Infinite families of congruences for p o (n)

+ (−1)^±p+13 q^p

2−1 3

∞

X

n=−∞

(−1)ⁿq^p2(3n2−2n)

=

p−1 2

X

k=−^p−1₂ k6=^±p+1₃

(−1)^kq^3k2−2k

∞

X

n=−∞

(−1)ⁿqpn(3pn+6k−2)

+ (−1)^±p+13 q^p

2−1

3 Ω(−q^p2).

Note that, if 3k²−2k≡ ^p2₃⁻¹ (mod p), thenk = ^±p+1₃ .

3.5 Infinite families of congruences for p

_o

(n)

We next prove certain infinite families of congruences for p_o(n) modulo 8 and 16 as stated in the following theorems.

Theorem 3.7. Let p ≥ 3 be a prime, n ≥ 0 and α ≥ 1. If −2

p

=−1, then we have

p_o 8p^2αn+ (3p+ 8j)p^2α−1

≡0 (mod 16), (3.33)

p_o 16p^2αn+ (6p+ 16j)p^2α−1

≡0 (mod 16), (3.34)

p_o 72p^2αn+ (9p+ 72j)p^2α−1

≡0 (mod 8), (3.35)

p_o 24p^2αn+ (17p+ 24j)p^2α−1

≡0 (mod 8), (3.36)

p_o 32p^2αn+ (4p+ 32j)p^2α−1

≡0 (mod 8), (3.37)

p_o 32p^2αn+ (12p+ 32j)p^2α−1

≡0 (mod 8). (3.38)

where j = 1,2, . . . , p−1.

Before we prove Theorem 3.7, we first prove the following results.

60 Overpartitions Into Odd Parts

Theorem 3.8. Let p≥3be a prime such that −2

p

=−1. Then, for all nonneg- ative integers n and α, we have

∞

X

n=0

p_o(8p^2αn+ 3p^2α)qⁿ≡4f(−q²)³ψ(q) (mod 16). (3.39)

Proof. From (3.16), we have that (3.39) is true for α = 0. We now use induction on α to complete the proof. Observe that (3.39) can also be written as

∞

X

n=0

p_o

8

p^2αn+ 3p^2α−1 8

+ 3

qⁿ≡4f(−q²)³ψ(q) (mod 16). (3.40)

We suppose that (3.40) holds for some α > 0. Substituting (1.20) and (1.23) in (3.40), we have, modulo 16

∞

X

n=0

p_o

8

p^2αn+ 3p^2α−1 8

+ 3

qⁿ (3.41)

≡4







 1 2

p−1

X

k=0 k6=^p−1₂

(−1)^kq^k2+k

∞

X

n=−∞

(−1)ⁿ(2pn+ 2k+ 1)qpn(pn+2k+1)

+(−1)^p−12 pq^p

2−1

4 f(−q^2p2)³

×





p−3 2

X

m=0

q^m2+m2 f

p²+ (2m+ 1)p

2 ,p²−(2m+ 1)p 2

+q^p

2−1

8 ψ(q^p2)



.

For a prime p≥3 and 0≤k ≤p−1, 0≤m≤ ^p−1₂ , we consider

(k²+k) + m²+m

2 ≡3p²−1

8 (mod p) (3.42)

which is equivalent to

2(2k+ 1)²+ (2m+ 1)² ≡0 (mod p).

3.5 Infinite families of congruences for p_o(n) 61

Since

−2 p

= −1, we have k = m = ^p−1₂ is the only solution of (3.42). There- fore, extracting the terms containing q^pn+3(p

2−1)

8 from both sides of (3.41), and then replacing q^p byq, we deduce that

∞

X

n=0

p_o

8

p^2α+1n+ 3p^2α+2−1 8

+ 3

qⁿ ≡4f(−q^2p)³ψ(q^p) (mod 16). (3.43)

Similarly, extracting the terms containing q^pn from both sides of (3.43), and then replacing q^p byq, we obtain

∞

X

n=0

p_o

8

p^2(α+1)n+ 3p^2(α+1)−1 8

+ 3

qⁿ≡4f(−q²)³ψ(q) (mod 16), (3.44)

proving the result for α+ 1. This completes the proof of the theorem.

Theorem 3.9. Let p≥3be a prime such that −2

p

=−1. Then, for all nonneg- ative integers n and α, we have

∞

X

n=0

p_o(72p^2αn+ 9p^2α)qⁿ≡6f(−q)f(−q²) (mod 8). (3.45)

Proof. Clearly, (3.45) is true whenα= 0 due to (3.23). We now use induction on α to complete the proof.

For a prime p≥5 and −^p−1₂ ≤k, m≤ ^p−1₂ , we consider the congruence 3k²+k

2 + 23m²+m

2 ≡3p²−1

24 (mod p), (3.46)

which is equivalent to

(6k+ 1)² + 2(6m+ 1)² ≡0 (mod p).

Since

−2 p

=−1, thereforek =m= ^±p−1₆ is the only solution of (3.46). By Lemma

62 Overpartitions Into Odd Parts

following congruence

∞

X

n=0

p_o

72

p^2α+1n+ 9p^2α+2−1 72

+ 9

qⁿ ≡6f(−q^p)f(−q^2p) (mod 8). (3.47)

We next extract the terms containing q^pn from both sides of the above congruence, and observe that (3.45) is true when α is replaced by α+ 1. This completes the

proof of the result.

Theorem 3.10. Let p ≥ 3 be a prime such that −2

p

= −1. Then, for all nonnegative integers n and α, we have

∞

X

n=0

p_o

24

p^2αn+ 17p^2α−1 24

+ 17

qⁿ≡4f(−q³)³Ω(−q) (mod 8). (3.48)

Proof. From (3.22) we can see that (3.48) is true when α= 0. Suppose that (3.48) holds for some α >0. Substituting (1.23) and (3.32) into (3.48), we have, modulo 8

∞

X

n=0

p_o

24

p^2αn+ 17p^2α−1 24

+ 17

qⁿ (3.49)

=







 1 2

p−1

X

m=0 m6=^p−1

2

(−1)^mq^3m2+m2

∞

X

n=−∞

(−1)ⁿ(2pn+ 2m+ 1)q3pn(pn+2m+1) 2

+(−1)^p−12 pq^3p

2−1

8 f(−q^3p2)³

×













p−1 2

X

k=−^p−1

2

k6=^±p+1₃

(−1)^kq^3k2−2k

∞

X

n=−∞

(−1)ⁿqpn(3pn+6k−2)+ (−1)^±p+13 q^p

2−1

3 Ω(−q^p2)











 .

For a prime p≥5, 0≤m≤p−1 and −^p−1₂ ≤k ≤ ^p−1₂ , we consider 3m(m+ 1)

2 + 3k²−2k≡17p²−1

24 (mod p), (3.50)

3.5 Infinite families of congruences for p_o(n) 63

which is equivalent to (6m+ 3)²+ 2(6k−2)² ≡0 (mod p). Since

−2 p

=−1, we havem = ^p−1₂ andk = ^±p+1₃ is the only solution of (3.50). Therefore, extracting the terms containing q^pn+17p

2−1

24 from both sides of (3.49), and then replacing q^p by q, we deduce that

∞

X

n=0

p_o

24

p^2α+1n+ 17p^2α+2−1 24

+ 17

qⁿ≡4f(−q^3p)³Ω(−q^p) (mod 8).

(3.51) Similarly, extracting the terms containing q^pn from both sides of (3.51), and then replacing q^p byq, we obtain

∞

X

n=0

p_o

24

p^2(α+1)n+ 17p^2(α+1)−1 24

+ 17

qⁿ≡4f(−q³)³Ω(−q) (mod 8).

(3.52)

This completes the proof of the result.

Proof of Theorem 3.7. From (3.43), it follows that

p_o(8p^2α+1(pn+j) + 3p^2α+2)≡0 (mod 16), (3.53) where j = 1,2, . . . , p−1. This completes the proof of (3.33) forα≥1.

Proof of (3.34) proceeds along similar lines to the proof of (3.33). Therefore, we omit the details for reasons of brevity.

Extracting the terms containing q^pn+j from (3.47), where j = 1,2, . . . , p−1, it follows that

p_o 72p^2αn+ (9p+ 72j)p^2α−1

≡0 (mod 8).

This completes the proof of (3.35) for α≥1.

64 Overpartitions Into Odd Parts

From (3.51), it follows that

p_o(24p^2α+1(pn+j) + 17p^2α+2)≡0 (mod 8), where j = 1,2, . . . , p−1. This completes the proof of (3.36) for α≥1.

We now substitute the p-dissection identities, namely (1.21), (1.22) and (3.31) into (3.19) and (3.20). For p ≥ 3, −^p−1₂ ≤ k, m ≤ ^p−1₂ and

−2 p

= −1, the congruences

3k²+k

2 + 3m²+m≡3p²−1

24 (mod p), 3k² +k

2 + 3m² + 2m≡9p²−1

24 (mod p)

have the only solutions k = m = ^±p−1₆ , and k = ^±p−1₆ , m = ^±p−1₃ , respectively.

Proceeding similarly as shown in the proof of (3.36), we obtain

∞

X

n=0

p_o

32

p^2α+1n+ 3p^2α+2−1 24

+ 4

qⁿ≡6(q^p;q^p)∞(q^2p;q^2p)⁵_∞

(q^4p;q^4p)²_∞ (mod 8), (3.54)

∞

X

n=0

p_o

32

p^2α+1n+ 9p^2α+2−1 24

+ 12

qⁿ≡4(q^p;q^p)³_∞(q^4p;q^4p)²_∞ (q^2p;q^2p)∞

(mod 8).

(3.55) Now, from (3.54) and (3.55), it follows that

p_o(32p^2α+1(pn+j) + 4p^2α+2)≡0 (mod 8), p_o(32p^2α+1(pn+j) + 12p^2α+2)≡0 (mod 8),

wherej = 1,2, . . . , p−1.This completes the proof of (3.37) and (3.38) forα≥1.

Theorem 3.11. Let p≡3 (mod 4)be a prime ,n ≥0and α≥1. If −2

p

=−1,

3.5 Infinite families of congruences for p_o(n) 65

then we have

p_o 16p^2αn+ (10p+ 16j)p^2α−1

≡0 (mod 16), (3.56)

where j = 1,2, . . . , p−1.

Before we prove Theorem 3.11, we first prove the following result.

Theorem 3.12. Let p ≥ 3 be a prime such that p ≡ 3 (mod 4). Then, for all nonnegative integers n and α, we have

∞

X

n=0

p_o

16

p^2αn+ 5p^2α−1 8

+ 10

qⁿ≡8f(−q⁴)³ψ(q) (mod 16). (3.57)

Proof. We use induction on α to proof the theorem. Clearly, (3.57) is true when α = 0 due to (3.18). Suppose that (3.57) holds for some α > 0. For a prime p≥ 5 and 0≤k ≤p−1, 0≤m≤ ^p−1₂ , the equation

2(k²+k) + m²+m

2 ≡5p² −1

8 (mod p),

which is equivalent to 4(2k+ 1)² + (2m+ 1)² ≡ 0 (mod p), has the only solution k = m = ^p−1₂ as p ≡ 3 (mod 4). Applying (1.20) and (1.23) in (3.57), and then proceeding similarly as shown in the proof of Theorem 3.8, we deduce that

∞

X

n=0

p_o

16

p^2α+1n+ 5p^2α+2−1 8

+ 10

qⁿ≡4f(−q^4p)³ψ(q^p) (mod 16). (3.58)

Extracting the terms containing q^pn from both sides of (3.58), we find that

∞

X

n=0

p_o

16

p^2(α+1)n+ 5p^2(α+1)−1 8

+ 10

qⁿ≡4f(−q⁴)³ψ(q) (mod 16),

completing the proof of (3.57).

66 Overpartitions Into Odd Parts

Proof of Theorem 3.11. From (3.58), it follows that

p_o(16p^2α+1(pn+j) + 10p^2α+2)≡0 (mod 16),

where j = 1,2, . . . , p−1. This completes the proof of the Theorem 3.11.

4

Andrews’ Singular Overpartitions

4.1 Introduction

In [4], Andrews defined the partition function C_k,i(n), called singular overparti- tion, which counts the number of overpartitions of n in which no part is divisible by k and only parts ≡ ±i (mod k) may be overlined. For example,C_3,1(4) = 10 with the relevant partitions being 4,4,2 + 2,2 + 2,2 + 1 + 1,2 + 1 + 1,2 + 1 + 1,2 + 1 + 1,1 + 1 + 1 + 1,1 + 1 + 1 + 1. For k ≥ 3 and 1≤ i≤_k

2

, the generating function

1The contents of this chapter have been published inThe Ramanujan J.(2018) and some parts are under review.

68 Singular Overpartitions

for Ck,i(n) is given by

∞

X

n=0

C_k,i(n)qⁿ= (q^k;q^k)∞(−qⁱ;q^k)∞(−q^k−i;q^k)∞

(q;q)∞

. (4.1)

Andrews proved the following Ramanujan-type congruences C_3,1(9n+ 3)≡C_3,1(9n+ 6)≡0 (mod 3)

using some q-series identities. Chen, Hirschhorn and Sellers [17] later showed that Andrews’ congruences modulo 3 are two examples of an infinite family of congruences modulo 3 which hold for the function C_3,1(n). Recently, Ahmed and Baruah [2]

found congruences for C_3,1(n) modulo 4, 18 and 36, infinite families of congruences modulo 2 and 4 for C_8,2(n), congruences modulo 2 and 3 forC_12,2(n) and C_12,4(n);

and congruences modulo 2 for C_28,8(n) and C_48,16(n).

In a very recent work, Naika and Gireesh [40] prove that two congruences for C_3,1(n) modulo 36 proved by Ahmed and Baruah hold for modulo 72. They also prove congruences modulo 6, 12, 16, 18, and 24 for C_3,1(n) and infinite families of congruences modulo 12, 18, 48, and 72 for C_3,1(n). They further conjecture that

C_3,1(12n+ 11)≡0 (mod 144) for all n≥0. In this chapter we prove the conjecture.