Congruences modulo prime for fractional colour partition function

Rahman, Riyajur; Saikia, Nipen

doi:10.1108/AJMS-09-2021-0235

Purpose

Let p_[1,r;t] be defined by $\sum_{n = 0}^{\infty} p_{[1, r; t]} (n) q^{n} = {(E_{1} E_{r})}^{t},$ where t is a non-zero rational number, r ≥ 1 is an integer and $E_{r} = \prod_{n = 0}^{\infty} (1 - q^{r (n + 1)})$ for |q| < 1. The function p_[1,r;t](n) is the generalisation of the two-colour partition function p_[1,r;−1](n). In this paper, the authors prove some new congruences modulo odd prime ℓ by taking r = 5, 7, 11 and 13, and non-integral rational values of t.

Design/methodology/approach

Using q-series expansion/identities, the authors established general congruence modulo prime number for two-colour partition function.

Findings

In the paper, the authors study congruence properties of two-colour partition function for fractional values. The authors also give some particular cases as examples.

Originality/value

The partition functions for fractional value is studied in 2019 by Chan and Wang for Ramanujan's general partition function and then extended by Xia and Zhu in 2020. In 2021, Baruah and Das also proved some congruences related to fractional partition functions previously investigated by Chan and Wang. In this sequel, some congruences are proved for two-colour partitions in this paper. The results presented in the paper are original.

1. Introduction

A partition of a positive integer n is a non-increasing sequence of positive integers called parts, whose sum equals n. The number of partition of a non-negative integer n is usually denoted by p(n) with p(0) = 1. The generating function of p(n) (due to Euler) is given by

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

(1.1)

where for any complex number a,

{(a; q)}_{\infty} = \prod_{n = 0}^{\infty} (1 - a q^{n}), | q | < 1 .

(1.2)

For convenience, for any positive integer r, we write

E_{r} = {(q^{r}; q^{r})}_{\infty} .

Ramanujan [1, 2] found following three beautiful congruences modulo 5, 7 and 11 for the partition function p(n):

p (5 n + 4) \equiv 0 (\mod 5),

(1.3)

p (7 n + 5) \equiv 0 (\mod 7),

(1.4)

and

p (11 n + 6) \equiv 0 (\mod 11) .

(1.5)

In a letter to Hardy written from Fitzroy House late in 1918 [3, pp. 192–193], Ramanujan introduced the general partition function p_t(n) for any non-negative integer n and non-zero integer t as

\sum_{n = 0}^{\infty} p_{t} (n) q^{n} = \frac{1}{E_{1}^{t}} .

(1.6)

For t = 1, p₁(n) is the usual unrestricted partition function p(n) defined in (1.1). Ramanujan [3] further claimed that, if λ is a positive integer and $\overset{̅}{w}$ is a prime of the form 6λ − 1, then

p_{- 4} (n \overset{̅}{w} - \frac{(\overset{̅}{w} + 1)}{6}) \equiv 0 (\mod \overset{̅}{w}) .

(1.7)

After Ramanujan, the congruence properties of the partition function p_t(n) have been studied by many authors. For some recent works on p_t(n) , we refer to [4–13] and references therein.

Chan and Wang [14] studied the partition function p_t(n) for non-integral rational values of t and found many congruences. This paper closely follows the techniques of Chan and Wang. Xia and Zhu [15] proved many of the congruences conjectured in [14]. Recently, Baruah and Das [16] proved some new families of congruences modulo powers of primes for p_t(n) for non-integral rational values of t. Baruah and Das [16] also investigated another general partition function p_[1,r;t](n) which is defined by

\sum_{n = 0}^{\infty} p_{[1, r; t]} (n) = {(E_{1} E_{r})}^{t}

(1.8)

for non-integral fractional values of t. The partition function p_[1,r;t](n) is the generalisation of the two-colour partition function p_[1,r;−1](n) , one of the colours appears only in parts that are multiples of r. The partition function p_[1,r;−1](n) and other related functions have been studied by many authors. We refer [17–24] and references therein for details.

In [16], Baruah and Das established three congruences for p_[1,r;t](n) by taking r = 2, 3 and 4, and t = −a/b with gcd(a, b) = 1. For example, Baruah and Das [16, Theorem 1.13] proved that:

Theorem.

(Baruah & Das) [16, Theorem 1.13] Suppose $a, b, d \in Z$ , b ≥ 1 and gcd(a, b) = 1. Let ℓ be an odd prime divisor of a + db and 0 ≤ r < ℓ. Suppose d, ℓ and r satisfy the following conditions: d = 3, ℓ ≡ 5 or 11 (mod 12) and 2r + 1 ≡ 0 (mod ℓ). Then for all n ≥ 0,

p_{[1,3; - a / b]} (ℓ n + r) \equiv 0 (\mod ℓ) .

In this paper, we will prove four new congruences for p_[1,r;t](n) modulo an odd prime ℓ by taking r = 5, 7, 11 and 13 in (1.8) which are analogous to the above theorem. In order to verify some of our congruences, we list following q-series expansions:

{(E_{1} E_{5})}^{- 1 / 4} = 1 + \frac{q}{4} + \frac{13}{32} q^{2} + \frac{55}{128} q^{3} + \frac{1235}{2048} q^{4} + \frac{6663}{8192} q^{5} + \frac{59897}{65536} q^{6} + \frac{226967}{262144} q^{7} + \frac{9878595}{8388608} q^{8} + \frac{41862795}{33554432} q^{9} + \frac{514093347}{268435456} q^{10} + O {[q]}^{11},

(1.9)

{(E_{1} E_{7})}^{- 1 / 3} = 1 + \frac{q}{3} + \frac{5}{9} q^{2} + \frac{50}{81} q^{3} + \frac{215}{243} q^{4} + \frac{646}{729} q^{5} + \frac{8711}{6561} q^{6} + \frac{32036}{19683} q^{7} + \frac{112486}{59049} q^{8}

+ \frac{3386515}{1594323} q^{9} + \frac{12675397}{4782969} q^{10} + O {[q]}^{11},

(1.10)

{(E_{1} E_{11})}^{- 1 / 2} = 1 + \frac{q}{2} + \frac{7}{8} q^{2} + \frac{17}{16} q^{3} + \frac{203}{128} q^{4} + \frac{455}{256} q^{5} + \frac{2723}{1024} q^{6} + \frac{6001}{2048} q^{7} + \frac{133107}{32768} q^{8}

+ \frac{312011}{65536} q^{9} + \frac{1613529}{262144} q^{10} + O {[q]}^{11},

(1.11)

{(E_{1} E_{13})}^{1 / 2} = 1 - \frac{q}{2} - \frac{5}{8} q^{2} - \frac{5}{16} q^{3} - \frac{45}{128} q^{4} + \frac{33}{256} q^{5} - \frac{209}{1024} q^{6} + \frac{755}{2048} q^{7} + \frac{1155}{32768} q^{8}

+ \frac{15045}{65536} q^{9} + \frac{45077}{262144} q^{10} + O {[q]}^{11}

(1.12)

and

{(E_{1} E_{13})}^{- 5 / 6} = 1 + \frac{5}{6} q + \frac{115}{72} q^{2} + \frac{2915}{1296} q^{3} + \frac{112585}{31104} q^{4} + \frac{889249}{186624} q^{5} + \frac{49269935}{6718464} q^{6} + \frac{381199585}{40310784} q^{7} + \frac{26238866555}{1934917632} q^{8} + \frac{1853700203575}{104485552128} q^{9} + \frac{30342040805197}{1253826625536} q^{10} + O {[q]}^{11},

(1.13)

where O[q]¹¹ contains the terms involving q¹¹ or higher powers of q.

2. New congruences for p_[1,r;t](n)

In this section, we prove four new congruence modulo an odd prime ℓ. To prove our congruences, we employ the following q-series identity from [25]:

E_{1}^{3} = \sum_{n = - \infty}^{\infty} (4 n + 1) q^{({(4 n + 1)}^{2} - 1) / 8}

(2.1)

We also require the following congruence which follows from binomial theorem (or see [26]): For prime ℓ and integer k ≥ 1,

E_{k}^{ℓ} \equiv E_{k ℓ} (\mod ℓ) .

(2.2)

Theorem 2.1.

Suppose a and b are relatively prime integers with b ≥ 1 . Let ℓ be an odd prime divisor of a + 3b , and let r be an integer with 0 ≤ r < ℓ. Suppose ℓ and r satisfy any of the following two conditions:

4r + 3 ≡ 0 (mod ℓ), ℓ ≡ 2 or 3 (mod 5) and ℓ ≡ 1 (mod 4),
4r + 3 ≡ 0 (mod ℓ), ℓ ≡ 1 or 4 (mod 5) and ℓ ≡ 3 (mod 4).

Then for alln ≥ 0,

p_{[1,5; - a / b]} (ℓ n + r) \equiv 0 (\mod ℓ) .

(2.3)

Proof. Since ℓ|(a + db), so we can write a + db = ℓm for some integer m. Also, since gcd(a, b) = 1, it follows that gcd(b, ℓ) = 1. Setting t = −a/b in (1.8), we find that

\sum_{n = 0}^{\infty} p_{[1,5; - a / b]} (n) q^{n} = {(E_{1} E_{5})}^{- a / b} = \frac{{(E_{1} E_{5})}^{d}}{{(E_{1} E_{5})}^{ℓ m / b}} .

(2.4)

Employing () in , we obtain

\sum_{n = 0}^{\infty} p_{[1,5; - a / b]} (n) q^{n} \equiv \frac{{(E_{1} E_{5})}^{d}}{{(E_{ℓ} E_{5 ℓ})}^{m / b}} (\mod ℓ) .

(2.5)

For d = 3, from , we find that

{(E_{1} E_{5})}^{3} = \sum_{n = - \infty}^{\infty} \sum_{m = - \infty}^{\infty} (4 n + 1) (4 m + 1) q^{({(4 n + 1)}^{2} + 5 {(4 m + 1)}^{2} - 6) / 8} .

(2.6)

We note that

N = ({(4 n + 1)}^{2} + 5 {(4 m + 1)}^{2} - 6) / 8

is equivalent to

8 N + 6 = {(4 n + 1)}^{2} + 5 {(4 m + 1)}^{2} .

If

ℓ \equiv 2 or 3 (\mod 5) and ℓ \equiv 1 (\mod 4),

or

ℓ \equiv 1 or 4 (\mod 5) and ℓ \equiv 3 (\mod 4) .

then

(\frac{- 5}{ℓ}) = - 1

⁠, where here and throughout the paper

(\frac{\cdot}{\cdot})

denotes the Legendre symbol. So it follows that

8 N + 6 \equiv 0 (\mod ℓ),

or

4 N + 3 \equiv 0 (\mod ℓ),

if and only if 4n + 1 ≡ 0 (mod ℓ) and 4m + 1 ≡ 0 (mod ℓ). So the congruences (2.3) now follows by employing (2.6) and then comparing the coefficient of q^ℓn + r. □

Corollary 2.2.

We have

\begin{align} (i) & p_{[1,5; - 1 / 4]} (13 n + 9) \equiv 0 (\mod 13), \\ (i i) & p_{[1,5; - 2 / 3]} (11 n + 2) \equiv 0 (\mod 11), \\ (i i i) & p_{[1,5; - 2 / 5]} (17 n + 12) \equiv 0 (\mod 17), \\ (i v) & p_{[1,5; - 7 / 8]} (31 n + 7) \equiv 0 (\mod 31) . \end{align}

Proof. (i) For ℓ ≡ 3 (mod 5) and ℓ ≡ 1 (mod 4), take ℓ = 13.From [27, p. 176, Theorem 9.2(c)], we note that if p is an odd prime and a is a integer relatively prime to p, then

(\frac{a}{p}) \equiv a^{\frac{p - 1}{2}} (\mod p) .

(2.7)

From , it follows that

(\frac{- 5}{13}) \equiv {(- 5)}^{\frac{13 - 1}{2}} (\mod 13) .

Since

\frac{13 - 1}{2}

is an even integer, we have

\begin{align} (\frac{- 5}{13}) & \equiv {(5)}^{\frac{13 - 1}{2}} (\mod 13), \end{align}

(2.8)

\begin{align} \equiv (\frac{5}{13}) (\mod 13), \end{align}

(2.9)

As Legendre symbol take values 1 and −1 only, (2.16) gives

(\frac{- 5}{13}) = (\frac{5}{13}) .

(2.10)

Using Quadratic Reciprocity Law for Legendre symbol [27, p. 186], we find that

(\frac{5}{13}) (\frac{13}{5}) = {(- 1)}^{\frac{13 - 1}{2} \cdot \frac{5 - 1}{2}} = 1 .

Since 13 ≡ 3 (mod 5),

(\frac{13}{5}) = (\frac{3}{5}) \equiv 3^{\frac{5 - 1}{2}} \equiv 1 (\mod 5) .

So from ,

(\frac{- 5}{13}) = - 1

⁠. Now set a = 1 and b = 4 so that a + 3b is divisible by 13. Also, if r = 9, then 4r + 3 ≡ 0 (mod 13). Thus, (i) follows from Theorem 2.1.

Similarly, (ii)-(iv) follows from Theorem 2.1 with {a = 2, b = 3, ℓ = 11, r = 2}, { a = 1, b = 4, ℓ = 13, r = 9}, {a = 2, b = 5, ℓ = 17, r = 12} and {a = 7, b = 8, ℓ = 31, r = 7}, respectively. □

The congruence presented in Corollary 2.2(i) can be easily verified from the series expansion of ${(E_{1} E_{5})}^{- 1 / 4}$ in (1.9).

Theorem 2.3.

Suppose a and b are relatively prime integers with b ≥ 1 . Let ℓ be an odd prime divisor of a + 3b and let r be an integer with 0 ≤ r < ℓ. Suppose ℓ and r satisfy any of the following two conditions:

r + 1 ≡ 0 (mod ℓ), ℓ ≡ 3, 5, 6 (mod 7) and ℓ ≡ 1 (mod 4),
r + 1 ≡ 0 (mod ℓ), ℓ ≡ 3, 5, 6 (mod 7) and ℓ ≡ 3 (mod 4).

Then for alln ≥ 0,

p_{[1,7, - a / b]} (ℓ n + r) \equiv 0 (\mod ℓ) .

(2.11)

Proof. Since ℓ|(a + db), so we can write a + db = ℓm for some integer m. Also, since gcd(a, b) = 1, it follows that gcd(b, ℓ) = 1. Setting t = −a/b in (1.8), we find that

\sum_{n = 0}^{\infty} p_{[1,7, - a / b]} (n) q^{n} = {(E_{1} E_{7})}^{- a / b} = \frac{{(E_{1} E_{7})}^{d}}{{(E_{1} E_{7})}^{ℓ m / b}} .

(2.12)

Employing () in , we obtain

\sum_{n = 0}^{\infty} p_{[1,7, - a / b]} (n) q^{n} \equiv \frac{{(E_{1} E_{7})}^{d}}{{(E_{ℓ} E_{7 ℓ})}^{m / b}} (\mod ℓ) .

(2.13)

For d = 3, from , we find that

{(E_{1} E_{7})}^{3} = \sum_{n = - \infty}^{\infty} \sum_{m = - \infty}^{\infty} (4 n + 1) (4 m + 1) q^{({(4 n + 1)}^{2} + 7 {(4 m + 1)}^{2} - 6) / 8} .

(2.14)

We note that

N = ({(4 n + 1)}^{2} + 7 {(4 m + 1)}^{2} - 8) / 8,

is equivalent to

8 N + 8 = {(4 n + 1)}^{2} + 7 {(4 m + 1)}^{2} .

If

ℓ \equiv 3,5,6 (\mod 7) and ℓ \equiv 1 (\mod 4),

or

ℓ \equiv 3,5,6 (\mod 7) and ℓ \equiv 3 (\mod 4) .

then

(\frac{- 7}{ℓ}) = - 1

⁠. So it follows that

8 N + 8 \equiv 0 (\mod ℓ),

or

N + 1 \equiv 0 (\mod ℓ),

if and only if 4n + 1 ≡ 0 (mod ℓ) and 4m + 1 ≡ 0 (mod ℓ). So the congruences () now follow by employing () and then comparing the coefficient of q^ℓn + r. □

Corollary 2.4.

We have

\begin{align} (i) & p_{[1,7; - 1 / 3]} (5 n + 4) \equiv 0 (\mod 5), \\ (i i) & p_{[1,7; - 1 / 4]} (13 n + 12) \equiv 0 (\mod 13), \\ (i i i) & p_{[1,7; - 2 / 5]} (17 n + 16) \equiv 0 (\mod 17), \\ (i v) & p_{[1,7; - 1 / 6]} (19 n + 18) \equiv 0 (\mod 19) . \end{align}

Proof. For ℓ ≡ 5 (mod 7) and ℓ ≡ 1 (mod 4), set ℓ = 5. From (2.7), it follows that

(\frac{- 7}{5}) \equiv {(- 7)}^{\frac{5 - 1}{2}} \equiv {(7)}^{\frac{5 - 1}{2}} \equiv (\frac{7}{5}) (\mod 5) .

(2.15)

Using Quadratic Reciprocity Law, we have

(\frac{5}{7}) (\frac{7}{5}) = {(- 1)}^{\frac{5 - 1}{2} \cdot \frac{7 - 1}{2}} = 1 .

(2.16)

Now,

(\frac{5}{7}) \equiv 5^{\frac{7 - 1}{2}} \equiv - 1 (\mod 7) .

(2.17)

Therefore, by

(\frac{- 7}{5}) \equiv - 1

⁠. Take a = 1 and b = 3 so that a + 3b is divisible by 5, and also if r = 4, then r + 1 ≡ 0 (mod 5). So (i) follows immediately from Theorem 2.3.

Similarly, to prove (ii)–(iv) we set {a = 1, b = 4, ℓ = 13, r = 12}, {a = 2, b = 5, ℓ = 17, r = 16} and {a = 1, b = 6, ℓ = 19, r = 18} in Theorem 2.3, respectively. □

The congruence presented in Corollary 2.4(i) can be easily verified from the series expansion of ${(E_{1} E_{7})}^{(- 1 / 3)}$ in (1.10).

Theorem 2.5.

Suppose a and b are relatively prime integers with b ≥ 1 . Let ℓ be an odd prime divisor of a + 3b and let r be an integer with 0 ≤ r < ℓ. Suppose ℓ and r satisfy any of the following two conditions:

2r + 3 ≡ 0 (mod ℓ), ℓ ≡ 2, 6, 7, 8, 10 (mod 11) and ℓ ≡ 1 (mod 4),
2r + 3 ≡ 0 (mod ℓ), ℓ ≡ 2, 6, 7, 8, 10 (mod 11) and ℓ ≡ 3 (mod 4).

Then for alln ≥ 0,

p_{[1,11; - a / b]} (ℓ n + r) \equiv 0 (\mod ℓ) .

(2.18)

Proof. Since ℓ|(a + db), so we can write a + db = ℓm for some integer m. Also, since gcd(a, b) = 1, it follows that gcd(b, ℓ) = 1. Setting t = −a/b in (1.8), we find that

\sum_{n = 0}^{\infty} p_{[1,11; - a / b]} (n) q^{n} = {(E_{1} E_{11})}^{- a / b} = \frac{{(E_{1} E_{11})}^{d}}{{(E_{1} E_{11})}^{ℓ m / b}} .

(2.19)

Employing () in , we obtain

\sum_{n = 0}^{\infty} p_{[1,11; - a / b]} (n) q^{n} \equiv \frac{{(E_{1} E_{11})}^{d}}{{(E_{ℓ} E_{11 ℓ})}^{m / b}} (\mod ℓ) .

(2.20)

For d = 3, from , we find that

{(E_{1} E_{11})}^{3} = \sum_{n = - \infty}^{\infty} \sum_{m = - \infty}^{\infty} (4 n + 1) (4 m + 1) q^{({(4 n + 1)}^{2} + 11 {(4 m + 1)}^{2} - 12) / 8} .

(2.21)

We note that

N = ({(4 n + 1)}^{2} + 11 {(4 m + 1)}^{2} - 12) / 8,

is equivalent to

8 N + 12 = {(4 n + 1)}^{2} + 11 {(4 m + 1)}^{2} .

If

ℓ \equiv 2,6,7,8,10 (\mod 11) and ℓ \equiv 1 (\mod 4),

or

ℓ \equiv 2,6,7,8,10 (\mod 11) and ℓ \equiv 3 (\mod 4) .

then

(\frac{- 11}{ℓ}) = - 1

⁠. So it follows that

8 N + 12 \equiv 0 (\mod ℓ),

or

2 N + 3 \equiv 0 (\mod ℓ),

if and only if 4n + 1 ≡ 0 (mod ℓ) and 4m + 1 ≡ 0 (mod ℓ). So the congruences (2.18) now follow by employing (2.21) and then comparing the coefficient of q^ℓn + r.□

Corollary 2.6.

We have

\begin{align} (i) & p_{[1,11; - 1 / 2]} (7 n + 2) \equiv 0 (\mod 7) . \\ (i i) & p_{[1,11; - 2 / 5]} (17 n + 7) \equiv 0 (\mod 17) . \\ (i i i) & p_{[1,11; - 5 / 8]} (29 n + 13) \equiv 0 (\mod 29) . \\ (i v) & p_{[1,11; - 8 / 11]} (41 n + 19) \equiv 0 (\mod 41) . \end{align}

Proof. (i) For ℓ ≡ 7 (mod 11) and ℓ ≡ 1 (mod 4), take ℓ = 7. From (2.7), it follows that

(\frac{- 11}{7}) \equiv {(- 11)}^{\frac{7 - 1}{2}} \equiv {(11)}^{\frac{7 - 1}{2}} \equiv (\frac{11}{7}) (\mod 7) .

Using Quadratic Reciprocity Law, we have

(\frac{11}{7}) (\frac{7}{11}) = {(- 1)}^{\frac{11 - 1}{2} \cdot \frac{7 - 1}{2}} = 1 .

(2.22)

Now,

(\frac{7}{11}) \equiv {(7)}^{\frac{11 - 1}{2}} \equiv - 1 (\mod 5) .

Therefore, by

(\frac{- 11}{7}) \equiv - 1

⁠. Take a = 1 and b = 2 so that a + 3b is divisible by 7, and also if r = 2, then 2r + 3 ≡ 0 (mod 7). So (i) follows immediately from Theorem 2.5.

Similarly, we set {a = 2, b = 5, ℓ = 17, r = 7}, {a = 5, b = 8, ℓ = 29, r = 13} and {a = 8, b = 11, ℓ = 41, r = 19} in Theorem 2.5 to arrive at (ii)–(iv), respectively. □

The congruence in Corollary 2.6(i) can be easily verified from the series expansion of ${(E_{1} E_{11})}^{- 1 / 2}$ in (1.11).

Theorem 2.7.

Suppose a and b are relatively prime integers with b ≥ 1 . Let ℓ be an odd prime divisor of a + 3b and let r be an integer with 0 ≤ r < ℓ. Suppose ℓ and r satisfy any of the following two conditions:

(i) 4 r + 7 \equiv 0 (\mod ℓ), ℓ \equiv 2,5,6,7,8,11 (\mod 13) a n d ℓ \equiv 1 (\mod 4),

(i i) 4 r + 7 \equiv 0 (\mod ℓ), ℓ \equiv 1,3,4,9,10,12 (\mod 13) a n d ℓ \equiv 3 (\mod 4) .

Then for alln ≥ 0,

p_{[1,13, - a / b]} (ℓ n + r) \equiv 0 (\mod ℓ) .

(2.23)

Proof. Since ℓ|(a + db), so we can write a + db = ℓm for some integer m. Also, since gcd(a, b) = 1, it follows that gcd(b, ℓ) = 1. Setting t = −a/b in (1.8), we find that

\sum_{n = 0}^{\infty} p_{[1,13; - a / b]} (n) q^{n} = {(E_{1} E_{13})}^{- a / b} = \frac{{(E_{1} E_{13})}^{d}}{{(E_{1} E_{13})}^{ℓ m / b}} .

(2.24)

Employing () in , we obtain

\sum_{n = 0}^{\infty} p_{[1,13; - a / b]} (n) q^{n} \equiv \frac{{(E_{1} E_{13})}^{d}}{{(E_{ℓ} E_{13 ℓ})}^{m / b}} (\mod ℓ) .

(2.25)

For d = 3, from , we find that

{(E_{1} E_{13})}^{3} = \sum_{n = - \infty}^{\infty} \sum_{m = - \infty}^{\infty} (4 n + 1) (4 m + 1) q^{({(4 n + 1)}^{2} + 13 {(4 m + 1)}^{2} - 14) / 8} .

(2.26)

We note that

N = ({(4 n + 1)}^{2} + 13 {(4 m + 1)}^{2} - 14) / 8,

is equivalent to

8 N + 14 = {(4 n + 1)}^{2} + 13 {(4 m + 1)}^{2} .

Now if

ℓ \equiv 2,5,6,7,8,11 (\mod 13) a n d ℓ \equiv 1 (\mod 4),

or

ℓ \equiv 1,3,4,9,10,12 (\mod 13) a n d ℓ \equiv 3 (\mod 4),

then

(\frac{- 13}{ℓ}) = - 1

⁠. So it follows that

8 N + 14 \equiv 0 (\mod ℓ),

or

4 N + 7 \equiv 0 (\mod ℓ),

if and only if 4n + 1 ≡ 0 (mod ℓ) and 4m + 1 ≡ 0 (mod ℓ). So the congruences () now follows by employing () and then comparing the coefficient of q^ℓn + r. □

Corollary 2.8.

We have

\begin{align} (i) & p_{[1,13; 1 / 2]} (5 n + 2) \equiv 0 (\mod 5), \\ (i i) & p_{[1,13; - 5 / 6]} (23 n + 4) \equiv 0 (\mod 23), \\ (i i i) & p_{[1,13; - 7 / 10]} (37 n + 26) \equiv 0 (\mod 37), \\ (i v) & p_{[1,13; - 8 / 11]} (41 n + 29) \equiv 0 (\mod 41) . \end{align}

Proof. (i) For ℓ ≡ 5 (mod 13) and ℓ ≡ 1 (mod 4), set ℓ = 5. From (2.7), it follows that

(\frac{- 13}{5}) \equiv {(- 13)}^{\frac{5 - 1}{2}} \equiv 1 3^{\frac{5 - 1}{2}} \equiv (\frac{13}{5}) (\mod 5) .

Using Quadratic Reciprocity Law, we have

(\frac{5}{13}) (\frac{13}{5}) = {(- 1)}^{\frac{13 - 1}{2} \cdot \frac{5 - 1}{2}} = 1 .

(2.27)

Now,

(\frac{5}{13}) \equiv 5^{\frac{13 - 1}{2}} \equiv - 1 (\mod 13) .

Therefore, by

(\frac{- 13}{5}) \equiv - 1

⁠. Now set a = −1 and b = 2 so that a + 3b is divisible by 5 and also if r = 2, then 4r + 7 ≡ 0 (mod 5). So (i) follows immediately from Theorem 2.7.

(ii) For ℓ ≡ 10 (mod 13) and ℓ ≡ 3 (mod 4), take ℓ = 23. From (2.7), it follows that

(\frac{- 13}{23}) \equiv {(- 13)}^{\frac{23 - 1}{2}} \equiv - {(13)}^{\frac{23 - 1}{2}} \equiv - (\frac{13}{23}) (\mod 23) .

Using Quadratic Reciprocity Law, we have

(\frac{23}{13}) (\frac{13}{23}) = {(- 1)}^{\frac{13 - 1}{2} \cdot \frac{23 - 1}{2}} = 1 .

(2.28)

Now,

(\frac{23}{13}) = (\frac{10}{13}) \equiv 1 0^{\frac{13 - 1}{2}} \equiv 1 (\mod 13) .

Therefore, () implies

(\frac{- 13}{23}) \equiv - 1

⁠. Take a = 5 and b = 6 so that a + 3b is divisible by 23, and also if r = 4, then 4r + 7 ≡ 0 (mod 23). So (ii) follows immediately from Theorem 2.7.

Similarly, we set {a = 7, b = 10, ℓ = 37, r = 26} and {a = 8, b = 11, ℓ = 41, r = 29} in Theorem 2.7 to arrive at (iii)–(iv), respectively.□

The congruences presented in Corollary 2.8(i) and (ii) can be verified from the series expansion of ${(E_{1} E_{13})}^{1 / 2}$ and ${(E_{1} E_{13})}^{- 5 / 6}$ in (1.12) and (1.13), respectively.

The authors would like to thank anonymous reviewers for their valuable comments which improved the quality of the paper.

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Riyajur Rahman and Nipen Saikia

Published in the Arab Journal of Mathematical Sciences. Published by Emerald Publishing Limited. This article is published under the Creative Commons Attribution (CC BY 4.0) license. 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 license may be seen at http://creativecommons.org/licences/by/4.0/legalcode

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