THE BG-RANK OF A PARTITION AND ITS APPLICATIONS

ALEXANDER BERKOVICH AND FRANK G. GARVAN

Abstract. Let π denote a partition into parts λ_{1} ≥ λ_{2} ≥ λ_{3} . . .. In a 2006 paper we defined BG-rank(π) as

BG-rank(π) = ∑_{j≥1}(−1)^{j+1} \frac{1 − (−1)^{λ_{j}}}{2} .

This statistic was employed to generalize and refine the famous Ramanujan modulo 5 partition congruence. Let p_{j}(n) denote the number of partitions of n with BG-rank = j. Here, we provide a combinatorial proof that

p_{j}(5n + 4) ≡ 0 (mod 5), j ∈ ℤ,

by showing that the residue of the 5-core crank mod 5 divides the partitions enumerated by p_{j}(5n + 4) into five equal classes. This proof uses the orbit construction from our previous paper and a new identity for the BG-rank. Let a_{t,j}(n) denote the number of t-cores of n with BG-rank = j. We find eta-quotient representations for

∑_{n≥0} a_{t,⌊ \frac{t+1}{4} ⌋}(n)q^{n} and ∑_{n≥0} a_{t,−⌊ \frac{t−1}{4} ⌋}(n)q^{n},

when t is an odd, positive integer. Finally, we derive explicit formulas for the coefficients a_{5,j}(n), j = 0, ±1.

1. Introduction

A partition π is a nonincreasing sequence

π = (λ_{1}, λ_{2}, λ_{3}, …)

of positive integers (parts) λ_{1} ≥ λ_{2} ≥ λ_{3} ≥ . . .. The norm of π, denoted |π|, is defined as

|π| = ∑_{i≥1} λ_{i}.

If |π| = n, we say that π is a partition of n. The (Young) diagram of π is a convenient way to represent π graphically: the parts of π are shown as rows of unit squares (cells). Given the diagram of π we label a cell in the i-th row and j-th column by the least nonnegative integer ≡ j − i (mod t). The resulting diagram is called a t-residue diagram [7]. We can also label cells in the infinite column 0 and the infinite row 0 in the same fashion and call the resulting diagram the extended t-residue diagram [5]. And so with each partition π and positive integer t we can

Date: April 27, 2007. 2000 Mathematics Subject Classification. Primary 11P81, 11P83; Secondary 05A17, 05A19. Key words and phrases. partitions, t-cores, BG-rank, eta-quotients, Lambert series, theta

series, even–odd dissections.

Research of both authors was supported in part by NSA grant MSPF-06G-150.

associate the t-dimensional vector

r⃗(π, t) = (r_{0}(π, t), r_{1}(π, t), …, r_{t−1}(π, t))

with

r_{i}(π, t) = r_{i}, 0 ≤ i ≤ t − 1

being the number of cells colored i in the t-residue diagram of π. If some cell of π shares a vertex or edge with the rim of the diagram of π, we call this cell a rim cell of π. A connected collection of rim cells of π is called a rim hook if (diagram of π)\(rim hook) represents a legitimate partition. We say that a partition is a t-core, denoted π_{t-core}, if its diagram has no rim hooks of length t [7].

The Durfee square of π is the largest square that fits inside the diagram of π. Reflecting the diagram of π about its main diagonal, one gets the diagram of π′ (the conjugate of π). More formally,

π′ = (λ′_{1}, λ′_{2}, λ′_{3}, …)

with λ′_{i} being the number of parts of π that are ≥ i. In [2] we defined a new partition statistic

(1.1)  BG-rank(π) := ∑_{j≥1}(−1)^{j} \frac{(−1)^{λ_{j}} − 1}{2}.

It is easy to verify that

(1.2)  BG-rank(π) = r_{0}(π, 2) − r_{1}(π, 2)

and

(1.3)  BG-rank(π) ≡ |π| (mod 2).

In [2] we proved the following (mod 5) congruences

(1.4)  p_{j}(5n) ≡ 0 (mod 5) if j ≡ 1, 2 (mod 5),
(1.5)  p_{j}(5n + 1) ≡ 0 (mod 5) if j ≢ 1, 2 (mod 5),
(1.6)  p_{j}(5n + 2) ≡ 0 (mod 5) if j ≢ 0, 3 (mod 5),
(1.7)  p_{j}(5n + 3) ≡ 0 (mod 5) if j ≡ 0, 3 (mod 5),
(1.8)  p_{j}(5n + 4) ≡ 0 (mod 5) for all j ∈ ℤ.

Here p_{j}(n) denotes the number of partitions of n with BG-rank = j. Clearly,

p(5n + 4) = ∑_{j} p_{j}(5n + 4)

with p(n) denoting the number of unrestricted partitions of n. And so (1.8) implies the famous Ramanujan congruence [11]

p(5n + 4) ≡ 0 (mod 5).

In this paper, we build on the developments in [2] to provide a combinatorial proof of (1.8).

For t-odd it is surprising that the BG-rank(π_{t−core}) assumes only finitely many values. In fact, we will show that if t is an odd, positive integer, then

(1.9)  −⌊\frac{t − 1}{4}⌋ ≤ BG-rank(π_{t−core}) ≤ ⌊\frac{t + 1}{4}⌋.

Here ⌊x⌋ denotes the integer part of x.

We will establish the following identities. For odd t > 1

(1.10)  C_{t,(−1)^{\frac{t−1}{2}} ⌊\frac{t−1}{4}⌋}(q) = q^{\frac{(t−1)(t−3)}{8}} F (t, q^{2}),
(1.11)  C_{t,(−1)^{\frac{t+1}{2}} ⌊\frac{t+1}{4}⌋}(q) = q^{\frac{t^{2}−1}{8}}  \frac{E^{t}(q^{4t})}{E(q^{4})},

where

C_{t,j}(q) = ∑_{n≥0} a_{t,j}(n)q^{n},

a_{t,j}(n) denotes the number of t-cores of n with BG-rank = j and

E(q) = ∏_{j=1}^{∞}(1 − q^{j}),

F (t, q) = \frac{E^{t−4}(q^{2t})E^{2}(q^{t})E^{3}(q^{2})}{E^{2}(q)}.

We observe that (1.3) suggests that C_{t,j}(q) is an even (odd) function of q if j is even (odd).

It is instructive to compare (1.10, 1.11) with the well-known identity [5] for unrestricted t-cores

(1.12)  ∑_{n≥0} a_{t}(n)q^{n} = \frac{E^{t}(q^{t})}{E(q)}.

Here a_{t}(n) denotes the number of t-cores of n.

The rest of this paper is organised as follows.

In Section 2 we discuss the Littlewood decomposition of π in terms of t-core and t-quotient of π. We describe the Garvan, Kim, Stanton bijection for t-cores and use a constant term technique to provide a simple proof of the Klyachko identity [8]

(1.13)  ∑_{\begin{matrix}n⃗∈ℤ^{t} \\ n⃗·1_{t}=0 \end{matrix}} q^{\frac{t}{2} n⃗·n⃗+b⃗_{t}·n⃗} = \frac{E^{t}(q^{t})}{E(q)}.

Here 1_{t} = (1, 1, …, 1) ∈ ℤ^{t}, b⃗_{t} = (0, 1, 2, …, t − 1).

In Section 3 we establish a fundamental identity connecting BG-rank and the Littlewood decomposition.

In Section 4 we discuss a combinatorial proof of (1.8).

Section 5 is devoted to the proof of the identities (1.10, 1.11).

Section 6 deals with 5-cores with prescribed BG-rank. There we derive the explicit formulas for the coefficients a_{5,j}(n), j = 0, ±1.

In Section 7 we give a generalization of the BG-rank and state a number of results.

2. Two Bijections

In this section we will follow closely the discussion in [4], [5] to recall some basic facts about t-cores and t-quotients. A region r in the extended t-residue diagram of π is the set of cells (i, j) satisfying t(r −1) ≤ j −i < tr. A cell of π is called exposed if it is at the end of a row. One can construct t bi-infinite words W_{0}, W_{1}, …, W_{t−1} of two letters N, E as

The r-th letter of W_{i} = \begin{cases}E, & if there is an exposed cell labelled i in the region r \\ N, & otherwise. \end{cases}

It is easy to see that the word set {W_{0}, W_{1}, …, W_{t−1}} fixes π uniquely.

Let P be the set of all partitions and P_{t-core} be the set of all t-cores. There is a well-known bijection

ϕ_{1} : P → P_{t-core} × P × P × P … × P

which goes back to Littlewood [9]

ϕ_{1}(π) = (π_{t-core}, π̂_{0}, π̂_{1}, …, π̂_{t−1})

such that

|π| = |π_{t-core}| + t ∑_{i=0}^{t−1} |π̂_{i}|.

Multipartition (π̂_{0}, π̂_{1}, …, π̂_{t−1}) is called the t-quotient of π. We remark that (1.12) is the immediate corollary of the Littlewood bijection. We describe ϕ_{1} in full detail a bit later.

The second bijection

ϕ_{2} : P_{t-core} → {n⃗ : n⃗ ∈ ℤ^{t}, n⃗ · 1_{t} = 0}

was introduced in [5]. It is for t-cores only

ϕ_{2}(π_{t-core}) = n⃗ = (n_{0}, n_{1}, …, n_{t−1})

where for 0 ≤ i ≤ t − 2

(2.1)  n_{i} = r_{i}(π_{t-core}, t) − r_{i+1}(π_{t-core}, t)

and

(2.2)  n_{t−1} = r_{t−1}(π_{t-core}, t) − r_{0}(π_{t-core}, t).

Clearly,

∑_{i=0}^{t−1} n_{i} = n⃗ · 1_{t} = 0.

Moreover,

(2.3)  |π_{t-core}| = \frac{t}{2}n⃗ · n⃗ + b⃗_{t} · n⃗,

as shown in [5]. And so

(2.4)  ∑_{n≥0} a_{t}(n)q^{n} = ∑_{\begin{matrix}n⃗∈ℤ^{t} \\ n⃗·1_{t}=0 \end{matrix}} q^{\frac{t}{2} n⃗·n⃗+b⃗_{t}·n⃗}.

Note that (1.12), (2.4) imply the Klyachko identity (1.13). The reader may wonder if (2.1, 2.2) can be used to define ϕ_{2}(π) = n⃗ for any partition π. This, of course, can be done. However, in general ϕ_{2} is not a 1 − 1 function and so ϕ_{2}^{−1} can’t be defined. Indeed, if π_{1} ≠ π_{2}, but π_{t-core} is a t-core of both π_{1} and π_{2} then

ϕ_{2}(π_{1}) = ϕ_{2}(π_{2}) = ϕ_{2}(π_{t-core}).

When a partition is a t-core, ϕ_{2} can be inverted. To do this we recall that the partition is a t-core iff for 0 ≤ i ≤ t − 1

\begin{aligned}Region & : & · · · · · · · · · & n_{i} − 1 & n_{i} & n_{i} + 1 & n_{i} + 2 & · · · · · · · · · \\ W_{i} & : & · · · · · · · · · & E & E & N & N & · · · · · · · · · \end{aligned}

as explained in [5]. For example, the word image of ϕ_{2}^{−1}((2, −1, −1)) is

Region : · · · · · · − 1 0 1 2 3 · · · · · ·
W_{0} : · · · · · · E E E E N · · · · · ·
W_{1} : · · · · · · E N N N N · · · · · ·
W_{2} : · · · · · · E N N N N · · · · · · .

This means that

(2.5)  ϕ_{2}^{−1}((2, −1, −1)) = (4, 2).

More generally, if

ϕ_{1}(π) = (π_{t-core}, π̂_{0}, π̂_{1}, …, π̂_{t−1})

with

π̂_{i} = (λ_{1}^{(i)}, λ_{2}^{(i)}, …, λ_{m_{i}}^{(i)}), 0 ≤ i ≤ t − 1,

then cells colored i are not exposed only in the regions

n_{i} + j − λ_{j}^{(i)}, 1 ≤ j ≤ m_{i}

and

n_{i} + m_{i} + k, k ≥ 1.

For example, if π̂_{i} = (λ_{1}, λ_{2}, λ_{3}) then

Region : · · · · · · n_{i} + 1 − λ_{1} · · · · · · n_{i} + 2 − λ_{2} · · · · · · n_{i} + 3 − λ_{3} · · · · · · n_{i} + 4 · · · · · ·
W_{i} : · · · · · · E N E · · · · · · E N E · · · · · · E N E · · · · · · E N · · · · · ·

Clearly, one can easily determine n⃗ and (π̂_{0}, π̂_{1}, …, π̂_{t−1}) from the word set {W_{0}, W_{1}, …, W_{t−1}}. And so

ϕ_{1}(π) = (ϕ_{2}^{−1}(n⃗), π̂_{0}, …, π̂_{t−1}).

We illustrate the above with the following example. If t = 3 and π = (7, 5, 4, 3, 2) then

Region : · · · · · · − 2 − 1 0 1 2 3 4 5 · · · · · ·
W_{0} : · · · · · · E E E N E E N N · · · · · ·
W_{1} : · · · · · · E N N E N N N N · · · · · ·
W_{2} : · · · · · · E N E N N N N N · · · · · · .

We have

n_{0} = 2,  π̂_{0} = (2),
n_{1} = −1,  π̂_{1} = (1, 1),
n_{2} = −1,  π̂_{2} = (1).

Using (2.5), we obtain

ϕ_{1}((7, 5, 4, 3, 2)) = ((4, 2), (2), (1, 1), (1)).

To proceed further we recall some standard q-hypergeometric notations [6]:

(a_{1}, a_{2}, a_{3}, …; q)_{N} = (a_{1}; q)_{N}(a_{2}; q)_{N}(a_{3}; q)_{N} …

where

(a; q)_{N} = (a)_{N} = \begin{cases}∏_{j=0}^{N−1}(1 − aq^{j}), & N > 0 \\ 1, & N = 0 \\ ∏_{j=1}^{−N}(1 − aq^{−j})^{−1}, & N < 0. \end{cases}

We shall also require the Jacobi triple product identity [6, (II.28)]

(2.6)  ∑_{n=−∞}^{∞} q^{n^{2}}z^{n} = (q^{2}, −zq, −\frac{q}{z}; q^{2})_{∞}.

We are now ready to prove the Klyachko identity (1.13). We will employ a so-called constant term technique. To this end we rewrite the left hand side of (1.13) as

LHS (1.13) = [z^{0}] ∑_{n⃗∈ℤ^{t}} q^{\frac{t}{2} n⃗·n⃗+b⃗_{t}·n⃗}z^{n⃗·1_{t}} = [z^{0}] ∏_{i=0}^{t−1} ∑_{n_{i}=−∞}^{∞} q^{\frac{t}{2} n_{i}^{2}+in_{i}}z^{n_{i}}

where [z^{i}]f(z) is the coefficient of z^{i} in the expansion of f(z) in powers of z. With the aid of (2.6) we derive

LHS (1.13) = [z^{0}] ∏_{i=0}^{t−1} (q^{t}, −q^{i+ \frac{t}{2}} z, −\frac{q^{\frac{t}{2}}}{q^{i}z}; q^{t})_{∞}

= [z^{0}]\frac{E^{t}(q^{t})}{E(q)} (q, −q^{\frac{t}{2}} z, −\frac{q}{q^{\frac{t}{2}} z}; q)_{∞}

= [z^{0}](\frac{E^{t}(q^{t})}{E(q)} ∑_{n=−∞}^{∞} q^{\frac{n^{2}}{2} + \frac{t−1}{2} n}z^{n})

= \frac{E^{t}(q^{t})}{E(q)},

as desired. The above proof is just a warm-up excercise to prepare the reader for a more sophisticated proof of (1.10) discussed in Section 5.

3. The Littlewood decomposition and BG-rank

The main goal of this section is to establish the following identities for BG-rank. If t is even and (n_{0}, …, n_{t−1}) = ϕ_{2}(π), then

(3.1)  BG-rank(π) = ∑_{i=0}^{\frac{t−2}{2}} n_{2i}.

If t is odd then

(3.2)  BG-rank(π_{t-core}) = bg(n⃗),

where n⃗ = ϕ_{2}(π_{t-core}) and

(3.3)  bg(n⃗) := \frac{1 − ∑_{j=0}^{t−1}(−1)^{j+n_{j}}}{4} .

Moreover, if t is odd and ϕ_{1}(π) = (π_{t-core}, π̂_{0}, …, π̂_{t−1}) then

(3.4)  BG-rank(π) = BG-rank(π_{t-core}) + ∑_{j=0}^{t−1}(−1)^{j+n_{j}}BG-rank(π̂_{j}).

The proof of (3.1) is straightforward. It is sufficient to observe that if some cell is colored i in the t-residue diagram of π, then it is colored \frac{1−(−1)^{i}}{2} in the 2-residue diagram of π. And so we obtain with the aid of (1.2)

BG-rank(π) = (r_{0} + r_{2} + r_{4} + ⋯ + r_{t−2}) − (r_{1} + r_{3} + r_{5} + ⋯ + r_{t−1})
= (r_{0} − r_{1}) + (r_{2} − r_{3}) + ⋯ + (r_{t−2} − r_{t−1})
= n_{0} + n_{2} + ⋯ + n_{t−2},

as desired. Next, let D(π) = D denote the size of the Durfee square of π. To prove (3.2) we begin by rewriting (1.1) as

(3.5)  BG-rank(π) = \frac{1}{2} (par(ν) + ∑_{j=1}^{ν}(−1)^{λ_{j}−j}) .

Here π = (λ_{1}, λ_{2}, …, λ_{ν}) and par(x) is defined as

par(x) := \frac{1 − (−1)^{x}}{2} .

Next, let π_{1}, π_{2} denote the partitions constructed from the first D = D(π_{t-core}) rows, columns of π_{t-core}, respectively. Let π_{3} denote a partition whose diagram is the Durfee square of π_{t-core}. It is plain that

BG-rank(π_{t-core}) = BG-rank(π_{1}) + BG-rank(π_{2}) − BG-rank(π_{3})

(3.6) = BG-rank(π_{1}) + BG-rank(π_{2}) − par(D).

We shall also require the following sets

P_{+} : = {i ∈ ℤ : 0 ≤ i ≤ t − 1, n_{i} > 0},
P_{−} : = {i ∈ ℤ : 0 ≤ i ≤ t − 1, n_{i} < 0}.

Here n_{i}’s are the components of ϕ_{2}(π_{t-core}). Note that if i ∈ P_{+}, then i is exposed in all positive regions ≤ n_{i} of π_{1}. This observation together with (3.5) implies that

BG-rank(π_{1}) = \frac{1}{2} (par(D) + ∑_{i∈P_{+}} ∑_{k=1}^{n_{i}}(−1)^{t(k−1)+i})

= \frac{1}{2} (par(D) + ∑_{i∈P_{+}}(−1)^{i}par(n_{i})) (3.7)

In [5], the authors showed that under conjugation ϕ_{2}(π_{t-core}) transforms as

(n_{0}, n_{1}, n_{2}, …, n_{t−1}) → (−n_{t−1}, −n_{t−2}, −n_{t−3}, …, −n_{0}).

Also it is easy to see that

BG-rank(π_{2}) = BG-rank(π′_{2}).

It follows that

(3.8)  BG-rank(π_{2}) = \frac{1}{2} (par(D) + ∑_{i∈P_{−}}(−1)^{i}par(n_{i})) .

Combining (3.6, 3.7, 3.8) and taking into account that par(0) = 0 we get

BG-rank(π_{t-core}) = \frac{1}{2} ∑_{i∈P_{−}∪P_{+}}(−1)^{i}par(n_{i})

= \frac{1}{2} ∑_{i=0}^{t−1}(−1)^{i}par(n_{i}) = \frac{1 − ∑_{i=0}^{t−1}(−1)^{i+n_{i}}}{4} ,

as desired. Note that formula (3.2) implies that BG-rank of odd t-core is bounded, as stated in (1.9). Next, let π^{∼}_{0,i}, π^{∼}_{2,i}, π^{∼}_{3,i}, … denote the partitions constructed from ϕ_{1}(π) = (π_{t-core}, π̂_{0}, π̂_{1}, …, π̂_{t−1}), for odd t as follows

π^{∼}_{0,i} = ϕ_{1}^{−1}(π_{t-core}, π̂_{0}, π̂_{1}, …, π̂_{i−1}, (0), π̂_{i+1}, …, π̂_{t−1}),
π^{∼}_{1,i} = ϕ_{1}^{−1}(π_{t-core}, π̂_{0}, π̂_{1}, …, π̂_{i−1}, (λ_{1}), π̂_{i+1}, …, π̂_{t−1}),
π^{∼}_{2,i} = ϕ_{1}^{−1}(π_{t-core}, π̂_{0}, π̂_{1}, …, π̂_{i−1}, (λ_{1}, λ_{2}), π̂_{i+1}, …, π̂_{t−1}),
· · · · · · · · · .

Here π̂_{i} = (λ_{1}, λ_{2}, …, λ_{ν}). Note that the W_{i} word of π^{∼}_{0,i} is

\begin{aligned}Region & : & · · · · · · · · · & n_{i} & n_{i} + 1 & · · · · · · · · · \\ W_{i} & : & · · · · · · · · · & E & N & · · · · · · · · · . \end{aligned}

To convert π^{∼}_{0,i} into π^{∼}_{1,i} we attach a rim hook of length tλ_{1} to π^{∼}_{0,i} so that W_{i} becomes

\begin{aligned}Region & : & · · · · · · · · · & n_{i} + 1 − λ_{1} & · · · · · · · · · & n_{i} + 2, & · · · · · · · · · \\ W_{i} & : & · · · · · · · · · & E N E & · · · · · · · · · & E N N & · · · · · · · · · . \end{aligned}

It is not hard to verify that the color of the head (north-eastern) cell of the added rim-hook in the 2-residue diagram of π^{∼}_{1,i} is given by par(tn_{i} + i) = par(n_{i} + i). Observe that zeros and ones alternate along the added hook rim. This means that BG-rank does not change if λ_{1} is even. If λ_{1} is odd then the change is determined by the color of the added head cell, i.e.

BG-rank(π^{∼}_{1,i}) = BG-rank(π^{∼}_{0,i}) + par(λ_{1})(1 − 2par(n_{i} + i))
= BG-rank(π^{∼}_{0,i}) + par(λ_{1})(−1)^{n_{i}+i},

Next, we convert π^{∼}_{1,i} into π^{∼}_{2,i} by adding the new hook rim of length tλ_{2} to π^{∼}_{1,i} so that W_{i} becomes

\begin{aligned}Region & : & · · · · · · & n_{i} + 1 − λ_{1} & · · · · · · & n_{i} + 2 − λ_{2} & · · · · · · & n_{i} + 3 & · · · · · · \\ W_{i} & : & · · · · · · & E N E & · · · · · · & E N E & · · · · · · & E N & · · · · · · . \end{aligned}

The color of the new head cell is given by

par(t(n_{i} + 1) + i) = par(n_{i} + 1 + i),

and so

BG-rank(π^{∼}_{2,i}) = BG-rank(π^{∼}_{1,i}) + par(λ_{2})(1 − 2par(n_{i} + 1 + i))
= BG-rank(π^{∼}_{0,i}) + (−1)^{n_{i}+i}(par(λ_{1}) − par(λ_{2})).

Proceeding as above we arrive at

BG-rank(π) = BG-rank(π^{∼}_{0,i}) + (−1)^{n_{i}+i} ∑_{j=1}^{ν}(−1)^{j+1}par(λ_{j})

(3.9) = BG-rank(π^{∼}_{0,i}) + (−1)^{n_{i}+i}BG-rank(π̂_{i}).

Formula (3.4) follows easily from (3.9). Let us now define B⃗_{t}, B̃_{t} ∈ ℤ^{t} as

B⃗_{t} = \begin{cases}∑i=0^{\frac{t−1}{2}} e⃗_{2i}, & if t ≡ 1 (mod 4) \\ ∑i=0^{\frac{t−3}{2}} e⃗_{1+2i}, & if t ≡ −1 (mod 4) \end{cases}

and

B̃_{t} = B⃗_{t} + ∑_{i=0}^{t−1} e⃗_{i} = \begin{cases}∑i=0^{\frac{t−3}{2}} e⃗_{1+2i}, & if t ≡ 1 (mod 4) \\ ∑i=0^{\frac{t−1}{2}} e⃗_{2i}, & if t ≡ −1 (mod 4) \end{cases}

Here e⃗_{i}’s are standard unit vectors in ℤ^{t} defined as e_{0} = (1, 0, …, 0), …, e⃗_{t−1} = (0, …, 0, 1).

We conclude this section with the following important observation. If odd t > 1, k = 0, 1, …, \frac{t−1}{2} and n⃗ ∈ ℤ^{t}, n⃗ · 1_{t} = 0, then

(3.10)  bg(n⃗) = (−1)^{\frac{t−1}{2}} (⌊\frac{t}{4}⌋ − k)

iff n⃗ ≡ B⃗_{t}+e⃗_{i_{0}} +e⃗_{i_{1}} +⋯+e⃗_{i_{2k}} (mod 2) for some 0 ≤ i_{0} < i_{1} < i_{2} < ⋯ < i_{2k} ≤ t−1. In particular, if n⃗ ∈ ℤ^{t}, n⃗ · 1_{t} = 0, then

(3.11)  bg(n⃗) = (−1)^{\frac{t+1}{2}} ⌊\frac{t + 1}{4}⌋

iff n⃗ ≡ B̃_{t} (mod 2). We leave the proof as an exercise for the interested reader.

4. Combinatorial proof of p_{j}(5n + 4) ≡ 0 (mod 5)

Throughout this section we assume that

|π| ≡ 4 (mod 5)

and

|π5-core| ≡ 4 (mod 5).

To prove (1.8) we shall require a few definitions. Following [5], we define the 5-core crank as

(4.1)  c_{5}(π) := 2(r_{0}(π, 5) − r_{4}(π, 5)) + (r_{1}(π, 5) − r_{3}(π, 5)) + 1 (mod 5).

Note that if |π_{5-core}| ≡ 4 (mod 5), then obviously

n_{0} + n_{1} + n_{2} + n_{3} + n_{4} = 0, (4.2)

n_{1} + 2n_{2} + 3n_{3} + 4n_{4} ≡ 4 (mod 5). (4.3)

10 ALEXANDER BERKOVICH AND FRANK G. GARVAN

Here, n⃗ = (n_{0}, n_{1}, n_{2}, n_{3}, n_{4}) = ϕ_{2}(π_{5-core}). Let’s introduce a new vector α⃗(n⃗) = (α_{0}, α_{1}, α_{2}, α_{3}, α_{4}), defined as

α_{0} = \frac{n_{0} − 3n_{1} − 2n_{2} − n_{3} + 1}{5}, (4.4)

α_{1} = \frac{−3n_{0} − n_{1} − 4n_{2} − 2n_{3} + 2}{5}, (4.5)

α_{2} = \frac{−3n_{0} − n_{1} + n_{2} − 2n_{3} + 2}{5}, (4.6)

α_{3} = \frac{n_{0} + 2n_{1} + 3n_{2} + 4n_{3} + 1}{5}, (4.7)

α_{4} = \frac{4n_{0} + 3n_{1} + 2n_{2} + n_{3} − 1}{5}. (4.8)

Using (4.2, 4.3) it is easy to verify that α⃗(n⃗) ∈ ℤ^{5} and that

(4.9)  (α_{0} + α_{1} + α_{2} + α_{3} + α_{4}) = 1.

Inverting (4.4–4.8) we find that n_{0} = α_{0} + α_{4}, (4.10) n_{1} = −α_{0} + α_{1} + α_{4}, (4.11) n_{2} = −α_{1} + α_{2}, (4.12) n_{3} = −α_{2} + α_{3} − α_{4}, (4.13) n_{4} = −α_{3} − α_{4}, (4.14) Note that in terms of these new variables we have

(4.15)  c_{5}(π) ≡ ∑_{i=0}^{4} iα_{i} (mod 5),
(4.16)  |π| = 5Q(α⃗) − 1 + 5 ∑_{i=0}^{4} |π̂_{i}|,

and

BG-rank(π) = \frac{1 − (−1)^{α_{0}+α_{1}} − (−1)^{α_{1}+α_{2}} − ⋯ − (−1)^{α_{4}+α_{0}}}{4}
+ (−1)^{α_{0}+α_{4}}BG-rank(π̂_{0})
+ (−1)^{α_{2}+α_{3}}BG-rank(π̂_{1})
+ (−1)^{α_{1}+α_{2}}BG-rank(π̂_{2})
+ (−1)^{α_{0}+α_{1}}BG-rank(π̂_{3})

+ (−1)^{α_{3}+α_{4}}BG-rank(π̂_{4}). (4.17) Here ϕ_{1}(π) = (π_{5-core}, π̂_{0}, …, π̂_{4}) and Q(α⃗) := α⃗ · α⃗ − (α_{0}α_{1} + α_{1}α_{2} + ⋯ + α_{4}α_{0}). It is convenient to combine ϕ_{1}, ϕ_{2}, α⃗ into a new invertible function Φ, defined as

Φ(π) = (α⃗(ϕ_{2}(π_{5-core})), π̂), where π̂ := (π̂_{0}, …, π̂_{4}). Following [2] we define

Ĉ_{1}(α⃗) = (α_{4}, α_{0}, α_{1}, α_{2}, α_{3}),
Ĉ_{2}(π̂) = (π̂_{4}, π̂_{2}, π̂_{3}, π̂_{0}, π̂_{1}),
Ô(π) = Φ^{−1}(Ĉ_{1}(α⃗), Ĉ_{2}(π̂)).

We observe that operator Ô has the following properties

|Ô(π)| = |π|,
Ô^{5}(π) = π,
BG-rank (Ô(π)) = BG-rank(π),
(4.18)  c_{5} (Ô(π)) ≡ 1 + c_{5}(π) (mod 5).

Clearly, Ô preserves the norm and the BG-rank of the partition. And so we can assemble all partitions of 5n + 4 with BG-rank = j into disjoint orbits:

π, Ô(π), Ô^{2}(π), Ô^{3}(π), Ô^{4}(π).

Here, π is some partition of 5n+4 with BG-rank = j. Formula (4.18) suggests that all five members of the same orbit are distinct. Clearly,

p_{j}(5n + 4) = 5 · (number of orbits).

Hence, p_{j}(5n + 4) ≡ 0 (mod 5), as desired. In fact, we have the following Theorem 4.1. Let j be any fixed integer. The residue of the 5-core crank mod 5 divides the partitions enumerated by p_{j}(5n + 4) into five equal classes.

We note that this theorem generalizes Theorem 4.1 [2, p.717].

5. Identities for odd t-cores with extreme BG-rank values

The main object of this section is to provide a proof of formulas (1.10) and (1.11). Thoughout this section t is presumed to be a positive odd integer. We will prove (1.11) first. To this end we employ the observation (3.10) together with (2.3) to rewrite it as

(5.1)  ∑_{\begin{matrix}n⃗∈ℤ^{t},n⃗·1_{t}=0 \\ n⃗≡ ^{}B̃_{t} (mod 2) \end{matrix}} q^{Q̃(n⃗)} = q^{\frac{t^{2}−1}{8}} \frac{E^{t}(q^{4t})}{E(q^{4})},

where

(5.2)  Q̃(n⃗) := \frac{t}{2}n⃗ · n⃗ + b⃗_{t} · n⃗.

Next we introduce new summation variables ñ = (ñ_{0}, …, ñ_{t−1}) ∈ ℤ^{t} as follows

(5.3)  n⃗ = 2ñ + ∑_{i=0}^{⌊ \frac{t−3}{4} ⌋} (e_{\frac{t−3}{2} −2i} − e_{\frac{t+1}{2} +2i}) .

Obviously, ñ is subject to the constraint

(5.4)  ñ · 1_{t} = 0.

Note that in terms of new variables we have

(5.5)  Q̃(n⃗) = Q̃(n⃗) + (t − 1)1_{t} · ñ = \frac{t^{2} − 1}{8} + 4\{\frac{t}{2}ñ · ñ + σ_{1} + σ_{2} + σ_{3}\},

where

σ_{1} = ∑_{i=0}^{⌊ \frac{t−3}{4} ⌋}(t − 1 − i)ñ_{\frac{t−3}{2} −2i},

σ_{2} = ∑_{i=0}^{⌊ \frac{t−3}{4} ⌋} iñ_{2i+ \frac{t+1}{2}} ,

σ_{3} = ∑_{i=−⌊ \frac{t−1}{4} ⌋}^{⌊ \frac{t−1}{4} ⌋}(\frac{t − 1}{2} + i)ñ_{\frac{t−1}{2} +2i}.

At this point it is natural to perform further changes:

ñ_{\frac{t−3}{2} −2i} → ñ_{t−1−i},  0 ≤ i ≤ ⌊\frac{t − 3}{4}⌋
ñ_{\frac{t+1}{2} +2i} → ñ_{i},  0 ≤ i ≤ ⌊\frac{t − 3}{4}⌋
ñ_{\frac{t−1}{2} +2i} → ñ_{\frac{t−1}{2} +i},  −⌊\frac{t − 1}{4}⌋ ≤ i ≤ ⌊\frac{t − 1}{4}⌋.

This way we obtain

Q̃(n⃗) = \frac{t^{2} − 1}{8} + 4Q̃(ñ),

ñ ∈ ℤ^{t}, ñ · 1_{t} = 0.

And so with the aid of the Klyachko identity (1.13) we find that

(5.6)  C_{t,(−1)^{\frac{t+1}{4}} ⌊ \frac{t+1}{4} ⌋}(q) = ∑_{\begin{matrix}^{}ñ∈ℤ^{t} \\ ^{}ñ·1_{t}=0 \end{matrix}} q^{\frac{t^{2}−1}{8} +4Q̃(^{}ñ)} = q^{\frac{t^{2}−1}{8}} \frac{E^{t}(q^{4t})}{E(q^{4})},

as desired. To prove (1.10) we shall require the following lemma. Lemma 5.1. For a positive odd t

(5.7)  ψ^{2}(q^{2}) = q^{\frac{t−1}{2}} ψ^{2}(q^{2t}) + \frac{E^{3}(q^{4t})}{f(−q^{t}, −q^{3t})} ∑_{i=0}^{\frac{t−3}{2}} q^{i} \frac{f(q^{t−1−2i}, −q^{1+2i})}{f(−q^{4i+2}, −q^{4t−2−4i})}

holds.

In the above we employed the Ramanujan notations

(5.8)  ψ(q) := \frac{E^{2}(q^{2})}{E(q)} = ∑_{n≥0} q^{\binom{n+1}{2}},
(5.9)  f(a, b) := (ab, −a, −b; ab)_{∞}.

Using (2.6) we can easily show that

(5.10)  f(a, b) = ∑_{n=−∞}^{∞} a^{\frac{n(n+1)}{2}} b^{\frac{n(n−1)}{2}} .

Setting a = q^{t−1−2i}, b = −q^{1+2i}, 0 ≤ i ≤ \frac{t−3}{2} in (5.10) and dissecting we obtain

f(q^{t−1−2i}, −q^{1+2i}) = f(−q^{2+t+4i}, −q^{3t−2−4i})

+ q^{t−1−2i}f(−q^{2−t+4i}, −q^{5t−2−4i}). (5.11)

To prove the above lemma we start with the Ramanujan _{1}ψ_{1}-summation formula [6, II.29]

(5.12)  ∑_{n=−∞}^{∞} \frac{(a)_{n}}{(b)_{n}} z^{n} = \frac{(az, \frac{q}{az} , q, \frac{b}{a}; q)_{∞}}{(z, \frac{b}{az} , b, \frac{q}{a}; q)_{∞}} , |\frac{b}{a}| < |z| < 1.

We set b = aq to obtain

(5.13) ∑_{n=−∞}^{∞} \frac{z^{n}}{1 − aq^{n}} = \frac{(az, \frac{q}{az} , q, q; q)_{∞}}{(z, \frac{q}{z} , a, \frac{q}{a}; q)_{∞}} = \frac{E^{3}(q)f(−az, −\frac{q}{az})}{f(−z, −\frac{q}{z})f(−a, −\frac{q}{a})}, |q| < |z| < 1.

If we replace q → q^{4}, z = q, a = q^{2} in (5.13) we find that

(5.14)  ∑_{n=−∞}^{∞} \frac{q^{n}}{1 − q^{2+4n}} = ψ^{2}(q^{2}).

Next we split the sum on the left of (5.14) as

(5.15)  ψ^{2}(q^{2}) = ∑_{\begin{matrix}i=0 \\ i≠ \frac{t−1}{2} \end{matrix}}^{t−1} ∑_{m_{i}=−∞}^{∞} q^{i} \frac{q^{tm_{i}}}{1 − q^{2+4i}q^{4tm_{i}}} + ∑_{m=−∞}^{∞} q^{\frac{t−1}{2}} \frac{q^{tm}}{1 − q^{2t}q^{4tm}} .

Using (5.14) with q → q^{t} it is easy to recognize the last sum in (5.15) as q^{\frac{t−1}{2}} ψ^{2}(q^{2t}). And so we have

(5.16) ψ^{2}(q^{2}) = q^{\frac{t−1}{2}} ψ^{2}(q^{2t}) + \frac{E^{3}(q^{4t})}{f(−q^{t}, −q^{3t})} ∑_{\begin{matrix}i=0 \\ i≠ \frac{t−1}{2} \end{matrix}}^{t−1} q^{i} \frac{f(−q^{2+4i+t}, −q^{3t−2−4i})}{f(−q^{2+4i}, −q^{4t−2−4i})},

where we have made a multiple use of (5.13). Finally, folding the last sum in half and using (5.11) we arrive at

ψ^{2}(q^{2}) = q^{\frac{t−1}{2}} ψ^{2}(q^{2t}) + ∑_{i=0}^{\frac{t−3}{2}} \frac{E^{3}(q^{4t})q^{i}}{f(−q^{t}, −q^{3t})f(−q^{2+4i}, −q^{4t−2−2i})}
× {f (−q^{2+4i+t}, −q^{3t−2−4i}) + q^{t−1−2i}f (−q^{5t−2−4i}, −q^{2−t+4i}) }

= q^{\frac{t−1}{2}} ψ^{2}(q^{2t}) + \frac{E^{3}(q^{4t})}{f(−q^{t}, −q^{3t})} ∑_{i=0}^{\frac{t−3}{2}} q^{i} \frac{f(q^{t−1−2i}, −q^{1+2i})}{f(−q^{2+4i}, −q^{4t−2−4i})}. (5.17)

This concludes the proof of Lemma 5.1.

We now move on to prove (1.10). Again, using the observation (3.10), we can rewrite it as

(5.18)  ∑_{j=0}^{t−1} ∑_{\begin{matrix}n⃗∈ℤ^{t},n⃗·1_{t}=0 \\ n⃗≡B⃗_{t}+e⃗_{j} (mod 2) \end{matrix}} q^{Q̃(n⃗)} = q^{\frac{(t−1)(t−3)}{8}} F (t, q^{2}).

Remarkably, (5.18) is just the constant term in z of the following more general identity

∑_{j=0}^{t−1} ∑_{\begin{matrix}n⃗∈ℤ^{t} \\ n⃗≡B⃗_{t}+e⃗_{j} (mod 2) \end{matrix}} q^{Q̃(n⃗)}z^{\frac{n⃗·1_{t}}{2}}

= q^{\frac{(t−1)(t−3)}{8}} F (t, q^{2}) ∑_{n=−∞}^{∞} q^{2n^{2}+(t−1)n}z^{n}. (5.19)

To prove (5.19) we observe that its right hand side satisfies the first order functional equation

(5.20)  D̂_{t,q}(f(z)) = f(z),

where

D̂_{t,q}(f(z)) := zq^{t+1}f(zq^{4}).

After a bit of labor one can verify that for 0 ≤ i ≤ t − 1

(5.21)  D̂_{t,q} (  ∑_{\begin{matrix}n⃗∈ℤ^{t} \\ n⃗≡B⃗_{t}+e⃗_{i} (mod 2) \end{matrix}} q^{Q̃(n⃗)}z^{\frac{n⃗·1_{t}}{2}} ) = ∑_{\begin{matrix}n⃗∈ℤ^{t} \\ n⃗≡B⃗_{t}+e⃗_{i+2} (mod 2) \end{matrix}} q^{Q̃(n⃗)}z^{\frac{n⃗·1_{t}}{2}} ,

where e⃗_{t} := e⃗_{0} and e⃗_{t+1} := e⃗_{1}. Clearly, (5.21) implies that the left hand side of (5.19) satisfies (5.20), as well. It remains to verify (5.19) at one nontrivial point. To this end we set

z = \begin{cases}1, & if t ≡ −1 (mod 4), \\ q^{2}, & if t ≡ 1 (mod 4) \end{cases}

in (5.19), and then replace q^{2} → q to get with the help of (2.6)

q^{\frac{t−1}{2}} ψ(q^{2t}) ∏_{j=0}^{\frac{t−3}{2}} f^{2} (q^{1+2j}, q^{2t−1−2j})

× \{1 + ∑_{i=1}^{\frac{t−1}{2}} q^{−i} \frac{f(q^{t}, q^{t})f(q^{2i}, q^{2t−2i})}{ψ(q^{2t})f(q^{t+2i}, q^{t−2i})}\} = ψ(q^{2})F (t, q). (5.22)

To proceed further we need to verify two product identities

ψ(q^{2}) ∏_{j=0}^{\frac{t−3}{2}} f^{2} (q^{1+2j}, q^{2t−1−2j}) = ψ(q^{2t})F (t, q)

and

ψ(q^{2t})\frac{f(q^{t}, q^{t})f(q^{2i}, q^{2t−2i})}{f(q^{t+2i}, q^{t−2i})} = E^{3}(q^{4t})\frac{f(q^{2i}, −q^{t−2i})}{f(−q^{t}, −q^{3t})f(−q^{2t+4i}, −q^{2t−4i})}, i ∈ ℕ.

Next, we multiply both sides of (5.22) by \frac{ψ(q^{2})}{F (t,q)} and simplify to arrive at

(5.23) q^{\frac{t−1}{2}} ψ^{2}(q^{2t}) + \frac{E^{3}(q^{4t})}{f(−q^{t}, −q^{3t})} ∑_{i=1}^{\frac{t−1}{2}} q^{\frac{t−1}{2} −i} \frac{f(q^{2i}, −q^{t−2i})}{f(−q^{2t+4i}, −q^{2t−4i})} = ψ^{2}(q^{2}),

which is essentially the identity in Lemma 5.1. This concludes our proof of (5.19). It follows that (5.18), (1.10) hold true.

6. 5-cores with prescribed BG-rank

Formula (1.9) suggests that BG-rank(π_{5-core}) can assume just three values: 0, ±1. This means that

(6.1)  a_{5}(n) = a_{5,−1}(n) + a_{5,0}(n) + a_{5,1}(n).

The generating function of version (6.1) is

(6.2)  \frac{E^{5}(q^{5})}{E(q)} = C_{5,−1}(q) + C_{5,0}(q) + C_{5,1}(q).

In the last section we proved (1.10), (1.11). These identities with t = 5 state that

(6.3)  C_{5,−1}(q) = q^{3} \frac{E^{5}(q^{20})}{E(q^{4})},
(6.4)  C_{5,1}(q) = qF (5, q^{2}).

By (1.3) we observe that C_{t,j}(q) is either an odd or an even functions of q with parity determined by the parity of j. Therefore, C_{5,0}(q) is an even function of q, and C_{5,±1}(q) are odd functions of q. Consequently, we see that

(6.5)  ep (\frac{E^{5}(q^{5})}{E(q)}) = C_{5,0}(q)

where

ep(f(x)) := \frac{f(x) + f(−x)}{2}.

In this section we will show that C_{5,0}(q) can be expressed as a sum of two infinite products

(6.6)  C_{5,0}(q) = R(q^{2}),

where

(6.7)  R(q) := \frac{E^{4}(q^{10})E(q^{5})E^{2}(q^{4})}{E^{2}(q^{20})E(q)} + q \frac{E^{2}(q^{20})E^{3}(q^{5})E^{6}(q^{2})}{E^{2}(q^{10})E^{2}(q^{4})E^{3}(q)}.

It is easy to rewrite (6.7) in a manifestly positive way as

R(q) = f(q, q^{4})f(q^{2}, q^{3}){φ(q^{5})ψ(q^{2}) + qφ(q)ψ(q^{10})},

where

φ(q) := f(q, q) = ∑_{n=−∞}^{∞} q^{n^{2}} = \frac{E^{5}(q^{2})}{E^{2}(q^{4})E^{2}(q)},

and ψ(q) is defined in (5.8). Formula (6.6) enabled us to discover and prove the new Lambert series identity

(6.8)  R(q) = ∑_{i=0}^{1} ∑_{n=−∞}^{∞}(−1)^{i}q^{5n+i} \frac{1 + q^{1+2i+10n}}{(1 − q^{1+2i+10n})^{2}} .

In what follows we will require three identities:

(6.9)  [ux, \frac{u}{x}, vy, \frac{v}{y}; q]_{∞} = [uy, \frac{u}{y}, vx, \frac{v}{x}; q]_{∞} + \frac{v}{x}[xy, \frac{x}{y}, uv, \frac{u}{v}; q]_{∞},

([6, ex. 5.21])

(6.10) f(a, b)f(c, d) = f(ac, bd)f(ad, bc) + af (\frac{b}{c}, \frac{c}{b}(abcd)) f (\frac{b}{d}, \frac{d}{b}(abcd)) ,

provided ab = cd ([1]) and

(6.11)  \frac{E^{5}(q^{5})}{E(q)} = ∑_{i=1}^{2} ∑_{n=−∞}^{∞}(−1)^{i+1} \frac{q^{5n+i−1}}{(1 − q^{5n+i})^{2}} ,

([6, ex. 5.7], [5, p.8]). Here

[a; q]_{∞} = (a, \frac{q}{a}; q)_{∞} ,
[a_{1}, a_{2}, …, a_{n}; q]_{∞} = ∏_{i=1}^{n}[a_{i}; q]_{∞}.

Next, we wish to establish the validity of

(6.12)  F (5, q) = \frac{E(q^{10})E^{2}(q^{5})E^{3}(q^{2})}{E^{2}(q)} = \frac{E^{5}(q^{5})}{E(q)} + q \frac{E^{5}(q^{10})}{E(q^{2})}.

To this end we multiply both sides of (6.12) by

\frac{[q, q^{3}; q^{10}]_{∞}^{2}[q^{2}, q^{4}; q^{10}]_{∞}}{E^{4}(q^{10})}

to obtain after simplification that

(6.13)  [q^{2}, q^{2}, q^{4}, q^{6}; q^{10}]_{∞} = [q, q^{3}, q^{5}, q^{5}; q^{10}]_{∞} + q[q, q, q^{3}, q^{3}; q^{10}]_{∞}.

But the last equation is nothing else but (6.9) with q replaced by q^{10} and u = q^{2}, v = q^{5}, x = 1, y = q. We now combine

ep (q \frac{E^{5}(q^{5})}{E(q)}) = qC_{5,−1}(q) + qC_{5,1}(q),

with (6.3), (6.5), and (6.12) to obtain

(6.14)  ep (q \frac{E^{5}(q^{5})}{E(q)}) = 2q^{4} \frac{E^{5}(q^{20})}{E(q^{4})} + q^{2} \frac{E^{5}(q^{10})}{E(q^{2})}.

This can be stated as the following eigenvalue problem

(6.15)  T_{2} (q \frac{E^{5}(q^{5})}{E(q)}) = q \frac{E^{5}(q^{5})}{E(q)},

where for prime p the Hecke operator T_{p} is defined by its action as

T_{p} (∑_{n≥0} a_{n}q^{n}) = ∑_{n≥0} a_{pn}q^{n} + p(\frac{p}{5}) ∑_{n≥0} a_{n}q^{pn},

with (\frac{a}{b}) being the Legendre symbol. We remark that (6.15) is the p = 2 case of the more general formula

(6.16)  T_{p} (q \frac{E^{5}(q^{5})}{E(q)}) = (p + (\frac{p}{5})) (q \frac{E^{5}(q^{5})}{E(q)}),

which can be deduced from (6.11). We shall not supply the details. Instead, we note that (6.16) together with (6.3, 6.4, 6.5) implies that

(6.17)  Tp̃(qC_{5,j}(q)) = (p̃ + (\frac{p̃}{5})) (qC_{5,j}(q)), j = 0 ± 1.

Here, p̃ is an odd prime.

To prove (6.6) we use (6.12) to deduce that

(6.18)  ep (\frac{E^{5}(q^{5})}{E(q)}) = ep(F (5, q)) = E(q^{10})E^{3}(q^{2}) · ep (\frac{E^{2}(q^{5})}{E^{2}(q)}) .

To proceed further we employ (6.10) with a = q, b = q^{9}, c = q^{3}, d = q^{7} to get

\frac{E(q^{5})}{E(q)} = \frac{E(q^{4})}{E(q^{20})E^{2}(q^{2})}f(q, q^{9})f(q^{3}, q^{7})

= \frac{E(q^{4})}{E(q^{20})E^{2}(q^{2})}\{f(q^{4}, q^{16})f(q^{8}, q^{12}) + qf(q^{6}, q^{14})f(q^{2}, q^{18})\}

= \frac{E^{2}(q^{20})E(q^{8})}{E(q^{40})E^{2}(q^{2})} + q \frac{E(q^{40})E(q^{10})E^{3}(q^{4})}{E(q^{20})E(q^{8})E^{3}(q^{2})}. (6.19)

It is clear that

(6.20)  ep (\frac{E^{2}(q^{5})}{E^{2}(q)}) = \frac{E^{4}(q^{20})E^{2}(q^{8})}{E^{2}(q^{40})E^{4}(q^{2})} + q^{2} \frac{E^{2}(q^{40})E^{2}(q^{10})E^{6}(q^{4})}{E^{2}(q^{20})E^{2}(q^{8})E^{6}(q^{2})}.

Combining (6.18) and (6.20) we find that

(6.21)  ep (\frac{E^{5}(q^{5})}{E(q)}) = R(q^{2}).

The last formula together with (6.5) implies (6.6). Next, we rewrite (6.11) as

\frac{E^{5}(q^{5})}{E(q)} = ∑_{i=1}^{2} ∑_{n=−∞}^{∞}(−1)^{i+1} \frac{q^{5n+i−1}(1 + 2q^{5n+i} + q^{10n+2i})}{(1 − q^{10n+2i})^{2}}.

Clearly,

ep (\frac{E^{5}(q^{5})}{E(q)}) = ∑_{i=1}^{2} ∑_{\begin{matrix}n=−∞ \\ n≡i−1 (mod 2) \end{matrix}}^{∞}(−1)^{i+1} \frac{q^{5n+i−1}(1 + q^{10n+2i})}{(1 − q^{10n+2i})^{2}}

= ∑_{i=0}^{1} ∑_{n=−∞}^{∞}(−1)^{i} \frac{q^{10n+i}(1 + q^{20n+4i+2})}{(1 − q^{20n+4i+2})^{2}}. (6.22)

Formula (6.8) with q → q^{2} follows easily from (6.21) and (6.22). Before we move on we wish to summarize some of the above observations in the formula below

\frac{E^{5}(q^{5})}{E(q)} = \{\frac{E^{4}(q^{20})E(q^{10})E^{2}(q^{8})}{E^{2}(q^{40})E(q^{2})} + q^{2} \frac{E^{2}(q^{40})E^{3}(q^{10})E^{6}(q^{4})}{E^{2}(q^{20})E^{2}(q^{8})E^{3}(q^{2})}\}

+ q\{\frac{E^{5}(q^{10})}{E(q^{2})} + 2q^{2} \frac{E^{5}(q^{20})}{E(q^{4})}\}. (6.23)

In [5], the authors used (6.11) to find explicit formulas for the coefficients

(6.24)  a_{5}(n) = \frac{2^{d+1} + (−1)^{d}}{3} · 5^{c} · ∏_{i=1}^{s} \frac{p_{i}^{a_{i}+1} − 1}{p_{i} − 1} ∏_{j=1}^{t} \frac{q_{j}^{b_{j}+1} + (−1)^{b_{j}}}{q_{j} + 1} .

Here

(6.25)  n + 1 = 2^{d}5^{c} ∏_{i=1}^{s} p_{i}^{a_{i}} ∏_{j=1}^{t} q_{j}^{b_{j}}

is the prime factorization of n + 1 and p_{i} ≡ ±1 (mod 5), 1 ≤ i ≤ s and q_{j} ≡ ±2 (mod 5), 1 ≤ j ≤ t are odd primes. Formulas (6.3), (6.4), (6.5) and (6.12) suggest the following relations. For n ∈ ℕ and r = 0, 1, 2, 3 one has

(6.26)  a_{5,0}(n) = \begin{cases}a_{5}(n), & if n ≡ 0 (mod 2), \\ 0, & otherwise, \end{cases}

(6.27)  a_{5,−1}(4n + r) = \begin{cases}a_{5}(n), & if r = 3, \\ 0, & otherwise, \end{cases}

(6.28)  a_{5,1}(4n + r) = \begin{cases}a_{5}(2n), & if r = 1, \\ a_{5}(n) + a_{5}(2n + 1), & if r = 3, \\ 0, & if r = 0, 2. \end{cases}

These relations together with (6.24) enabled us to derive explicit formulas for a_{5,j}(n) with −1 ≤ j ≤ 1. In particular, if the prime factorization of n+1 is given by (6.25), then

(6.29)  a_{5,1}(4n + 3) = 2^{d+1}5^{c} ∏_{i=1}^{s} \frac{p_{i}^{a_{i}+1} − 1}{p_{i} − 1} ∏_{j=1}^{t} \frac{q_{j}^{b_{j}+1} + (−1)^{b_{j}}}{q_{j} + 1} .

We would like to conclude this section with the following discussion. It is easy to check that (6.17) implies that

(6.30) a_{5,j}(pn+p−1)+p(\frac{p}{5})a_{5,j} (\frac{n + 1}{p} − 1) = (p + (\frac{p}{5})) a_{5,j}(n), j = 0, ±1,

where p is odd prime, n ∈ ℕ and a_{5,j}(x) = 0 if x ∉ ℤ. Setting p = 5 we find that

(6.31)  a_{5,j}(5n + 4) = 5a_{5,j}(n), j = 0, ±1.

This is a refinement of the well-known result

(6.32)  a_{5}(5n + 4) = 5a_{5}(n),

proven in [5]. We can prove (6.31) by adapting the combinatorial proof in [5].

Let’s define

n⃗ = (n_{0}, n_{1}, n_{2}, n_{3}, n_{4}) = ϕ_{2}(π_{5-core})

for some π_{5-core} with BG-rank(π_{5-core}) = j and |π_{5-core}| = n. Consider map n⃗ → ñ = (ñ_{0}, ñ_{1}, ñ_{2}, ñ_{3}, ñ_{4}) with

ñ_{0} = n_{1} + 2n_{2} + 2n_{4} + 1,
ñ_{1} = −n_{1} − n_{2} + n_{3} + n_{4} + 1,
ñ_{2} = 2n_{1} + n_{2} + 2n_{3},
ñ_{3} = −2n_{2} − 2n_{3} − n_{4} − 1,
ñ_{4} = −2n_{1} − n_{3} − 2n_{4} − 1.

Obviously ñ ∈ ℤ^{5} and ñ · 1_{5} = 0 and so we can define π̃_{5-core} = ϕ_{2}^{−1}(ñ). It is easy to check that

|π̃_{5-core}| = 5n + 4,

and that

BG-rank(π̃_{5-core}) = j,

and

c_{5}(π̃_{5-core}) ≡ 4 (mod 5).

Recall that the orbit {π̃_{5-core}, Ô(π̃_{5-core}), …, Ô^{4}(π̃_{5-core})} contains just one member with c_{5} ≡ 4 (mod 5). And so each 5-core of n with BG-rank j is in 1 − 1 correspondence with an appropriate 5-member orbit of t-cores of 5n + 4 with BG-rank j. This observation yields a combinatorial proof of (6.31).

7. Outlook

Given our combinatorial proof of

p_{j}(5n + 4) ≡ 0 (mod 5), j ∈ ℤ

one may wonder about a combinatorial proof of the other mod 5 congruences (1.4-1.7). We strongly suspect that such proof will be dramatically different from the one discussed in Section 4. In addition, one would like to have combinatorial insights into (6.30) for p ≠ 5.

In this paper we found “positive” eta-quotient representations for C_{5,j}(q), −1 ≤ j ≤ 1. In the general case (odd t, −⌊\frac{t−1}{4}⌋ ≤ j ≤ ⌊\frac{t+1}{4}⌋), we established such representation only for C_{t,±⌊ \frac{t±1}{4} ⌋}(q). Clearly, one wants to find “positive” etaquotient representations for other admissible values of BG-rank. (See [3] for a fascinating discussion of the t = 7 case).

Finally, we observe that (1.2) is the s = 2 case of the following more general definition

gbg-rank(π, s) = ∑_{j=0}^{s−1} r_{j}(π, s)ω_{s}^{j}

with

ω_{s} = e^{i \frac{2π}{s}} .

Many identities, proven here, can be generalized further. For example, we can prove that if (s, t) = 1 then

(7.1)  gbg-rank(π_{t-core}, s) = \frac{∑_{i=0}^{t−1} ω_{s}^{i+1}(ω_{s}^{tn_{i}} − 1)}{(1 − ω_{s}^{t})(1 − ω_{s})}

and for 1 ≤ i ≤ s − 1 that

(7.2)  ∑_{gbg-rank(π_{t-core},s)=g(i)} q^{|π_{t-core}|} = q^{a(i)}F_{i}(q^{s}).

Here,

(n_{0}, n_{1}, …, n_{t−1}) = ϕ_{2}(π_{t-core}),

a(i) = \frac{(t^{2} − 1)(s^{2} − 1)}{24} − \frac{(t − 1)(s − i)i}{2},

g(i) = \frac{1}{(1 − ω_{s})(1 − \frac{1}{ω_{s}})} − ωs^{\frac{t−1}{2}} \frac{1 + \frac{t−1}{ω_{s}^{i}}}{(1 − ω_{s}^{t})(1 − \frac{1}{ω_{s}})},

F_{i}(q) = E(q^{s})E(q^{st})^{t−2} \frac{[q^{it}; q^{st}]_{∞}}{[q^{i}; q^{s}]_{∞}} .

Setting s = 2 in (7.1), (7.2) we obtain (3.2), (1.10), respectively.

In addition we can show that

(7.3)  ∑_{gbg-rank(π_{t-core},s)=g(0)} q^{|π_{t-core}|} = q^{a(0)} \frac{E(q^{s^{2}t})^{t}}{E(q^{s^{2}})} .

Setting s = 2 in (7.3) we get (1.11).

In [10] Olsson and Stanton defined so-called (s, t)-good partitions. Surprisingly, t-cores with gbg-rank = g(0) coincide with (t, s)-good partitions.

Let ν(t, s) denote a number of distinct values that gbg-rank(π_{t-core}, s) may assume. Then it can be shown that

ν(s, t) ≤ \frac{\binom{t+s}{t}}{t + s},

provided that (s, t) = 1. Morever, if s is prime or if s is a composite number and t < 2p then

ν(s, t) = \frac{\binom{t+s}{t}}{t + s}.

Here, p is a smallest prime divisor of s and (s, t) = 1.

Details of these and related results will be left to a later paper.

Acknowledgement

We would like to thank Robin Chapman, Ole Warnaar, Herbert Wilf and Hamza Yesilyurt for their kind interest and stimulating discussions.

References

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6. G. Gasper, M. Rahman, Basic hypergeometric series, Encyclopedia of Mathematics and its applications v.35, Cambridge, 1990.

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11. S. Ramanujan, Some properties of p(n), the number of partitions on n, Proc. Cambridge Phil. Soc. 19 (1919), 207–210.

Department of Mathematics, University of Florida, Gainesville, Florida 32611-8105

  E-mail address: alexb@math.ufl.edu

Department of Mathematics, University of Florida, Gainesville, Florida 32611-8105

  E-mail address: frank@math.ufl.edu