Another Proof of a Conjecture by Hirschhorn and Sellers on Overpartitions

Liuquan Wang Department of Mathematics National University of Singapore Singapore, 119076 Singapore

                    wangliuquan@nus.edu.sg

Abstract We provide a new elementary proof for a Ramanujan-type congruence for the overpartition function modulo 40, which was previously conjectured by Hirschhorn and Sellers and later proved by Chen and Xia. We also find some new congruences for the overpartition function modulo 5 and 9.

1 Introduction and Main Results

An overpartition of an integer n is a partition in which the first occurrence of a part may be overlined. The number of overpartitions of n is denoted by \overline{p}(n). For example, \overline{p}(3) = 8 as there are 8 overpartitions of 3: 3, \overline{3}, 2 + 1, \overline{2} + 1, 2 + \overline{1}, \overline{2} + \overline{1}, 1 + 1 + 1, \overline{1} + 1 + 1. We define \overline{p}(0) = 1 for convenience. It is well known (see [5], for example) that the generating function of \overline{p}(n) is

∑_{n≥0} \overline{p}(n)q^{n} = \frac{(−q; q)_{∞}}{(q; q)_{∞}} = \frac{1}{φ(−q)},  (1)

where (a; q)_{∞} = (1 − a)(1 − aq) ⋯ (1 − aq^{n}) ⋯ is standard q-series notation, and φ(q) = ∑_{n=−∞}^{∞} q^{n^{2}} is one of Ramanujan’s theta functions.

Overpartitions were first introduced by MacMahon [14] and have drawn much attention during the past ten years. There are numerous results concerning the arithmetic properties of \overline{p}(n). For more details, we refer the reader to [3, 5, 8, 15].

In 2005, Hirschhorn and Sellers [9] proved numerous Ramanujan-type identities for \overline{p}(n). For example,

∑_{n≥0} \overline{p}(4n + 3)q^{n} = 8\frac{(q^{2}; q^{2})_{∞}(q^{4}; q^{4})_{∞}^{6}}{(q; q)_{∞}^{8}} ,  (2)

which clearly implies \overline{p}(4n+3) ≡ 0 (mod 8). Meanwhile, they proposed a curious conjecture:

Conjecture 1. For any integer n ≥ 0, we have

\overline{p}(40n + 35) ≡ 0 (mod 40).

In a recent paper, Chen and Xia [3] gave a proof of Conjecture 1 using (p, k)-parametrization of theta functions. Their proof is relatively long and complicated, and in this paper we give a shorter proof.

Let r_{k}(n) denote the number of representations of n as sum of k squares. We find the following arithmetic relation.

Theorem 2. For any integer n ≥ 1, we have

\overline{p}(5n) ≡ (−1)^{n}r_{3}(n) (mod 5).

We have two remarkable corollaries.

Corollary 3. For any integers n ≥ 0 and α ≥ 0, we have

\overline{p}(4^{α}(40n + 35)) ≡ 0 (mod 5)

and

\overline{p}(5 · 4^{α+1}n) ≡ (−1)^{n}\overline{p}(5n) (mod 5).

By letting α = 0 in Corollary 3, we get \overline{p}(40n + 35) ≡ 0 (mod 5). Together with (2), Conjecture 1 follows immediately.

Corollary 4. For any prime p ≡ −1 (mod 5), we have

\overline{p}(5p^{3}n) ≡ 0 (mod 5)

for all n coprime to p.

Corollary 4 was first proved by Treneer (see [17, Proposition 1.4]) in 2006 using the theory of modular forms, which is not elementary.

Furthermore, with Corollary 3 in mind, we are ready to generalize Conjecture 1 to the following result.

Theorem 5. For any integers n ≥ 0 and α ≥ 0, we have

\overline{p}(4^{α}(40n + 35)) ≡ 0 (mod 40).

Some miscellaneous congruences can be deduced from Theorem 2, and we list some of them here.

Theorem 6. For any integers α ≥ 1 and n ≥ 0, we have

\overline{p}(5^{2α+1}(5n + 1)) ≡ \overline{p}(5^{2α+1}(5n + 4)) ≡ 0 (mod 5).

Theorem 7. Let p ≥ 3 be a prime, and N a positive integer which is coprime to p. Let α be any nonnegative integer.

(1) If p ≡ 1 (mod 5), then \overline{p}(5p^{10α+9}N) ≡ 0 (mod 5).

(2) If p ≡ 2, 3, 4 (mod 5), then \overline{p}(5p^{8α+7}N) ≡ 0 (mod 5).

Using properties of r_{3}(n), we can establish some other congruences as corollaries of Theorem 2. Recently, Chen et al. [2] found some Ramanujan-type congruences mainly based on the theory of modular forms. Their work also contains a proof of Theorem 5. In addition, they proved some congruences such as \overline{p}(5n) ≡ (−1)^{n}\overline{p}(4 · 5n) (mod 5).

Finally, we mention some congruences for overpartitions modulo 3. In 2011, based on the generating function of \overline{p}(3n) discovered by Hirschhorn and Sellers [8], Lovejoy and Osburn [13] proved the following result. For any integer n ≥ 1, we have

\overline{p}(3n) ≡ (−1)^{n}r_{5}(n) (mod 3).

Using the same method as in the proof of Theorem 2, we are able to improve this congruence to the following one.

Theorem 8. For any integer n ≥ 1, we have

\overline{p}(3n) ≡ (−1)^{n}r_{5}(n) (mod 9).

Similar to Theorem 7, we can deduce the following interesting congruences from Theorem 8.

Theorem 9. Let p ≥ 3 be a prime and N a positive integer which is coprime to p.

(1) If p ≡ 1 (mod 3), then \overline{p}(3p^{6α+5}N) ≡ 0 (mod 3) and \overline{p}(3p^{18α+17}N) ≡ 0 (mod 9).

(2) If p ≡ 2 (mod 3), then \overline{p}(3p^{4α+3}N) ≡ 0 (mod 9).

2 Preliminaries

Lemma 10. (Cf. [16, Lemma 1.2].) Let p be a prime and α a positive integer. Then

(q; q)_{∞}^{p^{α}} ≡ (q^{p}; q^{p})_{∞}^{p^{α−1}} (mod p^{α}).

Lemma 11. (Cf. [1, Theorem 3.3.1, 3.5.4].) For any integer n ≥ 1, we have

r_{4}(n) = 8 ∑_{d|n,4∤d} d, r_{8}(n) = 16(−1)^{n} ∑_{d|n} (−1)^{d}d^{3}.

Lemma 12. For any prime p ≥ 3, we have

r_{4}(pn) ≡ r_{4}(n) (mod p), r_{8}(pn) ≡ r_{8}(n) (mod p^{3}).

Proof. By Lemma 11, we have

r_{4}(n) = 8 ∑_{d|n,4∤d} d = 8 ∑_{\begin{matrix}d|n \\ 4∤d,p∤d \end{matrix}} d + 8 ∑_{\begin{matrix}d|n \\ 4∤d,p|d \end{matrix}} d ≡ 8 ∑_{\begin{matrix}d|n \\ 4∤d,p∤d \end{matrix}} d (mod p),

and

r_{4}(pn) = 8 ∑_{d|pn,4∤d} d = 8 ∑_{\begin{matrix}d|pn \\ 4∤d,p∤d \end{matrix}} d + 8 ∑_{\begin{matrix}d|pn \\ 4∤d,p|d \end{matrix}} d = 8 ∑_{\begin{matrix}d|n \\ 4∤d,p∤d \end{matrix}} d + 8p ∑_{\begin{matrix}d|n \\ 4∤d \end{matrix}} d.

Combining them together, we deduce that r_{4}(pn) ≡ r_{4}(n) (mod p).

Similarly we can prove r_{8}(pn) ≡ r_{8}(n) (mod p^{3}).

Lemma 13. (Cf. [6, Theorem 1 in Chapter 4].) For any integers α ≥ 0 and n ≥ 0, we have r_{3}(4^{α}(8n + 7)) = 0 and r_{3}(4^{α}n) = r_{3}(n).

Lemma 14. (Cf. [7].) Let p ≥ 3 be a prime. For any integers n ≥ 1 and α ≥ 0, we have

r_{3}(p^{2α}n) = (\frac{p^{α+1} − 1}{p − 1} − (\frac{−n}{p})\frac{p^{α} − 1}{p − 1})r_{3}(n) − p\frac{p^{α} − 1}{p − 1}r_{3}(n/p^{2}).

where (\frac{·}{p}) denotes the Legendre symbol, and we take r_{3}(n/p^{2}) = 0 unless p^{2}|n .

Lemma 15. (Cf. [10, Theorem 3].) Let n be an integer which is neither a square nor twice a square. Then \overline{p}(n) ≡ 0 (mod 8).

Lemma 16. (Cf. [4].) Let p ≥ 3 be a prime, and n a positive integer such that p^{2} ∤ n. For any integer α ≥ 0, we have

r_{5}(p^{2α}n) = (\frac{p^{3α+3} − 1}{p^{3} − 1} − p(\frac{n}{p})\frac{p^{3α} − 1}{p^{3} − 1})r_{5}(n).

3 Proofs of The Theorems

Proof of Theorem 2. Replacing q by −q in (1), we get

∑_{n≥0} \overline{p}(n)(−q)^{n} = \frac{1}{φ(q)}.

Hence we have

φ(q)^{5} ∑_{n≥0} \overline{p}(n)(−q)^{n} = φ(q)^{4} = ∑_{n≥0} r_{4}(n)q^{n}.

By Lemma 10, we have φ(q)^{5} ≡ φ(q^{5}) (mod 5) and thus

φ(q^{5}) ∑_{n≥0} \overline{p}(n)(−q)^{n} ≡ ∑_{n≥0} r_{4}(n)q^{n} (mod 5).

Collecting all the terms of the form q^{5n} on both sides, we get

φ(q^{5}) ∑_{n≥0} \overline{p}(5n)(−q)^{5n} ≡ ∑_{n≥0} r_{4}(5n)q^{5n} (mod 5).

Replacing q^{5} by q and applying Lemma 12 with p = 5, we obtain

φ(q) ∑_{n≥0} \overline{p}(5n)(−q)^{n} ≡ ∑_{n≥0} r_{4}(5n)q^{n} ≡ ∑_{n≥0} r_{4}(n)q^{n} = φ(q)^{4} (mod 5).

Hence we have

∑_{n≥0} \overline{p}(5n)(−q)^{n} ≡ φ(q)^{3} = ∑_{n≥0} r_{3}(n)q^{n} (mod 5).

Theorem 1 follows by comparing the coefficients of q^{n} on both sides.

Proof of Corollary 3. This corollary follows immediately from Theorem 2 and Lemma 13.

Proof of Corollary 4. Let α = 1 and we replace n by np in Lemma 14. We have

r_{3}(p^{3}n) = (p + 1)r_{3}(n) − pr_{3}(n/p).

Since n is coprime to p and p + 1 ≡ 0 (mod 5), we get r_{3}(p^{3}n) ≡ 0 (mod 5). By Theorem 2, we deduce that \overline{p}(5p^{3}n) ≡ 0 (mod 5).

Proof of Theorem 5. Since 40n + 35 = 5(8n + 7) is an odd number, it cannot be twice a square. If 5(8n + 7) = x^{2} is a square where x is an odd number, then we know 5|x. Let x = 5y where y is an odd number, we get 8n + 7 = 5y^{2}. But 5y^{2} ≡ 5 (mod 8), which is a contradiction! Hence we know 4^{α}(40n + 35) is neither a square nor twice a square. By Lemma 15, we have \overline{p}(4^{α}(40n+35)) ≡ 0 (mod 8). Combining this with Corollary 3 completes the proof.

Proof of Theorem 6. Set p = 5 and n = 5m + r where r ∈ {1, 4} in Lemma 14. It is easy to deduce that r_{3}(5^{2α}(5m + r)) ≡ 0 (mod 5) for any integer α ≥ 1. By applying Theorem 2, we complete the proof.

Proof of Theorem 7. (1) Let n = pN and then replace α by 5α + 4 in Lemma 14. Since

\frac{p^{5α+5} − 1}{p − 1} = 1 + p + ⋯ + p^{5α+4} ≡ 0 (mod 5),

we have r_{3}(p^{10α+9}N) ≡ 0 (mod 5). By Theorem 2, we deduce that \overline{p}(5p^{10α+9}N) ≡ 0 (mod 5).

(2) Let n = pN and then replace α by 4α + 3 in Lemma 14. Since p^{4α+4} ≡ 1 (mod 5), we deduce that r_{3}(p^{8α+7}N) ≡ 0 (mod 5). By Theorem 2, we deduce that \overline{p}(5p^{8α+7}N) ≡ 0 (mod 5).

Proof of Theorem 8. We have

φ(q)^{9} ∑_{n≥0} \overline{p}(n)(−q)^{n} = φ(q)^{8} = ∑_{n≥0} r_{8}(n)q^{n}.

By Lemma 10, we have φ(q)^{9} ≡ φ(q^{3})^{3} (mod 9) and thus

φ(q^{3})^{3} ∑_{n≥0} \overline{p}(n)(−q)^{n} ≡ ∑_{n≥0} r_{8}(n)q^{n} (mod 9).

Collecting all the terms of the form q^{3n} on both sides, we get

φ(q^{3})^{3} ∑_{n≥0} \overline{p}(3n)(−q)^{3n} ≡ ∑_{n≥0} r_{8}(3n)q^{3n} (mod 9).

Replacing q^{3} by q and applying Lemma 12 with p = 3, we obtain

φ(q)^{3} ∑_{n≥0} \overline{p}(3n)(−q)^{n} ≡ ∑_{n≥0} r_{8}(3n)q^{n} ≡ ∑_{n≥0} r_{8}(n)q^{n} = φ(q)^{8} (mod 9).

Hence we have

∑_{n≥0} \overline{p}(3n)(−q)^{n} ≡ φ(q)^{5} = ∑_{n≥0} r_{5}(n)q^{n} (mod 9).

Theorem 8 follows by comparing the coefficients of q^{n} on both sides.

Proof of Theorem 9. (1) Let n = pN and then replace α by 3α + 2 in Lemma 16. Since

\frac{p^{9α+9} − 1}{p^{3} − 1} = 1 + p^{3} + ⋯ + p^{3(3α+2)} ≡ 0 (mod 3),

we have r_{5}(p^{6α+5}N) ≡ 0 (mod 3). By Theorem 8, we deduce that \overline{p}(3p^{6α+5}N) ≡ 0 (mod 3).

Similarly, let n = pN and replace α by 9α + 8 in Lemma 16. Since p ≡ 1 (mod 3) implies p^{3} ≡ 1 (mod 9), we have

\frac{p^{27α+27} − 1}{p^{3} − 1} = 1 + p^{3} + ⋯ + p^{3(9α+8)} ≡ 0 (mod 9).

Hence r_{5}(p^{18α+17}N) ≡ 0 (mod 9), and we deduce by Theorem 8 that \overline{p}(3p^{18α+17}N) ≡ 0 (mod 9).

(2) Let n = pN and replace α by 2α + 1 in Lemma 16. Note that p ≡ 2 (mod 3) implies p^{3} ≡ −1 (mod 9). Since p^{6α+6} ≡ 1 (mod 9), we have r_{5}(p^{4α+3}N) ≡ 0 (mod 9). By Theorem 8, we deduce that \overline{p}(3p^{4α+3}N) ≡ 0 (mod 9).

References

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2010 Mathematics Subject Classification: Primary 05A17; Secondary 11P83. Keywords: overpartition, congruence, sum of squares.

(Concerned with sequence \underline{A015128}.)

Received July 24 2014; revised version received August 25 2014; September 6 2014. Published in Journal of Integer Sequences, September 7 2014.

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