On the Andrews-Stanley Refinement of Ramanujan's Partition Congruence Modulo 5

0.21 MB·English

by  Alexander Berkovich

ON THE ANDREWS-STANLEY REFINEMENT OF
RAMANUJAN’S PARTITION CONGRUENCE MODULO 5
4
0
0 ALEXANDERBERKOVICHANDFRANKG.GARVAN
2

n
Abstract. In a recent study of sign-balanced, labelled posets Stanley [13],
a
introducedanewintegralpartitionstatistic
J
srank(π)=O(π)−O(π′),
3
where O(π)denotes the number ofoddparts of thepartition π and π′ is the

] conjugateofπ. In[1]AndrewsprovedthefollowingrefinementofRamanujan’s
O partitioncongruencemod5:
C p0(5n+4)≡p2(5n+4)≡0 (mod5),
h. p(n)=p0(n)+p2(n),
t where pi(n) (i = 0,2) denotes the number of partitions of n with srank ≡ i
a
(mod4) and p(n) is the number of unrestricted partitions of n. Andrews
m
askedforapartitionstatisticthatwoulddividethepartitions enumeratedby
[ pi(5n+4)(i=0,2)intofiveequinumerous classes.

In this paper we discuss two such statistics. The first one, while new, is
1
intimatelyrelatedtotheAndrews-Garvan[2]crank. Thesecondoneisinterms
v of the 5-core crank, introduced by Garvan, Kim and Stanton [9]. Finally, we
2 discusssomenewformulasforpartitionsthatare5-cores.
1
0
1
0
4 1. Introduction
0
Let p(n) be the number of unrestricted partitions of n. Ramanujan discovered
/
h and later proved that
t
a (1.1) p(5n+4) 0 (mod 5),
m ≡
(1.2) p(7n+5) 0 (mod 7),
: ≡
v
(1.3) p(11n+6) 0 (mod 11).
i ≡
X
Dyson[5]wasthefirsttoconsidercombinatorialexplanationsofthesecongruences.
r Hedefinedtherankofapartitionasthe largest part minus the numberof partsand
a
made the empirical observations that
p(5n+4)
(1.4) N(k,5,5n+4)= , 0 k 4,
5 ≤ ≤
p(7n+5)
(1.5) N(k,7,7n+5)= , 0 k 6,
7 ≤ ≤
where N(k,m,n) denotes the number of partitions of n with rank congruent to k
modulo m. Equation (1.4) means that the residue of the rank mod 5 divides the
partitionsof5n+4intofiveequalclasses. Similarly,(1.5)impliesthattheresidueof
Date:December26,2003.
1991 Mathematics Subject Classification. Primary11P81,11P83; Secondary05A17,05A19.
Key words and phrases. partitions, t-cores, ranks, cranks, Stanley’s statistic, Ramanujan’s
congruences.
1
2 ALEXANDERBERKOVICHANDFRANKG.GARVAN
therankmod7dividesthepartitionsof7n+5intosevenequalclasses. Dyson’srank
failed to explain (1.3), and so Dyson conjectured the existence of a hypothetical
statistic, called the crank, that would explain the Ramanujan congruence mod 11.
Identities(1.4)-(1.5)werelaterprovedbyAtkinandSwinnerton-Dyer[3]. Andrews
and Garvan [2] found a crank for all three Ramanujan congruences (1.1)-(1.3).
Their crank is defined as follows
ℓ(π), if µ(π)=0,
(1.6) crank(π)=
∼
(ν(π) µ(π), if µ(π)>0,
−
whereℓ(π) denotesthe largestpartofπ, µ(π) denotes the number ofonesinπ and
∼
ν(π) denotes the number of parts of π larger than µ(π).
Later,Garvan,KimandStanton[9]founddifferentcranks,whichalsoexplained
all three congruences (1.1)-(1.3). Their approach made essential use of t-cores of
partitions and led to explicit bijections between various equinumerous classes. In
particular, they provided what amounts to a combinatorial proof of the formula
(1 q5m)5
(1.7) p(5n+4)qn =5 − ,
(1 qm)6
n≥0 m≥1 −
X Y
considered by Hardy to be an example of Ramanujan’s best work.
The main results of [2] can be summarized as
p(5n+4)
(1.8) M(k,5,5n+4)= , 0 k 4,
5 ≤ ≤
p(7n+5)
(1.9) M(k,7,7n+5)= , 0 k 6,
7 ≤ ≤
p(11n+6)
(1.10) M(k,11,11n+6)= , 0 k 10,
11 ≤ ≤
and
∼
1+(x+x−1 1)q+ M(m,n)xmqn
−
n>1 m
XX
(1 qn)
(1.11) = − ,
(1 xqn)(1 x−1qn)
n≥1 − −
Y
∼
whereM(m,n)denotesthenumberofpartitionsofnwithcrankmandM(k,m,n)
denotes the number of partitions of n with crank congruent to k modulo m.
In [7] Garvan found a refinement of (1.1)
(1.12) M(k,2,5n+4) 0 (mod 5), k =0,1
≡
together with the combinatorial interpretation
M(α,2,5n+4)
(1.13) M(2k+α,10,5n+4)= , 0 k 4,
5 ≤ ≤
with α=0,1.
Recently,averydifferentrefinementof(1.1)wasgivenbyAndrews[1]. Building
on the work of Stanley [13], Andrews examined partitions π classified according
to (π) and (π′), where where (π) denotes the number of odd parts of the
O O O
THE ANDREWS-STANLEY REFINEMENT OF RAMANUJAN’S CONGRUENCE 3
partition π and π′ is the conjugate of π. He used recursive relations to show that
( zyq;q2)
(1.14) G(z,y,q):= S(n,r,s)qnzrys = − ∞ ,
(q4;q4) (z2q2;q4) (y2q2;q4)
∞ ∞ ∞
n,r,s≥0
X
where S(n,r,s) denotes the number of partitions π of n with (π)=r, (π′)=s,
O O
and
(1.15) (a;q) = lim (a;q) ,
∞ n
n→∞
1, if n=0,
(1.16) (a;q) =(a) =
n n ( jn=−01(1−aqj), if n>0.
Adirectcombinatorialproofof(1.14)QwaslatergivenbyA.Sills[12],A.J.Yee[14]
andC. Boulet[4]. Actually, C. Boulet proveda stronger versionof (1.14) with one
extra parameter. We define the Stanley rank of a partition π as
(1.17) srank(π)= (π) (π′).
O −O
It is easy to see that
(1.18) srank(π) 0 (mod 2),
≡
so that
(1.19) p(n)=p (n)+p (n),
0 2
wherep (n)(i=0,2)denotesthenumberofpartitionsofnwithsrank i (mod 4).
i
≡
We note that (1.14) with z =y−1 =√ 1 immediately implies the Stanley formula
−
[13, p.8]
( q;q2)
(1.20) (p (n) p (n))qn = − ∞ .
0 − 2 (q4;q4) ( q2;q4)2
n≥0 ∞ − ∞
X
Using (1.1), (1.19) and (1.20), Andrews proved the following refinement of (1.1)
(1.21) p (5n+4) p (5n+4) 0 (mod 5).
0 2
≡ ≡
His proof of (1.21) was analytic and so at the end of [1] he posed the problem of
findingapartitionstatisticthatwouldgiveacombinatorialinterpretationof(1.21).
The object of this paper is to provide a solution to the Andrews problem. It turns
outthattherearetwodistinctintegralpartitionstatistics,whoseresiduemod5split
the partitions enumeratedby p (5n+4) (with i=0,2)into five equal classes. The
i
first statistic, which we call the stcrank, is new. However, it is intimately related
to the Andrews-Garvan crank (1.6). Unexpectedly, the second statistic is the “5-
corecrank”,introducedby Garvan,Kim andStanton[9]. Thissecondstatistic not
onlyprovidesthe desiredcombinatorialinterpretation,butit alsoprovidesa direct
combinatorialproof of (1.21).
The restofthis paper is organizedas follows. In Section2 wedefine the stcrank
andshowthatisindeed,astatisticaskedforin[1]. InSection3webrieflyreviewthe
development in [9]. In Section 4 we state a number of new formulas for partitions
that are 5-cores. We sketch the “5-core crank” proof of (1.21) and we conclude
with some open problems.
4 ALEXANDERBERKOVICHANDFRANKG.GARVAN
2. The stcrank
We begin with some preliminaries about partitions and their conjugates. A
partition π is a nonincreasing sequence
(2.1) π =(λ ,λ ,λ ,...)
1 2 3
of nonnegative integers (parts)
(2.2) λ λ λ .
1 2 3
≥ ≥ ≥···
The weight of π, denoted by π is the sum of parts
| |
(2.3) π =λ +λ +λ + .
1 2 3
| | ···
If π = n, then we say that π is a partition of n. Often it is convenient to use
| |
another notation for π
(2.4) π =(1f1,2f2,3f3,...),
which indicates the number of times each integer occurs as a part. The number
f = f (π) is called the frequency of i in π. The conjugate of π is the partition
i i
π′ =(λ′,λ′,λ′,...) with
1 2 3
λ′ =f +f +f +f +
1 1 2 3 4 ···
(2.5) λ′ =f +f +f +
2 2 3 4 ···
λ′ =f +f +
3 3 4 ···
.
.
.
Next, we discuss two bijections. The first one relates π and bipartitions (π ,π ),
1 2
where π is a partition with no repeated even parts.
2
Bijection 1
1
π (π ,π ),
1 2
−→
where
π =(1f1,2f2,3f3,...),
π =(1⌊f2/2⌋,2⌊f4/2⌋,3⌊f6/2⌋,...),
1
π =(1f1,2{f2},3f3,4{f2},...),
2
x is the largest integer x, and
⌊ ⌋ ≤
x =x 2 x/2 .
{ } − ⌊ ⌋
Indeed, remove from π the maximum even number of even parts. The resulting
partition is π , The removed even parts can be organized into a new partition
2
(22⌊f2/2⌋,42⌊f4/2⌋,62⌊f6/2⌋,...), which can easily be mapped onto π . Clearly, we
1
have
(2.6) π =4π + π ,
1 2
| | | | | |
(2.7) srank(π)=srank(π ),
2
THE ANDREWS-STANLEY REFINEMENT OF RAMANUJAN’S CONGRUENCE 5
(cid:0)(cid:1)(cid:0)(cid:1)
(cid:0)(cid:1)(cid:0)(cid:1)
(cid:0)(cid:1)(cid:0)(cid:1)
π π
A 2 B
(cid:0)(cid:1)(cid:0)(cid:1)
(cid:0)(cid:1)(cid:0)(cid:1)
(cid:0)(cid:1)(cid:0)(cid:1)
(cid:0)(cid:1)(cid:0)(cid:1)
(cid:0)(cid:1)
(cid:0)(cid:1)
(cid:0)(cid:1)
(cid:0)(cid:1)
Figure 1. Graphical illustration of Bijection 2
so that
q|π|ysrank(π) = q4|π1| q|π2|ysrank(π2)
Xπ Xπ1 Xπ2
1
(2.8) = q|π2|ysrank(π2).
(q4;q4)
∞
Xπ2
Comparing (2.8) and (1.14) with zy =1, we see that
( q;q2)
(2.9) q|π2|ysrank(π2) = − ∞ ,
(y2q2;q4) (q2/y2;q4)
∞ ∞
Xπ2
where the sum is over all partitions with no repeated even parts.
To describe our second bijection we require a few definitions. We say that π
A
1
is a partition of type A iff π ((1),π ). We say that π = (λ ,λ ,λ ,...) is
A 2 B 1 2 3
−→
a partition of type B iff either π = 4, λ λ 2, λ′ λ′ 2, λ 2 and λ
| B| 6 1− 2 ≥ 1− 2 ≥ 1− 2
are not identical even integers and π has no repeated even parts, or π = (3,1).
B B
1
Obviously, π ((0),π ). Our second bijection relates partitions of type A and
B B
−→
B.
Bijection 2
2
π π ,
A B
−→
where
π =(1f1,2f2,3f3,...,mfm),
A
(1f1+2,2f2−2,3f3,4f4,...,(m 1)fm−1,mfm−1,(m+2)1), if m>2,
−
π = (1f1+2,41), if m=2, f =3,
B  2
(1f1+1,31), if m=2, f2 =2,
m≥2, f2 =2, 3, and f2i =0, 1 for i>1.
Clearly, we have
(2.10) π = π ,
A B
| | | |
(2.11) srank(π )=srank(π ).
A B
Next, we define a new partition statistic
1
(2.12) stcrank(π)=crank(π )+ srank(π)+Ψ(π),
1
2
6 ALEXANDERBERKOVICHANDFRANKG.GARVAN
1
where π is determined by π (π ,π ), and the correctionterm Ψ(π)=1 if π is
1 1 2
−→
of type B and zero, otherwise. We note that
1
(2.13) stcrank(π )= 1+ srank(π ),
A A
− 2
and
1
(2.14) stcrank(π )=1+ srank(π ).
B B
2
Equipped with the definitions above, we can now prove the following lemma.
Lemma 2.1. If
g(x,y,q):= q|π|xstcrank(π)ysrank(π),
π
X
then g(x,y,q) has the product representation
(q4;q4) ( q;q2)
∞ ∞
g(x,y,q)= − ,
(q4x,q4/x,q2y2x,q2/(y2x);q4)
∞
where
(a ,a ,a ,...;q) =(a ;q) (a ;q) (a ;q) .
1 2 2 ∞ 1 ∞ 2 ∞ 3 ∞
···
Proof. If π is not of type B and π 1 (π ,π ), then using (2.6)-(2.7), (2.12) we
1 2
−→
find that
(2.15) q|π|xstcrank(π)ysrank(π) =q4|π1|+|π2|xcrank(π1)(xy2)srank(π2)/2.
2
On the other hand, if π =π and π π , then
A A B
−→
q|πA|xstcrank(πA)ysrank(πA)+q|πB|xstcrank(πB)ysrank(πB)
(2.16) =q|πA|(x+x−1 1)(xy2)srank(πA)/2+q|πB|x0(xy2)srank(πB)/2.
−
Here we have used (2.6)-(2.7) and (2.10)-(2.14).
Equations (2.15), (2.16) imply that
(2.17) q|π|xstcrank(π)ysrank(π) = q4|π1|w(x,π ) q|π2|(xy2)srank(π2)/2
1
Xπ Xπ1 Xπ2
where
x+x−1 1, if π =(1),
1
(2.18) w(x,π )= −
1 (xcrank(π1), otherwise.
We note that in the first sum on the right side of (2.17) the summation is over
unrestrictedpartitions π , and in the secondsum the summation is overpartitions
1
π withnorepeatedevenparts. Finally,recalling(1.11)withq q4 and(2.9)with
2
→
y2 xy2, we obtain
→
(q4;q4) ( q;q2)
q|π|xstcrank(π)ysrank(π) = ∞ − ∞ ,
(xq4,q4/x;q4) · (xy2q2,q2/(xy2);q4)
∞ ∞
π
X
as desired. (cid:3)
Next we show that
(2.19) the coefficient of q5n+4 in g(ξ,1,q)=0,
(2.20) the coefficient of q5n+4 in g(ξ,√ 1,q)=0,
−
THE ANDREWS-STANLEY REFINEMENT OF RAMANUJAN’S CONGRUENCE 7
where ξ is a primitive fifth root of unity (ξ5 = 1). We use the method of [6]. We
need Jabobi’s triple product identity
∞
(2.21) znqn2 =(q2, qz q/z;q2) ,
∞
− −
n=−∞
X
which implies that
∞
(2.22) (q4;q4) ( q;q2) =(q4, q3, q;q4) = q2n2+n = qTk,
∞ ∞ ∞
− − −
n=−∞ k≥0
X X
and
1
(2.23) (q2ξ2,q2/ξ2,q2;q2) = ( 1)mq2Tmξ−2m(1 ξ4m+2).
∞ 1 ξ2 − −
− m≥0
X
Here T =k(k+1)/2. By Lemma 2.1 and equations (2.22) and (2.23) we have
k
qTk (q2ξ2,q2/ξ2,q2;q2)
g(ξ,1,q)= k≥0 = ∞ qTk
(q4ξ,q4/ξ,q2ξ,q2/ξ;q4) (q10;q10)
P ∞ ∞ k≥0
X
1 1
(2.24) = ( 1)mq2Tm+Tkξ−2m(1 ξ4m+2).
1 ξ2(q10;q10) − −
∞
− k,m≥0
X
Note that 2T +T 4 (mod 5) iff k m 2 (mod 5), but then 1 ξ4m+2 =0.
m k
≡ ≡ ≡ −
This proves (2.19). The proof of (2.20) is analogous.
LetP (k,m,n)denotethenumberofpartitionsofnwithsrank i (mod 4)and
i
≡
stcrank k (mod m). Clearly,
≡
4 g(ξ,1,q)+g(ξ,√ 1,q)
(2.25) ξk P (k,5,n)qn = − ,
0
2
k=0 n≥0
X X
4 g(ξ,1,q) g(ξ,√ 1,q)
(2.26) ξk P (k,5,n)qn = − − ,
2
2
k=0 n≥0
X X
Combining (2.19)-(2.20) and (2.25)-(2.26) we find that
4
(2.27) ξkP (k,5,5n+4)=0,(for i=0,2),
i
k=0
X
which implies that
(2.28) P (0,5,5n+4)=P (1,5,5n+4)= =P (4,5,5n+4).
i i i
···
On the other hand
4
(2.29) p (5n+4)= P (k,5,5n+4),
i i
k=0
X
so that
1
(2.30) P (k,5,5n+4)= p (5n+4),
i i
5
for i = 0,2 and k = 0,1,2,3,4. Thus, we have proved the main result of this
section.
8 ALEXANDERBERKOVICHANDFRANKG.GARVAN
Theorem 2.2. The residue of the partition statistic stcrank mod 5 divides the
partitions enumerated by p (5n+4) with i=0,2 into five equinumerous classes.
i
We illustrate this theorem in Table 1 below for the 30 partitions of 9. These
partitions are organized into five classes with six members each. In each class the
first 4 members have srank 0 (mod 4) and the remaining two members have
≡
srank 2 (mod 4).
≡
Table 1.
stcrank 0(mod 5) 1(mod 5) 2(mod 5) 3(mod 5) 4(mod 5)
≡
srank 0 (33) (15,22) (14,21,31) (11,24) (19)
≡
(mod 4) (13,21,41) (14,51) (13,32) (16,31) (12,22,31)
(11,31,51) (12,21,51) (11,42) (11,21,61) (23,31)
(41,51) (91) (22,51) (21,71) (12,71)
srank 2 (13,23) (11,21,32) (15,41) (17,21) (21,31,41)
≡
(mod 4) (13,61) (12,31,41) (11,81) (11,22,41) (31,61)
Finally, we note that the equation
(2.31) srank(π)= srank(π′)
−
implies that a partition π is self-conjugate only if srank(π) = 0. This means that
the involution π π′ has no fixed points if srank(π) 2 (mod 4). Hence,
−→ ≡
2 p (5n+4) and by (1.21) we have the stronger congruence
2
|
p (5n+4) 0 (mod 10).
2
≡
3. t-cores
In this section we recall some basic facts about t-cores and briefly review the
development in [9]. A partition π is a called a t-core, if it has no rim hooks of
length t [10]. We let a (n) denote the number of partitions of n which are t-cores.
t
Inwhatfollows,πt-coredenotesat-corepartition. Giventhediagramofapartition
π we label a cell in the i-th row and j-th column by the least nonnegative integer
congruenttoj i (mod t). Theresultingdiagramiscalledat-residuediagram[10,
−
p.84].
Let P be the set of all partitions and Pt-core be the set of all t-cores. There is
well-known bijection which goes back to Littlewood [11]. φ1 : P Pt-core P
→ × ×
P,
···×
(3.1) φ1(π)=(πt-core,π~t),
(3.2) π~ =(π ,π ,π ,...,π ),
t 0 1 2 t−1
b
such that
b b b b b
t−1
(3.3) π = πt-core +t πi .
| | | | | |
i=0
X
b
THE ANDREWS-STANLEY REFINEMENT OF RAMANUJAN’S CONGRUENCE 9
This bijection is described in moredetail in [10], [9] and[8]. The followingidentity
is an immediate corollary of this bijection.
1 1
(3.4) = p(n)qn = a (n)qn.
(q) (qt;qt)t t
∞ n≥0 ∞ n≥0
X X
It can be rewritten as
(qt;qt)t
(3.5) a (n)qn = ∞.
t
(q)
∞
n≥0
X
There is another bijection φ , introduced in [9]. It is for t-cores only. φ :
2 2
Pt-core ~n=(n0,n1,...,nt−1) : ni Z,n0+ +nt−1 =0 ,
→{ ∈ ··· }
(3.6) φ2(πt-core)=~n=(n0,n1,n2,...,nt−1).
We call~n an n-vector. It has the following properties.
(3.7) ~n Zt, ~n ~1 =0,
t
∈ ·
and
t−1 t−1
t
(3.8) |πt-core|= 2 n2i + ini,
i=0 u=0
X X
where the t-dimensional vector~1 has all components equal to 1. The generating
t
function identity that corresponds to this second bijection is
(3.9) at(n)qn = q2t||~n||2+~bt·~n.
nX≥0 ~nX∈Zt
~n·~1t=0
Here
t−1
(3.10) ~n 2 = n2, and ~b =(0,1,2,...,t 1).
|| || i t −
i=0
X
To construct the n-vector of πt-core in (3.6), we follow [8] and define
(3.11) ~r(πt-core)=(r0,r1,r2,...,rt−1),
where for 0 i t 1, ri(πt-core) denotes the number of cells labelled i (mod t)
≤ ≤ −
in the t-residue diagram of πt-core. Then (3.6) can be given explicitly as
(3.12) φ2(πt-core)=~n=(r0 r1,r1 r2,r2 r3,...,rt−1 r0).
− − − −
We note that t ~n 2 is a multiple of t since~n ~1 =0. Hence by (3.4) and (3.9) we
2|| || · t
have
(3.13) at(tn+δ)qtn+δ = q2t||~n||2+~bt·~n,
nX≥0 ~n∈ZtX, ~n·~1t=0
~n·~bt≡δ (modt)
and
1
(3.14) p(tn+δ)qn = a (tn+δ)qn,
(q)t t
n≥0 ∞ n≥0
X X
where δ =0,1,2, ..., t 1.
−
10 ALEXANDERBERKOVICHANDFRANKG.GARVAN
Wenowassumet=5. Forthecaseδ =4therightsideof(3.13)canbesimplified
using the the following change of variables.
n =α +α ,
0 0 4
n = α +α +α ,
1 0 1 4
−
(3.15) n = α +α ,
2 1 2
−
n = α +α α ,
3 2 3 4
− −
n = α α ,
4 3 4
− −
We find~n is an n-vector satisfying~n ~b 4 (mod 5) if and only if
5
· ≡
(3.16) α~ =(α ,α ,α ,α ,α ) Z5
0 1 2 3 4
∈
and
(3.17) α +α +α +α +α =1.
0 1 2 3 4
We call α~ and α-vector. Hence, by (3.13) and (3.14) we have
(3.18) a (5n+4)qn+1 = qQ(α~),
5
nX≥0 α~·X~15=1
α~∈Z5
and
1
(3.19) p(5n+4)qn+1 = qQ(α~),
(q)5
nX≥0 ∞ α~·X~15=1
α~∈Z5
where
(3.20) Q(α~)= α~ 2 (α α +α α + +α α ),
0 1 1 2 4 0
|| || − ···
If π 4 (mod 5) and t = 5, we can combine bijections φ and φ into a single
1 2
| | ≡
bijection
(3.21) Φ(π)=(α~,π~ ),
5
such that
b 4
(3.22) π =5Q(α~) 1+5 π .
i
| | − | |
i=0
X
Next, following [9] we define the 5-core crank of π wbhen π 4 (mod 5) as
| |≡
4 2
(3.23) c (π)=1+ iα 2(1+n n n +n ) 2+ ir (mod 5),
5 i 0 1 2 3 2−i
≡ − − ≡
i=0 i=−2
X X
where α is determined by (3.21).
It is easy to check that Q(α~) in (3.22) remains invariant under the following
cyclic permutation
(3.24) C (α~)=(α ,α ,α ,α ,α ),
1 4 0 1 2 3
while c (π) increases by 1 (mod 5) under the map
5
b
(3.25) O(π)=Φ−1(C (α~),~π ).
1 5
In other words, if π 4 (mod 5), then
| |≡ b b b
(3.26) π = O(π),
| | | |
b