ON THE ANDREWS-STANLEY REFINEMENT OF RAMANUJAN’S PARTITION CONGRUENCE MODULO 5 AND GENERALIZATIONS

ALEXANDER BERKOVICH AND FRANK G. GARVAN

Abstract. In a recent study of sign-balanced, labelled posets Stanley [14], introduced a new integral partition statistic

srank(π) = O(π) − O(π′),

where O(π) denotes the number of odd parts of the partition π and π′ is the conjugate of π. In [1] Andrews proved the following refinement of Ramanujan’s partition congruence mod 5:

p_{0}(5n + 4) ≡ p_{2}(5n + 4) ≡ 0 (mod 5),
p(n) = p_{0}(n) + p_{2}(n),

where p_{i}(n) (i = 0, 2) denotes the number of partitions of n with srank ≡ i (mod 4) and p(n) is the number of unrestricted partitions of n. Andrews asked for a partition statistic that would divide the partitions enumerated by p_{i}(5n + 4) (i = 0, 2) into five equinumerous classes.

In this paper we discuss three such statistics: the St-crank, the 2-quotient-rank and the 5-core-crank. The first one, while new, is intimately related to the Andrews-Garvan [2] crank. The second one is in terms of the 2-quotient of a partition. The third one was introduced by Garvan, Kim and Stanton in [9]. We use it in our combinatorial proof of the Andrews refinement. Remarkably, the Andrews result is a simple consequence of a stronger refinement of Ramanujan’s congruence mod 5. This more general refinement uses a new partition statistic which we term the BG-rank. We employ the BG-rank to prove new partition congruences modulo 5. Finally, we discuss some new formulas for partitions that are 5-cores and discuss an intriguing relation between 3-cores and the Andrews-Garvan crank.

1. Introduction

Let p(n) be the number of unrestricted partitions of n. Ramanujan discovered and later proved that

(1.1)  p(5n + 4) ≡ 0 (mod 5),
(1.2)  p(7n + 5) ≡ 0 (mod 7),
(1.3)  p(11n + 6) ≡ 0 (mod 11).

Dyson [5] was the first to consider combinatorial explanations of these congruences. He defined the rank of a partition as the largest part minus the number of parts and

Date: December 19, 2005. 1991 Mathematics Subject Classification. Primary 11P81, 11P83; Secondary 05A17, 05A19. Key words and phrases. partitions, t-cores, ranks, cranks, Stanley’s statistic, Ramanujan’s

congruences.

made the empirical observations that

(1.4)  N(k, 5, 5n + 4) = \frac{p(5n + 4)}{5}, 0 ≤ k ≤ 4,
(1.5)  N(k, 7, 7n + 5) = \frac{p(7n + 5)}{7}, 0 ≤ k ≤ 6,

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 partitions of 5n+4 into five equal classes. Similarly, (1.5) implies that the residue of the rank mod 7 divides the partitions of 7n+5 into seven equal classes. Dyson’s rank 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) were later proved by Atkin and Swinnerton-Dyer [3]. Andrews and Garvan [2] found a crank for all three Ramanujan congruences (1.1)–(1.3). Their crank is defined as follows

(1.6)  crank(π) = \begin{cases}ℓ(π), & if μ(π) = 0, \\ ν^{∼}(π) − μ(π), & if μ(π) > 0, \end{cases}

where ℓ(π) denotes the largest part of π, μ(π) denotes the number of ones in π and ν^{∼}(π) denotes the number of parts of π larger than μ(π).

Later, Garvan, Kim and Stanton [9] found different cranks, which also explained 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

(1.7)  ∑_{n≥0} p(5n + 4)q^{n} = 5 ∏_{m≥1} \frac{(1 − q^{5m})^{5}}{(1 − q^{m})^{6}} ,

considered by Hardy to be an example of Ramanujan’s best work.

The main results of [2] can be summarized as

(1.8)  M(k, 5, 5n + 4) = \frac{p(5n + 4)}{5}, 0 ≤ k ≤ 4,
(1.9)  M(k, 7, 7n + 5) = \frac{p(7n + 5)}{7}, 0 ≤ k ≤ 6,
(1.10)  M(k, 11, 11n + 6) = \frac{p(11n + 6)}{11}, 0 ≤ k ≤ 10,

and

1 + (x + x^{−1} − 1)q + ∑_{n>1} ∑_{m} M^{∼}(m, n)x^{m}q^{n}

= ∏_{n≥1} \frac{(1 − q^{n})}{(1 − xq^{n})(1 − x^{−1}q^{n})}, (1.11)

where M^{∼}(m, n) denotes the number of partitions of n with crank m and M(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

(1.13)  M(2k + α, 10, 5n + 4) = \frac{M(α, 2, 5n + 4)}{5}, 0 ≤ k ≤ 4,

with α = 0, 1.

Recently, a very different refinement of (1.1) was given by Andrews [1]. Building on the work of Stanley [14], Andrews examined partitions π classified according to O(π) and O(π′), where where O(π) denotes the number of odd parts of the partition π and π′ is the conjugate of π. He used recursive relations to show that

(1.14) G(z, y, q) := ∑_{n,r,s≥0} S(n, r, s)q^{n}z^{r}y^{s} = \frac{(−zyq; q^{2})_{∞}}{(q^{4}; q^{4})_{∞}(z^{2}q^{2}; q^{4})_{∞}(y^{2}q^{2}; q^{4})_{∞}} ,

where S(n, r, s) denotes the number of partitions π of n with O(π) = r, O(π′) = s, and

(1.15)  (a; q)_{∞} = lim_{n→∞}(a; q)_{n},
(1.16)  (a; q)_{n} = (a)_{n} = \begin{cases}1, & if n = 0, \\ ∏_{j=0}^{n−1}(1 − aq^{j}), & if n > 0. \end{cases}

A direct combinatorial proof of (1.14) was later given by A. Sills [13], A. J. Yee [15] and C. Boulet [4]. Actually, C. Boulet proved a stronger version of (1.14) with one extra parameter. We define the Stanley rank of a partition π as

(1.17)  srank(π) = O(π) − O(π′).

It is easy to see that

(1.18)  srank(π) ≡ 0 (mod 2),

so that

(1.19)  p(n) = p_{0}(n) + p_{2}(n),

where p_{i}(n) (i = 0, 2) denotes the number of partitions of n with srank ≡ i (mod 4). We note that (1.14) with z = y^{−1} = \sqrt{−1} immediately implies the Stanley formula [14, p.8]

(1.20)  ∑_{n≥0}(p_{0}(n) − p_{2}(n))q^{n} = \frac{(−q; q^{2})_{∞}}{(q^{4}; q^{4})_{∞}(−q^{2}; q^{4})_{∞}^{2}} .

Using (1.1), (1.19) and (1.20), Andrews proved the following refinement of (1.1)

(1.21)  p_{0}(5n + 4) ≡ p_{2}(5n + 4) ≡ 0 (mod 5).

His proof of (1.21) was analytic and so at the end of [1] he posed the problem of finding a partition statistic that would give a combinatorial interpretation of (1.21). The first goal of this paper is to provide such an interpretation. It turns out that there are several distinct integral partition statistics, whose residue mod 5 split the partitions enumerated by p_{i}(5n + 4) (with i = 0, 2) into five equal classes. The first statistic, which we call the St-crank, is new. However, it is intimately related to the Andrews-Garvan crank (1.6). The second statistic, which we call the 2-quotient-rank, is also new. Unexpectedly, the third statistic is the 5-core-crank, introduced by Garvan, Kim and Stanton [9]. This statistic not only provides the desired combinatorial interpretation, but it also leads to a direct combinatorial proof of (1.21).

Our second goal here is to show that Andrews’ result (1.21) is a straightforward corollary of the new refinement of (1.1). This stronger refinement uses a new partition statistic, which we term the BG-rank. Remarkably, the BG-rank enables us to discover and prove new partition congruences mod 5.

The rest of this paper is organized as follows. In Section 2 we define the St-crank and show that it is indeed, a statistic asked for in [1]. In Section 3 we give another combinatorial interpretation of (1.21), discuss a surprising relation between 3-cores and the Andrews-Garvan crank, and then briefly we review the development in [9]. In Section 4 we establish a number of new formulas for partitions that are 5-cores and outline a combinatorial proof of (1.21). The hardest parts of this proof are dealt with in Sections 5 and 6. Finally, in Section 7 we introduce the BG-rank and use it to prove new partition congruences mod 5.

2. The St-crank

We begin with some preliminaries about partitions and their conjugates. A partition π is a nonincreasing sequence

(2.1)  π = (λ_{1}, λ_{2}, λ_{3}, …)

of nonnegative integers (parts)

(2.2)  λ_{1} ≥ λ_{2} ≥ λ_{3} ≥ · · · .

The weight of π, denoted by |π| is the sum of parts

(2.3)  |π| = λ_{1} + λ_{2} + λ_{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 = (1^{f_{1}}, 2^{f_{2}}, 3^{f_{3}}, …),

which indicates the number of times each integer occurs as a part. The number f_{i} = f_{i}(π) is called the frequency of i in π. The conjugate of π is the partition π′ = (λ′_{1}, λ′_{2}, λ′_{3}, …) with

λ′_{1} = f_{1} + f_{2} + f_{3} + f_{4} + ⋯
(2.5)  λ′_{2} = f_{2} + f_{3} + f_{4} + ⋯
λ′_{3} = f_{3} + f_{4} + ⋯
⋮

Next, we discuss two bijections. The first one relates π and bipartitions (π_{1}, π_{2}), where π_{2} is a partition with no repeated even parts. Bijection 1

π ⟶^{1} (π_{1}, π_{2}),

where

π = (1^{f_{1}}, 2^{f_{2}}, 3^{f_{3}}, …),
π_{1} = (1^{⌊f_{2}/2⌋}, 2^{⌊f_{4}/2⌋}, 3^{⌊f_{6}/2⌋}, …),
π_{2} = (1^{f_{1}}, 2^{{f_{2}}}, 3^{f_{3}}, 4^{{f_{2}}}, …),

⌊x⌋ is the largest integer ≤ x, and

{x} = x − 2⌊x/2⌋.

Figure 1. Graphical illustration of Bijection 2

Indeed, remove from π the maximum even number of even parts. The resulting partition is π_{2}, The removed even parts can be organized into a new partition (2^{2⌊f_{2}/2⌋}, 4^{2⌊f_{4}/2⌋}, 6^{2⌊f_{6}/2⌋}, …), which can easily be mapped onto π_{1}. Clearly, we have

(2.6)  |π| = 4|π_{1}| + |π_{2}|,
(2.7)  srank(π) = srank(π_{2}),

so that

∑_{π} q^{|π|}y^{srank(π)} = ∑_{π_{1}} q^{4|π_{1}|} ∑_{π_{2}} q^{|π_{2}|}y^{srank(π_{2})}

= \frac{1}{(q^{4}; q^{4})_{∞}} ∑_{π_{2}} q^{|π_{2}|}y^{srank(π_{2})}. (2.8)

Comparing (2.8) and (1.14) with zy = 1, we see that

(2.9)  ∑_{π_{2}} q^{|π_{2}|}y^{srank(π_{2})} = \frac{(−q; q^{2})_{∞}}{(y^{2}q^{2}; q^{4})_{∞}(q^{2}/y^{2}; q^{4})_{∞}} ,

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} is a partition of type A iff π_{A} ⟶^{1} ((1), π_{2}). We say that π_{B} = (λ_{1}, λ_{2}, λ_{3}, …) is a partition of type B iff either |π_{B}| ≠ 4, λ_{1} − λ_{2} ≥ 2, λ′_{1} − λ′_{2} ≥ 2, λ_{1} − 2 and λ_{2} are not identical even integers and π_{B} has no repeated even parts, or π_{B} = (3, 1). Obviously, π_{B} ⟶^{1} ((0), π_{B}). Our second bijection relates partitions of type A and B. Bijection 2

π_{A} ⟶^{2} π_{B},

where

π_{A} = (1^{f_{1}}, 2^{f_{2}}, 3^{f_{3}}, …, m^{f_{m}}),

π_{B} = \begin{cases}(1^{f_{1}+2}, 2^{f_{2}−2}, 3^{f_{3}}, 4^{f_{4}}, …, (m − 1)^{f_{m−1}}, m^{f_{m}−1}, (m + 2)^{1}), & if m > 2, \\ (1^{f_{1}+2}, 4^{1}), & if m = 2, f_{2} = 3, \\ (1^{f_{1}+1}, 3^{1}), & if m = 2, f_{2} = 2, \end{cases}

m ≥ 2, f_{2} = 2, 3, and f_{2i} = 0, 1 for i > 1.

Clearly, we have

(2.10)  |π_{A}| = |π_{B}|,
(2.11)  srank(π_{A}) = srank(π_{B}).

Next, we define a new partition statistic

(2.12)  St-crank(π) = crank(π_{1}) + \frac{1}{2}srank(π) + Ψ(π),

where π_{1} is determined by π ⟶^{1} (π_{1}, π_{2}), and the correction term Ψ(π) = 1 if π is of type B and zero, otherwise. We note that

(2.13)  St-crank(π_{A}) = −1 + \frac{1}{2}srank(π_{A}),

and

(2.14)  St-crank(π_{B}) = 1 + \frac{1}{2}srank(π_{B}).

We give some examples. Let π = (1^{2}, 2^{4}, 3^{2}, 4^{1}, 5^{1}, 6^{2}). Then π ⟶^{1} ((1^{2}, 3^{1}), (1^{2}, 3^{2}, 4^{1}, 5^{1})) so that St-crank(π) = (1−2)+(5−1)/2+0 = 1. Next, we consider a partition of type B. Let π_{B} = (1^{2}, 3^{1}, 5^{1}). Then St-crank(π_{B}) = 0+(4−2)/2+1 = 2.

Equipped with the definitions above, we can now prove the following lemma. Lemma 2.1. If

g(x, y, q) := ∑_{π} q^{|π|}x^{St-crank(π)}y^{srank(π)},

then g(x, y, q) has the product representation

g(x, y, q) = \frac{(q^{4}; q^{4})_{∞}(−q; q^{2})_{∞}}{(q^{4}x, q^{4}/x, q^{2}y^{2}x, q^{2}/(y^{2}x); q^{4})_{∞}} ,

where

(a_{1}, a_{2}, a_{2}, …; q)_{∞} = (a_{1}; q)_{∞}(a_{2}; q)_{∞}(a_{3}; q)_{∞} · · · .

Proof. If π is not of type B and π ⟶^{1} (π_{1}, π_{2}), then using (2.6)–(2.7), (2.12) we find that

(2.15)  q^{|π|}x^{St-crank(π)}y^{srank(π)} = q^{4|π_{1}|+|π_{2}|}x^{crank(π_{1})}(xy^{2})^{srank(π_{2})/2}.

On the other hand, if π = π_{A} and π_{A} ⟶^{2} π_{B}, then

q^{|π_{A}|}x^{St-crank(π_{A})}y^{srank(π_{A})} + q^{|π_{B}|}x^{St-crank(π_{B})}y^{srank(π_{B})}

(2.16) = q^{|π_{A}|}(x + x^{−1} − 1)(xy^{2})^{srank(π_{A})/2} + q^{|π_{B}|}x^{0}(xy^{2})^{srank(π_{B})/2}.

Here we have used (2.6)–(2.7) and (2.10)–(2.14).

Equations (2.15), (2.16) imply that

(2.17) ∑_{π} q^{|π|}x^{St-crank(π)}y^{srank(π)} = ∑_{π_{1}} q^{4|π_{1}|}w(x, π_{1}) ∑_{π_{2}} q^{|π_{2}|}(xy^{2})^{srank(π_{2})/2}

where

(2.18)  w(x, π_{1}) = \begin{cases}x + x^{−1} − 1, & if π_{1} = (1), \\ x^{crank(π_{1})}, & otherwise. \end{cases}

We note that in the first sum on the right side of (2.17) the summation is over unrestricted partitions π_{1}, and in the second sum the summation is over partitions π_{2} with no repeated even parts. Finally, recalling (1.11) with q → q^{4} and (2.9) with y^{2} → xy^{2}, we obtain

∑_{π} q^{|π|}x^{St-crank(π)}y^{srank(π)} = \frac{(q^{4}; q^{4})_{∞}}{(xq^{4}, q^{4}/x; q^{4})_{∞}} · \frac{(−q; q^{2})_{∞}}{(xy^{2}q^{2}, q^{2}/(xy^{2}); q^{4})_{∞}} ,

as desired. □

Next we show that

(2.19) the coefficient of q^{5n+4} in g(ξ, 1, q) = 0,

(2.20) the coefficient of q^{5n+4} in g(ξ, \sqrt{−1}, q) = 0,

where ξ is a primitive fifth root of unity (ξ^{5} = 1). We use the method of [6]. We need Jacobi’s triple product identity

(2.21)  ∑_{n=−∞}^{∞} z^{n}q^{n^{2}} = (q^{2}, −qz, −q/z; q^{2})_{∞},

which implies that

(2.22)  (q^{4}; q^{4})_{∞}(−q; q^{2})_{∞} = (q^{4}, −q^{3}, −q; q^{4})_{∞} = ∑_{n=−∞}^{∞} q^{2n^{2}+n} = ∑_{k≥0} q^{T_{k}},

and

(2.23)  (q^{2}ξ^{2}, q^{2}/ξ^{2}, q^{2}; q^{2})_{∞} = \frac{1}{1 − ξ^{2}} ∑_{m≥0}(−1)^{m}q^{2T_{m}}ξ^{−2m}(1 − ξ^{4m+2}).

Here T_{k} = k(k + 1)/2. By Lemma 2.1 and equations (2.22) and (2.23) we have

g(ξ, 1, q) = \frac{∑_{k≥0} q^{T_{k}}}{(q^{4}ξ, q^{4}/ξ, q^{2}ξ, q^{2}/ξ; q^{4})_{∞}} = \frac{(q^{2}ξ^{2}, q^{2}/ξ^{2}, q^{2}; q^{2})_{∞}}{(q^{10}; q^{10})_{∞}} ∑_{k≥0} q^{T_{k}}

= \frac{1}{1 − ξ^{2}} \frac{1}{(q^{10}; q^{10})_{∞}} ∑_{k,m≥0}(−1)^{m}q^{2T_{m}+T_{k}}ξ^{−2m}(1 − ξ^{4m+2}). (2.24)

Note that 2T_{m} + T_{k} ≡ 4 (mod 5) iff k ≡ m ≡ 2 (mod 5), but then 1 − ξ^{4m+2} = 0. This proves (2.19). The proof of (2.20) is analogous.

Let P_{i}(k, m, n) denote the number of partitions of n with srank ≡ i (mod 4) and St-crank ≡ k (mod m). Clearly,

(2.25)  ∑_{k=0}^{4} ξ^{k} ∑_{n≥0} P_{0}(k, 5, n)q^{n} = \frac{g(ξ, 1, q) + g(ξ, \sqrt{−1}, q)}{2},

(2.26)  ∑_{k=0}^{4} ξ^{k} ∑_{n≥0} P_{2}(k, 5, n)q^{n} = \frac{g(ξ, 1, q) − g(ξ, \sqrt{−1}, q)}{2},

Combining (2.19)–(2.20) and (2.25)–(2.26) we find that

(2.27)  ∑_{k=0}^{4} ξ^{k}P_{i}(k, 5, 5n + 4) = 0, (for i = 0, 2),

which implies that

(2.28)  P_{i}(0, 5, 5n + 4) = P_{i}(1, 5, 5n + 4) = ⋯ = P_{i}(4, 5, 5n + 4).

On the other hand

(2.29)  p_{i}(5n + 4) = ∑_{k=0}^{4} P_{i}(k, 5, 5n + 4),

so that

(2.30)  P_{i}(k, 5, 5n + 4) = \frac{1}{5}p_{i}(5n + 4),

for i = 0, 2 and k = 0, 1, 2, 3, 4. Thus, we have proved the main result of this section. Theorem 2.2. The residue of the partition statistic St-crank mod 5 divides the partitions enumerated by p_{i}(5n + 4) with i = 0, 2 into five equinumerous classes.

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

St-crank ≡ 0(mod 5) 1(mod 5) 2(mod 5) 3(mod 5) 4(mod 5)
srank ≡ 0  (3^{3})  (1^{5}, 2^{2}) (1^{4}, 2^{1}, 3^{1}) (1^{1}, 2^{4})  (1^{9})
(mod 4)  (1^{3}, 2^{1}, 4^{1})  (1^{4}, 5^{1})  (1^{3}, 3^{2})  (1^{6}, 3^{1}) (1^{2}, 2^{2}, 3^{1})
(1^{1}, 3^{1}, 5^{1})  (1^{2}, 2^{1}, 5^{1}) (1^{1}, 4^{2}) (1^{1}, 2^{1}, 6^{1}) (2^{3}, 3^{1})
(4^{1}, 5^{1})  (9^{1})  (2^{2}, 5^{1})  (2^{1}, 7^{1})  (1^{2}, 7^{1})
srank ≡ 2  (1^{3}, 2^{3})  (1^{1}, 2^{1}, 3^{2}) (1^{5}, 4^{1})  (1^{7}, 2^{1}) (2^{1}, 3^{1}, 4^{1})
(mod 4)  (1^{3}, 6^{1})  (1^{2}, 3^{1}, 4^{1}) (1^{1}, 8^{1}) (1^{1}, 2^{2}, 4^{1}) (3^{1}, 6^{1})

The analogs of (2.19)–(2.20) do not hold for 7n + 5 when ξ^{7} = 1 or for 11n + 6 when ξ^{11} = 1 so that the analog of Theorem 2.2 does not hold for Ramanujan’s congruence mod 7 or mod 11.

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_{2}(5n + 4) and by (1.21) we have the stronger congruence

p_{2}(5n + 4) ≡ 0 (mod 10).
3. t-cores

3.1. Preliminaries. In this section we recall some basic facts about t-cores and briefly review the development in [9]. A partition π is called a t-core, if it has no rim hooks of length t [11]. We let a_{t}(n) denote the number of partitions of n which are t-cores. In what follows, πt-core denotes a t-core partition. Given the diagram of a partition π we label a cell in the i-th row and j-th column by the least nonnegative integer congruent to j − i (mod t). The resulting diagram is called a t-residue diagram [11, p.84].

We also label cells in the infinite column 0 and in the infinite row 0 in the same way, and call the resulting diagram the extended t-residue diagram [9]. A region r in the extended diagram is the set of cells (i, j) satisfying t(r−1) ≤ j−i < tr. A cell 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 (not exposed) and E (exposed) as follows:

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

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 [12]. ϕ_{1} : P → Pt-core × P × ⋯ × P,

(3.1)  ϕ_{1}(π) = (πt-core, π̂_{t}),
(3.2)  π̂_{t} = (π̂_{0}, π̂_{1}, π̂_{2}, …, π̂_{t−1}),

such that

(3.3)  |π| = |πt-core| + t ∑_{i=0}^{t−1}|π̂_{i}|.

This bijection is described in more detail in [11], [9] and [8]. The following identity is an immediate corollary of this bijection.

(3.4)  \frac{1}{(q)_{∞}} = ∑_{n≥0} p(n)q^{n} = \frac{1}{(q^{t}; q^{t})_{∞}^{t}} ∑_{n≥0} a_{t}(n)q^{n}.

It can be rewritten as

(3.5)  ∑_{n≥0} a_{t}(n)q^{n} = \frac{(q^{t}; q^{t})_{∞}^{t}}{(q)_{∞}} .

There is another bijection ϕ_{2}, introduced in [9]. It is for t-cores only. ϕ_{2} : Pt-core → {n⃗ = (n_{0}, n_{1}, …, n_{t−1}) : n_{i} ∈ ℤ, n_{0} + ⋯ + n_{t−1} = 0},

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

We call n⃗ an n-vector. It has the following properties.

(3.7)  n⃗ ∈ ℤ^{t},  n⃗ · 1_{t} = 0,

and

(3.8)  |πt-core| = \frac{t}{2} ∑_{i=0}^{t−1} n_{i}^{2} + ∑_{i=0}^{t−1} in_{i},

where the t-dimensional vector 1_{t} has all components equal to 1. The generating function identity that corresponds to this second bijection is

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

Here

(3.10)  ||n⃗||^{2} = ∑_{i=0}^{t−1} n_{i}^{2}, and b⃗_{t} = (0, 1, 2, …, t − 1).

To construct the n-vector of πt-core in (3.6), we follow [8] and define

(3.11)  r⃗(πt-core) = (r_{0}, r_{1}, r_{2}, …, r_{t−1}),

where for 0 ≤ i ≤ t − 1, r_{i}(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)  ϕ_{2}(πt-core) = n⃗ = (r_{0} − r_{1}, r_{1} − r_{2}, r_{2} − r_{3}, …, r_{t−1} − r_{0}).

It was shown in [9] that a partition is a t-core with n-vector (n_{0}, n_{1}, …, n_{t−1}) iff for all i = 0, …, t − 1 the bi-infinite word W_{i} is of the form

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

We note that \frac{t}{2}||n⃗||^{2} is a multiple of t since n⃗ ·1_{t} = 0. Hence by (3.4) and (3.9) we have

(3.13)  ∑_{n≥0} a_{t}(tn + δ)q^{tn+δ} = ∑_{\begin{matrix}n⃗∈ℤ^{t}, n⃗·1_{t}=0 \\ n⃗·b⃗_{t}≡δ (mod t) \end{matrix}} q^{\frac{t}{2} ||n⃗||^{2}+b⃗_{t}·n⃗},

and

(3.14)  ∑_{n≥0} p(tn + δ)q^{n} = \frac{1}{(q)_{∞}^{t}} ∑_{n≥0} a_{t}(tn + δ)q^{n},

where δ = 0, 1, 2, …, t − 1.

3.2. The 2-quotient-rank. Having collected the necessary background on t-cores, we are now in a position to provide another combinatorial intrepretation of (1.21). To this end we introduce the following new partition statistic

(3.15)  2-quotient-rank(π) = ν(π̂_{0}) − ν(π̂_{1}),

where π̂_{0} and π̂_{1} are determined by

(3.16)  ϕ_{1}(π) = (π2-core, (π̂_{0}, π̂_{1})),

where ν(π) denotes the number of parts of π. Our main result here is Theorem 3.1. The residue of the 2-quotient-rank mod 5 divides the partitions enumerated by p_{i}(5n + 4) with i = 0, 2 into five equal classes.

Proof. We start by recalling Proposition 3.1(d) in [14]:

(3.17)  srank(π) ≡ |π| − |π2-core| (mod 4).

Using (3.3) with t = 2, we obtain from (3.17)

(3.18)  srank(π) ≡ 2(|π̂_{0}| + |π̂_{1}|) (mod 4).

Next, we define the generating function

(3.19)  G_{2}(x, y, q) := ∑_{π} q^{|π|}x^{2−quotient−rank(π)}y^{srank(π)}

It is possible to find a product representation for G_{2}(x, ω, q) when ω^{4} = 1, namely

(3.20) G_{2}(x, ω, q) = ∑_{\begin{matrix}π_{2−core} \\ π̂_{0},π̂_{1} \end{matrix}} q^{|π_{2−core}|+2(|π̂_{0}|+|π̂_{1}|)}x^{ν(π̂_{0})−ν(π̂_{1})}ω^{2(|π̂_{0}|+|π̂_{1}|)}

= \frac{∑_{k≥0} q^{T_{k}}}{(xq^{2}ω^{2}, q^{2}ω^{2}/x; q^{2}ω^{2})_{∞}} = \frac{(q^{4}; q^{4})_{∞}(−q; q^{2})_{∞}}{(xq^{2}ω^{2}, q^{2}ω^{2}/x; q^{2}ω^{2})_{∞}} .

Here we have used (2.22) along with the fact that a partition is a 2-core if and only if it is a staircase [11]. Hence using

(3.21)  (x^{±1}q^{2}ω^{2}; q^{2}ω^{2})_{∞} = (x^{±1}q^{2}ω^{2}, x^{±1}q^{4}; q^{4})_{∞},

we find that

(3.22)  G_{2}(x, \sqrt{±1}, q) = \frac{(q^{4}; q^{4})_{∞}(−q; q^{2})_{∞}}{(±xq^{2}, xq^{4}, ±q^{2}/x, q^{4}/x; q^{4})_{∞}} .

Comparing this with the product in Lemma 2.1, we see that

(3.23)  G_{2}(x, \sqrt{±1}, q) = g(x, \sqrt{±1}, q).

This means that

(3.24)  p^{∼}_{i}(m, n) = p_{i}^{∗}(m, n),  i = 0, 2.

Here p^{∼}_{i}(m, n) (resp. p_{i}^{∗}(m, n)) denotes the number of partitions of n with srank ≡ i (mod 4) and 2-quotient-rank= m (resp. St-crank= m). Theorem 3.1 follows easily from (3.24) and Theorem 2.2. □

We remark that the St-crank and the 2-quotient-rank are distinct statistics. For example, St-crank((5, 4, 1)) = 0 but 2-quotient-rank((5, 4, 1)) = 1. It would be interesting to find a direct combinatorial proof of (3.24).

3.3. 3-cores and the crank. It is natural to attempt to extend the construction of the previous section to (3-core, 3-quotient). To this end we define

(3.25)  G_{3}(x, q) := ∑_{π} q^{|π|}(x^{3n_{1}} + x^{3n_{2}+1} + x^{−3n_{2}−1})x^{3(ν(π̂_{1})−ν(π̂_{2}))},

where

(3.26)  ϕ_{1}(π) = (π3-core, (π̂_{0}, π̂_{1}, π̂_{2})),
(3.27)  ϕ_{2}(π3-core) = (−n_{1} − n_{2}, n_{1}, n_{2}).

Hence using (3.3), (3.8) with t = 3, we find that

(3.28)  |π| = Q_{3}(n_{1}, n_{2}) + 3(|π̂_{0}| + |π̂_{1}| + |π̂_{2}|),

where

(3.29)  Q_{3}(n_{1}, n_{2}) = 3(n_{1}^{2} + n_{1}n_{2} + n_{2}^{2}) + n_{1} + 2n_{2}.

Clearly,

G_{3}(x, q) (3.30)

= \frac{∑_{n_{1},n_{2}} q^{Q_{3}(n_{1},n_{2})}x^{3n_{1}} + ∑_{n_{1},n_{2}} q^{Q_{3}(n_{1},n_{2})}x^{3n_{2}+1} + ∑_{n_{1},n_{2}} q^{Q_{3}(n_{1},n_{2})}x^{−1−3n_{2}}}{(q^{3}, x^{3}q^{3}, q^{3}/x^{3}; q^{3})_{∞}} .

Next, we change summation variables in the first, second, and third sums respectively as

3n_{1} = n − m,  3n_{2} = n + 2m,
3n_{1} + 1 = −2n − m,  3n_{2} + 1 = n − m,
3n_{1} + 1 = −n − 2m,  3n_{2} + 1 = m − n,

respectively. In this way, we get

(3.31)  G_{3}(x, q) = \frac{∑_{n,m} q^{n^{2}+mn+m^{2}}x^{n−m}}{(q^{3}, x^{3}q^{3}, q^{3}/x^{3}; q^{3})_{∞}} .

Remarkably, the numerator on the right side of (3.31) has a product representation. By equation (1.23) in [10] we have (3.32)

∑_{n,m} q^{n^{2}+mn+m^{2}}x^{n−m} = (x + 1 + 1/x)(q; q)_{∞}(q^{3}; q^{3})_{∞} \frac{(x^{3}q^{3}; q^{3})_{∞}(q^{3}/x^{3}; q^{3})_{∞}}{(xq; q)_{∞}(q/x; q)_{∞}} .

We have

(3.33)  G_{3}(x, q) = (x + 1 + 1/x)\frac{(q)_{∞}}{(xq; q)_{∞}(q/x; q)_{∞}} .

Recalling (1.11), we see that (3.33) gives a surprising relation between 3-cores and the Andrews-Garvan crank. This certainly warrants further investigation.

3.4. 5-cores. We now assume t = 5. For the case δ = 4 the right side of (3.13) can be simplified using the the following change of variables.

n_{0} = α_{0} + α_{4},
n_{1} = −α_{0} + α_{1} + α_{4},
(3.34)  n_{2} = −α_{1} + α_{2},
n_{3} = −α_{2} + α_{3} − α_{4},
n_{4} = −α_{3} − α_{4},

We find n⃗ is an n-vector satisfying n⃗ · b⃗_{5} ≡ 4 (mod 5) if and only if

(3.35)  α⃗ = (α_{0}, α_{1}, α_{2}, α_{3}, α_{4}) ∈ ℤ^{5}

and

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

We call α⃗ an α-vector. Hence, by (3.13) and (3.14) we have

(3.37)  ∑_{n≥0} a_{5}(5n + 4)q^{n+1} = ∑_{\begin{matrix}α⃗·1_{5}=1 \\ α⃗∈ℤ^{5} \end{matrix}} q^{Q(α⃗)},

and

(3.38)  ∑_{n≥0} p(5n + 4)q^{n+1} = \frac{1}{(q)_{∞}^{5}} ∑_{\begin{matrix}α⃗·1_{5}=1 \\ α⃗∈ℤ^{5} \end{matrix}} q^{Q(α⃗)},

where

(3.39)  Q(α⃗) = ||α⃗||^{2} − (α_{0}α_{1} + α_{1}α_{2} + ⋯ + α_{4}α_{0}),

If |π| ≡ 4 (mod 5) and t = 5, we can combine bijections ϕ_{1} and ϕ_{2} into a single bijection

(3.40)  Φ(π) = (α⃗, π̂_{5}),

such that

(3.41)  |π| = 5Q(α⃗) − 1 + 5 ∑_{i=0}^{4}|π̂_{i}|.

Next, following [9] we define the 5-core crank of π when |π| ≡ 4 (mod 5) as

(3.42) c_{5}(π) = 1 + ∑_{i=0}^{4} iα_{i} ≡ 2(1 + n_{0} − n_{1} − n_{2} + n_{3}) ≡ 2 + ∑_{i=−2}^{2} ir_{2−i} (mod 5),

where α is determined by (3.40).

It is easy to check that Q(α⃗) in (3.41) remains invariant under the following cyclic permutation

(3.43)  Ĉ_{1}(α⃗) = (α_{4}, α_{0}, α_{1}, α_{2}, α_{3}),

while c_{5}(π) increases by 1 (mod 5) under the map

(3.44)  Ô(π) = Φ^{−1}(Ĉ_{1}(α⃗), π̂_{5}).

In other words, if |π| ≡ 4 (mod 5), then

(3.45)  |π| = |Ô(π)|,

and

(3.46)  c_{5}(π) + 1 ≡ c_{5}(Ô(π)) (mod 5).

This suggests that all partitions of 5n + 4 can be organized into orbits. Each orbit consists of five distinct members:

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

and each element of the orbit has a distinct 5-core crank (mod 5). Clearly, the total number of such orbits is \frac{1}{5}p(5n + 4), and so p(5n + 4) ≡ 0 (mod 5). This summarizes the combinatorial proof of (1.1) given in [9]. If we apply the map Ô (3.44) to the partitions of 5n + 4 that are 5-cores, we find that

(3.48)  a_{5}^{0}(5n + 4) = a_{5}^{1}(5n + 4) = ⋯ = a_{5}^{4}(5n + 4),

where, for 0 ≤ j ≤ 4, a_{5}^{j}(n) denotes the number of partitions of n that are 5-cores with 5-core crank congruent to j modulo 5. Hence,

(3.49)  a_{5}^{j}(5n + 4) = \frac{1}{5}a_{5}(5n + 4), j = 0, 1, …, 4,

which proves that

(3.50)  a_{5}(5n + 4) ≡ 0 (mod 5).

Actually, more is true. We have

(3.51)  a_{5}(5n + 4) = 5a_{5}(n).

We sketch the combinatorial proof of (3.51) given in [9]. See also [8]. The map θ : P5-core(n) ⟶ P5-core^{0}(5n + 4), defined in terms of n-vectors as

n⃗ ↦ n′ = (n_{1} + 2n_{2} + 2n_{4} + 1, −n_{1} − n_{2} + n_{3} + n_{4} + 1, 2n_{1} + n_{2} + 2n_{3},

− 2n_{2} − 2n_{3} − n_{4} − 1, −2n_{1} − n_{3} − 2n_{4} − 1), (3.52) is a bijection. Here P5-core(n) is the set of all 5-cores of n, and P5-core^{0}(n) is set of all 5-cores of n with 5-core crank congruent to zero modulo 5. Since θ is a bijection, we have

(3.53)  a_{5}(n) = a_{5}^{0}(5n + 4).

The proof of (3.51) easily follows from (3.49) and (3.53). Finally, we remark that Ramanujan’s result (1.7) is a straightforward consequence of (3.14) with(t, δ) = (5, 4), (3.51), and (3.5) with t = 5.

4. Refinement of Ramanujan’s mod 5 congruence, the srank and the 5-core crank

In the previous section we discussed the combinatorial proof in [9] of Ramanujan’s congruence (1.1) using the the 5-core crank (3.42). It is somewhat unexpected that the 5-core crank can be employed to prove the refinement (1.21) as well.

In fact, we were amazed to discover the following elegant formulas

(4.1)  srank(π5-core) ≡ ∑_{i=0}^{4}(n_{i} + i)^{3} (mod 4),

(4.2)  srank(π) ≡ srank(π5-core) + ∑_{i=0}^{4} srank(π̂_{i})

+ 2 ∑_{i=0}^{4}|π̂_{i}|(n_{i} + i) (mod 4),

where π5-core, π̂ = (π̂_{0}, π̂_{1}, π̂_{2}, π̂_{3}, π̂_{4}) are determined by (3.1) with t = 5, and

n⃗ = (n_{0}, n_{1}, …, n_{4}) = ϕ_{2}(π5-core).

In spite of their simple appearance, the above formulas are far from obvious. In Sections 4 and 5 we prove generalizations of (4.1)–(4.2). Here we restrict our attention to some implications of (4.1)–(4.2).

First, we note that if |π5-core| ≡ 4 (mod 5), then (4.1) can be written in terms of an α-vector (3.34) as (4.3) srank(π5-core) ≡ α_{0}α_{1}(α_{0} − α_{1}) + α_{1}α_{2}(α_{1} − α_{2}) + ⋯ + α_{4}α_{0}(α_{4} − α_{0}) (mod 4).

Similarly, if |π| ≡ 4 (mod 5), then

srank(π) ≡ α_{0}α_{1}(α_{0} − α_{1}) + ⋯ + α_{4}α_{0}(α_{4} − α_{0}) + ∑_{i=0}^{4} srank(π̂_{i})

+ 2{(α_{0} + α_{4})|π̂_{0}| + (α_{2} + α_{3})|π̂_{1}| + (α_{1} + α_{2})|π̂_{2}| (4.4)

+ (α_{0} + α_{1})|π̂_{3}| + (α_{3} + α_{4})|π̂_{4}|} (mod 4).

Remarkably, (4.3) suggests that srank(π5-core) with |π5-core| ≡ 4 (mod 5) remains invariant mod 4 under the cyclic permutation (3.43), and we have the following refinement of (3.49):

(4.5)  a_{5,i}^{j}(5n + 4) = \frac{1}{5}a_{5,i}(5n + 4),

where j = 0,…,4 and i = 0, 2. Here a_{5,i}(n) denotes the number of 5-cores of n with srank ≡ i (mod 4), and a_{5,i}^{j}(n) denotes the number of 5-cores of n with srank ≡ i (mod 4) and 5-core crank ≡ j (mod 5). Moreover, it is not difficult to verify that the map θ, given by (3.52), preserves the srank mod 4. Indeed, recalling that n_{0} + n_{1} + n_{2} + n_{3} + n_{4} = 0 we find after some simplication that

∑_{i=0}^{4}((n_{i} + i)^{3} − (n′_{i} + i)^{3}) ≡ 2(n_{0}n_{2}(n_{0} + n_{2}) + n_{1}n_{3}(n_{1} + n_{3}) + n_{2}n_{3}(n_{2} + n_{3})

+ n_{1}(n_{1} + 1) + n_{2}(n_{2} + 1) + n_{3}(n_{3} + 1)) (4.6)

≡ 0 (mod 4),

where n′ is defined in (3.52). Hence, (3.53) and (3.51) can be refined as

(4.7)  a_{5,i}(n) = a_{5,i}^{0}(5n + 4),  (i = 0, 2),

and

(4.8)  a_{5,i}(5n + 4) = 5a_{5,i}(n),  (i = 0, 2),

respectively.

It is less trivial to prove the 5-core crank analogue of Theorem 2.2. Namely, Theorem 4.1. The residue of the 5-core crank mod 5 divides the partitions enumerated by p_{i}(5n + 4) with i = 0, 2 into five equal classes.

Proof. We sketch a proof using (4.4) and (4.5). We define the cyclic shift operator Ĉ_{2} by

(4.9)  Ĉ_{2}(π̂_{5}) = (π̂_{4}, π̂_{2}, π̂_{3}, π̂_{0}, π̂_{1}),

Next, we use (4.9) to modify (3.44) as

(4.10)  Ô_{s}(π) = Φ^{−1}(Ĉ_{1}(α⃗), Ĉ_{2}(π̂_{5})),

where Φ(π) = (α⃗, π̂_{5}). Fix i = 0, 2. By (4.4) we see that Ô_{s} preserves the srank mod 4, and we may assemble all partitions of 5n + 4 with srank ≡ i (mod 4) into orbits:

π, Ô_{s}(π), Ô_{s}^{2}(π), Ô_{s}^{3}(π), Ô_{s}^{4}(π),

where π is some partition of 5n + 4 with srank(π) ≡ i (mod 4). As before, each orbit contains exactly five members and the 5-core crank increases by 1 mod 5 along the orbit. The number of these orbits is \frac{1}{5}p_{i}(5n + 4), consequently p_{i}(5n + 4) ≡ 0 (mod 5) and the result follows. □

Theorem 4.1 is illustrated below in Table 2, which contains all 30 partitions of 9, organized into 6 orbits. Each row in this table represents an orbit, and the first row lists all partitions of 9 that are 5-cores. In the table we have also included the image of each partition under the bijection ϕ_{1}. Instead of giving the full 5-quotient we have used a short-hand notation. Terms in the table have the form π → (π5-core, k) where k indicates that π̂_{5} = (π̂_{0}, π̂_{1}, π̂_{2}, π̂_{3}, π̂_{4}), where π̂_{i} = (1) if i = k, and (0) otherwise.

Now, we state some new formulas for a_{5,0}(n):

(4.11)  a_{5,0}(4n) = a_{5}(4n),
(4.12)  a_{5,0}(4n + 1) = a_{5}(4n + 1),
(4.13)  a_{5,0}(4n + 2) = 0,
(4.14)  a_{5,0}(4n + 3) = a_{5}(n).

Table 2

c_{5} ≡ 0(mod 5)  1(mod 5)  2(mod 5)  3(mod 5)  4(mod 5)
srank ≡ 0  (1^{4}, 5^{1})  (1^{3}, 3^{2})  (1^{4}, 2^{1}, 3^{1})  (1^{1}, 2^{1}, 6^{1})  (2^{2}, 5^{1})

(mod 4)

(1^{5}, 2^{2}) →  (2^{3}, 3^{1}) →  (1^{2}, 7^{1}) →  (4^{1}, 5^{1}) →  (1^{3}, 2^{1}, 4^{1}) →
((2^{2}), 3)  ((1^{4}), 2)  ((1^{2}, 2^{1}), 1)  ((1^{1}, 3^{1}), 4)  ((4^{1}), 0)
(3^{3}) →  (1^{9}) →  (1^{1}, 3^{1}, 5^{1}) → (1^{2}, 2^{2}, 3^{1}) →  (9^{1}) →
((2^{2}), 2)  ((1^{4}), 1)  ((1^{2}, 2^{1}), 4)  ((1^{1}, 3^{1}), 0)  ((4^{1}), 3)
(2^{1}, 7^{1}) →  (1^{2}, 2^{1}, 5^{1}) →  (1^{1}, 2^{4}) →  (1^{6}, 3^{1}) →  (1^{1}, 4^{2}) →
((2^{2}), 1)  ((1^{4}), 4)  ((1^{2}, 2^{1}), 0)  ((1^{1}, 3^{1}), 3)  ((4^{1}), 2)
srank ≡ 2  (1^{3}, 2^{3}) →  (1^{3}, 6^{1}) →  (2^{1}, 3^{1}, 4^{1}) →  (1^{1}, 8^{1}) →  (1^{2}, 3^{1}, 4^{1}) →
(mod 4)  ((2^{2}), 4)  ((1^{4}), 0)  ((1^{2}, 2^{1}), 3)  ((1^{1}, 3^{1}), 2)  ((4^{1}), 1)
(3^{1}, 6^{1}) →  (1^{1}, 2^{2}, 4^{1}) →  (1^{7}, 2^{1}) →  (1^{1}, 2^{1}, 3^{2}) →  (1^{5}, 4^{1}) →
((2^{2}), 0)  ((1^{4}), 3)  ((1^{2}, 2^{1}), 2)  ((1^{1}, 3^{1}), 1)  ((4^{1}), 4)

Formulas (4.11)–(4.13) follow from (4.1). Formula (4.14) is a consequence of the following bijective map, defined in terms of n-vectors by

(4.15)  n⃗ ↦ n′ = (2n_{1}, 1 + 2n_{4}, 2n_{2}, −1 + 2n_{0}, 2n_{3}).

To show that this is a bijection, one may easily verify that

(4.16)  |ϕ_{2}^{−1}(n′)| = 4|ϕ_{2}^{−1}(n⃗)| + 3,

and show that if |π5-core| ≡ 3 (mod 4) then srank(π5-core) ≡ 0 (mod 4) if and only if

(4.17)  ϕ_{2}(π5-core) ≡ (0, 1, 0, 1, 0) (mod 2).

5. The srank of t-cores

In this section we generalize equation (4.1) to t-cores for general t. Theorem 5.1. Let t ≥ 2, and

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

Let a = 0 or 1. If t ≡ 1 + 2a (mod 4), then

(5.1)  srank(πt-core) ≡ ∑_{i=0}^{t−1}(n_{i} + (1 − 2a)i + a)^{3} (mod 4);

and if t ≡ 2a (mod 4), then

(5.2)  srank(πt-core) ≡ ∑_{i=0}^{t−1} an_{i}^{2} + (i^{2} + i)n_{i} (mod 4).

Proof. For a partition π = (λ_{1}, λ_{2}, …, λ_{ν}),

(5.3)  srank(π) ≡ ∑_{j=1}^{ν}(λ_{j}^{2} + (1 − 2j)λ_{j}) (mod 4),

by Proposition 3.1(c) in [14]. Now let t ≥ 2, and suppose π is a t-core, π = (λ_{1}, λ_{2}, …, λ_{ν}), and

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

Suppose n_{i} > 0. Then i is exposed in each region k, 1 ≤ k ≤ n_{i}, and this exposed cell is to the right of the main diagonal. Each such exposed i in region k corresponds to a part of the partition in which the number of cells to the right of the main diagonal is t(k − 1) + i. Thus,

(5.4)  J = size of the Durfee square of π = ∑_{n_{i}>0} n_{i},

and

srank(π) ≡ ∑_{j=1}^{J}(λ_{j} − j)^{2} + (λ_{j} − j) + (j − j^{2}) + ∑_{j=J+1}^{ν} λ_{j}^{2} + (1 − 2j)λ_{j} (mod 4) (5.5)

≡ ∑_{n_{i}>0} ∑_{k=1}^{n_{i}}(t(k − 1) + i)^{2} + (t(k − 1) + i) + ∑_{j=1}^{J} j − j^{2}

+ ∑_{j=J+1}^{ν} λ_{j}^{2} + (1 − 2j)λ_{j} (mod 4)

≡ ∑_{n_{i}>0} g(t, n_{i}, i) + \frac{1}{3}(J − J^{3}) + ∑_{j=J+1}^{ν} λ_{j}^{2} + (1 − 2j)λ_{j} (mod 4),

where

(5.6)  g(t, n, i) = \frac{1}{3}t^{2}n^{3} + (ti − \frac{1}{2}t(t − 1))n^{2} + (i^{2} − i(t − 1) + \frac{1}{6}t^{2} − \frac{1}{2}t)n.

We note that for n > 0,

(5.7)  g(t, n, i) ≡ 0 (mod 2),

since g(t, n, i) is a sum of even integers. A calculation shows that

(5.8)  g(t, n, i) + g(t, −n, t − 1 − i) = 0.

It follows that g(t, n, i) is a sum of even integers for n < 0 and (5.7) holds for all n, so that

(5.9)  g(t, n, i) ≡ g(t, −n, t − 1 − i) (mod 4).

We have

(5.10)  ϕ_{2}(π′) = (−n_{t−1}, −n_{t−2}, …, −n_{0}).

See [9, p.3]. Let π_{1} be the partition consisting of the first J parts of π′. Then by (5.10), (5.7) and (5.9) we find that

(5.11)  srank(π_{1}) ≡ ∑_{n_{i}<0} g(t, −n_{i}, t − 1 − i) + \frac{1}{3}(J − J^{3})

≡ ∑_{n_{i}<0} g(t, n_{i}, i) + \frac{1}{3}(J − J^{3}) (mod 4).

Here we have used the fact that π′ is also a t-core. Now,

(5.12)  srank(π′_{1}) = −srank(π_{1}) ≡ srank(π_{1}) (mod 4),

and

(5.13)  srank(π′_{1}) ≡ ∑_{j=1}^{J} J^{2} + (1 − 2j)J + ∑_{j=J+1}^{ν} λ_{j}^{2} + (1 − 2j)λ_{j} (mod 4).

Hence

(5.14)  ∑_{j=J+1}^{ν} λ_{j}^{2} + (1 − 2j)λ_{j} ≡ ∑_{n_{i}<0} g(t, n_{i}, i) + \frac{1}{3}(J − J^{3}) (mod 4),

since

(5.15)  ∑_{j=1}^{J} J^{2} + (1 − 2j)J = 0.

Hence, by (5.5) and (5.14), we have

(5.16)  srank(π) ≡ ∑_{n_{i}>0} g(t, n_{i}, i) + ∑_{n_{i}<0} g(t, n_{i}, i) + \frac{2}{3}(J − J^{3}) (mod 4)

≡ ∑_{i=0}^{t−1} g(t, n_{i}, i) (mod 4),

since g(t, 0, i) = 0 and \frac{2}{3}(J − J^{3}) ≡ 2(J^{3} − J) ≡ 0 (mod 4).

We prove (5.1) and (5.2) by finding simplified forms for g(t, n, i) (mod 4). It is clear that the value of g(t, n, i) (mod 4) depends on the residue of t (mod 4). \underline{Case 1.} t ≡ 0 (mod 4). Then

g(t, n, i) ≡ g(0, n, i) ≡ (i^{2} + i)n (mod 4).

Thus (5.2) holds when a = 0. \underline{Case 2.} t ≡ 1 (mod 4). Then

g(t, n, i) ≡ g(1, n, i) ≡ −g(1, n, i) ≡ (n + i)^{3} − n − i^{3} (mod 4),

so that

srank(π) ≡ ∑_{i=0}^{t−1}(n_{i} + i)^{3} − ∑_{i=0}^{t−1} n_{i} − ∑_{i=0}^{t−1} i^{3} (mod 4)

≡ ∑_{i=0}^{t−1}(n_{i} + i)^{3} − \frac{t^{2}(t − 1)^{2}}{4} (mod 4)

≡ ∑_{i=0}^{t−1}(n_{i} + i)^{3} (mod 4),

since t ≡ 1 (mod 4). Here we have also used (3.7). Thus (5.1) holds when a = 0. \underline{Case 3.} t ≡ 2 (mod 4). Then

g(t, n, i) ≡ g(2, n, i) ≡ −g(2, n, i) ≡ (n^{2}+(i^{2}+i)n)−n+2in(n+i) ≡ (n^{2}+(i^{2}+i)n)−n (mod 4).

Thus (5.2) holds when a = 1. \underline{Case 4.} t ≡ 3 (mod 4). Then

g(t, n, i) ≡ g(3, n, i) ≡ −g(3, n, i) ≡ (n − i + 1)^{3} + n + (i − 1)^{3} (mod 4).

Figure 2. Attaching one cell to the rim

Since

∑_{i=0}^{t−1}(i − 1)^{3} = t\frac{(t − 3)}{4}(t(t − 3) + 4) ≡ 0 (mod 4),

for t ≡ 3 (mod 4), we see that (5.1) holds when a = 1. □

6. The srank and the t-quotient

In this section we prove that

(6.1)  srank(π) ≡ srank(πt-core) + 2a ∑_{i=0}^{t−1} |π̂_{i}| (mod 4),

provided t ≡ 2a (mod 4), and

(6.2)  srank(π) ≡ srank(πt-core) + 2 ∑_{i=0}^{t−1}(n_{i} + i + a)|π̂_{i}|

+ ∑_{i=0}^{t−1} srank(π̂_{i}) (mod 4),

if t ≡ 1 + 2a (mod 4). Here a = 0, 1 and

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

We note that when t = 5 equation (6.2) is formula (4.2). To prove (6.1), (6.2) we use (5.3) which we rewrite as

(6.3)  srank(π) ≡ ∑_{j=1}^{ν}(λ_{j}^{2} + (2j − 3)λ_{j}) (mod 4),

where π = (λ_{1}, λ_{2}, …, λ_{ν}).

Next, let π^{∗} be a partition obtained from π by attaching a single cell with coordinates (x, y) to the rim of the diagram of π as indicated in Figure 2.

It is easy to see that

(6.4) srank(π^{∗})−srank(π) ≡ (x^{2} −(x−1)^{2})−(y^{2} −(y −1)^{2}) ≡ 2(x+y) (mod 4).

Suppose, we create a new partition π^{∗∗} by attaching a border-strip of length ℓ to the diagram of π, such that the extreme North-East cell (head) of the strip has coordinates (x, y). Repeated use of (6.4) yields the following formula

(6.5) srank(π^{∗∗}) − srank(π) ≡ 2(x + y) + 2(x + y + 1) + ⋯ + 2(x + y + ℓ − 1)

≡ 2ℓ(x + y) + ℓ^{2} − ℓ (mod 4).

Here we have used the fact that a border-strip can be added to the diagram one cell at a time, in such a way that all intermediate diagrams correspond to partitions. It is straightforward to verify that the right side of (6.5) with ℓ = tλ becomes

(6.6)  2aλ (mod 4),  if t ≡ 2a (mod 4),

and

(6.7)  2aλ(x + y + a) + λ^{2} − λ (mod 4),  if t ≡ 1 + 2a (mod 4),

with a = 0, 1.

Note that (6.6) immediately implies (6.1). To prove (6.2) we need to work a little harder. Let’s consider a partition π^{∼}_{0}, such that

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

This partition has the following property. The rim cells with color i are exposed in all regions ≤ n_{i} and are not exposed in all regions > n_{i} of the extended t-residue diagram of π^{∼}_{0}. This means that the word W_{i} of π^{∼}_{0} has the form

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

Let us attach a border-strip of length ℓ = tλ to π^{∼}_{0} in such a way that the word W_{i} becomes

\begin{matrix}Region : & ⋯ & n_{i} − λ_{1} & n_{i} + 1 − λ_{1} & ⋯ & ⋯ & n_{i} + 1 & n_{i} + 2 & n_{i} + 3 & ⋯ \\ W_{i} : & ⋯ & E & N & E & ⋯ & E & N & N & ⋯ \end{matrix}

This way we create a partition π^{∼}_{1} such that

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

It is straightforward to verify that for t odd the coordinates (x, y) of the border-strip head satisfy

(6.8)  x + y ≡ n_{i} + i (mod 2).

Hence using (6.7), (6.8) we find that for t ≡ 1 + 2a (mod 4)

(6.9)  srank(π^{∼}_{1}) − srank(π^{∼}_{0}) ≡ 2λ_{1}(n_{i} + i + a) + λ_{1}^{2} − λ_{1} (mod 4),

where a = 0, 1.

Next, we add to the diagram of π^{∼}_{1} a new border-strip of length tλ_{2} with λ_{2} ≤ λ_{1}, such that W_{i} becomes

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

The new partition π^{∼}_{2} satisfies

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

Replacing λ_{1} by λ_{2} and n_{i} by n_{i} + 1 and repeating the argument that led us to (6.9) we obtain

(6.10)  srank(π^{∼}_{2}) − srank(π^{∼}_{1}) ≡ 2λ_{2}(n_{i} + 1 + i + a) + λ_{2}^{2} − λ_{2} (mod 4).

Let π^{∼}_{ν} denote the partition obtained from π^{∼}_{0}, such that

ϕ_{1}(π^{∼}_{ν}) = (πt-core, (π̂_{0}, …, π̂_{i}, …, π̂_{t−1})),

where π̂_{i} = (λ_{1}, λ_{2}, …, λ_{ν}). Proceeding as above and using (6.3) we find that for t ≡ 1 + 2a (mod 4)

(6.11) srank(π^{∼}_{ν}) − srank(π^{∼}_{0}) ≡ 2 ∑_{j=1}^{ν} λ_{j}(n_{i} + i + a + j − 1) + ∑_{j=1}^{ν}(λ_{j}^{2} − λ_{j})

≡ 2(n_{i} + i + a) ∑_{j=1}^{ν} λ_{j} + ∑_{j=1}^{ν}(λ_{j}^{2} + (2j − 3)λ_{j})

≡ 2(n_{i} + i + a)|π̂_{i}| + srank(π̂_{i}) (mod 4),

with a = 0, 1. Formula (6.2) follows easily from (6.11).

7. Generalization of Andrews’ refinement and new partition congruences modulo 5

In Section 3.2 we gave a new combinatorial interpretation of Andrews’ result (1.21) in terms of the 2-quotient of a partition. Further study of this development led us to a generalization of (1.21), which we now describe. We define the new partition statistic

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

where π = (λ_{1}, λ_{2}, …, λ_{ν}) and for an integer m, par(m) denotes the parity of m; i.e. par(m) = 1 if m is odd and 0, otherwise. If ϕ_{1}(π) = (π2-core, (π̂_{0}, π̂_{1})) and ϕ_{2}(π2-core) = (n_{0}, −n_{0}) then it is easy to verify that

(7.2)  BG-rank(π) = n_{0} = r_{0} − r_{1}.

Here r_{i} with i = 0, 1 denotes the number of cells colored i in the 2-residue diagram of π.

Next, we recall that

(7.3)  srank(π) ≡ |π| − |π2-core| ≡ |π| − n_{0}(2n_{0} − 1) (mod 4).

Here we have used (3.17) and (3.8). Thus, if |π| is given, then srank(π) (mod 4) is completely determined by BG-rank(π). Clearly, the converse is not true.

Let \overline{p}_{j}(m, n) denote the number of partitions of n with BG-rank = j and 2-quotient-rank = m. Then

(7.4)  f_{j}(x, q) = ∑_{\begin{matrix}n≥0, \\ m \end{matrix}} \overline{p}_{j}(m, n)x^{m}q^{n}

= q^{(2j−1)j} ∑_{π̂_{0},π̂_{1}} q^{2(|π̂_{0}|+|π̂_{1}|)}x^{ν(π̂_{0})−ν(π̂_{1})}

= \frac{q^{(2j−1)j}}{(q^{2}x, q^{2}/x; q^{2})_{∞}} .

Proceeding as in Section 2, we find that for ξ^{5} = 1, ξ ≠ 1

(7.5)  f_{j}(ξ, q) = \frac{1}{1 − ξ^{2}} \frac{q^{(2j−1)j}}{(q^{10}; q^{10})_{∞}} ∑_{n≥0}(−1)^{n}q^{n^{2}+n}ξ^{−2n}(1 − ξ^{4n+2}).

We have

(7.6)  f_{j}(ξ, q) = ∑_{\begin{matrix}n≥0 \\ 0≤m≤4 \end{matrix}} \overline{P}_{j}(m, 5, n)ξ^{m}q^{n},

where \overline{P}_{j}(m, 5, n) denotes the number of partitions of n with BG-rank = j and 2-quotient-rank ≡ m (mod 5).

Next, we observe that

(7.7)  n^{2} + n ≡ \begin{cases}0 (mod 5), & if n ≡ 0, 4 (mod 5), \\ 2 (mod 5), & if n ≡ 1, 3 (mod 5), \\ 1 (mod 5), & if n ≡ 2 (mod 5), \end{cases}

(7.8)  (2j − 1)j ≡ \begin{cases}0 (mod 5), & if j ≡ 0, 3 (mod 5), \\ 1 (mod 5), & if j ≡ 1, 2 (mod 5), \\ 3 (mod 5), & if j ≡ 4 (mod 5), \end{cases}

and (1 − ξ^{4n+2}) = 0 if n ≡ 2 (mod 5). This means that

(7.9)  ∑_{m=0}^{4} \overline{P}_{j}(m, 5, 5n)ξ^{m} = 0, if j ≡ 1, 2 (mod 5),

(7.10)  ∑_{m=0}^{4} \overline{P}_{j}(m, 5, 5n + 1)ξ^{m} = 0, if j ≢ 1, 2 (mod 5),

(7.11)  ∑_{m=0}^{4} \overline{P}_{j}(m, 5, 5n + 2)ξ^{m} = 0, if j ≢ 0, 3 (mod 5),

(7.12)  ∑_{m=0}^{4} \overline{P}_{j}(m, 5, 5n + 3)ξ^{m} = 0, if j ≡ 0, 3 (mod 5),

(7.13)  ∑_{m=0}^{4} \overline{P}_{j}(m, 5, 5n + 4)ξ^{m} = 0, for all j.

Hence we have the following

Theorem 7.1. For m = 0, 1, 2, 3, 4

(7.14)  \overline{P}_{j}(m, 5, 5n) = \frac{1}{5}\overline{p}_{j}(5n), if j ≡ 1, 2 (mod 5),
(7.15)  \overline{P}_{j}(m, 5, 5n + 1) = \frac{1}{5}\overline{p}_{j}(5n + 1), if j ≢ 1, 2 (mod 5),
(7.16)  \overline{P}_{j}(m, 5, 5n + 2) = \frac{1}{5}\overline{p}_{j}(5n + 2), if j ≢ 0, 3 (mod 5),
(7.17)  \overline{P}_{j}(m, 5, 5n + 3) = \frac{1}{5}\overline{p}_{j}(5n + 3), if j ≡ 0, 3 (mod 5),

and

(7.18)  \overline{P}_{j}(m, 5, 5n + 4) = \frac{1}{5}\overline{p}_{j}(5n + 4), for all j.

Corollary 7.2.

(7.19)  \overline{p}_{j}(5n) ≡ 0 (mod 5) if j ≡ 1, 2 (mod 5),
(7.20)  \overline{p}_{j}(5n + 1) ≡ 0 (mod 5) if j ≢ 1, 2 (mod 5),
(7.21)  \overline{p}_{j}(5n + 2) ≡ 0 (mod 5) if j ≢ 0, 3 (mod 5),
(7.22)  \overline{p}_{j}(5n + 3) ≡ 0 (mod 5) if j ≡ 0, 3 (mod 5),

and

(7.23)  \overline{p}_{j}(5n + 4) ≡ 0 (mod 5) for all j.

Recalling the comment after (7.3) we see that (7.23) gives an extension of Andrews’ result (1.21). However, congruences (7.19)–(7.22) appear to be new. It is possible to modify the construction in Section 4 in order to provide a direct combinatorial proof of (7.23). It is likely that a combinatorial proof of (7.19)–(7.22) would require significant new insights. To make this point plausible we note that the 5-core analog of (7.19)–(7.22) does not hold. In other words, it is not true that for r = 0, 1, 2, 3

(7.24)  \overline{a}_{5,j}(5n + r) ≡ 0 (mod 5).

On the other hand, it can be shown that

(7.25)  \overline{a}_{5,j}(5n + 4) ≡ 0 (mod 5).

Here \overline{a}_{5,j}(n) denotes the number of 5-core partitions of n with BG-rank = j.

As mentioned before, in [9], t-core cranks were given which combinatorially prove all three of Ramanujan’s congruences (1.1)–(1.3). It is natural to ask whether the methods of this paper and [9] can be extended to find refinements for Ramanujan’s partition congruences mod 7 and 11. In [9], there is a 7-cycle and an 11-cycle analogous to (3.43). However, this 7-cycle does not preserve the srank (mod 4) of 7-cores of 7n + 5 so that the srank does not work in this case. One could define a new srank for 7-cores that is analogous to (4.3); i.e. as a cyclically symmetric polynomial in α_{0}, α_{1}, …, α_{6}. The problem is to extend the definition of this statistic to all partitions of 7n + 5. The authors considered many such polynomials but were unable to identify a simple statistic for partitions of 7n + 5 which also gave a refinement to Ramanujan’s congruence mod 7 analogous to the srank.

It can be shown that the absolute value of the BG-rank of 7-cores of 7n + 5 is invariant under the 7-cycle. However, neither the BG-rank nor its absolute value gives a refinement of Ramanujan’s congruence mod 7 or 11. And so we would like to pose the Problem. Is there an analogue of the BG-rank, which gives a refinement of (1.2) and (1.3)?

Acknowledgements. We would like to thank Krishna Alladi, George Andrews, and Alain Lascoux for their interest and comments.

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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