CRANKS AND T -CORES

Frank Garvan,* Dongsu Kim,** and Dennis Stanton***

Abstract. New statistics on partitions (called cranks) are defined which combinatorially prove Ramanujan’s congruences for the partition function modulo 5, 7, 11, and 25. Explicit bijections are given for the equinumerous crank classes. The cranks are closely related to the t-core of a partition. Using q-series, some explicit formulas are given for the number of partitions which are t-cores. Some related questions for self-conjugate and distinct partitions are discussed.

1. Introduction.

Let p(n) be the number of partitions of n [1]. Dyson [7] proposed a combinatorial proof of Ramanujan’s congruence

p(5n + 4) ≡ 0 mod 5.

He defined an integral statistic on partitions, called the rank, whose value mod 5 split the set of partitions of 5n + 4 into 5 equal classes. He further conjectured that the rank also proved

p(7n + 5) ≡ 0 mod 7,

and hypothesized a statistic, called the crank, which would similarly prove

p(11n + 6) ≡ 0 mod 11.

Atkin and Swinnerton-Dyer [5] proved Dyson’s conjecture for 5 and 7. In [9], a crank for 11 as well as new cranks for 5 and 7 were found relative to vector partitions. Later, Andrews and Garvan [3] were able to define a crank, in terms of ordinary partitions, for 5, 7 and 11, thus completing the solution of Dyson’s crank conjecture. New relations for the crank of partitions mod 8, 9 and 10 have been found in [10].

Explicit bijections between these equinumerous classes have not been found in any of these three cases. This should be related to another possible approach to these problems: split the set of partitions of 5n + 4 into p(5n + 4)/5 classes, each of size 5. Hopefully, the bijections among the five classes should be realized by a 5-cycle which acts on all partitions of 5n + 4 with no fixed points. The orbits of

Typeset by AMS-TEX

1991 Mathematics Subject Classification. Primary 11P76.

Key words and phrases. partitions, congruences.

*School of Mathematics, Physics, Computing and Electronics, Macquarie University, Sydney, NSW 2109, Australia.

**School of Mathematics, University of Minnesota, Minneapolis, MN 55455.

***School of Mathematics, University of Minnesota, Minneapolis, MN 55455. This work was partially  supported  by  NSF  grant  DMS:8700995.

the 5 cycle are the p(5n + 4)/5 classes. A crank statistic should be defined which is {0, 1, 2, 3, 4} on an orbit. A 5-cycle, but no crank, was found by computer in [11]. In this paper, we complete this program by finding dihedral groups of size 10, 14, and 22 which act on the partitions of 5n+4, 7n+5, and 11n+6, respectively. The 5 (7 or 11) cycles in these groups have no fixed points, thus proving the congruences. We also define new statistics on partitions, i.e. new cranks, which split the set of partitions into equinumerous classes. A cycle in the dihedral groups increases the crank by 1, thus supplying an explicit bijection between the crank classes. Our definitions of the cycles and the cranks are completely combinatorial.

The key ingredients to our proof are two bijections for partitions and t-cores, which are given in §2. They allow us to find dihedral groups as symmetry groups of quadratic forms. There are interesting q-series attached to these forms, and these q-series imply some remarkable explicit formulas (Theorems 4 and 5 in §5) for the number of partitions which are t-cores, t = 5 or 7. The explicit cranks are given in Theorems 2 and 3, and in Proposition 1 of §3 and §4. A crank for 25n + 24 is given in §6. Similar bijections and generating functions for self-conjugate partitions and partitions with distinct parts are given in §7 and §8.

We shall use the notation for generating function from q-series, e.g.

(a; q)_{k} = ∏_{i=0}^{k−1}(1 − aq^{i}),

so

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

2. Two bijections.

In this section we give two bijections relating partitions and t-cores.

First we set the notation. Let P be the set of all partitions. For any λ ∈ P, let |λ| denote the number that λ partitions. Fix a positive integer t. Let P_{t−core} be the set of partitions which are t-cores. Recall that there are two equivalent definitions of a t-core: a partition λ is a t-core if, and only if, λ has no hook numbers that are multiples of t [13, 2.7.40]; or if, and only if, λ has no rim hooks that are multiples of t [13, p. 75]. We let a_{t}(n) be the number of partitions of n which are t-cores.

The first bijection is well-known [13, 2.7.17].

Bijection 1. There is a bijection ϕ_{1} : P → P_{t−core} × P × P × ⋯ × P,

ϕ_{1}(λ) = (λ̃, λ_{0}, λ_{1}, …, λ_{t−1}),

such that |λ| = |λ̃| + t ∑_{i=0}^{t−1} |λ_{i}|.

The generating function identity equivalent to Bijection 1 is

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

The second bijection is on t-cores only.

Bijection 2. There is a bijection ϕ_{2} : P_{t−core} → {n⃗ = (n_{0}, n_{1}, …, n_{t−1}) : n_{i} ∈ ℤ, n_{0} + ⋯ + n_{t−1} = 0}, where

|λ̃| = t∥n⃗∥^{2}/2 + b⃗ · n⃗,  b⃗ = (0, 1, ⋯, t − 1).

The generating function identity equivalent to Bijection 2 is

(2.2)  ∑_{n=0}^{∞} a_{t}(n)q^{n} = ∑_{\begin{matrix}n⃗·1=0 \\ n⃗∈ℤ^{t} \end{matrix}} q^{\frac{t}{2} ∥n⃗∥^{2}+b⃗·n⃗}.

Clearly (2.1) and (2.2) imply

(2.3)  ∑_{n=0}^{∞} p(n)q^{n} = \frac{1}{(q^{t}; q^{t})_{∞}^{t}} ∑_{\begin{matrix}n⃗·1=0 \\ n⃗∈ℤ^{t} \end{matrix}} q^{\frac{t}{2} ∥n⃗∥^{2}+b⃗·n⃗}.

We now give Bijection 2, and sketch Bijection 1.

Let λ̃ be a t-core. Define the vector n⃗ = ϕ_{2}(λ̃) in the following way. Label a cell in the i th row and j th column of λ̃ by j − i mod t. (This is called the t-residue diagram in [13, p. 84].) We also label the cells in column 0 in the same way, and call the resulting diagram the extended t-residue diagram. A cell is called exposed if it is at the end of a row. Region r of the extended t-residue diagram of λ̃ is the set of cells (i, j) satisfying t(r − 1) ≤ j − i < tr. We now define n_{i} to be the maximum region of λ̃ which contains an exposed cell labeled i. Since column 0 contains infinitely many exposed cells, n_{i} is well-defined.

If an exposed cell labeled i lies in region r, then there is an exposed cell labeled i in each region < r. For example, in region r − 1, if i is not exposed, then the rim hook whose northeast head is i in region r, and whose tail is i + 1 in region r − 1 has length t. (Note: If i = t − 1, the rim hook will lie entirely in region r.) This contradicts the assumption that λ̃ is a t-core.

Note that the size of the Durfee square of λ̃ is the sum of the positive n_{i}’s, since this is the number of exposed cells in positive regions.

We next verify that n_{0}+n_{1}+⋯+n_{t−1} = 0. Let λ̃′ be the conjugate of λ̃. First we show that if ϕ_{2}(λ̃) = (n_{0}, n_{1}, ⋯, n_{t−1}), then ϕ_{2}(λ̃′) = (−n_{t−1}, −n_{t−2}, ⋯, −n_{0}). Let a cell labeled i be exposed in region r of λ̃, so that i + 1 lies above i. In λ̃′, the cell corresponding to i+1 will be labeled t−i−1, lie in region 1−r, and not be the last cell in its row. Thus t − i − 1 is not exposed in region 1 − r of λ̃′. This shows that if i is exposed in region r of λ̃, r ≤ n_{i}; then t − 1 − i is not exposed in region r of λ̃′, r ≥ 1 − n_{i}. A similar argument shows that if a boundary cell labeled t − i − 1 is not exposed in region 1 − r, then i is exposed in region r. This verifies the claim about ϕ_{2}(λ̃′). Since the size of the Durfee square is unchanged by conjugation, the sum of the positive n_{i}’s must be equal to minus the sum of the negative n_{i}’s; that is, n_{0} + n_{1} + ⋯ + n_{t−1} = 0.

The inverse map to ϕ_{2} is easy to find: if (n_{0}, n_{1}, ⋯, n_{t−1}) is given, then the exposed cells in each region are known. This in turn gives the lengths of each row of the Ferrers diagram of λ̃, and thus λ̃.

Finally we show that if ϕ_{2}(λ̃) = n⃗, then λ̃ is a partition of the exponent in Bijection 2. The number of cells to the right of the main diagonal of the Ferrers diagram of λ is

∑_{n_{i}>0} in_{i} + t\binom{n_{i}}{2}.

For the cells below the main diagonal, we take λ̃′, to find

− ∑_{n_{j}<0}(t − 1 − j)n_{j} + t\binom{−n_{j}}{2}.

There are

∑_{n_{i}>0} n_{i}

cells on the main diagonal. The sum of these three terms is the exponent of q in (2.3).

We quickly sketch ϕ_{1} using the notation in Bijection 2. Given λ, we construct t biinfinite words in the letters N and E, w_{0}, …, w_{t−1}. The j th letter of w_{i} is E if i is exposed in region j of λ, otherwise the j th letter is N (not exposed). If λ is a t-core, then each word w_{i} is an infinite sequence of E’s followed by an infinite sequence of N’s. In this case, λ = λ̃, and λ_{i} = ∅ for all i. Otherwise there is an E to the right of some N. Find the rightmost E, say in position j. Find the rightmost N to the left of this E, say in position k < j. Delete a rim hook in λ whose northeast head is the cell corresponding to E, and whose southwest tail is adjacent to the cell corresponding to N. Place a part of size j − k in λ_{i}. The new partition has N’s in w_{i} from position j on, thus the E’s have been pushed to the left. This operation can be done in any order whatsoever, to finally obtain the t-core of λ, λ̃. The parts of each λ_{i} appear in increasing order.

3. Cranks.

In this section we give the cranks (Theorem 2) and the cycles which give bijections for these crank classes.

Let r be a residue class of t, and consider p(tn+r). On the right side of (2.3), all exponents of q which are equivalent to r mod t lie in the multisum. Since n⃗ ·1 = 0, we have \frac{t}{2}∥n⃗∥^{2} ≡ 0 mod t. Thus we have

(3.1)  ∑_{n=0}^{∞} p(tn + r)q^{tn+r} = \frac{1}{(q^{t}; q^{t})_{∞}^{t}} ∑_{\begin{matrix}n⃗·b⃗≡r mod t \\ n⃗·1=0 \\ n⃗∈ℤ^{t} \end{matrix}} q^{\frac{t}{2} ∥n⃗∥^{2}+b⃗·n⃗}.

The symmetry groups of the quadratic forms in (3.1) have been computed by computer for t ≤ 8 in [11]. Until now, we have assumed that the class t and residue r are arbitrary. We shall see that the choices of 5n + 4, 7n + 5, and 11n + 6 simplify (3.1).

For 5n + 4 we now change variables so that the symmetry group is obviously the dihedral group D_{5} of order 10. The five vectors n⃗ for ϕ_{2}(p(4)) are

(1) v⃗_{0} = (1, −1, 0, 0, 0),

(2) v⃗_{1} = (0, 1, −1, 0, 0), (3) v⃗_{2} = (0, 0, 1, −1, 0),

(4) v⃗_{3} = (0, 0, 0, 1, −1), and

(5) v⃗_{4} = (1, 1, 0, −1, −1).

They form a non-planar pentagon whose center c⃗ = (2/5, 1/5, 0, −1/5, −2/5).

We now change variables: write n⃗ = α_{0}v⃗_{0} + ⋯ + α_{4}v⃗_{4}. A vector n⃗ ∈ ℤ^{5} satisfies n⃗ ·1 = 0 and b⃗ · n⃗ ≡ 4 mod 5 if, and only if, n⃗ = α_{0}v_{0} + ⋯ + α_{4}v_{4} for some α⃗ ∈ ℤ^{5} such that α⃗ · 1 = 1.

It is easy to see that the quadratic form in (3.1) in terms of the vector α⃗ is

5∥α⃗∥^{2} − 5(α_{0}α_{1} + α_{1}α_{2} + ⋯ + α_{4}α_{0}) − 1.

Thus we obtain

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

where Q(α⃗) = ∥α⃗∥^{2} − (α_{0}α_{1} + α_{1}α_{2} + ⋯ + α_{4}α_{0}). Clearly the dihedral group D_{5} is the automorphism group.

It is also clear that the 5-cycle which cyclically permutes the α_{i}’s has no fixed points. Any fixed point α⃗ must have all entries equal. Since α⃗ · 1 = 1, this would imply that the entries are not integral.

The construction of the quadratic form and the group are completely analogous for 7n + 5 and 11n + 6. We collect these results in a theorem.

Theorem 1. For (t, r) = (5, 4), (7, 5) or (11, 6),

(3.2)  ∑_{n=0}^{∞} p(tn + r)q^{n+1} = \frac{1}{(q; q)_{∞}^{t}} ∑_{\begin{matrix}α⃗·1=1 \\ α⃗∈ℤ^{t} \end{matrix}} q^{Q(α⃗)},

where Q(α⃗) = ∥α⃗∥^{2} − ∑_{i=0}^{t−1} α_{i}α_{i+1}.

It should be noted that other identities for the generating functions of p(tn + r), which yield Ramanujan’s congruences, have been found by others. The most famous of these is Ramanujan’s [19, (17) p.213] identity for p(5n+4), which was considered by Hardy in agreement with MacMahon as Ramanujan’s most beautiful identity [19, p.xxxv]. Ramanujan [19, (18) p.213] also found an analog for p(7n+5). In fact for t = 5^{α}, 7^{α} similar identities have been found by Watson [20]. Using the theory of modular functions an analog for p(11n + 6) was found by Fine [8, (3.25)] and extended to higher powers of 11 by Atkin [4]. However, these other identities are of a different nature than (3.2) and do not yield obvious cranks. Further, Fine’s identity for t = 11 and Ramanujan’s identity for t = 5 or t = 7 are dissimilar so it is surprising that an identity like (3.2) holds for all three values t = 5, 7 and 11.

Finally we come to the cranks. An obvious statistic, which is increased by 1 mod 5 each time the α’s are cyclically permuted is ∑_{i=0}^{4} iα_{i}. So in terms of the α⃗ basis the crank is clear, but in terms of the n⃗ basis (given combinatorially in §2) the crank is not clear.

Theorem 2. A crank statistic for partitions λ of 5n + 4, 7n + 5, or 11n + 6 is given by the following algorithm.

(1) Find the t-core λ̃ of λ, (t = 5, 7, or 11) by Bijection 1.

(2) Find ϕ_{2}(λ̃) = n⃗ by Bijection 2.

(3) Let crank(λ) be the following mod t linear combination:

(t = 5) 4n_{0} + n_{1} + n_{3} + 4n_{4} for 5n + 4,

(t = 7) 4n_{0} + 2n_{1} + n_{2} + n_{4} + 2n_{5} + 4n_{6} for 7n + 5,

(t = 11) 4n_{0} +9n_{1} +5n_{2} +3n_{3} +n_{4} +n_{6} +3n_{7} +5n_{8} +9n_{9} +4n_{10} for 11n+6.

The new crank is not equal to the rank or the old crank. One can verify this on the 5 partitions of 9 which are 5-cores: 621, 522, 51111, 33111, and 321111.

4. More cranks.

The cranks in §3 depend only upon the t-core λ̃ of λ, and thus use Bijections 1 and 2. In this section we give two simpler ways (Proposition 1 and Theorem 3) of computing crank(λ), (one which is as a polynomial in the parts of λ), thus avoiding Bijections 1 and 2.

First, let r_{k}(λ) be the number of cells in the t-residue diagram of λ which are labeled k mod t. Suppose that λ is a t-core. A relation between the r_{k}’s and the n_{i}’s can be found by counting the cells in a row ending with i in region n_{i}. It is

(4.1)  r_{k} = ∑_{i=0}^{t−1}(n_{i}^{2}/2 + n_{i}χ(i ≥ k))

where

χ(S) = \{ \begin{matrix}1/2  if S is true, \\ −1/2  if S is false. \end{matrix}

It is easy to see that (4.1) implies

(4.2)  n_{k} = r_{k} − r_{k+1},

so that the relation in (4.1) for r_{0} for t-cores becomes

r_{0} = ∑_{i=0}^{t−1}(r_{i}^{2} − r_{i}r_{i+1}).

The cranks in Theorem 2 are defined by a linear combination of the n_{i}’s. Using (4.2), we can find mod t equivalent linear combinations of the r_{k}’s.

Proposition 1. The following mod t linear combinations of the t-residue numbers r_{k} are crank statistics:

(1) r_{1} + 2r_{2} − 2r_{3} − r_{4} for 5n + 4,

(2) 2r_{1} + r_{2} + r_{3} − r_{4} − r_{5} − 2r_{6} for 7n + 5,

(3) −6r_{1} − 4r_{2} − 2r_{3} − 2r_{4} − r_{5} + r_{6} + 2r_{7} + 2r_{8} + 4r_{9} + 6r_{10} for 11n + 6.

Proof. We already know from Theorem 2 that certain linear combinations give the crank for t-cores, e.g. (2, −1, 1, −2) for t = 5. If we multiply this linear combination by 3, we obtain (1) for t-cores. We reverse the sign to obtain (2), and (3) is immediate from Theorem 2. Now consider (1)-(3) as definitions of a crank for any partition λ, not just t-cores. Removing a rim hook of length t from λ reduces each r_{k} by one. The above linear combinations do not change, since the sum of all of the coefficients is zero. Thus the crank can be defined by Proposition 1 for all partitions λ. □

Next, we give a definition of a crank based upon the parts of λ : λ_{1} ≥ λ_{2} ≥ ⋯ ≥ λ_{m}.

Theorem 3. For t = 5, 7 or 11, let p_{t}(x) = (x − (t − 1)/2)^{t−3} mod t. A crank statistic for partitions λ : λ_{1} ≥ λ_{2} ≥ ⋯ ≥ λ_{m} of tn + r (where r = 4, 5, and 6, respectively) is given by

crank(λ) = ∑_{i=1}^{m}(p_{t}(λ_{i} − i) − p_{t}(i − 1)) mod t.

Proof. Let a_{i} be the coefficient of r_{i} in Proposition 1. Recall that a_{1} + a_{2} + ⋯ + a_{t−1} = 0.

First we evaluate the contribution to the crank in Proposition 1 from the first row of λ. The t-residue diagram of λ has an x ≡ λ_{1} − 1 mod t at the end of the first row. Then the contribution of this row to the crank is

(4.3)  a_{1} + a_{2} + ⋯ + a_{x}.

It remains to find a polynomial whose values at x = 0, 1, …, t − 1 agree with (4.3) mod t. Here are such polynomials:

(1) q_{5}(x) = 3(p_{5}(x) − 4) for t = 5,

(2) q_{7}(x) = −(p_{7}(x) − 4) for t = 7, and

(3) q_{11}(x) = p_{11}(x) − 4 for t = 11. Thus we have shown that the first row contributes q_{t}(x) ≡ q_{t}(λ_{1} −1)−q_{t}(0) mod t.

Consider row i of the t-residue diagram of λ: the first entry is 1 − i mod t, and the final entry is λ_{i} − i mod t. The contribution of this row to the crank is

a_{1} + a_{2} + ⋯ + a_{λ_{i}−i} + (a_{t+1−i} + ⋯ + a_{t−1}) ≡
a_{1} + a_{2} + ⋯ + a_{λ_{i}−i} − (a_{1} + ⋯ + a_{t−i}) ≡ q_{t}(λ_{i} − i) − q_{t}(t − i) mod t.

Thus

crank(λ) = ∑_{i=1}^{m}(q_{t}(λ_{i} − i) − q_{t}(t − i)) mod t.

Since q_{t}(t − 1 − i) ≡ q_{t}(i) mod t, and q_{t}(i) is a linear function of p_{t}(i), the result follows. □

5. Associated q-series.

From (2.1) there is an explicit generating function for the partitions of n which are t-cores. We investigate these functions for t = 5, 7, and 11, and give explicit formulas for a_{5}(n) and a_{7}(n) in Theorems 4 and 5.

The five cycle in §3 implies that a_{5}(5n + 4) ≡ 0 mod 5. However, much more is true,

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

we give two proofs of (5.1): one which is analytic, and proves more (Theorem 4), and another which is a bijection.

Theorem 4. Let n + 1 = 5^{c}p_{1}^{a_{1}} ⋯ p_{s}^{a_{s}}q_{1}^{b_{1}} ⋯ q_{t}^{b_{t}} be the prime factorization of n + 1 into primes p_{i} ≡ 1, 4 mod 5, and q_{j} ≡ 2, 3 mod 5. Then

a_{5}(n) = 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} .

Proof. Clearly (2.1) implies

∑_{n=0}^{∞} a_{5}(n)q^{n} = \frac{(q^{5} ; q^{5})_{∞}^{5}}{(q; q)_{∞}} .

However an identity of Ramanujan ([2, (3.46)] or [6]) is

(5.2)  q \frac{(q^{5} ; q^{5})_{∞}^{5}}{(q; q)_{∞}} = ∑_{n=0}^{∞} (\frac{n}{5})\frac{q^{n}}{(1 − q^{n})^{2}} .

where (\frac{a}{b}) is the Legendre symbol. Clearly the right side of (5.2) is

∑_{n=1}^{∞} ∑_{d|n} (\frac{d}{5})\frac{n}{d}q^{n}.

so

(5.3)  a_{5}(n − 1) = ∑_{d|n} (\frac{d}{5})\frac{n}{d}.

Since (\frac{d}{5})/d is multiplicative, (5.3) and [12, Theorem 265] implies that a_{5}(n − 1) is multiplicative. It remains to evaluate a_{5}(p^{a} − 1) for primes p, which can be done by (5.3). □

The combinatorial proof of (5.1) is explicit: given a partition λ̃ of n which is a 5-core, find five partitions θ_{0}, θ_{1}, θ_{2}, θ_{3}, θ_{4}, of 5n + 4 which are also 5-cores. Let ϕ_{2}(λ̃) = m⃗. We now give the image under ϕ_{2} of the five partitions of 5n + 4.

(1) ϕ_{2}(θ_{0}) = (m_{1} + 2m_{2} + 2m_{4} + 1, −m_{1} − m_{2} + m_{3} + m_{4} + 1, 2m_{1} + m_{2} + 2m_{3}, −2m_{2} − 2m_{3} − m_{4} − 1, −2m_{1} − m_{3} − 2m_{4} − 1),

(2) ϕ_{2}(θ_{1}) = (2m_{1}+2m_{4}+1+m_{3}, −1−m_{2}−2m_{3}−2m_{4}, −2m_{1}−2m_{2}−m_{4}, m_{1}+ 2m_{2} + 2m_{3}, −m_{1} + m_{2} + m_{4} − m_{3}),

(3) ϕ_{2}(θ_{2}) = (m_{1} − m_{2} − m_{4} + m_{3}, 2m_{2} + m_{4} + 1 + 2m_{3}, −m_{1} − 2m_{4} − 1 − 2m_{3}, −2m_{1} − 2m_{2} − m_{3}, 2m_{1} + m_{2} + 2m_{4}),

(4) ϕ_{2}(θ_{3}) = (−2m_{1} − m_{2} − 2m_{4}, −m_{1} − 2m_{2} − 2m_{3}, 2m_{2} + 2m_{4} + 1 + m_{3}, m_{1} + m_{2} − m_{4} − 1 − m_{3}, 2m_{1} + 2m_{3} + m_{4}),

(5) ϕ_{2}(θ_{4}) = (−2m_{1} − m_{4} − 2m_{3}, 2m_{1} + 2m_{2} + m_{3}, m_{1} − m_{2} + m_{4} − m_{3}, m_{2} + 2m_{4} + 1 + 2m_{3}, −m_{1} − 2m_{2} − 2m_{4} − 1).

We delete the verification of this claim. In fact, these five partitions θ_{i} form an orbit under the five cycle, and

crank(θ_{i}) ≡ 3i mod 5.

We denote by γ_{n} the 5 − to − 1 map from the set of 5-cores of 5n + 4 to the set of 5-cores of n, γ_{n}(θ_{i}) = λ̃ for 0 ≤ i ≤ 4.

For t = 7, we have a_{7}(7n + 5) ≡ 0 mod 7. It is not true that a_{7}(7n + 5) = 7a_{7}(n). There are however two multiplicative functions here. The identity which is analogous to (5.2) is due to Ramanujan [8, p. 159]

(5.4)  8q^{2} \frac{(q^{7} ; q^{7})_{∞}^{7}}{(q; q)_{∞}} + q(q^{7} ; q^{7})_{∞}^{3}(q; q)_{∞}^{3} = ∑_{n=1}^{∞} (\frac{n}{7})\frac{q^{n}(1 + q^{n})}{(1 − q^{n})^{3}}.

Let a′(n), b(n), c(n) denote the coefficient of q^{n} in the expansion of each of the three functions in (5.4) so that

8a_{7}(n) = a′(n + 2),
a′(n) + b(n) = c(n).

We will give in Lemmas 1 and 2 explicit formulas for b(n) and c(n), resulting in an explicit formula for a_{7}(n) in Theorem 5.

The proof of Theorem 4 analogously establishes the multiplicativity of c(n). We delete the details. The resulting explicit formula is given in Lemma 1.

Lemma 1. If n = 7^{c}p_{1}^{a_{1}} ⋯ p_{s}^{a_{s}}q_{1}^{b_{1}} ⋯ q_{t}^{b_{t}} is the prime factorization of n into primes p_{i} ≡ 1, 2, 4 mod 7, and q_{j} ≡ 3, 5, 6 mod 7, then

c(n) = 7^{2c} ∏_{i=1}^{s} \frac{p_{i}^{2a_{i}+2} − 1}{p_{i}^{2} − 1} ∏_{j=1}^{t} \frac{q_{j}^{2b_{j}+2} + (−1)^{b_{j}}}{q_{j}^{2} + 1} .

To establish the multiplicativity of the sequence b(n), we apply the theory of modular forms.

Proposition 2. The sequence b(n) is multiplicative, moreover

b(p^{m}) = b(p)b(p^{m−1}) − (\frac{p}{7})p^{2}b(p^{m−2})

for any prime p and integer m ≥ 2.

Proof. If q = e^{2πiz}, the generating function B(z) for b(n) (see (5.4)) can be considered a function of z in the upper half plane. In terms of the classical η-function it is B(z) = (η(z)η(7z))^{3}, which is a cusp form of level 7, weight 3, and character χ(n) = (\frac{n}{7}) [14, Prob. 12, 13, p.145]. Since this space of cusp forms is onedimensional, B(z) must be an eigenform for the Hecke operators T_{n}. This proves the multiplicativity of b(n), and the three term recurrence follows from the Euler product for the associated Dirichlet series [14, p. 160 (5.11)]. □

To find an explicit formula for b(n), we need only find b(p) for all primes p. The three term recurrence in Proposition 2 gives b(p^{m}), and multiplicativity gives b(n) for all n. The next proposition gives the values of b(p).

Proposition 3. Suppose p is an odd prime. If p ≡ 3, 5 or 6 mod 7, then b(p) = 0. If p ≡ 1, 2 or 4 mod 7, let p = x^{2} + 7y^{2}, where x and y are positive integers. Then b(p) = 2(x^{2} − 7y^{2}).

Remark. The result also holds for p = 2 except in this case x = y = \frac{1}{2}.

Proof. From Jacobi’s identity for (q; q)_{∞}^{3} [12, Thm 357] we find that

b(p) = ∑_{\begin{matrix}m,n≥0 \\ (2m+1)^{2}+7(2n+1)^{2}=8p \end{matrix}}(−1)^{m+n}(2m + 1)(2n + 1).

We need to determine all of the solutions (m, n) to

(p1)  8p = (2m + 1)^{2} + 7(2n + 1)^{2},

where m, n ≥ 0. Clearly the right side is a quadratic residue mod 7, so if p ≡ 3, 5 or 6 mod 7, there are no solutions (m, n) to (p1) and b(p) = 0.

Thus we can assume that p ≡ 1, 2 or 4 mod 7. We first show in this case that p has a unique representation p = x^{2} + 7y^{2} for positive integers x, y. We work in K = ℚ(\sqrt{−7}), whose ring of algebraic integers, O_{K} = ℤ[\frac{1+\sqrt{−7}}{2}], is a unique factorization domain [12, Th. 246]. Since

1 = (\frac{p}{7}) = (\frac{−7}{p}) (by quadratic reciprocity),

p divides a^{2} +7 for some integer a. Since p does not divide a±\sqrt{−7}, p is not prime. Thus p splits

p = ββ̄,

where β = x + y\sqrt{−7} is prime in O_{K} and x, y > 0. Now, x, y ∈ ℤ, otherwise x = m + \frac{1}{2}, y = n + \frac{1}{2} (for some m, n ∈ ℤ), p = m(m + 1) + 7n(n + 1) + 2 which is even and contradicts the assumption that p was odd. Hence p = ββ̄ = x^{2} + 7y^{2} for some positive integers x, y. The uniqueness of the representation follows from unique factorization, and the only units being ±1.

We next show that (p1) always has two solutions (m, n), whose contribution to b(p) is 2(x^{2} − 7y^{2}). (p1) is equivalent to 2p = γγ̄, where

γ = ^{\frac{(2m+1)+(2n+1)\sqrt{−7}}{2}} = \frac{x_{1}+y_{1}\sqrt{−7}}{2}, x_{1}, y_{1} > 0.

Let

2 = (\frac{1+\sqrt{−7}}{2}) (\frac{1−\sqrt{−7}}{2}) = ππ̄.

It follows that 2p factors as

2p = (πβ)(π̄β̄) = (πβ̄)(π̄β) = γγ̄.

By unique factorization we must have γ = ±πβ, ±π̄β, ±π̄β̄, or γ = ±πβ̄. This allows eight possible solutions (x_{1}, y_{1}), six of which violate x_{1}, y_{1} > 0. The two solutions can be explicitly given, depending upon which of the three inequalities 0 < y < 7y < x, 0 < x < y < 7y, or 0 < y < x < 7y occur. In each of these three cases we find b(p) = 2(x^{2} − 7y^{2}) as required. □

For p ≡ 1, 2 or 4 mod 7 note that Proposition 3 implies that b(p) ≠ 0, since b(p) = 0 would imply that x^{2} = 7y^{2}, and thus p = 14y^{2} ≡ 0 mod 7. We also note that b(p) = 2(x^{2} −7y^{2}) = β^{2} +β̄^{2} where p = ββ̄, β = x+y\sqrt{−7} and x, y > 0. Using the three term recurrence in Proposition 2 and the value of b(p) in Proposition 3, the value of b(p^{m}) is easily found. It is given in the next lemma.

Lemma 2. Let n = 7^{c}p_{1}^{a_{1}} ⋯ p_{s}^{a_{s}}q_{1}^{b_{1}} ⋯ q_{t}^{b_{t}} be the prime factorization of n into primes p_{i} ≡ 1, 2, 4 mod 7, and q_{j} ≡ 3, 5, 6 mod 7. Let p_{i} = x_{i}^{2} + 7y_{i}^{2} for some positive integers x_{i}, y_{i} in the case when p_{i} is odd and x_{1} = y_{1} = \frac{1}{2} for the case p_{1} = 2. Then

b(n) = 0 if some b_{j} is odd,

b(n) = (−7)^{c} ∏_{i=1}^{s} \frac{β_{i}^{2a_{i}+2} − β̄_{i}^{2a_{i}+2}}{β_{i}^{2} − β̄_{i}^{2}} ∏_{j=1}^{t} q_{j}^{b_{j}} otherwise,

where β_{i} = x_{i} + y_{i}\sqrt{−7}.

Note that Lemma 2 implies that b(p^{m}) = 0 if, and only if, p ≡ 3, 5 or 6 mod 7 and m is odd. We can now state the explicit formula for a_{7}(n).

Theorem 5. Let n + 2 = 7^{c}p_{1}^{a_{1}} ⋯ p_{s}^{a_{s}}q_{1}^{b_{1}} ⋯ q_{t}^{b_{t}} be the prime factorization of n + 2 into primes p_{i} ≡ 1, 2, 4 mod 7, and q_{j} ≡ 3, 5, 6 mod 7, then

a_{7}(n) = \frac{1}{8}(c(n + 2) − b(n + 2)).

where c(n) and b(n) are given in Lemmas 1 and 2 respectively.

We state relations which are corollaries of Theorems 4 and 5.

Corollary 1. The following relations hold:

(1) a_{5}(5^{α}m − 1) = 5^{α}a_{5}(m − 1) ≡ 0 mod 5^{α},

(2) a_{7}(7^{α}m − 2) ≡ 0 mod 7^{α},

(3) a_{7}(49n + 19) = 49a_{7}(7n + 1),

(4) a_{7}(49n + 33) = 49a_{7}(7n + 3),

(5) a_{7}(49n + 40) = 49a_{7}(7n + 4).

Proof. (1)–(5) are immediate from Theorems 4 and 5. □

For a_{11}(n) there are identities for cusp forms which prove

(5.5)  a_{11}(11^{α}m − 5) = 0 mod 11^{α}

and

(5.6)  a_{11}(43n − 5) = 0 mod 44 for (n, 43) = 1.

We delete the proofs of (5.5) and (5.6).

6. A crank for p(25n + 24).

We can define a group G which acts on the set of partitions of tn+r: G = G_{1}×S_{t}, where G_{1} is the automorphism group of the quadratic form in Bijection 2, and S_{t} is the symmetric group on t letters. For example, for 5n + 4, G = D_{5} × S_{5}, where D_{5} is the dihedral group of order 10. We have concentrated on the action of G_{1} for congruences. We can also use S_{t} for congruences, and do so in this section to find a crank for

(6.1)  p(25n + 24) ≡ 0 mod 25.

The group G = D_{5} × S_{5} for 5n + 4 contains two commuting five cycles which generate a subgroup of order 25. If each five cycle has no fixed points, then G will have no fixed points, and the orbits of G will have size 25. We have seen that the five cycle inside D_{5} has no fixed points. Since the five cycle in S_{5} could have fixed points, we consider two cases to define the crank, which will be an ordered pair of integers, each between 0 and 4. This ordered pair could be considered as the base 5 expansion for a number between 0 and 24.

Theorem 6. A crank statistic for partitions λ of 25n+24 is given by the following algorithm.

(1) Find ϕ_{1}(λ) = (λ̃, λ_{0}, λ_{1}, λ_{2}, λ_{3}, λ_{4}) given by Bijection 1.

(2) Let crank(λ) be the following ordered pair of integers:

Suppose that λ_{0} = λ_{1} = λ_{2} = λ_{3} = λ_{4}, so that λ̃ is a partition of 25s+24 for some s. Let μ_{1} = γ_{5s+4}(λ̃) and μ_{2} = γ_{s}(μ_{1}) be partitions of 5s+4 and s respectively (see §5). Put

(i)  crank(λ) = (crank(μ_{1}), crank(μ_{2})).

Suppose that some λ_{i} ≠ λ_{j}, so that the five cyclic permutations of (λ_{0}, λ_{1}, λ_{2}, λ_{3},λ_{4}) are distinct. Order these five tuples in lex order, and define the crank of the i th tuple to be i − 1. Put

(ii)  crank(λ) = (crank(λ̃), crank((λ_{0}, λ_{1}, λ_{2}, λ_{3}, λ_{4}))).

Proof. The vectors γ_{n}(θ_{i}) = λ̃ in §5 form an orbit under C_{5} ≤ G_{1}, so that the cranks of these five partitions are {0, 1, 2, 3, 4}.

The second part of the definition similarly interprets C_{5} × C_{5}, as a subgroup of G. □

In Theorem 6(2)(ii) the lex order crank can be replaced by using following statistic. Find ϕ_{1}(λ), and if

|λ_{0}| + |λ_{1}| + |λ_{2}| + |λ_{3}| + |λ_{4}| ≡ 0 mod 5,

define the crank as in Theorem 6(2)(i). Otherwise define

crank(λ) = (crank(λ̃), |λ_{1}| + 2|λ_{2}| + 3|λ_{3}| + 4|λ_{4}| mod 5).

It is clear that the five cyclic permutations of the λ_{i} will give five distinct cranks for the second component.

7. Self-conjugate partitions.

In this section we consider the restriction of Bijections 1 and 2 to self-conjugate partitions. Let asc_{t}(n) be the number of t-cores which are self-conjugate.

In Bijection 1, suppose that λ is self-conjugate. Then λ̃ must be self-conjugate, and λ′_{i} = λ_{t−i−1}, 0 ≤ i ≤ t − 1. Since (−q; q^{2})_{∞} is the generating function for self-conjugate partitions, the self-conjugate version of (2.1) is

(7.1a)  (−q; q^{2})_{∞} = \frac{1}{(q^{2t}; q^{2t})_{∞}^{t/2}} ∑_{n=0}^{∞} asc_{t}(n)q^{n}  for t even,

and

(7.1b)  (−q; q^{2})_{∞} = \frac{(−q^{t}; q^{2t})_{∞}}{(q^{2t}; q^{2t})_{∞}^{(t−1)/2}} ∑_{n=0}^{∞} asc_{t}(n)q^{n}  for t odd.

While there are congruences for q_{0}(n), the number of self-conjugate partitions of n, e.g. q_{0}(125n + 99) ≡ 0 mod 5, it is not obvious how to prove these congruences from this approach. Nevertheless, for t = 5 there is a nice Lambert series, and a theorem analogous to Theorems 4 and 5.

The Lambert series is

(7.2)  q(−q; q^{2})_{∞} \frac{(q^{10} ; q^{10})_{∞}^{2}}{(−q^{5} ; q^{10})_{∞}} = ∑_{n=1}^{∞} χ_{20}(n)\frac{q^{n}}{1 − q^{n}} ,

where

χ_{20}(n) = \{ \begin{matrix}0 if (n, 10) > 1, \\ (−1)^{\frac{1}{2} (n−1)} otherwise. \end{matrix}

Thus only odd n occur in (7.2), and the appropriate signs for odd n mod 20 are + − 0 − + − +0 + −. (7.2) follows from the _{6}ψ_{6} summation formula [2, (3.1)]. The identity (7.2) also follows from an identity of Ramanujan [21, (1) p.60] and Jacobi’s formula for r_{2}(n), the number of representations of n as a sum of two squares [12, Thm 278]. The functions r_{2}(n) and asc_{5}(n) are related by

(7.3)  r_{2}(n) − r_{2}(\frac{n}{5}) = 4asc_{5}(n − 1).

As before we obtain from (7.2) the following theorem.

Theorem 7. Let n + 1 = 2^{c}5^{d}p_{1}^{a_{1}} ⋯ p_{s}^{a_{s}}q_{1}^{b_{1}} ⋯ q_{t}^{b_{t}} be the prime factorization of n + 1 into primes p_{i} ≡ 1 mod 4, and q_{j} ≡ 3 mod 4. Then

asc_{5}(n) = ∏_{i=1}^{s}(a_{i} + 1) ∏_{j=1}^{t}(1 + (−1)^{b_{j}})/2.

A surprising corollary results.

Corollary 2. The following relations hold:

(1) asc_{5}(2n + 1) = asc_{5}(n),

(2) asc_{5}(5n + 4) = asc_{5}(n),

(3) asc_{5}(n) = 0 if and only if there exists a prime q ≡ 3 mod 4 and an odd integer b such that q^{b} exactly divides n + 1.

There is a nice combinatorial proof of Corollary 2(2). Suppose λ̃ is a selfconjugate partition of 5n + 4 which is a 5-core. It is easy to see that only θ_{0} of §4 is self-conjugate. Thus, the restriction of the map γ_{n} to self-conjugate partitions is a bijection.

A similar bijection exists for Corollary 2(1). Just replace (n_{1}, n_{2}) in (7.4) by (−n_{1} + n_{2}, −n_{1} − n_{2} − 1). We do not have a combinatorial proof of Corollary 2(3).

In Bijection 2, clearly we have n_{i} = −n_{t−1−i}, so the generating function becomes

(7.4)  ∑_{n=0}^{∞} asc_{t}(n)q^{n} = ∑_{n⃗∈ℤ^{⌊t/2⌋}} q^{t∥n⃗∥^{2}+c⃗·n⃗}.

where

c⃗ = \{ \begin{matrix}(2, 4, ⋯, t − 1)  for t odd, \\ (1, 3, ⋯, t − 1)  for t even, \end{matrix}

and ⌊t/2⌋ is the integral part of t/2. Using (7.4) dissections of (−q; q^{2})_{∞} due to Kolberg [16] can be easily derived.

8. Partitions with distinct parts.

A bijection analogous to Bijection 1 for partitions with distinct parts has been given by Morris and Yaseen [17]. In this section we give the appropriate version of Bijections 1 and 2 for these partitions.

We begin with notation. Let pd(n) be the number of partitions λ of n with distinct parts. Given such a λ, the shifted Ferrers diagram of λ, S(λ), is the Ferrers diagram of λ with the i th row shifted to the right by i − 1 cells. The doubled partition of λ, denoted λλ, is defined by adding λ_{i} cells to the i − 1 st column of S(λ). For example, if λ = 3 1, then λλ = 4 3 1. We let DD denote the set of doubled distinct partitions λλ. We define the t-core of λλ as the usual t-core of λλ. Let DD_{t−core} be the set of partitions which are t-cores of doubled distinct partitions, and let add_{t}(n) be the number of these which are partitions of n. We state without proof that DD_{t−core} ⊂ DD, thus add_{t}(n) = 0 if n is odd.

By restricting Bijection 1, we find a bijection for doubled distinct partitions. We must take the conjugate of the λ_{0} in ϕ_{1} to find λ_{0} for ϕ_{3}.

Bijection 3. There is a bijection ϕ_{3} : DD → DD_{t−core} × DD × P × ⋯ × P,

ϕ_{1}(λλ) = (λ̃λ, λ_{0}, λ_{1}, …, λ_{t−1}),

such that 2|λ| = |λλ| = |λ̃λ| + t ∑_{i=0}^{t−1} |λ_{i}|, and λ_{t−i} = λ′_{i}, for 1 ≤ i ≤ t − 1.

The generating function identities equivalent to Bijection 3 is

(8.1a)  ∑_{n=0}^{∞} pd(n)q^{2n} = \frac{(−q^{2t}; q^{2t})_{∞}}{(q^{2t}; q^{2t})_{∞}^{(t−1)/2}} ∑_{n=0}^{∞} add_{t}(n)q^{n}  for t odd.

and

(8.1b)  ∑_{n=0}^{∞} pd(n)q^{2n} = \frac{(−q^{t}; q^{t})_{∞}}{(q^{2t}; q^{2t})_{∞}^{(t−2)/2}} ∑_{n=0}^{∞} add_{t}(n)q^{n}  for t even.

Bijection 4. There is a bijection ϕ_{4} : DD_{t−core} → {n⃗ = (n_{0}, n_{1}, …, n_{t−1}) : n_{i} ∈ ℤ, n_{0} = 0, n_{i} = −n_{t−i}, 1 ≤ i ≤ t − 1}, where

|λ̃λ| = t∥n⃗∥^{2}/2 + b⃗ · n⃗,  b⃗ = (0, 1, ⋯, t − 1).

The generating function identity equivalent to Bijection 4 is

(8.2)  ∑_{n=0}^{∞} add_{t}(n)q^{n} = ∑_{n⃗∈ℤ^{⌊(t−1)/2⌋}} q^{t∥n⃗∥^{2}+d⃗·n⃗}.

where

d⃗ = \{ \begin{matrix}(1, 3, ⋯, t − 2)  for t odd, \\ (2, 4, ⋯, t − 2)  for t even. \end{matrix}

9. Remarks.

Naturally it is desirable to prove the general congruence for powers of 5, 7, and 11 found in [4, 20]. It would be natural to assume that the symmetry group of the quadratic form for t = 5^{a}7^{b}11^{c} contains a cycle of the appropriate length. We have not been able to find this cycle. Our crank for p(25n + 24) was inspired by Bailey’s ([6], [2, p. 465]) proof of Ramanujan’s congruence for p(5n + 4). In fact Bailey’s method also yields the congruence for p(25n + 24). It may be possible to extend our methods to obtain a crank for p(49n + 47) and p(121n + 116). First we need to find a crank for a_{7}(49n + 47) and a_{11}(121n + 116). Our present method does not succeed in these cases since there is no analog of (5.1). It may be easier to find a mod 49 crank for p(49n + 19), p(49n + 33), p(49n + 40) in view of Corollary 1(3),(4),(5).

In [11] the symmetry groups of the forms associated with p(tn + r) for t ≤ 8 and all r were computed. The symmetry groups for p(11n + r) for r ≠ 6 need to be found. There is reason to believe these groups are larger than C_{2}. It seems that an unusually high proportion of the coefficients of a_{11}(n) are divisible by 11 so there may be more 11-cycles. A higher proportion of the coefficients are divisible by 5. In fact, using modular forms which appear in an analog of (5.4) for t = 11, we can prove the following theorem.

Theorem 8. If (\frac{n}{11}) = 1 and n not divisble by 5, then a_{11}(n − 5) = 0 mod 5.

We can prove by a bijection from (7.4) that

(9.1)  asc_{7}(4n + 6) = asc_{7}(n).

Further, by utilizing Gauss’ result for sums of three squares [12, p.311] it can be shown that

(9.2)  asc_{7}(n) = 0  if and only if n + 2 = 4^{a}(8m + 1).

(This observation was made by Doug McDoniel.)

Kolberg [15, 16] has given several infinite products for dissections of p(n), q(n) and q_{0}(n). Many of these results (for q(n) and q_{0}(n)) follow easily from (7.4), (8.2) and the Jacobi triple product formula.

John Stembridge informed the authors of the existence of [17, 18] for distinct partitions. It should be noted that the definition of t-core given in §8 for t-even does not agree with [17, 18].

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