On 3- and 9-Regular Cubic Partitions

D. S. Gireesh and C. Shivashankar

Department of Mathematics M. S. Ramaiah University of Applied Sciences

Bengaluru-560 058 Karnataka India

  gireeshdap@gmail.com
 shankars224@gmail.com

M. S. Mahadeva Naika Department of Mathematics Bengaluru Central University

Central College Campus

Bengaluru-560 001 Karnataka India

msmnaika@rediffmail.com

Abstract

Let a_{3}(n) and a_{9}(n) be 3- and 9-regular cubic partitions of n. In this paper, we establish several infinite families of congruences modulo powers of 3. For example, for all non-negative integers n and α we find that

a_{3} (3^{2α}n + \frac{3^{2α} − 1}{4}) ≡ 0 (mod 3^{α})

and a_{9} (3^{α+1}n + 3^{α+1} − 1) ≡ 0 (mod 3^{α+1}).

1 Introduction

A partition of a positive integer n is a non-increasing sequence of positive integers whose sum is n. Let p(n) denote the number of partitions of n and the generating function is

∑_{n≥0} p(n)q^{n} = \frac{1}{f_{1}} ,

where, here and throughout the paper, we set

f_{k} = (q^{k}; q^{k})_{∞} = ∏_{m=1}^{∞}(1 − q^{km}).

Chan [1] studied the cubic partition function denoted by a(n), whose generating function is

∑_{n≥0} a(n)q^{n} = \frac{1}{f_{1}f_{2}} .

He found the generating function

∑_{n≥0} a(3n + 2)q^{n} = 3\frac{f_{3}^{3}f_{6}^{3}}{f_{1}^{4}f_{2}^{4}} ,

which readily implies that

a(3n + 2) ≡ 0 (mod 3).

Chan [2] established infinite family of congruence modulo powers of 3 for a(n). For each n, k ≥ 1, he proved that

a(3^{k}n + c_{k}) ≡ 0 (mod 3^{k+δ(k)}),  (1)

where c_{k} is the reciprocal modulo 3^{k} of 8 and

δ(k) := \begin{cases}1, & if k is even; \\ 0, & k is odd. \end{cases}

Zhao and Zhong [9] studied cubic partition pairs denoted by b(n), the generating function satisfied by b(n) is

∑_{n≥0} b(n)q^{n} = \frac{1}{f_{1}^{2}f_{2}^{2}} .  (2)

For each n ≥ 0, they found Ramanujan’s type congruences for b(n), namely

b(5n + 4) ≡ 0 (mod 5),
b(7n + i) ≡ 0 (mod 7),
b(9n + 7) ≡ 0 (mod 9),

where i ∈ {2, 3, 4, 6}.

Lin [6] studied the cubic partition pairs and established the following congruences modulo 27:

b(27n + 16) ≡ 0 (mod 27),  (3)
b(27n + 25) ≡ 0 (mod 27),  (4)
b(81n + 61) ≡ 0 (mod 27).  (5)

Also, he proposed the following conjectures:

Conjecture 1. For each n ≥ 0,

b(81n + 61) ≡ 0 (mod 81).  (6)

Conjecture 2.

∑_{n≥0} b(81n + 7)q^{n} ≡ 9\frac{f_{2}f_{3}^{2}}{f_{6}} (mod 81),  (7)

∑_{n≥0} b(81n + 34)q^{n} ≡ 36\frac{f_{1}f_{6}^{2}}{f_{3}} (mod 81).  (8)

The above conjectures were proved by Gireesh and Naika [4], Chern [3], and Lin et al. [7].

Motivated by the above results, in this paper, we study 3- and 9-regular cubic partitions, which are defined as follow:

• Let a_{3}(n) denote the number of 3-regular cubic partitions of n, whose generating func-

tion is

∑_{n≥0} a_{3}(n)q^{n} = \frac{f_{3}f_{6}}{f_{1}f_{2}} .  (9)

• Let a_{9}(n) denote the number of 9-regular cubic partitions of n, whose generating func-

tion is

∑_{n≥0} a_{9}(n)q^{n} = \frac{f_{9}f_{18}}{f_{1}f_{2}} .  (10)

We shall show that

∑_{n≥0} a_{3}(3n + 2)q^{n} = 3\frac{f_{3}^{3}f_{6}^{3}}{f_{1}^{3}f_{2}^{3}}  (11)

and

∑_{n≥0} a_{9}(3n + 2)q^{n} = 3\frac{f_{3}^{4}f_{6}^{4}}{f_{1}^{4}f_{2}^{4}} .  (12)

These are analogous to Ramanujan’s most beautiful identities [8, pp. 239, 243]

∑_{n≥0} p(5n + 4)q^{n} = 5\frac{f_{5}^{5}}{f_{1}^{6}}  (13)

and

∑_{n≥0} p(7n + 5)q^{n} = 7\frac{f_{7}^{3}}{f_{1}^{4}} + 49q \frac{f_{7}^{7}}{f_{1}^{8}} .  (14)

We also obtain infinite families of congruences modulo powers of 3 for a_{3}(n) and a_{9}(n), which are stated in the following theorems:

Theorem 3. For each n, α ≥ 0,

a_{3} (3^{2α}n + \frac{3^{2α} − 1}{4}) ≡ 0 (mod 3^{α}),  (15)
a_{3} (3^{2α+1}n + \frac{3^{2α+2} − 1}{4}) ≡ 0 (mod 3^{α+1}),  (16)
a_{3} (3^{2α+2}n + \frac{7 × 3^{2α+1} − 1}{4}) ≡ 0 (mod 3^{α+2}),  (17)
a_{3} (3^{2α+2}n + \frac{11 × 3^{2α+1} − 1}{4}) ≡ 0 (mod 3^{α+2}).  (18)

Theorem 4. For each n, α ≥ 0,

a_{9} (3^{α+1}n + 3^{α+1} − 1) ≡ 0 (mod 3^{α+1}).  (19)

The results (15)–(19) are analogous to Ramanujan’s congruences modulo powers of 5 [5], for n, α ≥ 0,

p (5^{2α+1}n + \frac{19 × 5^{2α+1} + 1}{24}) ≡ 0 (mod 5^{2α+1})  (20)

and

p (5^{2α+2}n + \frac{23 × 5^{2α+2} + 1}{24}) ≡ 0 (mod 5^{2α+2}).  (21)

2 Preliminaries

Define

ζ = \frac{f_{1}f_{2}}{qf_{9}f_{18}}

and

T = \frac{f_{3}^{4}f_{6}^{4}}{q^{3}f_{9}^{4}f_{18}^{4}} .

Let H be the “huffing” operator defined by

H (∑ a_{n}q^{n}) = ∑ a_{3n}q^{3n}.

From Chan [2, (11)–(19)], for each i ≥ 1, we have

H (\frac{1}{ζ^{i}} ) = ∑_{j=1}^{i} \frac{m_{i,j}}{T^{j}} ,  (22)

where m_{i,j}’s are defined in the following matrix. The m_{i,j} form a matrix M, the first nine rows of which are

M = (\begin{matrix}3 & 0 & 0 & 0 & 0 & 0 & 0 & ⋯ \\ 2 & 3^{3} & 0 & 0 & 0 & 0 & 0 & ⋯ \\ 1 & 3^{3} & 3^{5} & 0 & 0 & 0 & 0 & ⋯ \\ 0 & 2 · 3^{2} & 2^{2} · 3^{4} & 3^{7} & 0 & 0 & 0 & ⋯ \\ 0 & 5 & 2 · 3^{3} · 5 & 3^{6} · 5 & 3^{9} & 0 & 0 & ⋯ \\ 0 & 1 & 2 · 3^{2} · 7 & 3^{6} · 5 & 2 · 3^{9} & 3^{11} & 0 & ⋯ \\ 0 & 0 & 2 · 3 · 7 & 2^{2} · 3^{4} · 7 & 3^{8} · 7 & 3^{10} · 7 & 3^{13} & ⋯ \\ 0 & 0 & 2^{3} & 2 · 3^{3} · 19 & 2^{4} · 3^{7} & 2^{2} · 3^{9} · 7 & 2^{3} · 3^{12} & ⋯ \\ 0 & 0 & 1 & 2^{2} · 3^{4} & 3^{9} & 3^{9} · 5^{2} & 2^{2} · 3^{13} & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix})  (23)

and for i ≥ 4, m_{i,1} = 0, and for j ≥ 2,

m_{i,j} = 9m_{i−1,j−1} + 3m_{i−2,j−1} + m_{i−3,j−1}.  (24)

In fact m_{4i−3,j} = 0 for j ≤ i − 1, so we can write

H (\frac{1}{ζ^{4i−3}} ) = ∑_{j=i}^{4i−3} \frac{m_{4i−3,j}}{T^{j}} = ∑_{j=1}^{3i−2} \frac{m_{4i−3,i+j−1}}{T^{i+j−1}} = ∑_{j=1}^{3i−2} \frac{a_{i,j}}{T^{i+j−1}} ,  (25)

where

a_{i,j} = m_{4i−3,i+j−1}.  (26)

Similarly, m_{4i−1,j} = 0 if j ≤ i − 1, so we can write

H (\frac{1}{ζ^{4i−1}} ) = ∑_{j=i}^{4i−1} \frac{m_{4i−1,j}}{T^{j}} = ∑_{j=1}^{3i} \frac{m_{4i−1,i+j−1}}{T^{i+j−1}} = ∑_{j=1}^{3i} \frac{b_{i,j}}{T^{i+j−1}} ,  (27)

where

b_{i,j} = m_{4i−1,i+j−1}.  (28)

And m_{4i,j} = 0 if j ≤ i, so we can write

H (\frac{1}{ζ^{4i}} ) = ∑_{j=1+i}^{4i} \frac{m_{4i,j}}{T^{j}} = ∑_{j=1}^{3i} \frac{m_{4i,i+j}}{T^{i+j}} = ∑_{j=1}^{3i} \frac{c_{i,j}}{T^{i+j}} ,  (29)

where

c_{i,j} = m_{4i,i+j}.  (30)

We can write (25) as

H ((q \frac{f_{9}f_{18}}{f_{1}f_{2}} )^{4i−3}) = ∑_{j=1}^{3i−2} a_{i,j} (q^{3} \frac{f_{9}^{4}f_{18}^{4}}{f_{3}^{4}f_{6}^{4}} )^{i+j−1} ,  (31)

which can be rearranged to

H (q^{i−3} (\frac{f_{3}f_{6}}{f_{1}f_{2}} )^{4i−3}) = ∑_{j=1}^{3i−2} a_{i,j}q^{3j−3} (\frac{f_{9}f_{18}}{f_{3}f_{6}} )^{4j−1} .  (32)

The equation (27) can be written as

H ((q \frac{f_{9}f_{18}}{f_{1}f_{2}} )^{4i−1}) = ∑_{j=1}^{3i} b_{i,j} (q^{3} \frac{f_{9}^{4}f_{18}^{4}}{f_{3}^{4}f_{6}^{4}} )^{i+j−1} ,  (33)

which implies that

H (q^{i−1} (\frac{f_{3}f_{6}}{f_{1}f_{2}} )^{4i−1}) = ∑_{j=1}^{3i} b_{i,j}q^{3j−3} (\frac{f_{9}f_{18}}{f_{3}f_{6}} )^{4j−3} .  (34)

Similarly (29) is

H ((q \frac{f_{9}f_{18}}{f_{1}f_{2}} )^{4i}) = ∑_{j=1}^{3i} c_{i,j} (q^{3} \frac{f_{9}^{4}f_{18}^{4}}{f_{3}^{4}f_{6}^{4}} )^{i+j} ,  (35)

and this can be rearranged to

H (q^{i} (\frac{f_{3}f_{6}}{f_{1}f_{2}} )^{4i}) = ∑_{j=1}^{3i} c_{i,j}q^{3j} (\frac{f_{9}f_{18}}{f_{3}f_{6}} )^{4j} .  (36)

3 Generating functions

In this section, we deduce generating functions that are useful in proving our main results.

Theorem 5. For each α ≥ 0,

∑_{n≥0} a_{3} (3^{2α}n + \frac{3^{2α} − 1}{4}) q^{n} = ∑_{i≥1} x_{2α,i}q^{i−1} (\frac{f_{3}f_{6}}{f_{1}f_{2}} )^{4i−3}  (37)

and

∑_{n≥0} a_{3} (3^{2α+1}n + \frac{3^{2α+2} − 1}{4}) q^{n} = ∑_{i≥1} x_{2α+1,i}q^{i−1} (\frac{f_{3}f_{6}}{f_{1}f_{2}} )^{4i−1} ,  (38)

where the coefficient vectors x_{α} = (x_{α,1}, x_{α,2}, …) are given by

x_{0} = (x_{0,1}, x_{0,2}, x_{0,3}, …) = (1, 0, 0, …),  (39)

and

x_{α+1} = x_{α}A if α is even,  (40)
x_{α+1} = x_{α}B if α is odd,  (41)

where A = (a_{i,j})_{i,j≥1} and B = (b_{i,j})_{i,j≥1}. Proof. The identity (9) is the α = 0 case of (37).

Suppose (37) holds for some α ≥ 0. Then

∑_{n≥0} a_{3} (3^{2α}n + \frac{3^{2α} − 1}{4}) q^{n} = ∑_{i≥1} x_{2α,i}q^{i−1} (\frac{f_{3}f_{6}}{f_{1}f_{2}} )^{4i−3} ,  (42)

which is equivalent to

∑_{n≥0} a_{3} (3^{2α}n + \frac{3^{2α} − 1}{4}) q^{n−2} = ∑_{i≥1} x_{2α,i}q^{i−3} (\frac{f_{3}f_{6}}{f_{1}f_{2}} )^{4i−3} .  (43)

Applying the operator H to (43), we find that

∑_{n≥0} a_{3} (3^{2α}(3n + 2) + \frac{3^{2α} − 1}{4}) q^{3n} = ∑_{i≥1} x_{2α,i}H (q^{i−3} (\frac{f_{3}f_{6}}{f_{1}f_{2}} )^{4i−3})

= ∑_{i≥1} x_{2α,i} ∑_{j=1}^{3i−2} a_{i,j}q^{3j−3} (\frac{f_{9}f_{18}}{f_{3}f_{6}} )^{4j−1}

= ∑_{j≥1} (∑_{i≥1} x_{2α,i}a_{i,j}) q^{3j−3} (\frac{f_{9}f_{18}}{f_{3}f_{6}} )^{4j−1}

= ∑_{j≥1} x_{2α+1,j}q^{3j−3} (\frac{f_{9}f_{18}}{f_{3}f_{6}} )^{4j−1} ,

which implies equation (38).

Now suppose (38) holds for some α ≥ 0. Then

∑_{n≥0} a_{3} (3^{2α+1}n + \frac{3^{2α+2} − 1}{4}) q^{n} = ∑_{i≥1} x_{2α+1,i}q^{i−1} (\frac{f_{3}f_{6}}{f_{1}f_{2}} )^{4i−1} .  (44)

Applying the operator H to (44), we find that

∑_{n≥0} a_{3} (3^{2α+1}(3n) + \frac{3^{2α+2} − 1}{4}) q^{3n} = ∑_{i≥1} x_{2α+1,i}H (q^{i−1} (\frac{f_{3}f_{6}}{f_{1}f_{2}} )^{4i−1})

= ∑_{i≥1} x_{2α+1,i} ∑_{j=1}^{3i} b_{i,j}q^{3j−3} (\frac{f_{9}f_{18}}{f_{3}f_{6}} )^{4j−3}

= ∑_{j≥1} (∑_{i≥1} x_{2α+1,i}b_{i,j}) q^{3j−3} (\frac{f_{9}f_{18}}{f_{3}f_{6}} )^{4j−3}

= ∑_{j≥1} x_{2α+2,j}q^{3j−3} (\frac{f_{9}f_{18}}{f_{3}f_{6}} )^{4j−3} .

After simplification, we obtain (37) with α + 1 in place of α. This completes the proof of (37) and (38) by induction.

Theorem 6. For each α ≥ 0,

∑_{n≥0} a_{9} (3^{α+1}n + 3^{α+1} − 1) q^{n} = ∑_{i≥1} y_{α,i}q^{i−1} (\frac{f_{3}f_{6}}{f_{1}f_{2}} )^{4i}  (45)

where the coefficient vectors Y_{α} = (y_{α,1}, y_{α,2}, …) are given by

Y_{0} = (y_{0,1}, y_{0,2}, y_{0,3}, …) = (3, 0, 0, …),  (46)

and

Y_{α+1} = Y_{α}C,  (47)

where C = (c_{i,j})_{i,j≥1}.

Proof. We prove this by induction on α. The identity (12) is the α = 0 case of (45). Suppose (45) holds for some α ≥ 0. Then

∑_{n≥0} a_{9} (3^{α+1}n + 3^{α+1} − 1) q^{n} = ∑_{i≥1} y_{α,i}q^{i−1} (\frac{f_{3}f_{6}}{f_{1}f_{2}} )^{4i} ,  (48)

which is equivalent to

∑_{n≥0} a_{9} (3^{α+1}n + 3^{α+1} − 1) q^{n−2} = q^{−3} ∑_{i≥1} y_{α,i}q^{i} (\frac{f_{3}f_{6}}{f_{1}f_{2}} )^{4i} .  (49)

Applying the operator H to (49), we find that

∑_{n≥0} a_{9} (3^{α+1}(3n + 2) + 3^{α+1} − 1) q^{3n} = q^{−3} ∑_{i≥1} y_{α,i}H (q^{i} (\frac{f_{3}f_{6}}{f_{1}f_{2}} )^{4i})

= ∑_{i≥1} y_{α,i} ∑_{j=1}^{3i} c_{i,j}q^{3j−3} (\frac{f_{9}f_{18}}{f_{3}f_{6}} )^{4j}

= ∑_{j≥1} (∑_{i≥1} y_{α,i}c_{i,j}) q^{3j−3} (\frac{f_{9}f_{18}}{f_{3}f_{6}} )^{4j}

= ∑_{j≥1} y_{α+1,j}q^{3j−3} (\frac{f_{9}f_{18}}{f_{3}f_{6}} )^{4j} ,

which implies that

∑_{n≥0} a_{9} (3^{α+2}n + 3^{α+2} − 1) q^{n} = ∑_{j≥1} y_{α+1,j}q^{j−1} (\frac{f_{3}f_{6}}{f_{1}f_{2}} )^{4j} ,  (50)

which is (45) with α + 1 for α .

4 Congruences

Let ν(N) be the largest power of 3 that divides N. Note that ν(0) = +∞. Proof of Theorem 3. It follows from (23) and (24) that

ν(m_{i.j}) ≥ 3j − i − 1,  (51)

and from (26), (28) and (51),

ν(a_{i.j}) ≥ 3(i + j − 1) − (4i − 3) − 1 = 3j − i − 1  (52)

and

ν(b_{i.j}) ≥ 3(i + j − 1) − (4i − 1) − 1 = 3j − i − 3.  (53)

It is not hard to show that

ν(x_{2α,j}) ≥ α + 3j − 4  (54)

and

ν(x_{2α+1,j}) ≥ α + 1 + 3(j − 1).  (55)

The identity (54) is true for α = 0, by (39).

Suppose (54) is true for some α ≥ 0. Then

ν(x_{2α+1,j}) ≥ min_{i≥1} (ν(x_{2α,i}) + ν(a_{i,j}))
= ν(x_{2α,1}) + ν(a_{1,j})
≥ α + 3j − 2
≥ α + 1 + 3(j − 1),

which is (55).

Now suppose (55) is true for all α ≥ 0. Then

ν(x_{2α+2,j}) ≥ min_{i≥1} (ν(x_{2α+1,i}) + ν(b_{i,j}))
= ν(x_{2α+1,1}) + ν(b_{1,j})
≥ α + 1 + 3j − 4,

which is (54) with α+1 in place of α. This completes the proof of (54) and (55) by induction.

The congruence (15) follows from (37) together with (54), and the congruence (16) follows from (38) together with (55).

From (38) and (55), we have

∑_{n≥0} a_{3} (3^{2α+1}n + \frac{3^{2α+2} − 1}{4}) q^{n} ≡ 3^{α+1} \frac{f_{3}^{3}f_{6}^{3}}{f_{1}^{3}f_{2}^{3}} (mod 3^{α+4}).  (56)

By the binomial theorem, it is easy to see that

f_{1}^{3} ≡ f_{3} (mod 3).  (57)

In view of (57), the congruence (56) can be expressed as

∑_{n≥0} a_{3} (3^{2α+1}n + \frac{3^{2α+2} − 1}{4}) q^{n} ≡ 3^{α+1} \frac{f_{9}f_{18}}{f_{3}f_{6}} (mod 3^{α+2}).  (58)

Equating the coefficients of q^{3n+1} and q^{3n+2} in (58), we obtain (17) and (18), respectively.

Proof of Theorem 4. It follows from (30) and (51) that

ν(c_{i.j}) ≥ 3(i + j) − 4i − 1 = 3j − i − 1.  (59)

It is not hard to show that

ν(y_{α,j}) ≥ α + 1 + 3(j − 1).  (60)

The identity (60) is true for α = 0, by (46).

Suppose (60) is true for some α ≥ 0. Then

ν(y_{α+1,j}) ≥ min_{i≥1} (ν(y_{α,i}) + ν(c_{i,j}))
= ν(y_{α,1}) + ν(c_{1,j})
≥ α + 1 + 3j − 2
≥ α + 2 + 3(j − 1),

which is (60) with α + 1 for α.

The congruence (19) follows from (45) together with (60).

5 Acknowledgments

The authors would like to thank the anonymous referee for his/her helpful suggestions and comments.

References

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[5] M. D. Hirschhorn and D. C. Hunt, A simple proof of the Ramanujan conjecture for powers of 5, J. Reine Angew. Math. 326 (1981), 1–17.

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2010 Mathematics Subject Classification: Primary 05A17; Secondary 11P83. Keywords: partition, 3-regular cubic partition, 9-regular cubic partition, congruence.

(Concerned with sequences \underline{A335602}, \underline{A335604}.)

Received November 27 2019. Revised versions received June 4 2020; June 17 2020; June 18 2020. Published in Journal of Integer Sequences, June 24 2020.

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