma2.4andtelescoping,wehave
n
X
k=1(1)k+1n
k2qank(q)= n
X
k=1(1)k+1 X
(π,ρ)B(k)
nB(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=.
permutationsof[n],whereσj=σj,1σj,2···σj,n foreachj[d]. Fixi[n1].Letusformawordw=w1w2···wdinthe
alphabet{R,F},wherewjisRorFdependingonwhetherσj,iσj,i+1 isariseorafallinthepermutationσjforeach
j[d].Wewillsaythatσcontainsthewordwifthewordwarisesasdescribedforsomeiandthatσavoidswifit
doesnot.
LetT(d)
nSd
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
σT(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,dB(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
σB(r)
n,dB(r+1)
n,dqinv(σ)=n
rd
qa(d)
nr(q), n>r>1.
ReasoningasintheproofofTheorem2.5thenshowsthat
a(d)
n(q)= n
X
k=1(1)k+1n
kd
qa(d)
nk(q), n>1,
whencew(d)
n(q)=a(d)
n(q)foralln>0,uponcomparisonwith(3).
Remark.
AnalgebraicproofofTheorem2.6whichmakesuseofseveralrecurrencesisgivenin[3].
800
T. Mansour, M. Shattuck
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)
nSd
ncomprisethosemembersσ=(σ1,σ2,...,σd)whichavoidFF···F. Deﬁnethepolynomial
b(d)
n(q)by b(d)
n(q)= X
σU(d)
nqinv(σ), n>1,
withb(d)
0(q)=1.Ifn>r>1,thenletC(r)
n,dSd
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
σC(r)
n,dC(r+1)
n,dqinv(σ)=qd(r2)n
rd
qb(d)
nr(q), n>r>1.
n,dC(r+1)
n,d
foratotal ofdr2inversionsin allsincetheﬁrstrlettersofeachcomponentareindescendingorder,whencethe
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).
themainonebeingtoDeﬁnition2.3. There,condition(b)shouldnowbeρ1>ρ2>...> ρr,with RF replacing
RR incondition(c). NotealsoinLemma2.4thattherewouldbeafactorofq(r2)ontheright-handsideof(4)dueto
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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(1x)kn
k2qank(q,x)=xan(q,x), n>1.(5)
1x+x
X
n=0an(q,x)zn
nq!nq!=fq(z(1x))
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!=1x
fq(z(1x))x.
Remark.
involution.
Lettingx=0inTheorem3.1yieldsTheorem2.5.Takingthelimitasx1inTheorem3.1implies
X
n=0an(q,1) zn
nq!nq!=lim
x11x
fq(z(1x))x=lim
x11
zf0q(0)1=1
1z,
byL0Hôpital’srule,whichisinaccordwiththefactthatan(q,1)=nq!nq!.Weremarkthattheq=1caseofTheorem3.1
isgivenas(2.15)of[2].
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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···π2nS2nsuchthatπ1>π2<π3>π4<...>π2n),aresultattributedtoAndré.Inthissection,we
considertwofurtherexamplesinthisdirectioninwhichbothf(z)andf1(z)areexponentialgeneratingfunctions(egf’s)
withintegercoeﬃcientsandsupplycombinatorialinterpretationsforthecoeﬃcientsofthelatter,giventheformer.These
resultsmaythenbeq-generalizedbyargumentssimilartothoseofthesecondsectionabove.
Inwhatfollows,ifw=w1w2··· denotesaword,thenamaximalsequenceofconsecutiveletterswiwi+1···wi+k1inw
suchthatwi<wi+1 <...<wi+k1willbecalledak-run.Runswillbesaidtobeevenorodddependingontheparity
ofk. Forexample,withinthewordw=12434385471[8]11,therearetwo1-runs(the5andthelast1),three2-runs
(34,38,and47),andone3-run(124).
LetFn,nN,denotetheFibonaccisequencedeﬁnedbytherecurrenceFn=Fn1+Fn2ifn>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)k1Fkn
kdnk, n>1.(6)
Todoso,ﬁxk,16k6n,andformpermutationsof[n]byﬁrstselectingak-subsetSof[n],writingitselementsin
increasingorder,andthenwritingtheremainingnkelementsof[n]aftertheelementsofSaccordingtoapermutation
ρhavingnoevenruns. Firstnotethatifkisodd,thenn
kdnkcountsallthemembersofDnstartingwithak-runas
wellasallmembersofSnDnwhichstartwithanevenrunoflengthatleastk+1andcontainnootherevenruns.
Toseethis,notethattheformercaseoccurswhentheﬁrstletterofρissmallerthanthelargest(=last)elementofS,
whenceweobtainaninitialrunoflengthk,andthelattercaseoccurswhentheﬁrstletterofρexceedsthelargest
elementofS,whenceweobtainaninitialevenrunwithlengthexceedingk. Bysimilarreasoning,ifkiseven,then
n
kdnkcountsallmembersofDnstartingwitharunhavinglengthatleastk+1aswellasallmembersofSnDn
whichstartwithak-runandcontainnootherevenruns.
Tocompletetheproof,wewilluseacharacteristicfunctionargument. SupposeαSnDnstartswitha2m-runfor
somem,16m6bn/2c,andcontainsnootherevenruns.Thentheabovereasoningimpliesthatαiscounted
(F1+F3+···+F2m1)F2m=0
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A combinatorial proof of a result for permutation pairs
timesbytherightsideof(6)sinceitiscountedF2i1timesinapositivewaybytheF2i1n
2i1dn2i+1 termforeach
i[m]andF2mtimes inanegativewaybytheF2mn
2mdn2mterm. Onthe otherhand, ifαDnstarts witha
(2m1)-runforsomem,thenαiscounted
F2m1(F2+F4+···+F2m2)=1
timebytherightsideof(6),byasimilarreasoning.Thus,allmembersofDn(andonlythey)arecountedonce,which
implies(6),asdesired.
Remark.
Wenowconsidertheproblemofavoidingrunsofacertainﬁxedlengthormore.Suppose>2isaﬁxedintegerand
letλn,nN,denotethesequencedeﬁnedby
λn=
1ifn0(mod);
1ifn1(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
1dn1n
dn+n
+1dn1n
2dn2+n
2+1dn21··· (7)
Todoso,ﬁrstdeﬁnethesetsA(r)
nfor16r6nby
A(r)
n=πSn:πstartswithani-runforsomei,r6i6r+1,withallotherrunsoflengthatmost1.
ThenthecardinalityofA(r)
nisn
rdnr. Tosee this,note thatmembersofA(r)
nmaybeformedbyﬁrstselectingan
r-elementsubsetSof[n],writingtheelementsofSinincreasingorder,andthenarrangingthenrmembersof[n]S
aftertheelementsofSaccordingtoapermutationρcontainingnorunsoflengthormore. Theresultingmemberof
A(r)
nwillstartwitharunoflengthriftheﬁrstletterofρissmallerthanthelargestelementofSandwillstartwitha
runwhoselengthisstrictlybetweenrandr+otherwise.
Tocompletetheproof,weagainuseacharacteristicfunctionargument. SupposeπSn
r=1A(r)
nstartswithani-run,
wherei>. Ifdoesnotdivideiandm=bi/c,thenπbelongstobothA(m)
nandA(m+1)
nandisthereforecountedonce
inanegativewaybythen
mdnmtermontherightsideof(7)aboveandonceinapositivewaybythen
m+1dnm1
804
T. Mansour, M. Shattuck
term(andsothesecontributionscancel). Ifdividesi,thenπiscountedinapositivewaybythen
i+1dni+1term
andinanegativewaybythenidniterm.Thus,ifπSn
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)=ezandthustheegfenumeratingthepermutationsof[n]avoidingallrunsof
lengthtwoormoreisj1(z)=ezasanticipatedsinceonlyπ=n(n1)···1ispermitted.If=4inTheorem4.2,then
theegfwhichenumeratesallthepermutationsof[n]havingnorunsoflengthfourormoreisj1(z),where
j(z)=12ez+coszsinz.
Notethatforingeneral,thefunctionj(z)maybeexpressedusingth rootsofunity. Forexample,letting=3in
Theorem4.2impliesthattheegfforthenumberofpermutationsof[n]whichavoidallrunsoflengththreeormoreis
j1(z),where j(z)=(1ω2)
3eωz+(1ω)
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
πA(r)
nqinv(π)=n
rqdnr(q),(8)
whereA(r)
nisasintheproofofTheorem4.2.Toshowthis,ﬁrstnotethatn
rqaccountsforboththechoiceofthesubsetS
comprisingtheﬁrstrlettersofπ=π1π2···πnA(r)
naswellastheinversionsthattheycreatewithinπwiththeletters
in[n]S.Toseethis,letw=w1w2···wnbethebinarywordwherewi=0ifiSand1otherwise.Thenthenumber
ofinversionscausedbytheelementsofSwithinπisthesameasthenumberofinversionsofw,withn
rqthegenerating
functionforthenumberofinversionsinbinarywordsoflengthncontainingexactlyrzeros,see[7,Proposition1.3.17].
Sincednr(q)accountsfortheorderingoftheﬁnalnrlettersofπoncetheelementsofShavebeenchosenaswell
astheinversionswithintheseletters,equation(8)follows.
Theproofof(7)nowappliesandshowsfurtherthat
dn(q)=n
1qdn1(q)n
qdn(q)+n
+1qdn`1(q)···, n>1,
whichimpliestn(q)=dn(q)foralln,uponcomparingtherecurrences.
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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)k1Fkn
kqdnk(q), n>1.(9)
Todo so,we generalizetheproofof (6). Observeinitially thatthe ﬁrstparagraphintheproof of(6)carriesover
withn
kdnkreplacedbyn
kqdnk(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
[5]SloaneN.J.,TheOn-LineEncyclopediaofIntegerSequences,http://oeis.org
[6]StanleyR.P.,Binomialposets,Möbiusinversion,andpermutationenumeration,J.CombinatorialTheorySer.A,1976,
20(3),336–356
bridge,1997
806
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