Arithmetic properties of (2, 3)-regular overcubic bipartitions

S. Shivaprasada Nayaka can be contacted at: shivprasadnayaks@gmail.com

Purpose

Let ${\bar{b}}_{2,3} (n)$ ⁠, which enumerates the number of (2, 3)-regular overcubic bipartition of n. The purpose of the paper is to describe some congruences modulo 8 for ${\bar{b}}_{2,3} (n)$ ⁠. For example, for each α ≥ 0 and n ≥ 1, ${\bar{b}}_{2,3} (8 n + 5) \equiv 0 (\mod 8),$ ${\bar{b}}_{2,3} (2 \cdot 3^{α + 3} n + 4 \cdot 3^{α + 2}) \equiv 0 (\mod 8) .$

Design/methodology/approach

H.C. Chan has studied the congruence properties of cubic partition function a(n), which is defined by $\sum_{n = 0}^{\infty} a (n) q^{n} = \frac{1}{{(q; q)}_{\infty} {(q^{2}; q^{2})}_{\infty}}$ ⁠.

Findings

To establish several congruence modulo 8 for ${\bar{b}}_{2,3} (n)$ ⁠, here the author keeps to the classical spirit of q-series techniques in the proofs.

Originality/value

The results established in the work are extension to those proved in ℓ-regular cubic partition pairs.

1. Introduction

A partition λ of a natural number n is a finite non-increasing sequence of positive integer parts λ_i (1 ≤ i ≤ m) such that

n = λ_{1} + λ_{2} + λ_{3} + \dots + λ_{m} .

In this case, we write |λ| = n. The number of partitions of n is denoted by p(n) and the generating function is given by as follows:

\sum_{n = 0}^{\infty} p (n) q^{n} = \frac{1}{{(q; q)}_{\infty}} .

Ramanujan’s three famous congruences of p(n) are as follows:

\begin{array}{l} p (5 n + 4) & \equiv 0 (\mod 5), \\ p (7 n + 5) & \equiv 0 (\mod 7), \\ p (11 n + 6) & \equiv 0 (\mod 11) . \end{array}

In [1–3], H.C. Chan has studied the congruence properties of cubic partition function a(n), which is defined by as follows:

\sum_{n = 0}^{\infty} a (n) q^{n} = \frac{1}{{(q; q)}_{\infty} {(q^{2}; q^{2})}_{\infty}} .

B. Kim [4] studied its overpartition analog, the overcubic partition function $\bar{a} (n)$ ⁠, which is defined by as follows:

\sum_{n = 0}^{\infty} \bar{a} (n) q^{n} = \frac{{(- q; q)}_{\infty} {(- q^{2}; q^{2})}_{\infty}}{{(q; q)}_{\infty} {(q^{2}; q^{2})}_{\infty}} .

In [5], M.D. Hirschhorn obtained the results satisfied by $\bar{a} (n)$ ⁠, which appeared in Kim’s paper [4], and Sellers [6] has proved a number of arithmetic properties of $\bar{a} (n)$ by employing elementary generating function methods. Zhao and Zhong [7] studied cubic partition pairs, which are denoted by b(n), and the generating function is as follows:

\sum_{n = 0}^{\infty} b (n) q^{n} = \frac{1}{{(q; q)}_{\infty}^{2} {(q^{2}; q^{2})}_{\infty}^{2}} .

Recently, Kim [8] studied congruence properties of $\bar{b} (n)$ ⁠, which denotes overcubic partition pairs of n, whose generating function is given by as follows:

\sum_{n = 0}^{\infty} \bar{b} (n) q^{n} = \frac{{(- q; q)}_{\infty}^{2} {(- q^{2}; q^{2})}_{\infty}^{2}}{{(q; q)}_{\infty}^{2} {(q^{2}; q^{2})}_{\infty}^{2}} .

More recently, Lin [9] studied various arithmetic properties of $\bar{b} (n)$ modulo 3 and 5. For example, for any α ≥ 2, n ≥ 0,

\bar{b} (3^{α} (3 n + 2)) \equiv 0 (\mod 3),

for α ≥ 0,

\bar{b} (380 \cdot 5^{α}) \equiv 0 (\mod 3) .

In [10], Naika and Nayaka have established some congruences for ℓ-regular cubic partition pairs. Let ${\bar{b}}_{2,3} (n)$ denote the number of (2, 3)-regular overcubic bipartitions of n, whose generating function is given by as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (n) q^{n} = \frac{{(q^{2}; q^{2})}_{\infty}^{2} {(q^{3}; q^{3})}_{\infty}^{4} {(q^{4}; q^{4})}_{\infty}^{4} {(q^{24}; q^{24})}_{\infty}^{2}}{{(q; q)}_{\infty}^{4} {(q^{6}; q^{6})}_{\infty}^{2} {(q^{8}; q^{8})}_{\infty}^{2} {(q^{12}; q^{12})}_{\infty}^{4}} .

(1.1)

In this paper, we establish several congruences modulo 8 for ${\bar{b}}_{2,3} (n)$ ⁠. These results can be found in Theorems (3.1), and we keep to the classical spirit of q-series techniques in our proofs.

2. Preliminaries

For |ab| < 1, Ramanujan’s general theta function f(a, b) is defined as follows:

f (a, b) ≔ \sum_{n = - \infty}^{\infty} a^{n (n + 1) / 2} b^{n (n - 1) / 2} .

Some special cases of f(a, b) are as follows:

φ (q) ≔ f (q, q) = \sum_{n = - \infty}^{\infty} q^{n^{2}} = {(- q; q^{2})}_{\infty}^{2} {(q^{2}; q^{2})}_{\infty},

ψ (q) ≔ f (q, q^{3}) = \sum_{n = 0}^{\infty} q^{n (n + 1) / 2} = \frac{{(q^{2}; q^{2})}_{\infty}}{{(q; q^{2})}_{\infty}}

and

f (- q) ≔ f (- q, - q^{2}) = \sum_{n = - \infty}^{\infty} {(- 1)}^{n} q^{n (3 n - 1) / 2} = {(q; q)}_{\infty} .

Where the product representation of f(a, b) arises from Jacobi’s triple product identity [11, p. 35, Entry 19] as follows:

f (a, b) = {(- a; a b)}_{\infty} {(- b; a b)}_{\infty} {(a b; a b)}_{\infty} .

The following dissection formulas to prove our main results.

Lemma 2.1.

For each prime p and n ≥ 1,

{(q; q)}_{\infty}^{p^{n}} \equiv {(q^{p}; q^{p})}_{\infty}^{p^{n - 1}} (\mod p^{n}) .

(2.1)

Lemma 2.2.

The following 2-dissections holds:

\begin{align} {(q; q)}_{\infty}^{2} & = \frac{{(q^{2}; q^{2})}_{\infty} {(q^{8}; q^{8})}_{\infty}^{5}}{{(q^{4}; q^{4})}_{\infty}^{2} {(q^{16}; q^{16})}_{\infty}^{2}} - 2 q \frac{{(q^{2}; q^{2})}_{\infty} {(q^{16}; q^{16})}_{\infty}^{2}}{{(q^{8}; q^{8})}_{\infty}}, \end{align}

(2.2)

\begin{align} \frac{1}{{(q; q)}_{\infty}^{2}} & = \frac{{(q^{8}; q^{8})}_{\infty}^{5}}{{(q^{2}; q^{2})}_{\infty}^{5} {(q^{16}; q^{16})}_{\infty}^{2}} + 2 q \frac{{(q^{4}; q^{4})}_{\infty}^{2} {(q^{16}; q^{16})}_{\infty}^{2}}{{(q^{2}; q^{2})}_{\infty}^{5} {(q^{8}; q^{8})}_{\infty}} . \end{align}

(2.3)

Lemma (2.2) is a consequence of dissection formulas of Ramanujan, which is collected in Berndt’s book [11, p. 40, Entry 25].

Lemma 2.3.

The following 2-dissections holds:

\begin{align} \frac{{(q^{3}; q^{3})}_{\infty}^{3}}{{(q; q)}_{\infty}} & = \frac{{(q^{4}; q^{4})}_{\infty}^{3} {(q^{6}; q^{6})}_{\infty}^{2}}{{(q^{2}; q^{2})}_{\infty}^{2} {(q^{12}; q^{12})}_{\infty}} + q \frac{{(q^{12}; q^{12})}_{\infty}^{3}}{{(q^{4}; q^{4})}_{\infty}}, \end{align}

(2.4)

\begin{align} \frac{{(q; q)}_{\infty}^{3}}{{(q^{3}; q^{3})}_{\infty}} & = \frac{{(q^{4}; q^{4})}_{\infty}^{3}}{{(q^{12}; q^{12})}_{\infty}} - 3 q \frac{{(q^{2}; q^{2})}_{\infty}^{2} {(q^{12}; q^{12})}_{\infty}^{3}}{{(q^{4}; q^{4})}_{\infty} {(q^{6}; q^{6})}_{\infty}^{2}}, \end{align}

(2.5)

\begin{align} \frac{{(q^{3}; q^{3})}_{\infty}}{{(q; q)}_{\infty}^{3}} & = \frac{{(q^{4}; q^{4})}_{\infty}^{6} {(q^{6}; q^{6})}_{\infty}^{3}}{{(q^{2}; q^{2})}_{\infty}^{9} {(q^{12}; q^{12})}_{\infty}^{2}} + 3 q \frac{{(q^{4}; q^{4})}_{\infty}^{2} {(q^{6}; q^{6})}_{\infty} {(q^{12}; q^{12})}_{\infty}^{2}}{{(q^{2}; q^{2})}_{\infty}^{7}} . \end{align}

(2.6)

Hirschhorn, Garvan and Borwein [12] proved (2.4) and (2.5). For proof of (2.6), see [13].

Lemma 2.4.

The following 2-dissections holds:

\begin{array}{l} \frac{1}{{(q; q)}_{\infty} {(q^{3}; q^{3})}_{\infty}} & = \frac{{(q^{8}; q^{8})}_{\infty}^{2} {(q^{12}; q^{12})}_{\infty}^{5}}{{(q^{2}; q^{2})}_{\infty}^{2} {(q^{4}; q^{4})}_{\infty} {(q^{6}; q^{6})}_{\infty}^{4} {(q^{24}; q^{24})}_{\infty}^{2}} \end{array}

\begin{align} + q \frac{{(q^{4}; q^{4})}_{\infty}^{5} {(q^{24}; q^{24})}_{\infty}^{2}}{{(q^{2}; q^{2})}_{\infty}^{4} {(q^{6}; q^{6})}_{\infty}^{2} {(q^{8}; q^{8})}_{\infty}^{2} {(q^{12}; q^{12})}_{\infty}} \end{align}

(2.7)

\begin{array}{l} {(q; q)}_{\infty} {(q^{3}; q^{3})}_{\infty} & = \frac{{(q^{2}; q^{2})}_{\infty} {(q^{8}; q^{8})}_{\infty}^{2} {(q^{12}; q^{12})}_{\infty}^{4}}{{(q^{4}; q^{4})}_{\infty}^{2} {(q^{6}; q^{6})}_{\infty} {(q^{24}; q^{24})}_{\infty}^{2}} \end{array}

\begin{align} - q \frac{{(q^{4}; q^{4})}_{\infty}^{4} {(q^{6}; q^{6})}_{\infty} {(q^{24}; q^{24})}_{\infty}^{2}}{{(q^{2}; q^{2})}_{\infty} {(q^{8}; q^{8})}_{\infty}^{2} {(q^{12}; q^{12})}_{\infty}^{2}} . \end{align}

(2.8)

Eqn (2.7) was proved by Baruah and Ojah [14]. Replacing q by − q in (2.7) and using the fact that ${(- q; - q)}_{\infty} = \frac{f_{2}^{3}}{f_{1} f_{4}}$ ⁠, we get (2.8).

Lemma 2.5.

The following 3-dissection hold:

\begin{align} {(q; q)}_{\infty} {(q^{2}; q^{2})}_{\infty} & = \frac{{(q^{6}; q^{6})}_{\infty} {(q^{9}; q^{9})}_{\infty}^{4}}{{(q^{3}; q^{3})}_{\infty} {(q^{18}; q^{18})}_{\infty}^{2}} - q {(q^{9}; q^{9})}_{\infty} {(q^{18}; q^{18})}_{\infty} \\ - 2 q^{2} \frac{{(q^{3}; q^{3})}_{\infty} {(q^{18}; q^{18})}_{\infty}^{4}}{{(q^{6}; q^{6})}_{\infty} {(q^{9}; q^{9})}_{\infty}^{2}} . \end{align}

(2.9)

One can see this identity in [15].

Lemma 2.6.

[11, p. 345, Entry 1 (iv)]. We have the following 3-dissection

{(q; q)}_{\infty}^{3} = {(q^{9}; q^{9})}_{\infty}^{3} (ζ^{- 1} - 3 q + 4 q^{3} ζ^{2}),

(2.10)

where

ζ = \frac{{(q^{3}; q^{3})}_{\infty} {(q^{18}; q^{18})}_{\infty}^{3}}{{(q^{6}; q^{6})}_{\infty} {(q^{9}; q^{9})}_{\infty}^{3}} .

(2.11)

3. Congruences modulo 8 for ${\bar{b}}_{2,3} (n)$

Theorem 3.1.

For each α ≥ 0 and n ≥ 1, we have

{\bar{b}}_{2,3} (6 n + 5) \equiv 0 (\mod 8),

(3.1)

{\bar{b}}_{2,3} (8 n + 5) \equiv 0 (\mod 8),

(3.2)

{\bar{b}}_{2,3} (12 n + 7) \equiv 0 (\mod 8),

(3.3)

{\bar{b}}_{2,3} (18 n + 15) \equiv 0 (\mod 8),

(3.4)

{\bar{b}}_{2,3} (36 n + 21) \equiv 0 (\mod 8),

(3.5)

{\bar{b}}_{2,3} (72 n + 38) \equiv 0 (\mod 8),

(3.6)

{\bar{b}}_{2,3} (72 n + 51) \equiv 0 (\mod 8),

(3.7)

{\bar{b}}_{2,3} (72 n + 57) \equiv 0 (\mod 8),

(3.8)

{\bar{b}}_{2,3} (216 n + 99) \equiv 0 (\mod 8),

(3.9)

{\bar{b}}_{2,3} (216 n + 171) \equiv 0 (\mod 8),

(3.10)

{\bar{b}}_{2,3} (8 \cdot 9^{α + 2} n + 57 \cdot 9^{α + 1}) \equiv 0 (\mod 8),

(3.11)

{\bar{b}}_{2,3} (8 \cdot 9^{α + 2} n + 35 \cdot 9^{α + 2}) \equiv 0 (\mod 8),

(3.12)

{\bar{b}}_{2,3} (2 \cdot 3^{α + 3} n + 4 \cdot 3^{α + 2}) \equiv 0 (\mod 8),

(3.13)

{\bar{b}}_{2,3} (72 n + 3) \equiv b_{2,3} (24 n + 1) (\mod 8),

(3.14)

{\bar{b}}_{2,3} (36 n + 3) \equiv b_{2,3} (12 n + 1) (\mod 8) .

(3.15)

Proof.

Employing (2.4) and (2.5) in (1.1), we have

\begin{array}{l} \sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (n) q^{n} & = \frac{{(q^{4}; q^{4})}_{\infty}^{13} {(q^{6}; q^{6})}_{\infty}^{3} {(q^{24}; q^{24})}_{\infty}^{2}}{{(q^{2}; q^{2})}_{\infty}^{9} {(q^{8}; q^{8})}_{\infty}^{2} {(q^{12}; q^{12})}_{\infty}^{7}} \end{array}

\begin{align} + 4 q \frac{{(q^{4}; q^{4})}_{\infty}^{9} {(q^{6}; q^{6})}_{\infty} {(q^{24}; q^{24})}_{\infty}^{2}}{{(q^{2}; q^{2})}_{\infty}^{7} {(q^{8}; q^{8})}_{\infty}^{2} {(q^{12}; q^{12})}_{\infty}^{3}} \end{align}

(3.16)

\begin{align} + 3 q^{2} \frac{{(q^{4}; q^{4})}_{\infty}^{5} {(q^{12}; q^{12})}_{\infty} {(q^{24}; q^{24})}_{\infty}^{2}}{{(q^{2}; q^{2})}_{\infty}^{5} {(q^{6}; q^{6})}_{\infty} {(q^{8}; q^{8})}_{\infty}^{2}}, \end{align}

(3.17)

which implies the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (2 n + 1) q^{n} = 4 \frac{{(q^{2}; q^{2})}_{\infty}^{9} {(q^{3}; q^{3})}_{\infty} {(q^{12}; q^{12})}_{\infty}^{2}}{{(q; q)}_{\infty}^{7} {(q^{4}; q^{4})}_{\infty}^{2} {(q^{6}; q^{6})}_{\infty}^{3}} .

(3.18)

Invoking (2.1) in (3.18), we obtain the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (2 n + 1) q^{n} \equiv 4 {(q; q)}_{\infty} {(q^{3}; q^{3})}_{\infty} {(q^{2}; q^{2})}_{\infty} {(q^{6}; q^{6})}_{\infty} (\mod 8) .

(3.19)

Substituting (2.8) into (3.19), we get the generating function as follows:

\begin{align} \sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (2 n + 1) q^{n} \equiv & 4 \frac{{(q^{2}; q^{2})}_{\infty}^{2} {(q^{8}; q^{8})}_{\infty}^{2} {(q^{12}; q^{12})}_{\infty}^{4}}{{(q^{4}; q^{4})}_{\infty}^{2} {(q^{24}; q^{24})}_{\infty}^{2}} \\ + 4 q \frac{{(q^{6}; q^{6})}_{\infty}^{2} {(q^{4}; q^{4})}_{\infty}^{4} {(q^{24}; q^{24})}_{\infty}^{2}}{{(q^{8}; q^{8})}_{\infty}^{2} {(q^{12}; q^{12})}_{\infty}^{2}} (\mod 8) . \end{align}

(3.20)

Extracting the terms in which powers of q are congruent to 1 modulo 2 from (3.20), we have the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (4 n + 3) q^{n} \equiv 4 \frac{{(q^{3}; q^{3})}_{\infty}^{2} {(q^{2}; q^{2})}_{\infty}^{4} {(q^{12}; q^{12})}_{\infty}^{2}}{{(q^{4}; q^{4})}_{\infty}^{2} {(q^{6}; q^{6})}_{\infty}^{2}} (\mod 8) .

(3.21)

Invoking () in , we obtain as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (4 n + 3) q^{n} \equiv 4 {(q^{3}; q^{3})}_{\infty}^{2} {(q^{6}; q^{6})}_{\infty}^{2} (\mod 8) .

(3.22)

Extracting the terms involving q³ⁿ from (3.22), replacing q³ by q, we have the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (12 n + 3) q^{n} \equiv 4 {(q; q)}_{\infty}^{2} {(q^{2}; q^{2})}_{\infty}^{2} (\mod 8) .

(3.23)

Employing (2.2) into (3.23), we find the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (12 n + 3) q^{n} \equiv 4 \frac{{(q^{2}; q^{2})}_{\infty}^{3} {(q^{8}; q^{8})}_{\infty}^{5}}{{(q^{4}; q^{4})}_{\infty}^{2} {(q^{16}; q^{16})}_{\infty}^{2}} (\mod 8) .

(3.24)

Extracting the terms involving q²ⁿ from (3.24), replacing q² by q, we have the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (24 n + 3) q^{n} \equiv 4 \frac{{(q; q)}_{\infty}^{3} {(q^{4}; q^{4})}_{\infty}^{5}}{{(q^{2}; q^{2})}_{\infty}^{2} {(q^{8}; q^{8})}_{\infty}^{2}} (\mod 8) .

(3.25)

Invoking (2.1) in (3.25), we get the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (24 n + 3) q^{n} \equiv 4 \frac{{(q^{2}; q^{2})}_{\infty}^{2}}{{(q; q)}_{\infty}} (\mod 8) .

(3.26)

Ramanujan recorded the following identity in his third note book; for proof, one can see [11, p. 49].

ψ (q) = \frac{{(q^{2}; q^{2})}_{\infty}^{2}}{{(q; q)}_{\infty}} = \frac{{(q^{6}; q^{6})}_{\infty} {(q^{9}; q^{9})}_{\infty}^{2}}{{(q^{3}; q^{3})}_{\infty} {(q^{18}; q^{18})}_{\infty}} + q \frac{{(q^{18}; q^{18})}_{\infty}^{2}}{{(q^{9}; q^{9})}_{\infty}} .

(3.27)

Substituting (3.27) into (3.26), we obtain the generating function as follows

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (24 n + 3) q^{n} \equiv 4 \frac{{(q^{6}; q^{6})}_{\infty} {(q^{9}; q^{9})}_{\infty}^{2}}{{(q^{3}; q^{3})}_{\infty} {(q^{18}; q^{18})}_{\infty}} + 4 q \frac{{(q^{18}; q^{18})}_{\infty}^{2}}{{(q^{9}; q^{9})}_{\infty}} (\mod 8) .

(3.28)

Congruence (3.7) follows from (3.28).

Extracting the terms in which powers of q are congruent to 1 modulo 3 from (3.28), we have the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (72 n + 27) q^{n} \equiv 4 \frac{{(q^{6}; q^{6})}_{\infty}^{2}}{{(q^{3}; q^{3})}_{\infty}} (\mod 8) .

(3.29)

The results (3.9) and (3.10) follow from (3.29).

From (3.29), we obtain the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (216 n + 27) q^{n} \equiv 4 \frac{{(q^{2}; q^{2})}_{\infty}^{2}}{{(q; q)}_{\infty}} (\mod 8) .

(3.30)

Using the congruences (3.30) and (3.26), we can see that

{\bar{b}}_{2,3} (216 n + 27) \equiv {\bar{b}}_{2,3} (24 n + 3) (\mod 8) .

By mathematical induction on α, we find that

{\bar{b}}_{2,3} (216 \cdot 9^{α} n + 27 \cdot 9^{α}) \equiv {\bar{b}}_{2,3} (24 n + 3) (\mod 8) .

(3.31)

Using (3.7) in (3.31), we get (3.12).

Extracting the terms involving q³ⁿ from (3.28), replacing q³ by q, we have the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (72 n + 3) q^{n} \equiv 4 \frac{{(q^{2}; q^{2})}_{\infty} {(q^{3}; q^{3})}_{\infty}^{2}}{{(q; q)}_{\infty} {(q^{6}; q^{6})}_{\infty}} (\mod 8) .

(3.32)

Invoking (2.1) in (3.32), we get the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (72 n + 3) q^{n} \equiv 4 {(q; q)}_{\infty} (\mod 8) .

(3.33)

From (3.20), we can see that

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (4 n + 1) q^{n} \equiv 4 \frac{{(q; q)}_{\infty}^{2} {(q^{4}; q^{4})}_{\infty}^{2} {(q^{6}; q^{6})}_{\infty}^{4}}{{(q^{2}; q^{2})}_{\infty}^{2} {(q^{12}; q^{12})}_{\infty}^{2}} (\mod 8) .

(3.34)

Invoking (2.1) in (3.34), we have the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (4 n + 1) q^{n} \equiv 4 {(q^{2}; q^{2})}_{\infty} {(q^{4}; q^{4})}_{\infty} (\mod 8) .

(3.35)

Congruence (3.2) follows from (3.34).

Extracting the terms involving q²ⁿ from (3.35), replacing q² by q, we have the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (8 n + 1) q^{n} \equiv 4 {(q; q)}_{\infty} {(q^{2}; q^{2})}_{\infty} (\mod 8) .

(3.36)

Employing (2.9) into (3.36), we obtain the generating function as follows:

\begin{align} \sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (8 n + 1) q^{n} \equiv & 4 \frac{{(q^{6}; q^{6})}_{\infty} {(q^{9}; q^{9})}_{\infty}^{4}}{{(q^{3}; q^{3})}_{\infty} {(q^{18}; q^{18})}_{\infty}^{2}} \\ + 4 q {(q^{9}; q^{9})}_{\infty} {(q^{18}; q^{18})}_{\infty} (\mod 8) . \end{align}

(3.37)

Extracting the terms in which powers of q are congruent to 1 modulo 3 from (3.37), we have the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (24 n + 9) q^{n} \equiv 4 {(q^{3}; q^{3})}_{\infty} {(q^{6}; q^{6})}_{\infty} (\mod 8) .

(3.38)

The results (3.6) and (3.8) follow from (3.38).

From (3.38), we have the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (24 n + 9) q^{n} \equiv 4 {(q; q)}_{\infty} {(q^{2}; q^{2})}_{\infty} (\mod 8) .

(3.39)

Using the congruences (3.39) and (3.36), we can see that

{\bar{b}}_{2,3} (72 n + 9) \equiv {\bar{b}}_{2,3} (8 n + 1) (\mod 8) .

By mathematical induction on α, we obtain the generating function as follows:

{\bar{b}}_{2,3} (8 \cdot 9^{α + 1} n + 9^{α + 1}) \equiv {\bar{b}}_{2,3} (8 n + 1) (\mod 8) .

(3.40)

Using (3.8) in (3.40), we get (3.11).

Extracting the terms involving q³ⁿ from (3.37), replacing q³ by q, we have the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (24 n + 1) q^{n} \equiv 4 \frac{{(q^{2}; q^{2})}_{\infty} {(q^{3}; q^{3})}_{\infty}^{4}}{{(q; q)}_{\infty} {(q^{6}; q^{6})}_{\infty}^{2}} (\mod 8) .

(3.41)

Invoking (2.1) in (3.41), we get the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (24 n + 1) q^{n} \equiv 4 {(q; q)}_{\infty} (\mod 8) .

(3.42)

Using the congruences (3.33) and (3.42), we obtain (3.14).

From (3.19), it can be rewritten as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (2 n + 1) q^{n} \equiv 4 {(q; q)}_{\infty} {(q^{2}; q^{2})}_{\infty} {(q^{3}; q^{3})}_{\infty}^{3} (\mod 8) .

(3.43)

Employing (2.9) into (3.43), we obtain the generating function as follows:

\begin{align} \sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (2 n + 1) q^{n} \equiv & 4 \frac{{(q^{3}; q^{3})}_{\infty}^{2} {(q^{6}; q^{6})}_{\infty} {(q^{9}; q^{9})}_{\infty}^{4}}{{(q^{18}; q^{18})}_{\infty}^{2}} \\ + 4 q {(q^{3}; q^{3})}_{\infty}^{3} {(q^{9}; q^{9})}_{\infty} {(q^{18}; q^{18})}_{\infty} (\mod 8) . \end{align}

(3.44)

Congruence (3.1) follows from (3.44).

Extracting the terms in which powers of q are congruent to 1 modulo 3 from (3.44), we have the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (6 n + 3) q^{n} \equiv 4 {(q; q)}_{\infty}^{3} {(q^{3}; q^{3})}_{\infty} {(q^{6}; q^{6})}_{\infty} (\mod 8),

(3.45)

which implies as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (6 n + 3) q^{n} \equiv 4 {(q; q)}_{\infty}^{3} {(q^{3}; q^{3})}_{\infty}^{3} (\mod 8) .

(3.46)

Substituting (2.10) into (3.46), we obtain the generating function as follows:

\begin{align} \sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (6 n + 3) q^{n} \equiv & 4 \frac{{(q^{3}; q^{3})}_{\infty}^{2} {(q^{6}; q^{6})}_{\infty} {(q^{9}; q^{9})}_{\infty}^{6}}{{(q^{18}; q^{18})}_{\infty}^{3}} \\ + 4 q {(q^{3}; q^{3})}_{\infty}^{3} {(q^{9}; q^{9})}_{\infty}^{3} (\mod 8) . \end{align}

(3.47)

Congruence (3.4) follows from (3.47).

Extracting the terms in which powers of q are congruent to 1 modulo 3 from (3.47), we get the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (18 n + 9) q^{n} \equiv 4 {(q; q)}_{\infty}^{3} {(q^{3}; q^{3})}_{\infty}^{3} (\mod 8) .

(3.48)

Using the congruences (3.48) and (3.46), we find that

{\bar{b}}_{2,3} (18 n + 9) \equiv {\bar{b}}_{2,3} (6 n + 3) (\mod 8) .

By mathematical induction on α, we obtain the generating function as follows:

{\bar{b}}_{2,3} (2 \cdot 3^{α + 2} n + 3^{α + 2}) \equiv {\bar{b}}_{2,3} (6 n + 3) (\mod 8) .

(3.49)

Using (3.4) in (3.49), we get (3.13).

From (3.47), we have the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (18 n + 3) q^{n} \equiv 4 \frac{{(q; q)}_{\infty}^{2} {(q^{2}; q^{2})}_{\infty} {(q^{3}; q^{3})}_{\infty}^{6}}{{(q^{6}; q^{6})}_{\infty}^{3}} (\mod 8) .

(3.50)

Invoking (2.1) in (3.50), we find that

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (18 n + 3) q^{n} \equiv 4 {(q^{2}; q^{2})}_{\infty}^{2} (\mod 8) .

(3.51)

Congruence (3.5) follows from (3.51).

Extracting the terms involving q³ⁿ from (3.43), replacing q³ by q, we have the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (6 n + 1) q^{n} \equiv 4 \frac{{(q; q)}_{\infty}^{2} {(q^{2}; q^{2})}_{\infty} {(q^{3}; q^{3})}_{\infty}^{4}}{{(q^{6}; q^{6})}_{\infty}^{2}} (\mod 8) .

(3.52)

Invoking (2.1) in (3.52), we obtain the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (6 n + 1) q^{n} \equiv 4 {(q^{2}; q^{2})}_{\infty}^{2} (\mod 8) .

(3.53)

Congruence (3.3) easily follows from (3.53).

From (3.51) and (3.53), we have the generating function as follows:

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (36 n + 3) q^{n} \equiv 4 {(q; q)}_{\infty}^{2} (\mod 8)

(3.54)

and

\sum_{n = 0}^{\infty} {\bar{b}}_{2,3} (12 n + 1) q^{n} \equiv 4 {(q; q)}_{\infty}^{2} (\mod 8) .

(3.55)

Using the congruences (3.54) and (3.55), we get internal congruence (3.15).

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S. Shivaprasada Nayaka

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Arithmetic properties of (2, 3)-regular overcubic bipartitions

1. Introduction

2. Preliminaries

3. Congruences modulo 8 for ${\bar{b}}_{2,3} (n)$

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Arithmetic properties of (2, 3)-regular overcubic bipartitions Open Access

1. Introduction

2. Preliminaries

3. Congruences modulo 8 for b¯2,3(n)

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Arithmetic properties of (2, 3)-regular overcubic bipartitions

3. Congruences modulo 8 for ${\bar{b}}_{2,3} (n)$