Some New Congruences for l-Regular Partitions Modulo 13, 17, and 23

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Some New Congruences for l-Regular Partitions Modulo 13, 17, and 23

S Abinash, T Kathiravan, K Srilakshmi

To cite this version:

S Abinash, T Kathiravan, K Srilakshmi. Some New Congruences for l-Regular Partitions Modulo 13, 17, and 23. Hardy-Ramanujan Journal, Hardy-Ramanujan Society, 2020, Volume 42 - Special Commemorative volume in honour of Alan Baker. �hal-02301897v2�

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Some new congruences for l-regular partitions modulo 13, 17, and 23

S. Abinash, T. Kathiravan and K. Srilakshmi

Dedicated to the memory of Alan Baker

Abstract. A partition ofnisl-regular if none of its parts is divisible byl. Letbl(n) denote the number ofl-regular partitions ofn. In this paper, using theta function identities due to Ramanujan, we establish some new infinite families of congruences for bl(n) modulol, wherel= 17,23, and forb65(n) modulo 13.

Keywords. congruences,l-regular partitions, theta function identities.

2010 Mathematics Subject Classification. 11P83, 05A17

1. Introduction

We first recall a few terminologies and notations. Let n be a positive integer. A partition ofn is a non-increasing sequence of positive integers whose sum is n, and the members of the sequence are called parts. For a positive integer l ≥2, a partition of n is said to bel-regular if none of its parts is divisible by l. Letb_l(n) denote the number ofl-regular partitions ofn. By convention, we assume thatbl(0) = 1. The generating function for bl(n) is given by

∞

X

n=0

b_l(n)qⁿ= f_l

f₁, (1.1)

where|q|<1 and for any positive integerk,f_k is defined by f_k :=

∞

Y

i=1

1−q^ki

. It can be easily verified that for any prime number l,

f_l≡f₁^l (modl). (1.2)

In recent years, various congruences satisfied by l-regular partitions for different values of l has been established. For instance, Lovejoy and Penniston [LoPe01] established criteria for 3-divisibility ofb3(n). Dandurand and Penniston [DaPe09] employed the theory of complex multiplication to give a precise description of thosensuch thatl|b_l(n) forl∈ {5,7,11}. Cui and Gu [CuiGu13] derived infinite families of congruences modulo 2 forb_l(n) wherel∈ {2,4,5,8,13,16}. Webb [Web11] established an infinite family of congruences modulo 3 for b13(n). Furcy and Penniston [FuPe12] obtained families of congruences modulo 3 for other values of l which are congruent to 1 modulo 3. Xia and Yao [XiYa14a] found some infinite families of congruences forb₉(n) modulo 2, and Cui and Gu [CuiGu15]

derived congruences for b9(n) modulo 3. Xia [Xia15] established infinite families of congruences for b_l(n) modulol wherel∈ {13,17,19}, for example, for anyn≥0,k≥0,

b13 5^12kn+ 1 2

5^12k−1

!

≡b13(n) (mod 13).

We thankepisciences.orgfor providing open access hosting of the electronic journalHardy-Ramanujan Journal

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74 2. Preliminaries

For a complete literature review, see [Xia15]. For more related works, see [AdDa18,AhLo01,AHS10, CaWe14,CuiGu18,DLY14,Kei14,LiWa14,Pen02,Pen08,Ran17,Tan18,Wan17,Wan18a,Wan18b, Xia14,XiYa14b,Yao14,ZJY18].

In this paper, we establish infinite families of Ramanujan-type congruences for bl(n) modulo l, where l ∈ {17,23}, and for b₆₅(n) modulo 13. In fact, no such congruence has been established for 23-regular partitions till now, to the best of the authors’ knowledge. The following are the main results.

Theorem 1.1. For any n≥0, k≥0, b17 5^4kn+2

3

5^4k−1

!

≡2^kb17(n) (mod 17), (1.3) and for r∈ {0,1,2,4},

b₁₇ 5^4k+4n+ 1 3

5^4k+3(3r+ 1)−2

!

≡0 (mod 17). (1.4)

Theorem 1.2. For any n≥0, k≥0, b₂₃ 5^24kn+ 11

12

5^24k−1

!

≡14^kb₂₃(n) (mod 23), (1.5) and for r∈ {0,1,2,3},

b₂₃ 5^24k+24n+ 1 12

5^24k+23(12r+ 7)−11

!

≡0 (mod 23). (1.6)

Theorem 1.3. For any n≥0, k≥0, b₆₅ 5^12kn+ 8

3

5^12k−1

!

≡12^kb₆₅(n) (mod 13), (1.7) and for r∈ {2,4},

b65 5^12k+12n+ 1 3

5^12k+11(3r+ 1)−8

!

≡0 (mod 13). (1.8)

2. Preliminaries

In this section we shall state a preliminary lemma.

Lemma 2.1. For

R(q) :=

∞

Y

n=1

1−q⁵ⁿ⁻⁴

1−q⁵ⁿ⁻¹ 1−q⁵ⁿ⁻³

1−q⁵ⁿ⁻², we have

A. ([Ram62], [Wat38])

f1 =f25

1

R(q⁵) −q−q²R q⁵

!

. (2.9)

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B.([Wat29a],[Wat29b])

1

R(q)⁵ −11q−q²R(q)⁵ = f₁⁶

f₅⁶. (2.10)

C.

1 f₁ = f₂₅⁵

f₅⁶ 1

R(q⁵)⁴+ q

R(q⁵)³+ 2q²

R(q⁵)²+ 3q³

R(q⁵)+ 5q⁴−3q⁵R q⁵

+ 2q⁶R q⁵2

−q⁷R q⁵3

+q⁸R q⁵4

! . (2.11) Note that the identity in (2.11) can be easily established by first substituting q by q⁵ in (2.10) and then using (2.9).

3. Congruences for 17-Regular Partitions

Proof of Theorem 1.1. Taking l= 17 in (1.1), we get

∞

X

n=0

b₁₇(n)qⁿ= f₁₇ f1

.

Now by (1.2),

∞

X

n=0

b₁₇(n)qⁿ≡f₁¹⁶ (mod 17). (3.12)

Replacing f₁ using (2.9) and then extracting the terms involving q⁵ⁿ⁺¹, we get

∞

X

n=0

b₁₇(5n+ 1)q⁵ⁿ⁺¹≡f₂₅¹⁶ q

R(q⁵)¹⁵ + 13q⁶

R(q⁵)¹⁰ + 8q¹¹

R(q⁵)⁵ −q¹⁶−8q²¹R q⁵5

+ 13q²⁶R q⁵10

−q³¹R q⁵15

!

(mod 17).

If we divide by q and substitute q by q^1/5, then we get

∞

X

n=0

b17(5n+ 1)qⁿ≡f₅¹⁶

1

R(q)⁵ −11q−q²R(q)⁵ 3

+ 12q 1

R(q)⁵ −11q−q²R(q)⁵ 2

+ 14q² 1

R(q)⁵ −11q−q²R(q)⁵

+ 7q³

!

(mod 17).

By (2.10), we get

∞

X

n=0

b17(5n+ 1)qⁿ≡f₅¹⁶ f₁¹⁸

f₅¹⁸ + 12qf₁¹²

f₅¹² + 14q²f₁⁶ f₅⁶ + 7q³

!

(mod 17).

In view of (3.12), we get

∞

X

n=0

b₁₇(5n+ 1)qⁿ≡ f₁¹⁸

f₅² + 12qf₁¹²f₅⁴+ 14q²f₁⁶f₅¹⁰+ 7

∞

X

n=0

b₁₇(n)q⁵ⁿ⁺³ (mod 17). (3.13)

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76 3. Congruences for 17-Regular Partitions

Replacing f1 using (2.9), we get

∞

X

n=0

b₁₇(5n+ 1)qⁿ≡ f₂₅¹⁸ f₅²

1

!18

+ 12qf₅⁴f₂₅¹² 1

!12

+ 14q²f₅¹⁰f₂₅⁶ 1

!6

+ 7

∞

X

n=0

b17(n)q⁵ⁿ⁺³ (mod 17).

If we extract the terms involvingq⁵ⁿ⁺³, divide byq³, and substituteq byq^1/5, then by (2.10) we get

∞

X

n=0

b17(5²n+ 16)qⁿ≡ f₅¹⁸ f₁² q³

!

+ 12f₁⁴f₅¹² 3f₁¹² f₅¹² + 3q²

!

+ 14f₁¹⁰f₅⁶ 11f₁⁶ f₅⁶ + 9q

!

+ 7

∞

X

n=0

b₁₇(n)qⁿ (mod 17), which, after rearranging the similar terms, yields

∞

X

n=0

b₁₇ 5²n+ 16

qⁿ≡7qf₁¹⁰f₅⁶+ 2q²f₁⁴f₅¹²+q³f₅¹⁸ f₁² + 10

∞

X

n=0

b₁₇(n)qⁿ (mod 17).

Replacing f₁ and 1/f₁ using (2.9) and (2.11), we get

∞

X

n=0

b₁₇ 5²n+ 16

qⁿ≡7qf₅⁶f₂₅¹⁰ 1

R(q⁵)−q−q²R q⁵

!10

+ 2q²f₅¹²f₂₅⁴ 1

!4

+q³f₅⁶f₂₅¹⁰ 1

R(q⁵)⁴ + q

R(q⁵)³ + 2q²

R(q⁵)² + 3q³

R(q⁵) + 5q⁴−3q⁵R q⁵ + 2q⁶R q⁵2

−q⁷R q⁵3

+q⁸R q⁵4

!2

+ 10

∞

X

n=0

b₁₇(n)qⁿ (mod 17).

If we extract the terms involvingq⁵ⁿ⁺¹, divide byq, and substitute q by q^1/5, then by (2.10) we get

∞

X

n=0

b₁₇ 5³n+ 41

qⁿ≡7f₁⁶f₅¹⁰ f₁¹²

f₅¹² + 12qf₁⁶ f₅⁶ +q²

!

+ 2f₁¹²f₅⁴ 12q

!

+f₁⁶f₅¹⁰ 10qf₁⁶ f₅⁶ + 6q²

!

+ 10

∞

X

n=0

b₁₇(5n+ 1)qⁿ (mod 17), which, after rearranging the similar terms, yields

∞

X

n=0

b17 5³n+ 41

qⁿ≡7f₁¹⁸

f₅² + 16qf₁¹²f₅⁴+ 13q²f₁⁶f₅¹⁰+ 10

∞

X

n=0

b17(5n+ 1)qⁿ (mod 17).

Now by (3.13), we get

∞

X

n=0

b17 5³n+ 41 qⁿ≡2

∞

X

n=0

b17(n)q⁵ⁿ⁺³ (mod 17). (3.14) Comparing the coefficients of the terms of the formq⁵ⁿ⁺³ in (3.14), we can conclude that for any n≥0,

b17 5⁴n+ 416

≡2b17(n) (mod 17),

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or,

b₁₇ 5⁴n+ 2 3

5⁴−1

!

≡2b₁₇(n) (mod 17). (3.15)

Now (1.3) follows from (3.15), by mathematical induction.

On the other hand, comparing the coefficients of the terms of the form q^5n+rwherer∈ {0,1,2,4}

in (3.14), we can conclude that for anyn≥0,

b₁₇ 5³(5n+r) +5³−2 3

!

≡0 (mod 17), or,

b₁₇ 5⁴n+ 1 3

5³(3r+ 1)−2

!

≡0 (mod 17). (3.16)

Now (1.4) follows from (1.3) and (3.16).

4. Congruences for 23-Regular Partitions

∞

X

n=0

b₂₃(n)qⁿ= f23

f₁ . Now by (1.2),

∞

X

n=0

b₂₃(n)qⁿ≡f₁²² (mod 23). (4.17)

Replacing f1 using (2.9) and then extracting the terms involving q⁵ⁿ⁺², we get

∞

X

n=0

b23(5n+ 2)q⁵ⁿ⁺² ≡f₂₅²² 2q²

R(q⁵)²⁰ + 21q⁷

R(q⁵)¹⁵ + 3q¹²

R(q⁵)¹⁰ + 8q¹⁷

R(q⁵)⁵ −q²²−8q²⁷R q⁵5

+ 3q³²R q⁵10

−21q³⁷R q⁵15

+ 2q⁴²R q⁵20

!

(mod 23).

If we divide by q² and substitute q by q^1/5, then we get

∞

X

n=0

b₂₃(5n+ 2)qⁿ≡f₅²² 2 1

R(q)⁵ −11q−q²R(q)⁵ 4

+ 17q 1

R(q)⁵ −11q−q²R(q)⁵ 3

+ 17q² 1

R(q)⁵ −11q−q²R(q)⁵ 2

+ 14q⁴

!

(mod 23).

By (2.10), we get

∞

X

n=0

b23(5n+ 2)qⁿ≡f₅²² 2f₁²⁴

f₅²⁴+ 17qf₁¹⁸

f₅¹⁸+ 17q²f₁¹²

f₅¹² + 14q⁴

!

(mod 23).

In view of (4.17), we get

∞

X

n=0

b23(5n+ 2)qⁿ≡2f₁²⁴

f₅² + 17qf₁¹⁸f₅⁴+ 17q²f₁¹²f₅¹⁰+ 14

∞

X

n=0

b23(n)q⁵ⁿ⁺⁴ (mod 23). (4.18)

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78 4. Congruences for 23-Regular Partitions

Replacing f1 using (2.9), we get

∞

X

n=0

b₂₃(5n+ 2)qⁿ≡2f₂₅²⁴ f₅²

1

!24

+ 17qf₅⁴f₂₅¹⁸ 1

!18

+ 17q²f₅¹⁰f₂₅¹² 1

R(q⁵)−q−q²R q⁵

!12

+ 14

∞

X

n=0

b23(n)q⁵ⁿ⁺⁴ (mod 23).

If we extract the terms involvingq⁵ⁿ⁺⁴, divide byq⁴, and substituteq byq^1/5, then by (2.10) we get

∞

X

n=0

b₂₃(5²n+ 22)qⁿ≡2f₅²⁴ f₁² q⁴

!

+ 17f₁⁴f₅¹⁸ 19f₁¹⁸ f₅¹⁸+ 7q³

!

+ 17f₁¹⁰f₅¹² 8f₁¹² f₅¹² + 3q²

!

+ 14

∞

X

n=0

b23(n)qⁿ (mod 23), which, after rearranging the similar terms, yields

∞

X

n=0

b23 5²n+ 22

qⁿ≡5q²f₁¹⁰f₅¹²+ 4q³f₁⁴f₅¹⁸+ 2q⁴f₅²⁴ f₁² + 13

∞

X

n=0

b23(n)qⁿ (mod 23). (4.19) Replacing f₁ and 1/f₁ using (2.9) and (2.11), we get

∞

X

n=0

b23 5²n+ 22

qⁿ≡5q²f₅¹²f₂₅¹⁰ 1

!10

+ 4q³f₅¹⁸f₂₅⁴ 1

!4

+ 2q⁴f₅¹²f₂₅¹⁰ 1

R(q⁵)⁴ + q

R(q⁵)³ + 2q²

R(q⁵)² + 3q³

R(q⁵)+ 5q⁴−3q⁵R q⁵ + 2q⁶R q⁵2

−q⁷R q⁵3

+q⁸R q⁵4

!2

+ 13

∞

X

n=0

b₂₃(n)qⁿ (mod 23).

If we extract the terms involvingq⁵ⁿ⁺², divide byq², and substituteq byq^1/5, then by (2.10) we get

∞

X

n=0

b₂₃ 5³n+ 72

qⁿ≡5f₁¹²f₅¹⁰ f₁¹²

f₅¹² + 20qf₁⁶

f₅⁶ + 16q²

!

+ 4f₁¹⁸f₅⁴ 18q

!

+ 2f₁¹²f₅¹⁰ 10qf₁⁶

f₅⁶ + 10q²

!

+ 13

∞

X

n=0

b23(5n+ 2)qⁿ (mod 23), which, after rearranging the similar terms, yields

∞

X

n=0

b₂₃ 5³n+ 72

qⁿ≡5f₁²⁴

f₅² + 8qf₁¹⁸f₅⁴+ 8q²f₁¹²f₅¹⁰+ 13

∞

X

n=0

b₂₃(5n+ 2)qⁿ (mod 23).

∞

X

n=0

b₂₃ 5³n+ 72

qⁿ≡8f₁²⁴

f₅² + 22qf₁¹⁸f₅⁴+ 22q²f₁¹²f₅¹⁰+ 21

∞

X

n=0

b₂₃(n)q⁵ⁿ⁺⁴ (mod 23).

Replacing f₁ using (2.9), we get

∞

X

n=0

b₂₃(5³n+ 72)qⁿ≡8f₂₅²⁴ f₅²

1

!24

+ 22qf₅⁴f₂₅¹⁸ 1

!18

+ 22q²f₅¹⁰f₂₅¹² 1

!12

+ 21

∞

X

n=0

b23(n)q⁵ⁿ⁺⁴ (mod 23).

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If we extract the terms involvingq⁵ⁿ⁺⁴, divide byq⁴, and substituteq byq^1/5, then by (2.10) we get

∞

X

n=0

b₂₃(5⁴n+ 572)qⁿ≡8f₅²⁴ f₁² q⁴

!

+ 22f₁⁴f₅¹⁸ 19f₁¹⁸ f₅¹⁸+ 7q³

!

+ 22f₁¹⁰f₅¹² 8f₁¹² f₅¹² + 3q²

!

+ 21

∞

X

n=0

b23(n)qⁿ (mod 23), which, after rearranging the similar terms, yields

∞

X

n=0

b₂₃ 5⁴n+ 572

qⁿ≡20q²f₁¹⁰f₅¹²+ 16q³f₁⁴f₅¹⁸+ 8q⁴f₅²⁴ f₁² + 17

∞

X

n=0

b₂₃(n)qⁿ (mod 23).

∞

X

n=0

b23 5⁴n+ 572 qⁿ≡4

∞

X

n=0

b23 5²n+ 22

qⁿ+ 11

∞

X

n=0

b23(n)qⁿ (mod 23), or,

∞

X

n=0

b₂₃ 5⁴n+11 12

5⁴−1

! qⁿ≡4

∞

X

n=0

b₂₃ 5²n+ 22

qⁿ+ 11

∞

X

n=0

b₂₃(n)qⁿ (mod 23).

Substituting nby 5²n+ 22, we get

∞

X

n=0

b₂₃ 5⁶n+11 12

5⁶−1

! qⁿ

≡4

∞

X

n=0

b23 5⁴n+ 11 12

5⁴−1

!

qⁿ+ 11

∞

X

n=0

b23 5²n+ 22

qⁿ (mod 23),

≡4 4

∞

X

n=0

b23 5²n+ 22

qⁿ+ 11

∞

X

n=0

b23(n)qⁿ

! + 11

∞

X

n=0

b23 5²n+ 22

qⁿ (mod 23),

≡4

∞

X

n=0

b23 5²n+ 22

qⁿ+ 21

∞

X

n=0

b23(n)qⁿ (mod 23).

Repeating the same process as above, it follows that

∞

X

n=0

b₂₃ 5²²n+11 12

5²²−1

!

qⁿ≡18

∞

X

n=0

b₂₃ 5²n+ 22

qⁿ+ 20

∞

X

n=0

b₂₃(n)qⁿ (mod 23).

∞

X

n=0

b₂₃ 5²²n+ 11 12

5²²−1

!

qⁿ≡21q²f₁¹⁰f₅¹²+ 3q³f₁⁴f₅¹⁸+ 13q⁴f₅²⁴ f₁² +

∞

X

n=0

b₂₃(n)qⁿ (mod 23).

Replacing f₁ and 1/f₁ using (2.9) and (2.11), we get

∞

X

n=0

b₂₃ 5²²n+11 12

5²²−1

! qⁿ

≡21q²f₅¹²f₂₅¹⁰ 1

!10

+ 3q³f₅¹⁸f₂₅⁴ 1

!4

+ 13q⁴f₅¹²f₂₅¹⁰ 1

R(q⁵)⁴ + q

R(q⁵)³ + 2q²

R(q⁵)² + 3q³

R(q⁵) + 5q⁴−3q⁵R q⁵

+ 2q⁶R q⁵2

−q⁷R q⁵3

+q⁸R q⁵4

!2

+

∞

X

n=0

b23(n)qⁿ (mod 23).

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80 5. Congruences for 65-Regular Partition

If we extract the terms involvingq⁵ⁿ⁺², divide byq², and substituteq byq^1/5, then by (2.10) we get

∞

X

n=0

b₂₃ 5²³n+7·5²³−11 12

!

qⁿ≡21f₁¹²f₅¹⁰ f₁¹²

f₅¹² + 20qf₁⁶

f₅⁶ + 16q²

!

+ 3f₁¹⁸f₅⁴ 18q

!

+ 13f₁¹²f₅¹⁰ 10qf₁⁶

f₅⁶ + 10q²

! +

∞

X

n=0

b₂₃(5n+ 2)qⁿ (mod 23), which, after rearranging the similar terms, yields

∞

X

n=0

b23 5²³n+7·5²³−11 12

!

qⁿ≡21f₁²⁴

f₅² + 6qf₁¹⁸f₅⁴+ 6q²f₁¹²f₅¹⁰+

∞

X

n=0

b23(5n+ 2)qⁿ (mod 23).

∞

X

n=0

b23 5²³n+ 7·5²³−11 12

!

qⁿ≡14

∞

X

n=0

b23(n)q⁵ⁿ⁺⁴ (mod 23). (4.20) Comparing the coefficients of the terms of the formq⁵ⁿ⁺⁴ in (4.20), we can conclude that for any n≥0,

b23 5²⁴n+11 12

5²⁴−1

!

≡14b23(n) (mod 23). (4.21)

Now (1.5) follows from (4.21), by mathematical induction.

On the other hand, comparing the coefficients of the terms of the form q^5n+rwherer∈ {0,1,2,3}

in (4.20), we can conclude that for anyn≥0,

b₂₃ 5²³(5n+r) +7·5²³−11 12

!

≡0 (mod 23), or,

b23 5²⁴n+ 1 12

5²³(12r+ 7)−11

!

≡0 (mod 23). (4.22)

Now (1.6) follows from (1.5) and (4.22).

5. Congruences for 65-Regular Partition

∞

X

n=0

b₆₅(n)qⁿ= f₆₅

f₁ . (5.23)

Replacing 1/f₁ using (2.11), we get

∞

X

n=0

b₆₅(n)qⁿ=f₆₅·f₂₅⁵ f₅⁶

1

R(q⁵)⁴ + q

R(q⁵)³ + 2q²

R(q⁵)² + 3q³

R(q⁵) + 5q⁴

−3q⁵R q⁵

+ 2q⁶R q⁵2

−q⁷R q⁵3

+q⁸R q⁵4

! .