Arithmetic properties of singular overpartition pairs without multiples of k

Nayaka, S. Shivaprasada; Sreelakshmi, T.K.; Kumar, Santosh

doi:10.1108/AJMS-01-2021-0013

Purpose

In this paper, the author defines the function ${\bar{B}}_{i, j}^{δ, k} (n)$ ⁠, the number of singular overpartition pairs of n without multiples of k in which no part is divisible by δ and only parts congruent to ± i, ± j modulo δ may be overlined.

Design/methodology/approach

Andrews introduced to combinatorial objects, which he called singular overpartitions and proved that these singular overpartitions depend on two parameters δ and i can be enumerated by the function ${\bar{C}}_{δ, i} (n)$ , which gives the number of overpartitions of n in which no part divisible by δ and parts ≡ ± i(Mod δ) may be overlined.

Findings

Using classical spirit of q-series techniques, the author obtains congruences modulo 4 for ${\bar{B}}_{2,4}^{8,3} (n)$ ⁠, ${\bar{B}}_{2,4}^{8,5}$ and ${\bar{B}}_{2,4}^{12,3}$ ⁠.

Originality/value

The results established in this work are extension to those proved in Andrews’ singular overpatition pairs of n.

1. Introduction

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

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

(1.1)

where the product representations arise from Jacobi's triple product identity [1, p. 35, Entry 19].

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

(1.2)

Throughout the paper, we use the standard q-series notation, and f_k is defined as

f_{k} ≔ {(q^{k}; q^{k})}_{\infty} = \lim_{n \to \infty} \prod_{m = 1}^{n} (1 - q^{m k}) .

The special cases of f(a, b) are

φ (q) ≔ f (q, q) = \sum_{n = - \infty}^{\infty} q^{n^{2}} = {(- q; q^{2})}_{\infty}^{2} {(q^{2}; q^{2})}_{\infty} = \frac{f_{2}^{5}}{f_{1}^{2} f_{4}^{2}},

(1.3)

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}} = \frac{f_{2}^{2}}{f_{1}}

(1.4)

and

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

(1.5)

A partition of a positive integer n is a nonincreasing sequence of positive integers whose sum is n. An overpartition, introduced by Corteel and Lovejoy [2], of a nonnegative integer n is a nonincreasing sequence of natural numbers whose sum is n in which the first occurrence of a number may be overlined.

Recently, G. E. Andrews [3] defined combinatorial objects, which he called singular overpartitions and proved that these singular overpartitions depend on two parameters δ and i can be enumerated by the function ${\bar{C}}_{δ, i} (n)$ , which gives the number of overpartitions of n in which no part divisible by δ and parts ≡ ±i (mod δ) may be overlined. The generating function of ${\bar{C}}_{δ, i} (n)$ is

\sum_{n = 0}^{\infty} {\bar{C}}_{δ, i} (n) q^{n} = \frac{{(q^{δ}; q^{δ})}_{\infty} {(- q^{i}; q^{δ})}_{\infty} {(- q^{δ - i}; q^{δ})}_{\infty}}{{(q; q)}_{\infty}} .

(1.6)

He also proved that

{\bar{C}}_{3,1} (9 n + 3) \equiv {\bar{C}}_{3,1} (9 n + 6) \equiv 0 (\mod 3) .

(1.7)

Andrews [3] proves that, for all n ≥ 0, ${\bar{C}}_{3,1} (n) = {\bar{A}}_{3} (n)$ ⁠, where ${\bar{A}}_{3} (n)$ is the number of overpartitions of n into parts not divisible by 3. The function ${\bar{A}}_{ℓ} (n)$ ⁠, which counts the number of overpartitions of n into parts not divisible by ℓ, plays a key role in the work of Lovejoy [4].

Chen et al. [5] have generalized (1.7) and proved some congruences modulo 2, 3, 4 and 8 for ${\bar{C}}_{3,1} (n)$ ⁠. They also proved some congruence for ${\bar{C}}_{4,1} (n)$ ⁠, ${\bar{C}}_{6,1} (n)$ and ${\bar{C}}_{6,2} (n)$ modulo powers of 2 and 3. More recently, Ahmed and Baruah [6] have found some new congruences for ${\bar{C}}_{3,1} (n)$ ⁠, ${\bar{C}}_{8,2} (n)$ ⁠, ${\bar{C}}_{12,4} (n)$ ⁠, ${\bar{C}}_{24,8} (n)$ and ${\bar{C}}_{48,16} (n)$ modulo 18, 36. Chen [7] has also found some congruences modulo powers of 2 for ${\bar{C}}_{3,1} (n)$ ⁠, ${\bar{C}}_{4,1} (n)$ ⁠. Yao [8] has proved congruences modulo 16, 32, 64 for ${\bar{C}}_{3,1} (n)$ ⁠. Naika and Gireesh [9] have found some congruences modulo 6, 12, 16, 18, 24, 48 and 72 for ${\bar{C}}_{3,1} (n)$ ⁠. Naika and Nayaka [10] have proved some congruences for ${\bar{C O}}_{3,1} (n)$ modulo powers of 2 and 3. They have also proved in a paper [11] modulo 4 for ${\bar{C}}_{4,1}^{3} (n)$ and ${\bar{C}}_{4,1}^{5} (n)$ ⁠.

In [12, 13], Naika et al. have defined the Andrews' singular overpartition pairs of n. Let ${\bar{C}}_{i, j}^{δ} (n)$ denote the number of Andrews' singular overpartition pairs of n in which no part is divisible by δ and only parts congruent to ± i, ± j modulo δ may be overlined. Andrews' singular overpartition pair π of n is a pair of Andrews' singular overpartitions (λ, μ) such that the sum of all of the parts is n. They also established Ramanujan-like congruences for ${\bar{A}}_{1,2}^{6} (n)$ modulo 3,9, 27 and infinite families of congruences for ${\bar{A}}_{1,5}^{12} (n)$ modulo 4, 6 and 9.

In this paper, we define the function ${\bar{B}}_{i, j}^{δ, k} (n)$ ⁠, the number of singular overpartition pairs of n without multiples of k in which no part is divisible by δ and only parts congruent to ± i, ± j modulo δ may be overlined. The generating function of ${\bar{B}}_{i, j}^{δ, k} (n)$ is given by

\sum_{n = 0}^{\infty} {\bar{B}}_{i, j}^{δ, k} (n) q^{n} = \frac{f (q^{i}, q^{δ - i}) f (q^{j}, q^{δ - j}) {(q^{k}; q^{k})}_{\infty}^{2}}{f (q^{k i}, q^{k (δ - i)}) f (q^{k j}, q^{k (δ - j)}) {(q; q)}_{\infty}^{2}} .

(1.8)

In this paper, we establish some congruences modulo 4 for ${\bar{B}}_{2,4}^{8,3} (n)$ ⁠, ${\bar{B}}_{2,4}^{8,5}$ and ${\bar{B}}_{2,4}^{12,3}$ ⁠. The main results of this paper can be stated as follows.

Theorem 1.1.

For all integers α ≥ 0 and n ≥ 0,

\begin{align} {\bar{B}}_{2,4}^{8,3} (16 n + 9) \equiv 0 (\mod 4), \end{align}

(1.9)

\begin{align} {\bar{B}}_{2,4}^{8,3} (16 n + 13) \equiv 0 (\mod 4), \end{align}

(1.10)

\begin{align} {\bar{B}}_{2,4}^{8,3} (32 n + 23) \equiv 0 (\mod 4), \end{align}

(1.11)

\begin{align} {\bar{B}}_{2,4}^{8,3} (32 n + 31) \equiv 0 (\mod 4), \end{align}

(1.12)

\begin{align} {\bar{B}}_{2,4}^{8,3} (64 n + 35) \equiv 0 (\mod 4), \end{align}

(1.13)

\begin{align} {\bar{B}}_{2,4}^{8,3} (64 n + 51) \equiv 0 (\mod 4), \end{align}

(1.14)

\begin{align} {\bar{B}}_{2,4}^{8,3} (256 n + 139) \equiv 0 (\mod 4), \end{align}

(1.15)

\begin{align} {\bar{B}}_{2,4}^{8,3} (256 n + 203) \equiv 0 (\mod 4), \end{align}

(1.16)

\begin{align} {\bar{B}}_{2,4}^{8,3} (4 n + 3) \equiv {\bar{B}}_{2,4}^{8,3} (64 n + 43) (\mod 4), \end{align}

(1.17)

\begin{align} {\bar{B}}_{2,4}^{8,3} (8 n + 5) \equiv {\bar{B}}_{2,4}^{8,3} (32 n + 19) \equiv {\bar{B}}_{2,4}^{8,3} (128 + 75) (\mod 4) . \end{align}

(1.18)

Theorem 1.2.

Let p be a prime ≥ 5, $(\frac{- 6}{p}) = - 1$ ⁠. Then for all integers α ≥ 1, and n ≥ 0,

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (16 p^{2 α} n + \frac{14 p^{2 α} + 1}{3}) q^{n} \equiv 2 f_{1} f_{6} (\mod 4) .

(1.19)

Theorem 1.3.

Let p be a prime ≥ 5, $(\frac{- 6}{p}) = - 1$ ⁠. Then for all integers α ≥ 0, and n ≥ 0,

{\bar{B}}_{2,4}^{8,3} (16 p^{2 α + 2} n + 16 p^{2 α + 1} i + \frac{14 p^{2 α + 2} + 1}{3}) \equiv 0 (\mod 4),

(1.20)

whereiis an integer and 1 ≤ i ≤ p − 1.

Theorem 1.4.

For all integers α ≥ 0 and n ≥ 0,

\begin{align} {\bar{B}}_{2,4}^{8,5} (4 n + 3) & \equiv 0 (\mod 4), \end{align}

(1.21)

\begin{align} {\bar{B}}_{2,4}^{8,5} (8 n + 5) & \equiv 0 (\mod 4) . \end{align}

(1.22)

Theorem 1.5.

For any prime p ≥ 5, α ≥ 0 and n ≥ 0, we have

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,5} (8 p^{2 α} n + \frac{p^{3 α} + 2}{3}) q^{n} \equiv 2 f_{1} (\mod 4) .

(1.23)

Theorem 1.6.

For any prime p ≥ 5, α ≥ 0, n ≥ 0 and l = 1, 2, …p − 1, we have

{\bar{B}}_{2,4}^{8,5} (8 p^{2 α + 1} (p n + l) + \frac{p^{3 α} + 2}{3}) \equiv 0 (\mod 4) .

(1.24)

Theorem 1.7.

For all integers α ≥ 0 and n ≥ 0,

\begin{align} {\bar{B}}_{2,4}^{12,3} (6 n + 5) \equiv 0 (\mod 4), \end{align}

(1.25)

\begin{align} {\bar{B}}_{2,4}^{12,3} (12 n + 9) \equiv 0 (\mod 4), \end{align}

(1.26)

\begin{align} {\bar{B}}_{2,4}^{12,3} (36 n + 27) \equiv 0 (\mod 4), \end{align}

(1.27)

\begin{align} {\bar{B}}_{2,4}^{12,3} (108 n + 51) \equiv 0 (\mod 4), \end{align}

(1.28)

\begin{align} {\bar{B}}_{2,4}^{12,3} (108 n + 87) \equiv 0 (\mod 4), \end{align}

(1.29)

\begin{align} {\bar{B}}_{2,4}^{12,3} (4 \cdot 3^{2 α + 3} n + \frac{32 \cdot 3^{2 α + 3} + 3}{2}) \equiv 0 (\mod 4) . \end{align}

(1.30)

Theorem 1.8.

For any prime p ≥ 5, α ≥ 0 and n ≥ 0, we have

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{12,3} (36 p^{2 α} n + \frac{3 (p^{3 α} + 1)}{2}) q^{n} \equiv 2 f_{1} (\mod 4) .

(1.31)

Theorem 1.9.

For any prime p ≥ 5, α ≥ 0, n ≥ 0 and l = 1, 2, …p − 1, we have

{\bar{B}}_{2,4}^{12,3} (36 p^{2 α + 1} (p n + l) + \frac{3 (p^{3 α} + 1)}{2}) \equiv 0 (\mod 4) .

(1.32)

2. Preliminary results

We need the following few dissection formulas to prove our main results,

Lemma 2.1.

The following two dissections hold:

\begin{align} \frac{f_{3}^{3}}{f_{1}} = \frac{f_{4}^{3} f_{6}^{2}}{f_{2}^{2} f_{12}} + q \frac{f_{12}^{3}}{f_{4}}, \end{align}

(2.1)

\begin{align} \frac{f_{3}}{f_{1}^{3}} = \frac{f_{4}^{6} f_{6}^{3}}{f_{2}^{9} f_{12}^{2}} + 3 q \frac{f_{4}^{2} f_{6} f_{12}^{2}}{f_{2}^{7}} . \end{align}

(2.2)

Hirschhorn, Garvan and Borwein [14] have proved Eqn (2.1). For proof of (2.2), see [15].

Lemma 2.2.

The following two dissections hold:

\begin{align} \frac{1}{f_{1} f_{3}} = \frac{f_{8}^{2} f_{12}^{5}}{f_{2}^{2} f_{4} f_{6}^{4} f_{24}^{2}} + q \frac{f_{4}^{5} f_{24}^{2}}{f_{2}^{4} f_{6}^{2} f_{8}^{2} f_{12}}, \end{align}

(2.3)

\begin{align} f_{1} f_{3} = \frac{f_{2} f_{8}^{2} f_{12}^{4}}{f_{4}^{2} f_{6} f_{24}^{2}} - q \frac{f_{4}^{4} f_{6} f_{24}^{2}}{f_{2} f_{8}^{2} f_{12}^{2}} . \end{align}

(2.4)

Eqn (2.3) was proved by Baruah and Ojah [16]. Replacing q by − q in (2.3) and using the fact that ${(- q; - q)}_{\infty} = \frac{f_{2}^{3}}{f_{1} f_{4}}$ ⁠, we get (2.4).

Lemma 2.3.

The following two dissections hold:

\begin{align} \frac{f_{3}}{f_{1}} = \frac{f_{4} f_{6} f_{16} f_{24}^{2}}{f_{2}^{2} f_{8} f_{12} f_{48}} + q \frac{f_{6} f_{8}^{2} f_{48}}{f_{2}^{2} f_{16} f_{24}}, \end{align}

(2.5)

\begin{align} \frac{f_{1}}{f_{3}} = \frac{f_{2} f_{16} f_{24}^{2}}{f_{6}^{2} f_{8} f_{48}} - q \frac{f_{2} f_{8}^{2} f_{12} f_{48}}{f_{4} f_{6}^{2} f_{16} f_{24}} . \end{align}

(2.6)

Xia and Yao [17] gave a proof of Lemma (2.3). Replacing q by − q in (2.5) and using the fact that ${(- q; - q)}_{\infty} = \frac{f_{2}^{3}}{f_{1} f_{4}}$ ⁠, we get (2.6).

Lemma 2.4.

The following two dissections hold:

\begin{align} \frac{f_{3}^{2}}{f_{1}^{2}} = \frac{f_{4}^{4} f_{6} f_{12}^{2}}{f_{2}^{5} f_{8} f_{24}} + 2 q \frac{f_{4} f_{6}^{2} f_{8} f_{24}}{f_{2}^{4} f_{12}}, \end{align}

(2.7)

\begin{align} \frac{f_{1}^{2}}{f_{3}^{2}} = \frac{f_{2} f_{4}^{2} f_{12}^{4}}{f_{6}^{5} f_{8} f_{24}} - 2 q \frac{f_{2}^{2} f_{8} f_{12} f_{24}}{f_{4} f_{6}^{4}} . \end{align}

(2.8)

Xia and Yao [18] proved (2.7) by employing an addition formula for theta functions.

Lemma 2.5.

The following two dissections hold:

\frac{f_{5}}{f_{1}} = \frac{f_{8} f_{20}^{2}}{f_{2}^{2} f_{40}} + q \frac{f_{4}^{3} f_{10} f_{40}}{f_{2}^{3} f_{8} f_{20}} .

(2.9)

Eqn (2.9) was proved by Hirschhorn and Sellers [19].

Lemma 2.6.

The following three dissections hold:

f_{1} f_{2} = \frac{f_{6} f_{9}^{4}}{f_{3} f_{18}^{2}} - q f_{9} f_{18} - 2 q^{2} \frac{f_{3} f_{18}^{4}}{f_{6} f_{9}^{2}} .

(2.10)

One can see this identity in [20].

Lemma 2.7.

(Cui and Gu [21, Theorem 2.2]). For any prime p ≥ 5,

f_{1} = \sum_{k = \frac{1 - p}{2} k \neq \frac{\pm p - 1}{6}}^{\frac{p - 1}{2}} {(- 1)}^{k} q^{\frac{3 k^{2} + k}{2}} f (- q^{\frac{3 p^{2} + (6 k + 1) p}{2}}, - q^{\frac{3 p^{2} - (6 k + 1) p}{2}}) + {(- 1)}^{\frac{\pm p - 1}{6}} q^{\frac{p^{2} - 1}{24}} f_{p^{2}},

where

\frac{\pm p - 1}{6} ≔ \{\begin{cases} \frac{p - 1}{6}, & if p \equiv 1 (\mod 6), \\ \frac{- p - 1}{6}, & if p \equiv - 1 (\mod 6) . \end{cases}

Lemma 2.8.

For any prime p and positive integer n,

f_{1}^{p^{n}} \equiv f_{p}^{p^{n - 1}} (\mod p^{n}) .

(2.11)

3. Proof of Theorem (1.1)

Setting i = 2, j = 4, δ = 8 and k = 3 in (1.8), we see that

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (n) q^{n} = \frac{f (q^{2}, q^{6}) f (q^{4}, q^{4}) {(q^{3}; q^{3})}_{\infty}^{2}}{f (q^{6}, q^{18}) f (q^{12}, q^{12}) {(q; q)}_{\infty}^{2}} .

(3.1)

By the definition of f(a, b) and the well-known Jacobi triple product identity, we get

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (n) q^{n} = \frac{f_{3}^{2} f_{6} f_{8}^{5} f_{48}^{2}}{f_{1}^{2} f_{2} f_{16}^{2} f_{24}^{5}} .

(3.2)

Substituting (2.7) into (3.2), we have

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (n) q^{n} = \frac{f_{6}^{2} f_{8}^{4} f_{48}^{2} f_{4}^{4} f_{12}^{2}}{f_{2}^{6} f_{16}^{2} f_{24}^{6}} + 2 q \frac{f_{6}^{3} f_{8}^{6} f_{4} f_{48}^{2}}{f_{2}^{5} f_{16}^{2} f_{24}^{4} f_{12}} .

(3.3)

Equating odd parts of the aforementioned equation, we obtain

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (2 n + 1) q^{n} = 2 \frac{f_{3}^{3} f_{4}^{6} f_{2} f_{24}^{2}}{f_{1}^{5} f_{8}^{2} f_{12}^{4} f_{6}} .

(3.4)

Involving (2.11) in (3.4), we get

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

(3.5)

Employing (2.2) into (3.5), we arrive at

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (2 n + 1) q^{n} \equiv 2 \frac{f_{4}^{8} f_{6}^{3}}{f_{2}^{9} f_{12}^{2}} + 2 q \frac{f_{4}^{4} f_{6} f_{12}}{f_{2}^{7}} (\mod 4) .

(3.6)

Extracting the terms involving q²ⁿ from both sides of (3.6), we have

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (4 n + 1) q^{n} \equiv 2 \frac{f_{2}^{8} f_{3}^{3}}{f_{1}^{9} f_{6}^{2}} (\mod 4) .

(3.7)

Using (2.11) in the aforementioned equation, we obtain

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

(3.8)

Substituting (2.1) into (3.8), we find that

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

(3.9)

which implies

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

(3.10)

Invoking (2.11) in (3.10), the equation reduces to

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

(3.11)

Employing (2.8) into (3.11), we obtain

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (8 n + 1) q^{n} \equiv 2 \frac{f_{2}^{4} f_{4}^{2} f_{12}^{4}}{f_{6}^{5} f_{8} f_{24}} (\mod 4) .

(3.12)

Congruence (1.9) easily follows from the aforementioned equation.

From (3.6), we have

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (4 n + 3) q^{n} \equiv 2 \frac{f_{2}^{4} f_{3} f_{6}^{2}}{f_{1}^{7}} (\mod 4) .

(3.13)

Using (2.11) in (3.6), we found

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

(3.14)

Substituting (2.5) into (3.14), we obtain

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (4 n + 3) q^{n} \equiv 2 \frac{f_{6}^{3} f_{4} f_{16} f_{24}^{2}}{f_{2} f_{8} f_{12} f_{48}} + 2 q \frac{f_{6}^{3} f_{8}^{2} f_{48}}{f_{2} f_{16} f_{24}} (\mod 4) .

(3.15)

Extracting the terms involving q²ⁿ⁺¹ from (3.15), dividing by q and replacing q² by q, we arrive at

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (8 n + 7) q^{n} \equiv 2 \frac{f_{3}^{3} f_{4}^{2} f_{24}}{f_{1} f_{8} f_{12}} (\mod 4) .

(3.16)

Invoking (2.11) in (3.16), we get

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (8 n + 7) q^{n} \equiv 2 \frac{f_{3}^{3} f_{12}}{f_{1}} (\mod 4) .

(3.17)

Employing (2.1) into (3.17), we obtain

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (8 n + 7) q^{n} \equiv 2 \frac{f_{4}^{3} f_{6}^{2}}{f_{2}^{2}} + 2 q \frac{f_{12}^{4}}{f_{4}} (\mod 4) .

(3.18)

Extracting the terms involving q²ⁿ from both sides of (3.18), we have

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (16 n + 7) q^{n} \equiv 2 \frac{f_{2}^{3} f_{3}^{2}}{f_{1}^{2}} (\mod 4) .

(3.19)

Using (2.11) in (3.19), we arrive at

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (16 n + 7) q^{n} \equiv 2 f_{2} f_{6} (\mod 4) .

(3.20)

Congruence (1.11) easily follows from (3.20).

From (3.18), we have

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (16 n + 15) q^{n} \equiv 2 \frac{f_{6}^{4}}{f_{2}} (\mod 4) .

(3.21)

Congruence (1.12) follows by extracting the terms involving q²ⁿ⁺¹ from (3.21).

From (3.15), we get

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (8 n + 3) q^{n} \equiv 2 \frac{f_{3}^{3} f_{2} f_{8} f_{12}^{2}}{f_{1} f_{4} f_{6} f_{24}} (\mod 4) .

(3.22)

Using (2.11) in (3.22), we obtain

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

(3.23)

Employing (2.1) into (3.23), we reduce that

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

(3.24)

Extracting the terms involving q²ⁿ from both sides of (3.24), we find that

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (16 n + 3) q^{n} \equiv 2 \frac{f_{3} f_{2}^{4}}{f_{1} f_{6}} (\mod 4) .

(3.25)

Substituting (2.5) into (3.25), we arrive at

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (16 n + 3) q^{n} \equiv 2 \frac{f_{2}^{2} f_{4} f_{16} f_{24}^{2}}{f_{8} f_{12} f_{48}} + 2 q \frac{f_{2}^{2} f_{8}^{2} f_{48}}{f_{16} f_{24}} (\mod 4),

(3.26)

which implies

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (32 n + 3) q^{n} \equiv 2 \frac{f_{1}^{2} f_{2} f_{8} f_{12}^{2}}{f_{4} f_{6} f_{24}} (\mod 4) .

(3.27)

Invoking (2.11) in (3.27), we obtain

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (32 n + 3) q^{n} \equiv 2 \frac{f_{8}}{f_{6}} (\mod 4) .

(3.28)

Congruence (1.13) follows by extracting the terms involving q²ⁿ⁺¹ from (3.28).

Extracting the terms involving q²ⁿ⁺¹ from (3.26), dividing by q and replacing q² by q, we arrive at

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (32 n + 19) q^{n} \equiv 2 \frac{f_{1}^{2} f_{4}^{2} f_{24}}{f_{8} f_{12}} (\mod 4) .

(3.29)

Using (2.11) in (3.29), we get

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (32 n + 19) q^{n} \equiv 2 f_{2} f_{12} (\mod 4) .

(3.30)

Congruence (1.14) follows from (3.30).

Extracting the terms involving q²ⁿ from both sides of (3.24), we have

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (16 n + 11) q^{n} \equiv 2 \frac{f_{1} f_{6}^{3}}{f_{3}} (\mod 4) .

(3.31)

Employing (2.6) into (3.31), we found

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (16 n + 11) q^{n} \equiv 2 \frac{f_{2} f_{6} f_{16} f_{24}^{2}}{f_{8} f_{48}} + 2 q \frac{f_{2} f_{6} f_{8}^{2} f_{12} f_{48}}{f_{4} f_{16} f_{24}} (\mod 4),

(3.32)

which implies

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (32 n + 11) q^{n} \equiv 2 \frac{f_{1} f_{3} f_{8} f_{12}^{2}}{f_{4} f_{24}} (\mod 4) .

(3.33)

Substituting (2.4) into (3.33), we get

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (32 n + 11) q^{n} \equiv 2 \frac{f_{2} f_{8}^{3} f_{12}^{6}}{f_{4}^{3} f_{6} f_{24}^{3}} + 2 q \frac{f_{4}^{3} f_{6} f_{24}}{f_{2} f_{8}} (\mod 4),

(3.34)

which implies

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (64 n + 11) q^{n} \equiv 2 \frac{f_{1} f_{4}^{3} f_{6}^{6}}{f_{2}^{3} f_{3} f_{12}^{3}} (\mod 4) .

(3.35)

Using (2.11) in (3.35), we find that

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (64 n + 11) q^{n} \equiv 2 \frac{f_{1}}{f_{3} f_{2}^{3}} (\mod 4) .

(3.36)

Employing (2.6) into (3.36), we get

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (64 n + 11) q^{n} \equiv 2 \frac{f_{2}^{4} f_{16} f_{24}^{2}}{f_{6}^{2} f_{8} f_{48}} + 2 q \frac{f_{2}^{4} f_{8}^{2} f_{12} f_{48}}{f_{4} f_{6}^{2} f_{16} f_{24}} (\mod 4) .

(3.37)

Extracting the terms involving q²ⁿ from (3.37) and replacing q² by q, we obtain

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (128 n + 11) q^{n} \equiv 2 \frac{f_{1}^{4} f_{8} f_{12}^{2}}{f_{3}^{2} f_{4} f_{24}} (\mod 4) .

(3.38)

Invoking (2.11) in (3.38), we arrive at

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (128 n + 11) q^{n} \equiv 2 \frac{f_{2}^{2} f_{8}}{f_{4} f_{6}} (\mod 4) .

(3.39)

Congruence (1.15) follows by extracting the terms involving q²ⁿ⁺¹ from (3.39).

From (3.37), we have

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (128 n + 75) q^{n} \equiv 2 \frac{f_{1}^{4} f_{4}^{2} f_{6} f_{24}}{f_{2} f_{3}^{2} f_{8} f_{12}} (\mod 4) .

(3.40)

Invoking (2.11) in (3.40), we obtain

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (128 n + 75) q^{n} \equiv 2 f_{2} f_{12} (\mod 4) .

(3.41)

Congruence (1.16) follows from (3.41).

From (3.34), we arrive at

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (64 n + 43) q^{n} \equiv 2 \frac{f_{2}^{3} f_{3} f_{12}}{f_{1} f_{4}} (\mod 4) .

(3.42)

Using (2.11) in (3.42), we get

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (64 n + 43) q^{n} \equiv 2 \frac{f_{3} f_{2} f_{12}}{f_{1}} (\mod 4) .

(3.43)

From the equations (3.14) and (3.43), we obtain (1.17).

Extracting the terms involving q²ⁿ⁺¹ from (3.9), dividing by q and replacing q² by q, we obtain

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (8 n + 5) q^{n} \equiv 2 \frac{f_{1}^{4} f_{6}^{3}}{f_{2} f_{3}^{2}} (\mod 4) .

(3.44)

Invoking (2.11) in (3.44), we get

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

(3.45)

Congruence (1.10) follows by extracting the terms involving q²ⁿ⁺¹ from (3.45).

From equations (3.45), (3.30) and (3.40), we obtain (1.18).

4. Proof of Theorem (1.2)

Extracting the terms involving q²ⁿ from (3.45) and replacing q² by q, we have

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,3} (16 n + 5) q^{n} \equiv 2 f_{1} f_{6} (\mod 4) .

(4.1)

define

\sum_{n = 0}^{\infty} g (n) q^{n} = f_{1} f_{6} .

(4.2)

Combining (4.1) and (4.2), we find that

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

(4.3)

For a prime, p ≥ 5 or $\frac{- (p - 1)}{2} \leq k, m \leq \frac{p - 1}{2}$ ⁠, consider

\frac{3 k^{2} + k}{2} + 6 \cdot \frac{3 m^{2} + m}{2} \equiv \frac{7 p^{2} - 7}{24} (\mod p),

(4.4)

therefore,

{(6 k + 1)}^{2} + 6 \cdot {(6 m + 1)}^{2} \equiv 0 (\mod p),

Since $(\frac{- 6}{p}) = - 1$ the congruence relation (4.4) holds if and only if both $k = m = \frac{\pm p - 1}{6}$ ⁠. Therefore, if we substitute Lemma (2.7) into (4.2) and then extract the terms in which the powers of q are congruent to $\frac{7 p^{2} - 7}{24}$ modulo p and then divide by $q^{\frac{7 p^{2} - 7}{24}}$ ⁠, we find that

\sum_{n = 0}^{\infty} g (p n + \frac{7 p^{2} - 7}{24}) q^{p n} = f_{p^{2}} f_{6 p^{2}},

which implies that

\sum_{n = 0}^{\infty} g (p^{2} n + \frac{7 p^{2} - 7}{24}) q^{n} = f_{1} f_{6}

(4.5)

and for n ≥ 0,

g (p^{2} n + p i + \frac{7 p^{2} - 7}{24}) = 0,

(4.6)

where i is an integer and 1 ≤ i ≤ p − 1. By induction, we see that for n ≥ 0 and α ≥ 0,

g (p^{2 α} n + \frac{7 p^{2 α} - 7}{24}) = g (n) .

(4.7)

Replacing n by $p^{2 α} n + \frac{7 p^{2 α} - 7}{24}$ in (4.3), we arrive at (1.19).

5. Proof of Theorem (1.3)

Replacing n by $p^{2} n + p i + \frac{7 p^{2} - 7}{24}$ in (4.7) and using (4.6), we find that for n ≥ 0 and α ≥ 0,

g (p^{2 α + 2} n + p^{2 α + 1} i + \frac{7 p^{α + 2} - 7}{24}) = 0 .

(5.1)

Comparing coefficients of qⁿ from both sides of (4.3), we see that for n ≥ 0,

{\bar{B}}_{2,4}^{8,3} (16 n + 5) \equiv 2 g (n) (\mod 4) .

(5.2)

The required result follows from (5.1) and (5.2).

6. Proof of Theorem (1.4)

Setting i = 2, j = 4, δ = 8 and k = 5 in (1.8), we see that

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,5} (n) q^{n} = \frac{f (q^{2}, q^{6}) f (q^{4}, q^{4}) {(q^{5}; q^{5})}_{\infty}^{2}}{f (q^{10}, q^{30}) f (q^{20}, q^{20}) {(q; q)}_{\infty}^{2}} .

(6.1)

By the definition of f(a, b) and the well-known Jacobi triple product identity, we get

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,5} (n) q^{n} = \frac{f_{5}^{2} f_{8}^{5} f_{10} f_{80}^{2}}{f_{1}^{2} f_{2} f_{16}^{2} f_{40}^{5}} .

(6.2)

Substituting (2.9) into (6.2), we have

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,5} (n) q^{n} = \frac{f_{8}^{7} f_{10} f_{20}^{4} f_{80}^{2}}{f_{2}^{5} f_{16}^{2} f_{40}^{7}} + 2 q \frac{f_{4}^{3} f_{8}^{5} f_{10}^{2} f_{20} f_{80}^{2}}{f_{2}^{6} f_{16}^{2} f_{40}^{5}} + q^{2} \frac{f_{4}^{6} f_{8}^{3} f_{10}^{3} f_{80}^{2}}{f_{2}^{7} f_{16}^{2} f_{40}^{3} f_{20}^{2}} .

(6.3)

Equating odd parts of the aforementioned equation, we obtain

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,5} (2 n + 1) q^{n} = 2 \frac{f_{2}^{3} f_{4}^{5} f_{5}^{2} f_{10} f_{40}^{2}}{f_{1}^{6} f_{8}^{2} f_{20}^{5}} .

(6.4)

Involving (2.11) in (6.4), we get

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,5} (2 n + 1) q^{n} \equiv 2 f_{4} (\mod 4) .

(6.5)

Congruences (1.21) and (1.22) follow from the aforementioned equation.

7. Proof of Theorem (1.5)

From (6.5), we have

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,5} (8 n + 1) q^{n} \equiv 2 f_{1} (\mod 4) .

(7.1)

Employing Lemma (2.7) into (7.1), it can be see that

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,5} (8 (p n + \frac{p^{2} - 1}{24}) + 1) q^{n} \equiv 2 f_{p} (\mod 4),

(7.2)

which implies

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,5} (8 p^{2} n + \frac{p^{3} + 2}{3}) q^{n} \equiv 2 f_{1} (\mod 4) .

(7.3)

Therefore,

{\bar{B}}_{2,4}^{8,5} (8 p^{2} n + \frac{p^{3} + 2}{3}) \equiv {\bar{B}}_{2,4}^{8,5} (8 n + 1) (\mod 4) .

Using the aforementioned relation and by induction on α, we arrive at (1.23).

8. Proof of Theorem (1.6)

Combining (7.2) with Theorem (1.5), we derive that for α ≥ 0,

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,5} (8 p^{2 α + 1} n + \frac{p^{3 α} + 2}{3}) \equiv 2 f_{p} (\mod 4) .

Therefore, it follows that

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{8,5} (8 p^{2 α + 1} (p n + l) + \frac{p^{3 α} + 2}{3}) \equiv 0 (\mod 4) .

where l = 1, 2, …, p − 1, we obtain (1.24).

9. Proof of Theorem (1.7)

Setting i = 2, j = 4, δ = 12 and k = 3 in (1.8), we see that

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{12,3} (n) q^{n} = \frac{f (q^{2}, q^{10}) f (q^{4}, q^{8}) {(q^{3}; q^{3})}_{\infty}^{2}}{f (q^{6}, q^{30}) f (q^{12}, q^{24}) {(q; q)}_{\infty}^{2}} .

(9.1)

By the definition of f(a, b) and the well-known Jacobi triple product identity, we get

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{12,3} (n) q^{n} = \frac{f_{3}^{2} f_{4} f_{6}^{2}}{f_{1}^{2} f_{2} f_{18} f_{36}} .

(9.2)

Substituting (2.7) into (9.2), we have

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{12,3} (n) q^{n} = \frac{f_{4}^{5} f_{6}^{3} f_{12}^{2}}{f_{2}^{6} f_{8} f_{18} f_{24} f_{36}} + 2 q \frac{f_{4}^{2} f_{6}^{4} f_{8} f_{24}}{f_{2}^{5} f_{12} f_{18} f_{36}} .

(9.3)

Equating odd parts of the aforementioned equation, we obtain

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{12,3} (2 n + 1) q^{n} = 2 \frac{f_{2}^{2} f_{3}^{4} f_{4} f_{12}}{f_{1}^{5} f_{6} f_{9} f_{18}} .

(9.4)

Involving (2.11) in (9.4), we get

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{12,3} (2 n + 1) q^{n} \equiv 2 \frac{f_{2}^{2} f_{6} f_{12}}{f_{1} f_{9} f_{18}} (\mod 4) .

(9.5)

Ramanujan recorded the following identity in his third note book:

ψ (q) = \frac{f_{2}^{2}}{f_{1}} = \frac{f_{6} f_{9}^{2}}{f_{3} f_{18}} + q \frac{f_{18}^{2}}{f_{9}} .

(9.6)

Substituting (9.6) into (9.5), we deduce that

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{12,3} (2 n + 1) q^{n} \equiv 2 \frac{f_{6}^{2} f_{9} f_{12}}{f_{3} f_{18}^{2}} + 2 q \frac{f_{6} f_{12} f_{18}}{f_{9}^{2}} (\mod 4) .

(9.7)

Congruence (1.25) easily follows from the aforementioned equation.

Extracting the terms involving q³ⁿ⁺¹ from (9.7), dividing by q and replacing q³ by q, we arrive at

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

(9.8)

Using (2.11) in (9.8), we get

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

(9.9)

Congruence (1.26) follows by extracting the terms involving q²ⁿ⁺¹ from (9.9).

From (9.9), we can reduce that

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

(9.10)

Employing (2.10) into (9.10), we get

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{12,3} (12 n + 3) q^{n} \equiv 2 \frac{f_{6} f_{9}^{4}}{f_{3} f_{18}^{2}} + 2 q f_{9} f_{18} (\mod 4) .

(9.11)

Congruence (1.27) follows from (9.11).

Extracting the terms involving q³ⁿ⁺¹ from (9.11), dividing by q and replacing q³ by q, we have

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{12,3} (36 n + 15) q^{n} \equiv 2 f_{3} f_{6} (\mod 4) .

(9.12)

Congruences (1.28) and (1.29) follow by extracting the terms involving q³ⁿ⁺¹ and q³ⁿ⁺² from (9.9).

From (9.12), we find that

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{12,3} (108 n + 15) q^{n} \equiv 2 f_{1} f_{2} (\mod 4) .

(9.13)

Combining (9.10) and (9.13), we get

{\bar{B}}_{2,4}^{12,3} (108 n + 15) \equiv {\bar{B}}_{2,4}^{12,3} (12 n + 3) (\mod 4) .

(9.14)

Using the aforementioned relation and by induction on α, we have

{\bar{B}}_{2,4}^{12,3} (4 \cdot 3^{2 α + 3} n + \frac{3 \cdot (3^{2 α + 2} + 1)}{2}) \equiv {\bar{B}}_{2,4}^{12,3} (12 n + 3) (\mod 4) .

(9.15)

Using (1.28) in (9.15), we obtain (1.30).

10. Proof of Theorem (1.8)

From (9.11), we find that

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{12,3} (36 n + 3) q^{n} \equiv 2 \frac{f_{2} f_{3}^{4}}{f_{1} f_{6}^{2}} (\mod 4) .

(10.1)

Invoking (2.11) in (10.1), we obtain

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{12,3} (36 n + 3) q^{n} \equiv 2 f_{1} (\mod 4) .

(10.2)

Employing Lemma (2.7) into (10.2), it can be see that

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{12,3} (36 (p n + \frac{p^{2} - 1}{24}) + 3) q^{n} \equiv 2 f_{p} (\mod 4),

(10.3)

which implies

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{12,3} (36 p^{2} n + \frac{3 (p^{3} + 1)}{2}) q^{n} \equiv 2 f_{1} (\mod 4) .

(10.4)

Therefore,

{\bar{B}}_{2,4}^{12,3} (36 p^{2} n + \frac{3 (p^{3} + 1)}{2}) \equiv {\bar{B}}_{2,4}^{12,3} (36 n + 3) (\mod 4) .

Using the aforementioned relation and by induction on α, we arrive at (1.31).

11. Proof of Theorem (1.9)

Combining (10.3) with Theorem (1.8), we derive that for α ≥ 0,

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{12,3} (36 p^{2 α + 1} n + \frac{3 (p^{3 α} + 1)}{2}) \equiv 2 f_{p} (\mod 4) .

Therefore, it follows that

\sum_{n = 0}^{\infty} {\bar{B}}_{2,4}^{12,3} (36 p^{2 α + 1} (p n + l) + \frac{3 (p^{3 α} + 1)}{2}) \equiv 0 (\mod 4) .

where l = 1, 2, …, p − 1, we obtain (1.32).

Dedicated to Prof. M. S. Mahadeva Naika on his 62nd birthday.

Acknowledgement:

The authors would like to thank the anonymous referee for his valuable suggestions to improve the quality of our paper.

Funding: This research received no specific grants from any funding agency in the public, commercial or not for profit sectors.

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S. Shivaprasada Nayaka, T.K. Sreelakshmi and Santosh Kumar

Published in Arab Journal of Mathematical Sciences. Published by Emerald Publishing Limited. This article is published under the Creative Commons Attribution (CC BY 4.0) licence. Anyone may reproduce, distribute, translate and create derivative works of this article (for both commercial and non-commercial purposes), subject to full attribution to the original publication and authors. The full terms of this licence may be seen at http://creativecommons.org/licences/by/4.0/legalcode

Arithmetic properties of singular overpartition pairs without multiples of k

1. Introduction

2. Preliminary results

3. Proof of Theorem (1.1)

4. Proof of Theorem (1.2)

5. Proof of Theorem (1.3)

6. Proof of Theorem (1.4)

7. Proof of Theorem (1.5)

8. Proof of Theorem (1.6)

9. Proof of Theorem (1.7)

10. Proof of Theorem (1.8)

11. Proof of Theorem (1.9)

Acknowledgement:

References

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Arithmetic properties of singular overpartition pairs without multiples of k Open Access

1. Introduction

2. Preliminary results

3. Proof of Theorem (1.1)

4. Proof of Theorem (1.2)

5. Proof of Theorem (1.3)

6. Proof of Theorem (1.4)

7. Proof of Theorem (1.5)

8. Proof of Theorem (1.6)

9. Proof of Theorem (1.7)

10. Proof of Theorem (1.8)

11. Proof of Theorem (1.9)

Acknowledgement:

References

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Arithmetic properties of singular overpartition pairs without multiples of k