A generalisation of a second partition theorem of Andrews to overpartitions

0.19 MB·English

by  Jehanne Dousse

A GENERALISATION OF A SECOND PARTITION THEOREM
OF ANDREWS TO OVERPARTITIONS
5
1
JEHANNEDOUSSE
0
2
n Abstract. In1968and1969, Andrewsprovedtwopartitiontheorems ofthe
a Rogers-RamanujantypewhichgeneraliseSchur’scelebratedpartitionidentity
J (1926). Andrews’ two generalisations of Schur’s theorem went on to become
two of the most influential results in the theory of partitions, finding appli-
9
cations in combinatorics, representation theory and quantum algebra. In a
2
recent paper, the author generalised the first of these theorems to overparti-

tions,usinganewtechniquewhichconsistsingoingbackandforthbetweenq-
]
O differenceequationsongeneratingfunctionsandrecurrenceequationsontheir
coefficients. Here, using a similar method, we generalise the second theorem
C
ofAndrewstooverpartitions.
.
h
t
a
m
1. Introduction
[
Apartitionofnisanon-increasingsequenceofnaturalnumberswhosesumisn.

1
An overpartition of n is a partition of n in which the first occurrence of a number
v
may be overlined. For example, there are 14 overpartitions of 4: 4, 4, 3+1, 3+1,
8
7 3+1, 3+1, 2+2, 2+2, 2+1+1, 2+1+1, 2+1+1, 2+1+1, 1+1+1+1 and
4 1+1+1+1.
7 In 1926, Schur [Sch26] proved the following partition identity.
0
. Theorem 1.1(Schur). Letnbeapositive integer. LetD1(n)denotethenumberof
1
partitions of n into distinct parts congruent to 1 or 2 modulo 3. Let E (n) denote
0 1
the number of partitions of n of the form n = λ +···+λ where λ −λ ≥ 3
5 1 s i i+1
1 with strict inequality if λi+1 ≡0 mod 3. Then D1(n)=E1(n).
:
v Forexample,forn=9,thepartitionscountedbyD1(9)are8+1,7+2and5+4
i andthepartitionscountedbyE (9)are9,8+1and7+2. ThusD (9)=E (9)=3.
X 1 1 1
Several proofs of Schur’s theorem have been given using a variety of different
r
a techniques such as bijective mappings [Bes91, Bre80], the method of weighted
words [AG93], and recurrences [And67, And68b, And71].
Schur’stheoremwassubsequentlygeneralisedtooverpartitionsbyLovejoy[Lov05]
using the method of weighted words. The case k =0 corresponds to Schur’s theo-
rem.
Theorem 1.2 (Lovejoy). Let D (k,n) denote the number of overpartitions of n
1
into parts congruent to 1 or 2 modulo 3 with k non-overlined parts. Let E (k,n)
1
denote the number of overpartitions of n with k non-overlined parts, where parts
differ by at least 3 if the smaller is overlined or both parts are divisible by 3, and
parts differ by at least 6 if the smaller is overlined and both parts are divisible by 3.
Then D (k,n)=E (k,n).
1 1
Theorem1.2was then provedbijectively by Raghavendraand Padmavathamma
[RP09], and using q-difference equations and recurrences by the author [Dou14b].
1
2 JEHANNEDOUSSE
Andrews extended the ideas of his proofs of Schur’s theorem to prove two much
more general theorems on partitions with difference conditions [And68a, And69].
Butbeforestatingtheseresultsintheirfullgeneralityweneedtointroducesomeno-
tation. LetA={a(1),...,a(r)}beasetofrdistinctintegerssuchthat k−1a(i)<
i=1
a(k) for all 1 ≤ k ≤ r and the 2r − 1 possible sums of distinct elements of A
are all distinct. We denote this set of sums by A′ = {α(1),...,α(2r −P1)}, where
α(1) < ··· < α(2r −1). Let us notice that α(2k) = a(k+1) for all 0 ≤ k ≤ r−1
and that any α between a(k) and a(k + 1) has largest summand a(k). Let N
be a positive integer with N ≥ α(2r −1) = a(1)+···+a(r). We further define
α(2r)=a(r+1)=N +a(1). Let A denote the set of positive integers congruent
N
to some a(i) mod N, −A the set of positive integers congruent to some −a(i)
N
mod N, A′ the set of positive integers congruent to some α(i) mod N and −A′
N N
the set of positive integers congruent to some −α(i) mod N. Let β (m) be the
N
least positive residue of m mod N. If α ∈ A′, let w(α) be the number of terms
appearing in the defining sum of α and v(α) the smallest a(i) appearing in this
sum.
To illustrate these notations in the remainder of this paper, it might be useful
to consider the example where a(k) = 2k−1 for 1 ≤ k ≤ r and α(k) = k for
1≤k ≤2r−1.
We are now able to state Andrews’ generalisations of Schur’s theorem.
Theorem 1.3 (Andrews). Let D(A ;n) denote the number of partitions of n into
N
distinct parts taken from A . Let E(A′ ;n) denote the number of partitions of n
N N
into parts taken from A′ of the form n=λ +···+λ , such that
N 1 s
λ −λ ≥Nw(β (λ ))+v(β (λ ))−β (λ ).
i i+1 N i+1 N i+1 N i+1
Then D(A ;n)=E(A′ ;n).
N N
Theorem1.4(Andrews). LetF(−A ;n)denotethenumberofpartitionsofninto
N
distinct parts taken from −A . Let G(−A′ ;n) denote the number of partitions of
N N
n into parts taken from −A′ of the form n=λ +···+λ , such that
N 1 s
λ −λ ≥Nw(β (−λ ))+v(β (−λ ))−β (−λ ),
i i+1 N i+1 N i+1 N i+1
and λ ≥N(w(β (−λ )−1). Then F(−A ;n)=G(−A′ ;n).
s N s N N
Not only have Andrews’ identities led to a number of important developments
in combinatorics [All97, CL06, Yee08] but they also play a natural role in group
representationtheory [AO91] and quantum algebra [Oh15].
The author generalised Theorem 1.3 to overpartitions in [Dou14a] by proving
the following.
Theorem 1.5. LetD(A ;k,n)denotethenumberof overpartitions of nintoparts
N
taken from A , having k non-overlined parts. Let E(A′ ;k,n) denote the number
N N
of overpartitions of n into parts taken from A′ of the form n = λ +···+λ ,
N 1 s
having k non-overlined parts, such that
λ −λ ≥Nw β (λ )−1+χ(λ ) +v(β (λ ))−β (λ ),
i i+1 N i+1 i+1 N i+1 N i+1
where χ(λ )=1 if λ(cid:0) is overlined and 0 oth(cid:1)erwise.
i+1 i+1
Then D(A ;k,n)=E(A′ ;k,n).
N N
Here we generalise Theorem 1.4 by showing the following.
A GENERALISATION OF A SECOND PARTITION THEOREM OF ANDREWS 3
Theorem 1.6. Let F(−A ;k,n) denote the number of overpartitions of n into
N
parts taken from −A , having k non-overlined parts. Let G(−A′ ;k,n) denote
N N
the number of overpartitions of n into parts taken from −A′ of the form n =
N
λ +···+λ , having k non-overlined parts, such that
1 s
λ −λ ≥Nw β (λ )−1+χ(λ ) +v(β (λ ))−β (λ ),
i i+1 N i+1 i+1 N i+1 N i+1
(cid:0) λ ≥N(w(β (−λ )(cid:1))−1).
s N s
Then F(−A ;k,n)=G(−A′ ;k,n).
N N
Theorem 1.1 (resp. Theorem 1.2) corresponds to N = 3, a(1) = 1, a(2) = 2
in Theorems 1.3 and 1.4 (resp. Theorems 1.5 and 1.6). Again, the case k = 0 of
Theorem 1.5 (resp. Theorem 1.6) gives Theorem 1.3 (resp. Theorem 1.4).
Let us illustrate Theorems 1.6 with an example. Let N = 7, r = 3, a(1) = 1,
a(2)=2, a(3)=4. The overpartitions of 8 counted by G(−A′;k,8) are 8, 8, 5+3
7
and 5+ 3. The overpartitions of 8 into parts congruent to 3, 5 or 6 modulo 7
(counted by F(−A ;k,8)) are 5+3, 5+3, 5+3 and 5+3. In both cases,we have
7
1 overpartition with 0 non-overlined parts, 2 overpartitions with 1 non-overlined
part, and 1 overpartition with 2 non-overlinedparts.
While the statements ofTheorems1.5 and 1.6resemble those of Andrews’theo-
rems,theproofsareconsiderablymoreintricateandinvolveanumberofnewideas.
The proofof Theorem1.6 uses ideas similar to the proofof Theorem 1.5 presented
in [Dou14a], in the sense that it relies in going back and forth from q-difference
equations to recurrence equations.
The remainder of this paper is devoted to the proof of Theorem 1.6. First, we
give the recurrence equation satisfied by the generating function for overpartitions
enumerated by G(−A′ ;k,n) having their largest part ≤ m, using some combina-
N
torialreasoningonthelargestpart. Thenweprovebyinductiononr thatthelimit
when m goes to infinity of a function satisfying this recurrence equation is equal
to r (−qN−a(j);qN)∞,whichisthegeneratingfunctionforoverpartitionscounted
j=1 (dqN−a(j);qN)∞
by F(−A ;k,n). Here we use the classical notation (a;q) := n−1(1−aqj).
Q N n j=0
Q
2. The recurrence equation
In this section, we establish the recurrence equation satisfied by the generating
function for overpartitions enumerated by G(−A′ ;k,n) having their largest part
N
≤m.
Letn,m∈N∗,k ∈N. Letπ (k,n)denotethenumberofoverpartitionscounted
m
by G(−A′ ;k,n) such that the largest part is ≤ m and overlined. Let φ (k,n)
N m
denotethe numberofoverpartitionscountedbyG(−A′ ;k,n)suchthatthelargest
N
partis≤mandnon-overlined. Thenψ (k,n):=π (k,n)+φ (k,n)isthenumber
m m m
of overpartitions counted by G(−A′ ;k,n) with largest part ≤m.
N
Then the following holds.
Lemma 2.1. We have
ψ (k,n)−ψ (k,n)
jN−α(m) jN−α(m+1)
(2.1) = ψjN−w(α(m))N−v(α(m))(k,n−jN +α(m))
+ψ (k,n−jN +α(m)).
jN−(w(α(m))−1)N−v(α(m))
4 JEHANNEDOUSSE
Proof: Let us first prove the following equation:
π (k,n)=π (k,n)
jN−α(m) jN−α(m+1)
(2.2) +πjN−w(α(m))N−v(α(m))(k,n−jN +α(m))
+φ (k,n−jN +α(m)).
jN−(w(α(m))−1)N−v(α(m))
We break the overpartitions counted by π (k,n) into two sets : those
jN−α(m)
with largest part < jN −α(m) and those with largest part equal to jN −α(m).
The first set is counted by π (k,n), and the second by
jN−α(m+1)
π (k,n−jN +a(m))
jN−w(α(m))N−v(α(m))
+φ (k,n−jN +a(m)).
jN−(w(α(m))−1)N−v(α(m))
To see this, let us consider an overparition n = λ + λ +··· + λ counted by
1 2 s
π (k,n) with largestpart equal to jN −α(m). Now removeits largestpart
jN−α(m)
λ =jN−α(m). Thenumberpartitionedbecomesn−jN+α(m). Thelargestpart
1
was overlined so the number of non-overlined parts is still k. If λ was overlined,
2
then we have
λ ≤λ −w(α(m))N −v(α(m))+α(m)
2 1
≤jN −w(α(m))N −v(α(m)),
andweobtainanoverpartitioncountedbyπ (k,n−jN+a(m)).
jN−w(α(m))N−v(α(m))
If λ was not overlined, then we have
2
λ ≤λ −(w(α(m))−1)N −v(α(m))+α(m)
2 1
≤jN −(w(α(m))−1)N −v(α(m)),
and we obtain an overpartition counted by φ (k,n−jN +
jN−(w(α(m))−1)N−v(α(m))
a(m)).
In the same way we can prove the following
φ (k,n)=φ (k,n)
jN−α(m) jN−α(m+1)
(2.3) +πjN−w(α(m))N−v(α(m))(k−1,n−jN +α(m))
+φ (k−1,n−jN +α(m)).
jN−(w(α(m))−1)N−v(α(m))
Adding equations (2.2) and (2.3) and noting that for all m,n,k, π (k−1,n)=
m
φ (k,n) (we can either overline the largest part or not), we obtain equation (2.1).
m
(cid:3)
We define, for m≥1, |q|<1, |d|<1,
∞ ∞
g =g (q,d):=1+ ψ (k,n)qndk,
m m m
n=1k=0
XX
andforall0≤k≤r−1,wesetg (q,d)=(−d)k forallkN ≤m≤(k+1)N. This
−m
definitionis consistentwith(2.1)andthe conditionthatλ ≥N(w(β (−λ ))−1).
s N s
We wantto find lim g , which is the generatingfunction for alloverpartitions
m
m→∞
counted by G(−A′ ;k,n). To do so, we establish a recurrence equation relating
N
g , for 0 ≤ j ≤ r. Let us start by giving some relations between gener-
(m−j)N−a(1)
ating functions.
Lemma 2.1 directly implies
A GENERALISATION OF A SECOND PARTITION THEOREM OF ANDREWS 5
Lemma 2.2. We have
g =g +qjN−α(m)g
jN−α(m) jN−α(m+1) jN−w(α(m))N−v(α(m))
(2.4)
+dqjN−α(m)g .
jN−(w(α(m))−1)N−v(α(m))
Let 1 ≤ k ≤ r+1. Adding equations (2.4) together for 1 ≤ m ≤ 2k−1−1, using
the fact that α 2k−1 =a(k), we obtain
(cid:0) (cid:1)
g =g
jN−a(1) jN−a(k)
(2.5) + qjN−αg +dqjN−αg .
(j−w(α))N−v(α) (j−w(α)+1)N−v(α)
α<Xa(k)(cid:0) (cid:1)
Let us now add equations (2.4) together for 2k−2 ≤m≤2k−1−1. This gives
(2.6)
g =g
jN−a(k) jN−a(k+1)
+ qjN−αg +dqjN−αg .
(j−w(α))N−v(α) (j−w(α)+1)N−v(α)
a(k)≤αX<a(k+1)(cid:0) (cid:1)
Every a(k)<α<a(k+1) is of the form α=a(k)+α′, with α′ <a(k). Hence we
can rewrite (2.6) as
g −g
jN−a(k) jN−a(k+1)
=qjN−a(k)g +dqjN−a(k)g
(j−1)N−a(k) jN−a(k)
+ qjN−a(k)−α′g(j−w(α′)−1)N−v(α′)+dqjN−a(k)−α′g(j−w(α′))N−v(α′)
α′X<a(k)(cid:16) (cid:17)
=qjN−a(k)g +dqjN−a(k)g
(j−1)N−a(k) jN−a(k)
+qN−a(k) g −g ,
(j−1)N−a(1) (j−1)N−a(k)
where the last(cid:0)equality follows from (2.5).(cid:1)
Thus
1−qjN−a(k) g =g +qN−a(k)g
jN−a(k) jN−a(k+1) (j−1)N−a(1)
(2.7)
(cid:16) (cid:17)
−qN−a(k) 1−q(j−1)N g .
(j−1)N−a(k)
(cid:16) (cid:17)
Wewanttofindtherecurrenceequationsatisfiedby(gℓN−a(1))ℓ∈N. Beforedoing
so, we must recall some facts about q-binomial coefficients defined by
m (1−qm)(1−qm−1)...(1−qm−r+1) if 0≤r ≤m,
:= (1−q)(1−q2)...(1−qr)
(cid:20)r(cid:21)q (0 otherwise.
They are q-analogues of the binomial coefficients and satisfy q-analogues of the
Pascaltriangle identity [Gas90].
Proposition 2.3. For all integers 0≤r≤m,
m m−1 m−1
(2.8) =qr + ,
r r r−1
(cid:20) (cid:21)q (cid:20) (cid:21)q (cid:20) (cid:21)q
6 JEHANNEDOUSSE
m m−1 m−1
(2.9) = +qm−r .
r r r−1
(cid:20) (cid:21)q (cid:20) (cid:21)q (cid:20) (cid:21)q
As q →1 these equations become Pascal’s identity.
Wearenowreadytostatethekeylemmawhichwillleadtothedesiredrecurrence
equation.
Lemma 2.4. For 1≤k ≤r+1, we have
k−1
1−dqℓN−a(j) g =g
ℓN−a(1) ℓN−a(k)
jY=1(cid:16) (cid:17)
k−1 k−j−1
j+m−1
+  dm qℓN−α (−1)m−1qℓ(m−1)N
(2.10) Xj=1 mX=0 α<Xa(k) (cid:20) m−1 (cid:21)q−N
 w(α)=j+m

j−1
j+m
+(−1)mqℓmN m  1−q(ℓ−h)N g(ℓ−j)N−a(1).
(cid:20) (cid:21)q−N!hY=1(cid:16) (cid:17)


Proof: Toprovethislemma,itissufficienttoreplaceq byq−1,thenxbyqℓN and
finally f qmN by g in the proof of Lemma 2.4 of [Dou14a]. (cid:3)
a(i) mN−a(i)
Writingu :=g andsettingk =r+1inLemma2.4,weobtainthedesired
(cid:0)ℓ (cid:1)ℓN−a(1)
recurrence equation
r
1−dqℓN−a(j) u =u
ℓ ℓ−1
jY=1(cid:16) (cid:17)
r r−j
j+m−1
+  dm qℓN−α (−1)m−1qℓ(m−1)N
m−1
(recN,r) Xj=1mX=0 α<Xa(r+1) (cid:20) (cid:21)q−N
 w(α)=j+m

j−1
j+m
+(−1)mqℓmN  1−q(ℓ−h)N uℓ−j,
m
(cid:20) (cid:21)q−N!hY=1(cid:16) (cid:17)


with the initial conditions u =(−d)k for all 0≤k ≤r−1.
−k
3. Evaluating lim u by induction
ℓ
ℓ→∞
Inthissection,weevaluate lim u ,whichisthegeneratingfunctionforpartitions
ℓ
ℓ→∞
counted by G(−A′ ;k,n). To do so, we prove the following theorem by induction
N
on r.
Theorem 3.1. Let r be a positive integer. Then for every N ≥α(2r−1), for every
sequence (um)m∈N satisfying (recN,r) and the initial condition u0 =1, we have
r (−qN−a(k);qN)
∞
lim u = .
ℓ→∞ ℓ (dqN−a(k);qN)∞
k=1
Y
A GENERALISATION OF A SECOND PARTITION THEOREM OF ANDREWS 7
The idea of the proof is to start from a function satisfying (rec ) and to do
N,r
some transformationsto obtain a function satisfying (rec ) in order to use the
N,r−1
inductionhypothesis. Inordertosimplifytheproof,wesplititintoseverallemmas.
Lemma 3.2. Let (u ) and (β ) be two sequences such that for all m∈N,
m m
m 1−dqjN−a(r)
β :=u
m m 1−qjN
j=1
Y
Then u = 1 and (u ) satisfies (rec ) if and only if β = 1 and (β ) satisfies
0 m N,r 0 m
the following recurrence equation
r
1+ dj−1 q−α+dj q−α(−1)jqjℓNβℓ
(rec′N,r)  Xj=1 wα(α<X)a=(jr−)1 wα<(Xαa)(=r)j  
   
r r min(j−1,h−1)
=β + c b (−1)h+1qhℓNβ ,
ℓ−1 k,j h−k,j ℓ−j
j=1h=1 k=0
XX X
where
ck,j :=q−Nk(k2+1)−ka(r) j−1 dk,
k
(cid:20) (cid:21)q−N
and
j+m−1
bm,j :=dm−1 q−α+dm q−α .
m−1
 α<Xa(r+1) α<Xa(r+1) (cid:20) (cid:21)q−N
 w(α)=j+m−1 w(α)=j+m 
 
Proof: Directly plugging the definition of (β ) into (rec ), we get
m N,r
r−1
(1−qℓN) 1−dqℓN−a(j) β =β
ℓ ℓ−1
jY=1(cid:16) (cid:17)
r r−j
j+m−1
+  dm qℓN−α (−1)m−1qℓ(m−1)N
m−1
Xj=1mX=0 α<Xa(r+1) (cid:20) (cid:21)q−N
 w(α)=j+m

j−1
j+m
+(−1)mqℓmN  1−dq(ℓ−h)N−a(r) βℓ−j.
m
(cid:20) (cid:21)q−N!hY=1(cid:16) (cid:17)


With the conventions that
q−α =0 for n≥r,
α<Xa(r)
w(α)=n
and
q−α =1,
α<Xa(r)
w(α)=0
8 JEHANNEDOUSSE
this can be reformulated as
r
1+ dj−1 q−α+dj q−α(−1)jqjℓNβℓ =βℓ−1
 Xj=1 α<Xa(r) α<Xa(r)  
  w(α)=j−1 w(α)=j  
   
r r−j+1
j+m−1
+  dm−1 q−α+dm q−α
m−1
Xj=1 mX=1  α<Xa(r+1) α<Xa(r+1) (cid:20) (cid:21)q−N
  w(α)=j+m−1 w(α)=j+m 
  
j−1
(−1)m−1qmℓN q−Nk(k2−1)−ka(r) j−1 dk(−1)kqkℓN βℓ−j,
k
k=0 (cid:20) (cid:21)q−N !
 X


because of the q-binomial theorem [Gas90]
n−1 n
(3.1) (1+qkt)= qk(k2−1) n tk,
k
k=0 k=0 (cid:20) (cid:21)q
Y X
in which we replace q by q−N, n by j−1 and t by −dq(ℓ−1)N−a(r). Finally, noting
that b = 0 if j +l−k −1 ≥ r, we can rewrite this as (rec′ ). Moreover,
l−k,j N,r
β =u =1 and the lemma is proved. (cid:3)
0 0
Wecandirectlytransform(rec′ )intoaq-differenceequationonthegenerating
N,r
function for (β ).
m
Lemma 3.3. Let (β ) be a sequence and f a function such that
m
∞
f(x):= β xn.
n
n=0
X
Then (β ) satisfies (rec′ ) and the initial condition β =1 if and only if f(0)=1
m N,r 0
and f satisfies the following recurrence equation
(eq )
N,r
r
(1−x)f(x)= dm−1 q−α+dm qα

mX=1 α<Xa(r) α<Xa(r)
w(α)=m−1 w(α)=m
r min(j−1,m−1)
+ c b xjqmjN (−1)m+1f(xqmN).
k,j m−k,j
!
j=1 k=0
X X
Proof: By the definition of f and (rec′ ), we have
N,r
r
(1−x)f(x)= dj−1 q−α+dj q−α(−1)j+1f xqjN
Xj=1 α<Xa(r) α<Xa(r)  (cid:0) (cid:1)
 w(α)=j−1 w(α)=j 
 
r r min(j−1,h−1)
+ c b (−1)h+1xjqhjNf xqhN .
k,j h−k,j
j=1h=1 k=0
XX X (cid:0) (cid:1)
A GENERALISATION OF A SECOND PARTITION THEOREM OF ANDREWS 9
Relabelling the summation indices and factorising leads to (eq ). Moreover,
N,r
f(0)=β =1. This completes the proof. (cid:3)
0
Let us now do some transformations starting from (rec ).
N,r−1
Lemma 3.4. Let (µ ) and (s ) be two sequences such that for all n,
n n
n
1
s :=µ .
n n 1−qNk
k=1
Y
Then (µ ) satisfies (rec ) and the initial condition µ =1 if and only if s =1
n N,r−1 0 0
and (s ) satisfies the following recurrence equation
n
(rec′′ )
N,r−1
r
1+ dj−1 q−α+dj q−α(−1)jqjℓNsℓ =sℓ−1
 Xj=1 α<Xa(r) α<Xa(r)  
  w(α)=j−1 w(α)=j  
   
r r−j
+ dm−1 q−α+dm q−α
Xj=1mX=1 α<Xa(r) α<Xa(r) 
 w(α)=j+m−1 w(α)=j+m 
 
j+m−1
× (−1)m+1qmℓNs .
ℓ−j
m−1
(cid:20) (cid:21)q−N
Proof: Using the definition of (s ) and (rec ), we get
n N,r−1
r−1
(1−qℓN) 1−dqℓN−a(j) s =s
ℓ ℓ−1
jY=1(cid:16) (cid:17)
r−1 r−j−1
j+m−1
+  dm qℓN−α (−1)m−1qℓ(m−1)N
m−1
Xj=1 mX=0 α<Xa(r+1) (cid:20) (cid:21)q−N
 w(α)=j+m

j−1
j+m
+(−1)mqℓmN  1−dq(ℓ−h)N−a(r) sℓ−j.
m
(cid:20) (cid:21)q−N!hY=1(cid:16) (cid:17)


Then, as in the proof of Lemma 3.2, this can be reformulated as (rec′′ ), and
N,r−1
s =µ =1. (cid:3)
0 0
Again,letustranslatethisintoarecurrenceequationonthegeneratingfunction
for (s ).
n
Lemma 3.5. Let (s ) be a sequence and G be a function such that
n
∞
G(x):= s xn.
n
n=0
X
10 JEHANNEDOUSSE
Then(s )satisfies(rec′′ )andtheinitialconditions =1ifandonlyifG(0)=1
n N,r−1 0
and G satisfies the following q-difference equation
(eq′′ )
N,r−1
r r−1
(1−x)G(x)= dm−1 q−α+dm q−α
mX=1Xj=0 α<Xa(r) α<Xa(r) 
 w(α)=j+m−1 w(α)=j+m 
 
j+m−1
× (−1)m+1xjqjmNG xqmN .
m−1
(cid:20) (cid:21)q−N
(cid:0) (cid:1)
Proof: By the definition of G and (rec′′ ), we have
N,r−1
r
(1−x)G(x)= dm−1 q−α+dm q−α(−1)m+1G xqmN
mX=1 α<Xa(r) α<Xa(r)  (cid:0) (cid:1)
 w(α)=m−1 w(α)=m 
 
r−1 r−1
+ dm−1 q−α+dm q−α
mX=1Xj=1 α<Xa(r) α<Xa(r) 
 w(α)=j+m−1 w(α)=j+m 
 
j+m−1
× (−1)m+1xjqjmNG xqmN .
m−1
(cid:20) (cid:21)q−N
(cid:0) (cid:1)
As the summand of the second term equals 0 when m = r, we can equivalently
write that the second sum is taken on m going from 1 to r. Then we observe
that the first term corresponds to j = 0 in the second term, and factorising gives
exactly (eq′′ ). Moreover, G(0)=s =1. This completes the proof. (cid:3)
N,r−1 0
Let us do a final transformationand obtain yet another q-difference equation.
Lemma 3.6. Let G and g be two sequences such that
∞
g(x):=G(x) 1+xqkN−a(r) .
kY=1(cid:16) (cid:17)
Then Gsatisfies (eq′′ )andtheinitial condition G(0)=1ifandonlyif g(0)=1
N,r−1
and g satisfies the following q-difference equation
(eq′ )
N,r−1
r r−1min(m−1,ν)
(1−x)g(x)= f e xνqνmN
m,µ m,ν−µ

m=1 ν=0 µ=0
X X X

r min(m−1,ν−1)
+ q−a(r) f e xνqνmN (−1)m+1g xqmN ,
m,µ m,ν−µ−1

ν=1 µ=0
X X (cid:0) (cid:1)

where
j+m−1
em,j :=dm−1 q−α+dm q−α ,
m−1
 α<Xa(r) α<Xa(r) (cid:20) (cid:21)q−N
 w(α)=j+m−1 w(α)=j+m 
 