Parity results for broken 11-diamond partitions

Yunjian Wu

doi:10.1515/math-2019-0031

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Parity results for broken 11-diamond partitions

Yunjian Wu

Veröffentlicht/Copyright: 16. Mai 2019

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Aus der Zeitschrift Open Mathematics Band 17 Heft 1

Abstract

Recently, Dai proved new infinite families of congruences modulo 2 for broken 11-diamond partition functions by using Hecke operators. In this note, we establish new parity results for broken 11-diamond partition functions. In particular, we generalize the congruences due to Dai by utilizing an identity due to Newman. Furthermore, we prove some strange congruences modulo 2 for broken 11-diamond partition functions. For example, we prove that if p is a prime with p ≠ 23 and Δ₁₁(2p + 1) ≡ 1 (mod 2), then for any k ≥ 0, Δ₁₁(2p^3k+3 + 1) ≡ 1 (mod 2), where Δ₁₁(n) is the number of broken 11-diamond partitions of n.

Keywords: broken k-diamond partition; congruence; parity results

MSC 2010: 11P83; 05A17

1 Introduction

A combinatorial study guided by MacMahon’s Partition Analysis led Andrews and Paule [1] to the construction of a new class of directed graphs called broken k-diamond partitions. For a fixed positive integer k, let Δ_k(n) denote the number of broken k-diamond partitions of n. Moreover, the following generating functions for Δ_k(n) was established by Andrews and Paule [1]:

∑n=0∞Δk(n)qn=(q2;q2)∞(q2k+1;q2k+1)∞(q;q)∞3(q4k+2;q4k+2)∞,

where

(q;q)∞:=∏n=1∞(1−qn).

Andrews and Paule [1] also proved that for n ≥ 0,

Δ1(2n+1)≡0(mod3)

by employing generating function manipulations. In addition, they also gave three conjectures modulo 2, 5 and 25 for Δ_k(n). Since then, a number of congruences for Δ_k(n) for small integers k have been proved. See, for example, [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12].

In 2010, by using theory of modular forms, Radu and Sellers [4] proved some parity results for Δ_k(n), where k ∈ {2, 3, 5, 6, 8, 9, 11}. For example, they proved that for n ≥ 0,

Δ11(46n+j)≡0(mod2).

where j ∈ {11, 15, 21, 23, 29, 31, 35, 39, 41, 43, 45}. Recently, Yao [12] proved several infinite families of congruences modulo 2 for Δ₁₁(n) which generalize Radu and Sellers’ congruences on Δ₁₁(n). Ahmed and Baruah [13] also proved some congruences modulo 2 for Δ₁₁(n). Very recently, Dai [2] established more parity results for Δ₁₁(n). Let p be a prime with p ≠ 23 and let n, k ≥ 0 be integers. He proved that if Δ₁₁(2p + 1) is even, then

Δ11(2p2k+2n+2rp2k+1+1)≡0(mod2) (1.1)

for 1 ≤ r ≤ p − 1 and if Δ₁₁(2p + 1) is odd, then

Δ11(2p3k+3n+2rp3k+2+1)≡0(mod2) (1.2)

for 1 ≤ r ≤ p − 1.

In this paper, we prove new congruences modulo 2 for Δ₁₁(n) by employing an identity due to Newman [14]. In particular, our results generalize (1.1) and (1.2). Moreover, we also prove some strange congruences modulo 2 for Δ₁₁(n). The main results of this paper can be stated as follows.

Theorem 1.1

Let k be a nonnegative integer and let p_i (1 ≤ i ≤ k + 1) be primes with p_i ≠ 23 (p_i may equal to 2). For n ≥ 0,

Δ11(2p1α(p1)⋯pkα(pk)pk+1α(pk+1)n+2rp1α(p1)⋯pkα(pk)pk+1α(pk+1)−1+1)≡0(mod2), (1.3)

where 1 ≤ r ≤ p_k+1 − 1 and

α(pi):=2,ifΔ11(2pi+1) is even,3,ifΔ11(2pi+1) is odd. (1.4)

Moreover,

Δ11(2p1α(p1)⋯pkα(pk)pk+1α(pk+1)+1)≡1(mod2). (1.5)

If we set p₁ = p₂ = ⋯ = p_k+1 = p in (1.3), we get (1.1) and (1.2). Moreover, if we set k = 1, p₁ = 3 and p₂ = 5 in (1.3), we find that for n ≥ 0,

Δ11(1350n+270r+1)≡0(mod2), (1.6)

where 1 ≤ r ≤ 4. Note that Dai’s congruences (1.1) and (1.2) do not imply (1.6). Therefore, (1.3) is a generalization of (1.1) and (1.2). Moreover, if set p₁ = p₂ = ⋯ = p_k+1 = p in (1.5), we deduce that if Δ₁₁(2p + 1) ≡ 0 (mod 2), then for k ≥ 0,

Δ11(2p2k+2+1)≡1(mod2)

and if Δ₁₁(2p + 1) ≡ 1 (mod 2), then for k ≥ 0,

Δ11(2p3k+3+1)≡1(mod2).

2 Two Lemmas

In order to prove Theorem 1.1, we prove the following two lemmas in this section.

Lemma 2.1

Let c(n) be defined by

∑n=0∞c(n)qn:=(q;q)∞(q23;q23)∞ (2.1)

and let p be an odd prime with p ≠ 23. If c(p − 1) ≡ 0 (mod 2), then for n ≥ 0,

c(p2n+p2−1)≡c(n)(mod2) (2.2)

and

c(p2n+rp−1)≡0(mod2), (2.3)

where r is an integer and 1 ≤ r ≤ p − 1.

Proof

Newman [14] proved that

c(pn+p−1)=c(p−1)c(n)−(−1)p−123pc((n−p+1)/p), (2.4)

where p is an odd prime with p ≠ 23 and ⋅p denotes the Legendre symbol. Identity (2.4) implies that

c(pn+p−1)≡c(p−1)c(n)+c((n−p+1)/p)(mod2). (2.5)

Therefore, if c(p − 1) ≡ 0 (mod 2), then

c(pn+p−1)≡c((n−p+1)/p)(mod2). (2.6)

Replacing n by pn + p − 1 in (2.6), we arrive at (2.2). Note that for 0 ≤ j ≤ p − 2, pn+j−p+1p is not an integer and

c((pn+j−p+1)/p)=0. (2.7)

Replacing n by pn + j (0 ≤ j ≤ p − 2) in (2.6) and using (2.7), we deduce that for n ≥ 0,

c(p2n+(j+1)p−1)≡0(mod2),0≤j≤p−2,

which is nothing but (2.3). This completes the proof. □

Lemma 2.2

Let p be an odd prime with p ≠ 23. If c(p − 1) ≡ 1 (mod 2), then for n ≥ 0,

c(p3n+p3−1)≡c(n)(mod2) (2.8)

and

c(p3n+rp2−1)≡0(mod2), (2.9)

where r is an integer and 1 ≤ r ≤ p − 1.

Proof

If c(p − 1) ≡ 1 (mod 2), then we can rewrite (2.5) as

c(pn+p−1)≡c(n)+c((n−p+1)/p)(mod2). (2.10)

Replacing n by pn + p − 1 in (2.10) yields

c(p2n+p2−1)≡c(pn+p−1)+c(n)(mod2). (2.11)

Substituting (2.10) into (2.11) yields

c(p2n+p2−1)≡c((n−p+1)/p)(mod2). (2.12)

Replacing n by pn + p − 1 in (2.12), we arrive at (2.8). Replacing n by pn + j (0 ≤ j ≤ p − 2) in (2.12) and employing (2.7), we deduce that for n ≥ 0,

c(p3n+(j+1)p2−1)≡0(mod2),0≤j≤p−2,

which yields (2.9). The proof is complete. □

3 Proof of Theorem 1.1

In [12], Yao established the generating function for Δ₁₁(2n + 1) modulo 2:

∑n=0∞Δ11(2n+1)qn≡1+q(q;q)∞(q23;q23)∞(mod2). (3.1)

In view of (2.1) and (3.1),

∑n=0∞Δ11(2n+1)qn≡1+∑n=0∞c(n)qn+1(mod2).

Therefore, for n ≥ 0,

Δ11(2n+3)≡c(n)(mod2). (3.2)

Yao [12] also proved that for n ≥ 0,

Δ11(16n+1)≡Δ11(2n+1)(mod2). (3.3)

and

Δ11(16n+9)≡0(mod2). (3.4)

Let k be a nonnegative integer and let p_i (1 ≤ i ≤ k + 1) be odd primes with p_i ≠ 23. By (3.2) and Lemmas 2.1 and 2.2, we see that for n ≥ 0

Δ11(2piα(pi)n+2piα(pi)+1)≡Δ11(2n+3)(mod2) (3.5)

and

Δ11(2pt+1α(pt+1)n+2rpt+1α(pt+1)−1+1)≡0(mod2), (3.6)

where 1 ≤ r ≤ p_t+1 − 1 and α(p_i) is defined by (1.4). By (3.3), (3.4) and the fact that Δ₁₁(5) ≡ 1 (mod 2), we find that (3.5) holds when p_i = 2 and (3.6) is also true when p_t+1 = 2. Replacing n by n − 1 in (3.5), we find that for n ≥ 1,

Δ11(2piα(pi)n+1)≡Δ11(2n+1)(mod2). (3.7)

Therefore, by (3.7) and iterative method, we see that for n ≥ 0,

Δ11(2p1α(pi)p2α(p2)⋯pkα(pk)n+1)≡Δ11(2n+1)(mod2), (3.8)

where p_i are primes with p_i ≠ 23 (p_i may equal to 2) for 1 ≤ i ≤ k. Replacing n by pk+1α(pk+1)+rpk+1α(pk+1)−1 in (3.8) and employing (3.6), we arrive at (1.3).

Setting n = 0 and i = k + 1 in (3.5) and using the fact that Δ₁₁(3) ≡ 1 (mod 2), we have

Δ11(2pk+1α(pk+1)+1)≡1(mod2). (3.9)

Setting n=pk+1α(pk+1) in (3.8) and using (3.9), we get (1.5). This completes the proof. □

Acknowledgments

The author is very grateful to the referee for his/her helpful comments. This work was supported by the National Natural Science Foundation of China (No. 11501101 and 61773115).

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Received: 2018-05-02

Accepted: 2019-03-06

Published Online: 2019-05-16

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broken k-diamond partition; congruence; parity results

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