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Arithmetic convolution sums derived from eta quotients related to divisors of 6

  • Nazli Yildiz Ikikardes , Jihyun Hwang and Daeyeoul Kim EMAIL logo
Published/Copyright: April 19, 2022
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Open Mathematics
From the journal Open Mathematics Volume 20 Issue 1

Abstract

The aim of this paper is to find arithmetic convolution sums of some restricted divisor functions. When divisors of a certain natural number satisfy a suitable condition for modulo 12, those restricted divisor functions are expressed by the coefficients of certain eta quotients. The coefficients of eta quotients are expressed by the sine function and cosine function, and this fact is used to derive formulas for the convolution sums of restricted divisor functions and of the number of divisors. In the sine function used to find the coefficients of eta quotients, the result is obtained by utilizing a feature with symmetry between the divisor and the corresponding divisor. Let N , r be positive integers and d be a positive divisor of N . Let e r ( N ; 12 ) denote the difference between the number of 2 N d − d congruent to r modulo 12 and the number of those congruent to − r modulo 12. The main results of this article are to find the arithmetic convolution identities for ∑ a 1 + ⋯ + a j = N ( ∏ i = 1 j e ˆ ( a i ) ) with e ˆ ( a i ) = e 1 ( a i ; 12 ) + 2 e 3 ( a i ; 12 ) + e 5 ( a i ; 12 ) and j = 1 , 2 , 3 , 4 . All results are obtained using elementary number theory and modular form theory.

Keywords: restricted divisor functions; eta quotient; convolution sums; q-series
MSC 2010: 11A07; 11A25

1 Introduction

Throughout this paper, N , N 0 , and Z will denote the set of natural numbers, the set of non-negative integers, and the ring of integers, respectively. For d , m , N ∈ N , r , s ∈ N 0 , and A , B ⊂ N 0 , we define some restricted divisor functions for our use in the sequel. Let

E r ( N ; m ) ≔ ∑ d ∣ N d ≡ r ( mod m ) 1 − ∑ d ∣ N d ≡ − r ( mod m ) 1 , E r , … , s ( N ; m ) ≔ E r ( N ; m ) + ⋯ + E s ( N ; m ) , e r ( N ; m ) ≔ ∑ d ∣ N 2 N d − d ≡ r ( mod m ) 1 − ∑ d ∣ N 2 N d − d ≡ − r ( mod m ) 1 , e r , … , s ( N ; m ) ≔ e r ( N ; m ) + ⋯ + e s ( N ; m ) , e ˆ ( N ) ≔ e 1 , 5 ( N ; 12 ) + 2 e 3 ( N ; 12 ) , A ⊗ B = { a ⋅ b ∣ a ∈ A , b ∈ B } , σ s ( N ) ≔ ∑ d ∣ N d s , σ ¯ 2 ( N ) ≔ ∑ d ∣ N d ≡ 1 ( mod 4 ) d 2 − ∑ d ∣ N d ≡ − 1 ( mod 4 ) d 2 , σ ˜ 2 ( N ) ≔ ∑ d ∣ N N d ≡ 1 ( mod 4 ) d 2 − ∑ d ∣ N N d ≡ − 1 ( mod 4 ) d 2 ,

and

σ ˆ ( N ) ≔ 2 ∑ d ∣ N d ≡ N d ( mod 6 ) d − ∑ d ∣ N d − N d ≡ 3 ( mod 6 ) d + ∑ d ∣ N d − N d ≡ ± 1 ( mod 6 ) d − ∑ d ∣ N d − N d ≡ ± 2 ( mod 6 ) d .

Here, d ∣ N means that d is a divisor of N . We also make use of the following convention:

σ s ( N ) = E r ( N ; m ) = e r ( N ; m ) = 0 if N ∉ Z or N ≤ 0 , σ ( N ) ≔ σ 1 ( N ) = ∑ d ∣ N d .

The exact evaluation of the basic convolution sum ∑ k = 1 N − 1 σ 1 ( k ) σ 1 ( N − k ) first appeared in a letter from Besge to Liouville in 1862 [1]. Much is known about the convolution sums of the divisor functions ∑ k = 1 N − 1 σ s ( k ) σ r ( N − k ) and ∑ k = 1 N − 1 σ s , r ( k ; m ) σ t , r ( N − k ; m ) , where σ s , r ( N ; m ) = ∑ d ∣ N d ≡ r ( mod m ) d s with r , s , t ∈ N 0 and m , N ∈ N . Among them, the beautiful results were found by Ramanujan [2,3] and Glaisher [4,5, 6,7] introduced interesting results. In recent years, related studies have been fulfilled in [8,9, 10,11]. The results of Cangul for special convolution sums related to a new graph invariant Ω can also be found in [12,13]. In the convolution sum of the restricted divisor functions ∑ k = 1 N − 1 E s ( k ; m ) E r ( N − k ; m ) , we introduce the results of Farkas [14,15], Williams [16], Guerzhoy and Raji [17], and Raji [18]. The convolution sum is characterized by having the same first and last term, and two symmetry structures in each term. Most of the aforementioned studies are the results of considering the divisors as modulo m . However, this paper attempts to calculate the convolution sums with a condition that considers the divisor of a given number and the corresponding divisor together. In other words, for a given natural number N and its divisor ω , we consider a convolution sum with the condition ω ≡ i ( mod m ) instead of the condition 2 N ω − ω ≡ i ( mod m ) with i ∈ N 0 and 0 ≤ i < m . Specifically, in this paper, we would like to deal with ∑ a 1 + ⋯ + a j = N ( ∏ i = 1 j e ˆ ( a i ) ) , a 1 , … , a j ∈ N , and j = 1 , 2 , 3 , 4 . Our results are different from those of Ramanujan and Farkas because the condition of the divisors of a given number is different. In order to derive these results, identities of infinite sums and infinite products given by eta quotients are needed. So we introduce eta quotients below.

Eta quotients are important subjects that are found in many fields of the theory of basic hyper-geometric series, partition functions, and modular forms [19]. An eta quotient is a function of the form f ( τ ) = ∏ ω ∣ T η b ω ( ω τ ) , where η is the Dedekind eta function defined by η ( τ ) ≔ q 1 24 ∏ n ≥ 1 ( 1 − q n ) . Here, q will denote a fixed complex number with ∣ q ∣ < 1 and b ω , T ∈ Z , so that we may write q = e 2 π i τ , where I m ( τ ) > 0 .

From here, we introduce the basic identity for infinite sums and infinite products through the work of Fine [20]. Let us define

(1.1) H ( q ) k ≔ η ( τ ) η ( 2 τ ) η ( 3 τ ) η ( 6 τ ) k = ∏ n ≥ 1 ( 1 − q n ) ( 1 − q 2 n ) ( 1 − q 3 n ) ( 1 − q 6 n ) k = ∑ N = 0 ∝ a k ( N ) q N

with k ∈ N . Here, a k ( 0 ) = 1 and a k ( N ) is the coefficient of q N in H ( q ) k with N ≥ 1 . In this article, the coefficients of H ( q ) k related to positive divisors of 6 are studied.

More precisely, we prove the following theorems.

Theorem 1

Let a 1 , … , a k , k , N ∈ N . If k − N ≡ 2 ( mod 4 ) , then

∑ a 1 + ⋯ + a k = N a 1 , … , a k odd e ˆ ( a 1 ) ⋯ e ˆ ( a k ) = ∑ a 1 + ⋯ + a k = N a 1 , … , a k : odd a 1 ( a 1 ) ⋯ a 1 ( a k ) = 0 .

For any N , does σ ( N ) become odd? The answer to this is well-known to [21, p. 28]:

“ σ ( N ) is odd if and only if N is a perfect square.”

Naturally, for other arithmetic functions, we can think of this question. Theorems 2 and 3 can give partial answers in terms of a i ( N ) with i = 1 , 2 .

Theorem 2

Let N be an odd positive integer. a 1 ( N ) is odd if and only if N ≡ 1 ( mod 12 ) is a perfect square. Furthermore,

a 1 ( N ) = − e ˆ ( N ) = − E 1 , 5 ( N ; 12 ) if N ≡ 1 ( mod 12 ) , 4 E 1 ( N ; 12 ) if N ≡ 5 ( mod 12 ) , − 4 E 1 , 5 ( N ; 12 ) if N ≡ 9 ( mod 12 ) , 0 if N ≡ 3 ( mod 4 ) .

Theorem 3

There does not exist an odd positive integer N satisfying a 2 ( N ) ≡ 1 ( mod 2 ) .

Theorem 4

Let N ( ≥ 2 ) be an integer with N = 2 n ⋅ 3 m ⋅ M ′ with gcd ( 6 , M ′ ) = 1 and n , m ∈ N 0 . Then

∑ k = 1 N − 1 e ˆ ( k ) e ˆ ( N − k ) = − 2 ( χ ( N ; 12 ) + 6 σ ( M ′ ) ) if N ≡ 0 ( mod 12 ) , 2 ( E 1 , 5 ( N ; 12 ) − σ ( N ) ) if N ≡ 1 ( mod 12 ) , 4 E 1 , 5 ( M ′ ; 12 ) − 3 σ ( M ′ ) if N ≡ 2 ( mod 12 ) , 4 σ ( M ′ ) if N ≡ 3 ( mod 12 ) , − 2 ( E 1 , 5 ( M ′ ; 12 ) − 3 σ ( M ′ ) ) if N ≡ 4 ( mod 12 ) , − 8 E 1 ( N ; 12 ) + σ ( N ) if N ≡ 5 ( mod 12 ) , − 2 ( χ ( N ; 12 ) + 6 σ ( M ′ ) ) if N ≡ 6 ( mod 12 ) , − 2 σ ( N ) if N ≡ 7 ( mod 12 ) , 4 E 1 , 5 ( M ′ ; 12 ) − 3 σ ( M ′ ) if N ≡ 8 ( mod 12 ) , 4 ( 2 E 1 , 5 ( N ; 12 ) + σ ( M ′ ) ) if N ≡ 9 ( mod 12 ) , − 2 ( E 1 , 5 ( M ′ ; 12 ) − 3 σ ( M ′ ) ) if N ≡ 10 ( mod 12 ) , σ ( N ) if N ≡ 11 ( mod 12 ) .

Here,

χ ( N ; 12 ) ≔ 0 if m ≡ 1 ( mod 2 ) or M ′ ≡ 7 , 11 ( mod 12 ) , 4 E 1 , 5 ( M ′ ; 12 ) if m ≡ 0 ( mod 2 ) and M ′ ≡ 1 , 5 ( mod 12 ) .

In particular, we obtain (Table 1).

Table 1

Values of Σ k = 1 N − 1 e ˆ ( k ) e ˆ ( N − k ) when N = 2 n ⋅ 3 m

N 2 2 n + 1 2 2 n + 2 3 2 m + 1 3 2 m + 2 2 n + 1 ⋅ 3 2 m + 1 2 n + 1 ⋅ 3 2 m
∑ k = 1 N − 1 e ˆ ( k ) e ˆ ( N − k ) 1 4 4 12 − 12 − 20

Using Theorems 1 and 4, we obtain:

Corollary 1

Let 2 N = 2 n ⋅ 3 m ⋅ M ′ be a positive integer with gcd ( 6 , M ′ ) = 1 , m ∈ N 0 , and n ∈ N . Then

∑ k = 1 N − 1 e ˆ ( 2 k ) e ˆ ( 2 N − 2 k ) = − 2 ( χ ( N ; 12 ) + 6 σ ( M ′ ) ) if 2 N ≡ 0 ( mod 12 ) , − 2 ( E 1 , 5 ( M ′ ; 12 ) − 3 σ ( M ′ ) ) if 2 N ≡ 4 ( mod 12 ) , 4 E 1 , 5 ( M ′ ; 12 ) − 3 σ ( M ′ ) if 2 N ≡ 8 ( mod 12 ) .

From now on, using modular form theory, we obtain the formulas ∑ a 1 + ⋯ + a j = N ( ∏ i = 1 j e ˆ ( a i ) ) , a 1 , … , a j ∈ N and j = 3 , 4 . Here, N is an odd integer. If N is an even integer, then the formula is very complex, so we will not deal with it in this article.

Let us define

T 1 ( q ) ≔ η ( 4 τ ) 2 η ( 8 τ ) 2 η ( 12 τ ) 4 η ( 24 τ ) 2 = q ∏ n ≥ 1 ( 1 − q 4 ) 2 ( 1 − q 8 ) 2 ( 1 − q 12 ) 4 ( 1 − q 24 ) 2 , T 2 ( q ) ≔ η ( 2 τ ) η ( 6 τ ) η ( 8 τ ) 4 η ( 12 τ ) η ( 4 τ ) = q 2 ∏ n ≥ 1 ( 1 − q 2 ) ( 1 − q 6 ) ( 1 − q 8 ) 4 ( 1 − q 12 ) ( 1 − q 4 ) , T 3 ( q ) ≔ η ( 4 τ ) 4 η ( 12 τ ) 2 η ( 24 τ ) 2 η ( 8 τ ) 2 = q 3 ∏ n ≥ 1 ( 1 − q 4 ) 4 ( 1 − q 12 ) 2 ( 1 − q 24 ) 2 ( 1 − q 8 ) 2 , T 4 ( q ) ≔ η ( 2 τ ) η ( 4 τ ) η ( 6 τ ) η ( 24 τ ) 4 η ( 12 τ ) = q 4 ∏ n ≥ 1 ( 1 − q 2 ) ( 1 − q 4 ) ( 1 − q 6 ) ( 1 − q 24 ) 4 ( 1 − q 12 ) ,

and

T ( q ) ≔ 18 7 ( T 1 ( q ) + 2 T 2 ( q ) + T 3 ( q ) + 2 T 4 ( q ) ) = ∑ N ≥ 1 t ( N ) q N .

Theorem 5

Let N ( ≥ 3 ) be an odd positive integer with N = 3 m ⋅ M ′ with gcd ( 3 , M ′ ) = 1 and m ∈ N 0 . Then

∑ a 1 + a 2 + a 3 = N a 1 , a 2 , a 3 ≥ 1 e ˆ ( a 1 ) e ˆ ( a 2 ) e ˆ ( a 3 ) = 3 E 1 , 5 ( N ; 12 ) − 6 σ ( N ) + 3 7 σ ¯ 2 ( N ) + t ( N ) if N ≡ 1 ( mod 12 ) , 12 σ ( M ′ ) − 5 7 σ ¯ 2 ( N ) − 135 7 σ ¯ 2 N 3 + t ( N ) if N ≡ 3 ( mod 12 ) , − 12 E 1 ( N ; 12 ) + 3 σ ( N ) + 3 7 σ ¯ 2 ( N ) + t ( N ) if N ≡ 5 ( mod 12 ) , − 6 σ ( N ) − 5 7 σ ¯ 2 ( N ) + t ( N ) if N ≡ 7 ( mod 12 ) , 12 E 1 , 5 ( N ; 12 ) + 12 σ ( M ′ ) + 3 7 σ ¯ 2 ( N ) + 81 7 σ ¯ 2 N 3 + t ( N ) if N ≡ 9 ( mod 12 ) , 3 σ ( N ) − 5 7 σ ¯ 2 ( N ) + t ( N ) if N ≡ 11 ( mod 12 ) .

To find the formula of ∑ a 1 + a 2 + a 3 + a 4 = N e ˆ ( a 1 ) e ˆ ( a 2 ) e ˆ ( a 3 ) e ˆ ( a 4 ) , we need the following eta quotients:

S i ( q ) ≔ η ( τ ) 10 − 2 i η ( 3 τ ) i − 1 η ( 6 τ ) 5 − i η ( 18 τ ) 2 i − 2 η ( 2 τ ) 5 − i η ( 9 τ ) i − 1 = q i ∏ n ≥ 1 ( 1 − q ) 10 − 2 i ( 1 − q 3 ) i − 1 ( 1 − q 6 ) 5 − i ( 1 − q 18 ) 2 i − 2 ( 1 − q 2 ) 5 − i ( 1 − q 9 ) i − 1

and

S ( q ) ≔ − 3 5 ( 7 S 1 ( q ) + 64 S 2 ( q ) + 192 S 3 ( q ) + 192 S 4 ( q ) ) + 4 T ( q ) = ∑ N ≥ 1 s ( N ) q N .

Theorem 6

Let N ( ≥ 4 ) be an odd positive integer with N = 3 m ⋅ M ′ with gcd ( 3 , M ′ ) = 1 and m ∈ N 0 . Then

∑ a 1 + a 2 + a 3 + a 4 = N a 1 , a 2 , a 3 , a 4 ≥ 1 e ˆ ( a 1 ) e ˆ ( a 2 ) e ˆ ( a 3 ) e ˆ ( a 4 ) = 4 E 1 , 5 ( N ; 12 ) − 12 σ ( N ) + 12 7 σ ¯ 2 ( N ) + 1 5 σ 3 ( N ) + s ( N ) if N ≡ 1 ( mod 12 ) , − 16 E 1 ( N ; 12 ) + 6 σ ( N ) + 12 7 σ ¯ 2 ( N ) − 2 5 σ 3 ( N ) + s ( N ) if N ≡ 5 ( mod 12 ) , − 12 σ ( N ) − 20 7 σ ¯ 2 ( N ) + 1 5 σ 3 ( N ) + s ( N ) if N ≡ 7 ( mod 12 ) , 6 σ ( N ) − 20 7 σ ¯ 2 ( N ) − 2 5 σ 3 ( N ) + s ( N ) if N ≡ 11 ( mod 12 ) , 24 σ ( M ′ ) − 20 7 σ ¯ 2 ( N ) − 540 7 σ ¯ 2 N 3 + 28 5 σ 3 N 3 + s ( N ) if N ≡ 3 , 15 ( mod 36 ) , 24 σ ( M ′ ) − 20 7 σ ¯ 2 ( N ) − 540 7 σ ¯ 2 N 3 − 1 10 σ 3 ( N ) + 42 5 σ 3 N 3 − 243 10 σ 3 N 9 + s ( N ) if N ≡ 27 ( mod 36 ) , 16 E 1 , 5 ( N ; 12 ) + 24 σ ( M ′ ) + 12 7 σ ¯ 2 ( N ) + 324 7 σ ¯ 2 N 3 − 1 10 σ 3 ( N ) + 42 5 σ 3 N 3 − 243 10 σ 3 N 9 + s ( N ) if N ≡ 9 ( mod 36 ) , 16 E 1 , 5 ( N ; 12 ) + 24 σ ( M ′ ) + 12 7 σ ¯ 2 ( N ) + 324 7 σ ¯ 2 N 3 + 28 5 σ 3 N 3 + s ( N ) if N ≡ 21 , 33 ( mod 36 ) .

This paper is organized as follows. In Section 2, some properties of certain infinite products and infinite sums are given. By using these equations, we derive computation formulas for the restricted divisor functions. In Section 3, we give values of a 1 ( N ) with N ≡ 1 (mod 2). Furthermore, we prove Theorems 1 and 2. In Section 4, we give values of a 1 ( N ) with N ≡ 1 ± 2 , ± 4 (mod 12). In Section 5, we give values of a 1 ( N ) with N ≡ 6 (mod 12). In Section 6, we give values of a 1 ( N ) with N ≡ 0 (mod 12). In Section 7, we give values of a 2 ( N ) and prove Theorems 3 and 4. In Section 8, we prove Theorems 5 and 6 using the theory of modular forms.

2 Preliminary

In [20, p. 10, 21], we find two curious identities

(2.1) ∏ n ≥ 1 ( 1 − q n ) 2 ( 1 − 2 q n cos 2 u + q 2 n ) = 1 − 4 sin u ∑ N ≥ 1 q N ∑ ω ∣ N sin 2 N ω − ω u

and

(2.2) ∏ n ≥ 1 ( 1 − q n ) 4 ( 1 − 2 q n cos u + q 2 n ) 2 = 1 − 8 sin 2 u 2 ∑ N ≥ 1 q N ∑ n k = N n , k ≥ 1 n cos ( k − n ) u .

Set u = π 6 in (2.1):

(2.3) ∏ n ≥ 1 ( 1 − q n ) ( 1 − q 2 n ) ( 1 − q 3 n ) ( 1 − q 6 n ) = 1 − 2 ∑ N ≥ 1 q N ∑ ω ∣ N sin 2 N ω − ω π 6 .

Now if we set u = π 3 in (2.2), we obtain

(2.4) ∏ n ≥ 1 ( 1 − q n ) ( 1 − q 2 n ) ( 1 − q 3 n ) ( 1 − q 6 n ) 2 = 1 − 2 ∑ N ≥ 1 q N ∑ n k = N n , k ≥ 1 n cos ( k − n ) π 3 .

Let ( a ¯ , b ¯ ) ≔ ( a , b ) ( mod 12 ) , that is, a ¯ ≡ a ( mod 12 ) and b ¯ ≡ b ( mod 12 ) . Table 2 shows values of sin 2 N ω − ω π 6 for each divisor ω of N . Table 3 is the operation table for the result of calculating the value of 2 N ω ⋅ ω (mod  12).

Table 2

Values of sin 2 N ω − ω π 6 when 2 N ω − ω ≡ i (mod 12)

2 N ω − ω ≡ i (mod 12) 0 1 2 3 4 5 6 7 8 9 10 11
sin 2 N ω − ω π 6 0 1 2 3 2 1 3 2 1 2 0 − 1 2 − 3 2 − 1 − 3 2 − 1 2
Table 3

Multiplicative operation table of ω ¯ , N ω ¯ satisfying ω ∣ N

N ω ¯ ω ¯ 0 1 2 3 4 5 6 7 8 9 10 11
0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 ♣ 1 2 ♠ 3 4 ⋆ 5 6 ♡ 7 8 ♢ 9 10 △ 11
2 0 † 2 4 ♮ 6 8 • 10 0 † 2 4 ♮ 6 8 • 10
3 0 ♠ 3 6 ♢ 9 0 ♠ 3 6 ♢ 9 0 ♠ 3 6 ♢ 9
4 0 □ 4 8 0 4 ∇ 8 0 □ 4 8 0 4 ∇ 8
5 0 ⋆ 5 10 ♠ 3 8 ♣ 1 6 △ 11 4 ♢ 9 2 ♡ 7
6 0 ♮ 6 0 ♮ 6 0 ♮ 6 0 ♮ 6 0 ♮ 6 0 ♮ 6
7 0 ♡ 7 2 ♢ 9 4 △ 11 6 ♣ 1 8 ♠ 3 10 ⋆ 5
8 0 ∇ 8 4 0 8 □ 4 0 ∇ 8 4 0 8 □ 4
9 0 ♢ 9 6 ♠ 3 0 ♢ 9 6 ♠ 3 0 ♢ 9 6 ♠ 3
10 0 • 10 8 ♮ 6 4 † 2 0 • 10 8 ♮ 6 4 † 2
11 0 △ 11 10 ♢ 9 8 ♡ 7 6 ⋆ 5 4 ♠ 3 2 ♣ 1

Lemma 1

Let N be a positive integer. Then

T even ( N ) ≔ ∑ ω ∣ N 2 N ω − ω ≡ 0 ( mod 2 ) sin 2 N ω − ω π 6 = 0 .

Proof

Let ω be a positive divisor of N . If N is an odd integer, then 2 N ω ≡ 0 ( mod 2 ) , ω ≡ 1 ( mod 2 ) , 2 N ω − ω ≡ 1 ( mod 2 ) , and thus T even ( N ) = 0 . So, we consider an even integer N and an even divisor ω satisfying 2 N ω − ω ≡ 0 ( mod 2 ) . It is easily checked that

(2.5) ∑ ω ∣ N 2 N ω − ω ≡ 0 ( mod 6 ) sin 2 N ω − ω π 6 = 0 .

Let S k ≔ 2 N ω , ω ∈ N × N ∣ 2 N ω − ω ≡ k ( mod 12 ) , ω ∣ N with k = 2 , 4 , 8 , 10 .

If 2 N ω − ω ≡ 2 (resp., 4) ( mod 12 ) , then N = 2 N ω ⋅ ω 2 and thus, 2 N ω ∣ N . Put ω ′ ≔ 2 N ω . Then

(2.6) 2 N ω − ω = ω ′ − 2 N ω ′ = − 2 N ω ′ − ω ′ and ω ≠ 2 N ω .

By (2.6), we directly see that

(2.7) 2 N ω − ω ≡ 2 ( resp. , 4 ) ( mod 12 ) if and only if 2 N ω ′ − ω ′ ≡ 10 ( resp. , 8 ) ( mod 12 ) .

Thus, by (2.6) and (2.7), f i : S 2 i → S 12 − 2 i with f i 2 N ω , ω = 2 N ω ′ , ω ′ is a bijective map and

(2.8) # S 2 i = # S 12 − 2 i with i = 1 , 2 .

Combining (2.8) with (2.5), we find

T even ( N ) = ∑ k = 0 5 ∑ ω ∣ N 2 N ω − ω ≡ 2 k ( mod 12 ) sin 2 N ω − ω π 6 = ∑ k = 1 2 ∑ ω ∣ N 2 N ω − ω ≡ ± 2 k ( mod 12 ) sin 2 N ω − ω π 6 = ( # S 2 + # S 4 ) 3 2 + ( # S 8 + # S 10 ) − 3 2 = 0 .

This completes the proof of Lemma 1.□

Using (2.1) and Table 2 we obtain the following result.

Remark 1

Let ω be a divisor of N satisfying 2 N ω − ω ≡ 0 ( mod 2 ) . It is easily verified that

(2.9) 2 N ω − ω ≡ 0 ( mod 2 ) if and only if ω ≡ 0 ( mod 2 ) .

Combining Lemma 1 with (2.9), we obtain

(2.10) T even ( N ) = ∑ ω ∣ N ω even sin 2 N ω − ω π 6 = 0 .

Lemma 2

Let N be a positive integer. Then

T odd ( N ) ≔ ∑ ω ∣ N 2 N ω − ω ≡ 1 ( mod 2 ) sin 2 N ω − ω π 6 = ∑ ω ∣ N ω odd sin 2 N ω − ω π 6 = 1 2 e ˆ ( N ) .

Proof

By (2.9) and (2.10), we can obtain

T odd ( N ) = ∑ ω ∣ N 2 N ω − ω ≡ 1 ( mod 2 ) sin 2 N ω − ω π 6 = ∑ ω ∣ N ω odd sin 2 N ω − ω π 6 .

Also by Table 2,

T odd ( N ) = ∑ k = 1 6 ∑ ω ∣ N 2 N ω − ω ≡ 2 k − 1 ( mod 12 ) sin 2 N ω − ω π 6 = 1 2 e ˆ ( N ) . □

Theorem 7

Let N ∈ N 0 . Then

a 1 ( N ) = 1 if N = 0 , − e ˆ ( N ) if N > 0 .

Proof

Equation (2.3) yields a 1 ( 0 ) = 1 . In fact, we can see that a 1 ( N ) = − 2 ( T odd ( N ) + T even ( N ) ) . Comparing Lemmas 1 and 2 with (2.3), we obtain the proof of Theorem 7.□

Remark 2

Comparing the infinite products for e ˆ ( N ) and E ˆ ( N ) ≔ E 1 , 5 ( N ; 12 ) + 2 E 3 ( N ; 12 ) , they have curious forms. E ˆ ( N ) is a simpler arithmetic function than e ˆ ( N ) , but from a perspective of infinite products, the infinite product for e ˆ ( N ) seems simpler than the infinite product for E ˆ ( N ) (Table 4).

Table 4

Infinite product forms of E ˆ ( N ) and e ˆ ( N )

E ˆ ( N ) e ˆ ( N )
Infinite product forms ∏ n ≥ 1 ( 1 − q 2 n ) ( 1 − q 4 n ) 2 ( 1 − q 6 n ) 3 ( 1 − q n ) ( 1 − q 3 n ) ( 1 − q 12 n ) 2 ∏ n ≥ 1 ( 1 − q n ) ( 1 − q 2 n ) ( 1 − q 3 n ) ( 1 − q 6 n )
References [20, p. 82] (1.1), Theorem 7

3 Coefficient of a 1 ( N ) with odd N

Lemma 3

If N ≡ 7 ( mod 12 ) is a natural number, then a 1 ( N ) = 0 .

Proof

By (2.3), a 1 ( N ) = − 2 ∑ ω ∣ N sin 2 N ω − ω π 6 . In Table 3, ♡ represents all cases N ω , ω satisfying N ω ⋅ ω ≡ 7 ( mod 12 ) and ω ∣ N . We note that ω 2 ≡ 1 ( mod 12 ) when ω ≡ 1 , 5 , 7 , 11 ( mod 12 ) . Hence,

(3.1) 2 N ω − ω ≡ 2 ω − ω ≡ 1 ω 2 ( 2 ω − ω 2 ⋅ ω ) ≡ 2 ω − ω ≡ ω ( mod 12 ) .

Here, 1 ω means that ω ⋅ 1 ω ≡ 1 ( mod 12 ) and 1 ω ∈ Z .

Using Table 2 and (3.1),

(3.2) a 1 ( N ) = − 2 ∑ ω ∣ N ω ≡ 1 ( 12 ) sin π 6 + ∑ ω ∣ N ω ≡ 5 ( 12 ) sin 5 π 6 + ∑ ω ∣ N ω ≡ 7 ( 12 ) sin 7 π 6 + ∑ ω ∣ N ω ≡ 11 ( 12 ) sin 11 π 6 .

Let F i ≔ { ω ∣ ω ≡ i ( mod 12 ) , ω ∣ N } with i = 1 , 5 , 7 , 11 . For each ω ∣ N , let f k : F k → F k + 6 be the maps defined by f k ( ω ) = N ω with k = 1 , 5 . It is easily verified that f k are bijective maps and # F i = # F i + 6 .

Therefore, a 1 ( N ) = − E 1 , 5 ( N ; 12 ) = − 2 ( # F 1 − # F 7 ) 1 2 + ( # F 5 − # F 11 ) 1 2 = 0 .□

Lemma 4

If N ≡ 11 ( mod 12 ) is a natural number, then a 1 ( N ) = 0 .

Proof

In Table 3, △ represents all cases N ω , ω satisfying N ω ⋅ ω ≡ 11 ( mod 12 ) and ω ∣ N . That is, possible ordered pairs N ω ¯ , ω ¯ are ( 1 ¯ , 11 ¯ ) , ( 11 ¯ , 1 ¯ ) , ( 5 ¯ , 7 ¯ ) , and ( 7 ¯ , 5 ¯ ) . Thus,

(3.3) a 1 ( N ) = − 2 ∑ ω ∣ N sin 2 N ω − ω π 6 = − 2 ∑ ω ∣ N ω ≡ 1 ( 12 ) sin 9 π 6 + ∑ ω ∣ N ω ≡ 5 ( 12 ) sin 9 π 6 + ∑ ω ∣ N ω ≡ 7 ( 12 ) sin 3 π 6 + ∑ ω ∣ N ω ≡ 11 ( 12 ) sin 3 π 6 = − 2 ∑ ω ∣ N ω ≡ 1 ( 12 ) ( − 1 ) + ∑ ω ∣ N ω ≡ 11 ( 12 ) 1 + ∑ ω ∣ N ω ≡ 5 ( 12 ) ( − 1 ) + ∑ ω ∣ N ω ≡ 7 ( 12 ) 1 .

It is trivial that ω ≠ N ω . If N ω , ω exists, then ω , N ω always exists. So, # { ω ∣ ω ≡ 1 ( mod 12 ) , ω ∣ N } = # { ω ∣ ω ≡ 11 ( mod 12 ) , ω ∣ N } and # { ω ∣ ω ≡ 5 ( mod 12 ) , ω ∣ N } = # { ω ∣ ω ≡ 7 ( mod 12 ) , ω ∣ N } .□

Lemma 5

If N ≡ 3 ( mod 12 ) is a natural number, then a 1 ( N ) = 0 .

Proof

In Table 3, ♠ represents all cases N ω , ω satisfying N ω ⋅ ω ≡ 3 ( mod 12 ) and ω ∣ N . In equation (3.4), N ω ¯ , ω ¯ = ( a ¯ , b ¯ ) is abbreviated as ( a ¯ , b ¯ ) . Thus,

(3.4) a 1 ( N ) = − 2 ∑ ω ∣ N ( 1 ¯ , 3 ¯ ) sin 11 π 6 + ∑ ω ∣ N ( 3 ¯ , 1 ¯ ) sin 5 π 6 + ∑ ω ∣ N ( 3 ¯ , 5 ¯ ) sin π 6 + ∑ ω ∣ N ( 5 ¯ , 3 ¯ ) sin 7 π 6 + ∑ ω ∣ N ( 3 ¯ , 9 ¯ ) sin 9 π 6 + ∑ ω ∣ N ( 9 ¯ , 3 ¯ ) sin 3 π 6 + ∑ ω ∣ N ( 7 ¯ , 9 ¯ ) sin 5 π 6 + ∑ ω ∣ N ( 9 ¯ , 7 ¯ ) sin 11 π 6 + ∑ ω ∣ N ( 9 ¯ , 11 ¯ ) sin 7 π 6 + ∑ ω ∣ N ( 11 ¯ , 9 ¯ ) sin π 6 = 0 .

The last identity in (3.4) is derived from the following identities. # { N ω ¯ , ω ¯ ∣ N ω ¯ , ω ¯ = ( 1 ¯ , 3 ¯ ) , ω ∣ N } = # { N ω ¯ , ω ¯ ∣ N ω ¯ , ω ¯ = ( 3 ¯ , 1 ¯ ) , ω ∣ N } , # { N ω ¯ , ω ¯ ∣ N ω ¯ , ω ¯ = ( 3 ¯ , 5 ¯ ) , ω ∣ N } = # { N ω ¯ , ω ¯ ∣ N ω ¯ , ω ¯ = ( 5 ¯ , 3 ¯ ) , ω ∣ N } , # { N ω ¯ , ω ¯ ∣ N ω ¯ , ω ¯ = ( 3 ¯ , 9 ¯ ) , ω ∣ N } = # { N ω ¯ , ω ¯ ∣ N ω ¯ , ω ¯ = ( 9 ¯ , 3 ¯ ) , ω ∣ N } , # { N ω ¯ , ω ¯ ∣ N ω ¯ , ω ¯ = ( 7 ¯ , 9 ¯ ) , ω ∣ N } = # { N ω ¯ , ω ¯ ∣ N ω ¯ , ω ¯ = ( 9 ¯ , 7 ¯ ) , ω ∣ N } , and # { N ω ¯ , ω ¯ ∣ N ω ¯ , ω ¯ = ( 9 ¯ , 11 ¯ ) ∣ , ω ∣ N } = # { N ω ¯ , ω ¯ ∣ N ω ¯ , ω ¯ = ( 11 ¯ , 9 ¯ ) , ω ∣ N } .□

Proof of Theorem 1

If N ≡ 3 ( mod 4 ) is a positive integer, then a 1 ( N ) = 0 by Lemmas 3–5. Similarly, using Lemmas 3–5, we obtain

∑ k = 1 N 2 a 1 ( 2 k − 1 ) a 1 ( N − 2 k + 1 ) = 0 if N ≡ 0 ( mod 4 ) , ∑ a + b + c = N a , b , c : odd a 1 ( a ) a 1 ( b ) a 1 ( c ) = 0 if N ≡ 1 ( mod 4 ) , ∑ a + b + c + d = N a , b , c , d : odd a 1 ( a ) a 1 ( b ) a 1 ( c ) a 1 ( d ) = 0 if N ≡ 2 ( mod 4 ) .

It is easily seen that k − N ≡ 0 ( mod 2 ) by a 1 ≡ … ≡ a k ≡ 1 ( mod 2 ) and ∑ i = 1 k a i = N . If a 1 ≡ … ≡ a k ≡ 1 ( mod 4 ) and N ≢ k ( mod 4 ) then ∑ i = 1 k a i ≡ k ≢ N ( mod 4 ) . Thus, there is at least one positive integer, a i ≡ 3 ( mod 4 ) . By Lemmas 3–5, a 1 ( a i ) = 0 . This completes the proof of Theorem 1.□

Lemma 6

If N ≡ 1 ( mod 12 ) is a natural number, then a 1 ( N ) = − E 1 , 5 ( N ; 12 ) .

Proof

( 1 ¯ , 1 ¯ ) , ( 5 ¯ , 5 ¯ ) , ( 7 ¯ , 7 ¯ ) , and ( 11 ¯ , 11 ¯ ) are ordered pairs of N ω ¯ , ω ¯ satisfying N ω ⋅ ω ≡ 1 ( mod 12 ) and ω ∣ N (see ♣ in Table 3). Thus, we have 2 N ω − ω ≡ ω ( mod 12 ) and

a 1 ( N ) = − 2 ∑ ω ∣ N sin 2 N ω − ω π 6 = − 2 ∑ i = 1 , 5 , 7 , 11 ∑ ω ∣ N ω ≡ i ( 12 ) sin ω π 6 = − E 1 , 5 ( N ; 12 ) .

This completes the proof of Lemma 6.□

Corollary 2

Let p 1 , … , p t be distinct primes and e 1 , … , e t ∈ N . If p i ( 1 ≤ i ≤ t ) are congruent to 1 or 5 modulo 12, then a 1 ( ∏ i = 1 t p i e i ) = − ∏ i = 1 t ( e i + 1 ) with ∏ i = 1 t p i e i ≡ 1 (mod 12). In particular, ∏ i = 1 t p i e i is square if and only if a 1 ( ∏ i = 1 t p i e i ) is odd.

Lemma 7

Let q 1 , … , q s be distinct primes, f 1 , … , f s ∈ N and N = ∏ j = 1 s q j f j . If q j ( 1 ≤ j ≤ s ) are congruent to 7 or 11 modulo 12, then

a 1 ( N ) = − 1 if f 1 ≡ f 2 ≡ … ≡ f s ≡ 0 ( mod 2 ) , 0 otherwise .

Proof

Let N = ∏ j = 1 s q j f j with ∑ j = 1 s f j ≡ 1 ( mod 2 ) . Then a 1 ( N ) = 0 by Lemmas 3 and 4. So, we consider N = ∏ j = 1 s q j f j with ∑ j = 1 s f j ≡ 0 ( mod 2 ) . Let us first consider the case where N is a square number. That is, say N = ∏ j = 1 s q j f j = ∏ j = 1 s q j 2 e j . Several sets are defined below for proof. For 1 ≤ i ≤ s , let

X i ≔ { 1 , q i 2 , q i 4 , … , q i 2 e i } ,

Y i ≔ { q i , q i 3 , … , q i 2 e i − 1 } ,

U i ≔ { d ∣ d ≡ 1 , 5 ( mod 12 ) , d ∣ ∏ j = 1 i q j 2 e j } ,

V i ≔ { d ∣ d ≡ 7 , 11 ( mod 12 ) , d ∣ ∏ j = 1 i q j 2 e j } .

We want to use mathematical induction to prove a 1 ( ∏ j = 1 s q j 2 e j ) = − 1 . First, consider the case of q 1 2 e 1 . Using # U 1 = # X 1 , # V 1 = # Y 1 and # U 1 − # V 1 = 1 , we obtain a 1 ( q 1 2 e 1 ) = − 1 by Lemma 6. For the ease of understanding, we show the case where s = 2 . Using 1 ⋅ 1 ≡ 1 ( mod 4 ) , 3 ⋅ 3 ≡ 1 ( mod 4 ) , and 1 ⋅ 3 ≡ 3 ( mod 4 ) , we obtain

(3.5) U 2 = ( U 1 ⊗ X 2 ) ∪ ( V 1 ⊗ Y 2 ) , V 2 = ( U 1 ⊗ Y 2 ) ∪ ( V 1 ⊗ X 2 ) , and # U 2 − # V 2 = ( e 1 + 1 ) ( e 2 + 1 ) + e 1 e 2 − ( e 1 + 1 ) e 2 − e 1 ( e 2 + 1 ) = 1 .

For 1 ≤ i ≤ s − 1 , we assume that # U i − # V i = 1 .

Similarly, with the same method as in (3.5), we obtain

U s = ( U s − 1 ⊗ X s ) ∪ ( V s − 1 ⊗ Y s ) , V s = ( U s − 1 ⊗ Y s ) ∪ ( V s − 1 ⊗ X s ) , and # U s − # V s = ( z ) ( e s + 1 ) + ( z − 1 ) e s − ( z ) e s − ( z − 1 ) ( e s + 1 ) = 1 .

Here, # U s − 1 ≔ z and # V s − 1 ≔ z − 1 . Therefore, by induction, a 1 ∏ j = 1 s q j 2 e j = − 1 .

Finally, we consider the case N ′ = ∏ j = 1 u q j 2 e j − 1 ∏ j = u + 1 s q j 2 e j with u ≥ 1 . Then # U i = # V i with i = 1 , … , u . Similarly, with the same method as in (3.5), we obtain

U u + 1 = ( U u ⊗ X u + 1 ) ∪ ( V u ⊗ Y u + 1 ) , V u + 1 = ( U u ⊗ Y u + 1 ) ∪ ( V u ⊗ X u + 1 ) , and # U u + 1 − # V u + 1 = z ′ ( e u + 1 + 1 ) + z ′ e u + 1 − z ′ e u + 1 − z ′ ( e u + 1 + 1 ) = 0 .

Here, # U u = # V u = z ′ . So, we obtain # U u + 1 − # V u + 1 = 0 . If we do this repeated calculation up to u + 2 , … , s , we obtain # U s − # V s = 0 and a 1 ( ∏ j = 1 u q j 2 e j − 1 ∏ j = u + 1 s q j 2 e j ) = 0 . After all the proof of Lemma 7 is completed.□

Using the method used in (3.5), Lemma 6, Corollary 2, and Lemma 7, we obtain the following corollary for general numbers.

Corollary 3

Let p 1 , … , p r ≡ 1 , 5 ( mod 12 ) and q 1 , … , q s ≡ 7 , 11 ( mod 12 ) be distinct primes and e 1 , … , e r , f 1 , … , f s ∈ N 0 . Then

a 1 ∏ j = 1 r p j e j ∏ i = 1 s q i f i = − ( e 1 + 1 ) ⋯ ( e r + 1 ) if f 1 ≡ … ≡ f s ≡ 0 ( mod 2 ) , − 1 if f 1 ≡ … ≡ f s ≡ 0 ( mod 2 ) , e 1 = … = e r = 0 , 0 otherwise .

Here, f 1 ≡ f 2 ≡ … ≡ f s ≡ 0 ( mod 2 ) includes f 1 = … = f s = 0 .

Lemma 8

If N ≡ 5 ( mod 12 ) is a natural number, then a 1 ( N ) = 4 E 1 ( N ; 12 ) .

Proof

Possible ordered pairs of N ω ¯ , ω ¯ satisfying N ω ⋅ ω ≡ 5 ( mod 12 ) and ω ∣ N are ( 1 ¯ , 5 ¯ ) , ( 5 ¯ , 1 ¯ ) , ( 7 ¯ , 11 ¯ ) , and ( 11 ¯ , 7 ¯ ) (see ⋆ in Table 3). Thus, # { ω ∣ ω ≡ 1 ( mod 12 ) , ω ∣ N } = # { ω ∣ ω ≡ 5 ( mod 12 ) , ω ∣ N } , # { ω ∣ ω ≡ 7 ( mod 12 ) , ω ∣ N } = # { ω ∣ ω ≡ 11 ( mod 12 ) , ω ∣ N } , and

a 1 ( N ) = − 2 ∑ ω ∣ N ( 1 ¯ , 5 ¯ ) sin 9 π 6 + ∑ ω ∣ N ( 5 ¯ , 1 ¯ ) sin 9 π 6 + ∑ ω ∣ N ( 7 ¯ , 11 ¯ ) sin 3 π 6 + ∑ ω ∣ N ( 11 ¯ , 7 ¯ ) sin 3 π 6 = − 2 ∑ ω ∣ N ω ≡ 1 , 5 ( 12 ) ( − 1 ) + ∑ ω ∣ N ω ≡ 7 , 11 ( 12 ) 1 = 4 ∑ ω ∣ N ω ≡ 1 ( 12 ) 1 + ∑ ω ∣ N ω ≡ 11 ( 12 ) ( − 1 ) = 4 E 1 ( N ; 12 ) . □

Lemma 9

If N ≡ 9 ( mod 12 ) is a natural number, then E 3 ( N ; 12 ) = 0 .

Proof

Write N = 3 m ⋅ M with gcd ( 3 , M ) = 1 and m ∈ N . Since N ≡ 1 ( mod 4 ) , we obtain M ≡ 1 ( mod 4 ) if m is even and M ≡ 3 ( mod 4 ) if m is odd. Let D N ≔ { d ∣ d ∈ N , d ∣ N } and D M ≔ { d ∣ d ∈ N , d ∣ M } . Then

D N = D M ⊔ 3 D M ⊔ ⋯ ⊔ 3 m D M .

Here, 3 k D M denotes the set { 3 k d ∣ d ∈ N , d ∣ M } and ⊔ is a symbol for disjoint union. Let d i be a positive divisor of M . Note that d i is not divisible by 2 and 3. If d i ≡ 1 ( mod 4 ) , then 3 2 l − 1 d i ≡ 3 ( mod 12 ) and 3 2 l d i ≡ 9 ( mod 12 ) for l ∈ N . If d i ≡ 3 ( mod 4 ) , then 3 2 l − 1 d i ≡ 9 ( mod 12 ) and 3 2 l d i ≡ 3 ( mod 12 ) for l ∈ N . This implies that # { d ∣ d ≡ 3 ( mod 12 ) , d ∈ 3 2 l − 1 D M ⊔ 3 2 l D M } = # { d ∣ d ≡ 9 ( mod 12 ) , d ∈ 3 2 l − 1 D M ⊔ 3 2 l D M } = # D M .

We assume that m is even. By the above observation, we have E 3 ( N ; 12 ) = E 3 ( M ; 12 ) . Since no divisors of M are congruent to ± 3 modulo 12, we conclude E 3 ( N ; 12 ) = 0 . Now let m be odd. Then

E 3 ( N ; 12 ) = E 3 ( M ; 12 ) + # { d ∣ d ≡ 3 ( mod 12 ) , d ∈ 3 D M } − # { d ∣ d ≡ 9 ( mod 12 ) , d ∈ 3 D M } .

Since M ≡ 3 ( mod 4 ) , M is not a perfect square. Thus, # D M is an even number. Let # D M = 2 r and write D M = d 1 , ⋯ , d r , M d 1 , ⋯ , M d r . We observe that d i ≡ ± 1 ( mod 4 ) if and only if M d i ≡ ∓ 1 ( mod 4 ) . Thus, 3 d i ≡ ± 3 ( mod 12 ) if and only if 3 M d i ≡ ∓ 3 ( mod 12 ) and then # { d ∣ d ≡ 3 ( mod 12 ) , d ∈ 3 D M } = # { d ∣ d ≡ 9 ( mod 12 ) , d ∈ 3 D M } . Finally, we obtain E 3 ( N ; 12 ) = E 3 ( M ; 12 ) = 0 .□

Lemma 10

If N ≡ 9 ( mod 12 ) is a natural number, then a 1 ( N ) = − 4 E 1 , 5 ( N ; 12 ) .

Proof

Possible ordered pairs of N ω ¯ , ω ¯ satisfying N ω ⋅ ω ≡ 9 ( mod 12 ) and ω ∣ N are ( 1 ¯ , 9 ¯ ) , ( 9 ¯ , 1 ¯ ) , ( 3 ¯ , 3 ¯ ) , ( 3 ¯ , 7 ¯ ) , ( 7 ¯ , 3 ¯ ) , ( 3 ¯ , 11 ¯ ) , ( 11 ¯ , 3 ¯ ) , ( 5 ¯ , 9 ¯ ) , ( 9 ¯ , 5 ¯ ) , and ( 9 ¯ , 9 ¯ ) (see ♢ in Table 3). Thus,

a 1 ( N ) = − 2 ∑ ( 1 ¯ , 9 ¯ ) ( 9 ¯ , 1 ¯ ) sin 5 π 6 + ∑ ( 3 ¯ , 7 ¯ ) ( 7 ¯ , 3 ¯ ) sin 11 π 6 + ∑ ( 3 ¯ , 11 ¯ ) ( 11 ¯ , 3 ¯ ) sin 7 π 6 + ∑ ( 5 ¯ , 9 ¯ ) ( 9 ¯ , 5 ¯ ) sin π 6 + ∑ ω ∣ N ( 3 ¯ , 3 ¯ ) sin 3 π 6 + ∑ ω ∣ N ( 9 ¯ , 9 ¯ ) sin 9 π 6 = − 2 ∑ ω ∣ N ( 9 ¯ , 1 ¯ ) 1 + ∑ ω ∣ N ( 3 ¯ , 7 ¯ ) ( − 1 ) + ∑ ω ∣ N ( 3 ¯ , 11 ¯ ) ( − 1 ) + ∑ ω ∣ N ( 9 ¯ , 5 ¯ ) 1 + ∑ ω ∣ N ( 3 ¯ , 3 ¯ ) 1 + ∑ ω ∣ N ( 9 ¯ , 9 ¯ ) ( − 1 ) = − 2 ∑ ω ∣ N ω ≡ 1 ( 12 ) 1 + ∑ ω ∣ N ω ≡ 7 ( 12 ) ( − 1 ) + ∑ ω ∣ N ω ≡ 11 ( 12 ) ( − 1 ) + ∑ ω ∣ N ω ≡ 5 ( 12 ) 1 + ∑ ω ∣ N ω ≡ 3 ( 12 ) 1 − ∑ ω ∣ N ω ≡ 7 ( 12 ) 1 − ∑ ω ∣ N ω ≡ 11 ( 12 ) 1

+ ∑ ω ∣ N ω ≡ 9 ( 12 ) ( − 1 ) − ∑ ω ∣ N ω ≡ 1 ( 12 ) ( − 1 ) − ∑ ω ∣ N ω ≡ 5 ( 12 ) ( − 1 ) = − 2 ( 2 E 1 , 5 ( N ; 12 ) + E 3 ( N ; 12 ) ) .

Now from Lemma 9 we have the result.□

Proof of Theorem 2

It is easy to see that a 1 ( 1 ) = − 1 . If n ( < − 1 ) is a negative and odd integer, then we find a natural number N satisfying N = p 1 − ( n + 1 ) with a prime p 1 ≡ 1 (mod 12). By Corollary 3, we obtain a 1 ( N ) = n . So, there is a natural number N where a 1 ( N ) = n for all negative and odd integers n . The remaining part of the proof of Theorem 2 is completed by Lemmas 3, 4, 5, 6, Corollary 2, Lemma 7, Corollary 3, and Lemmas 8, 10.□

4 Coefficient of a 1 ( N ) with N ≡ ± 2 , ± 4 (mod 12)

Lemma 11

When n is a natural number,

a 1 ( 2 n ) = − 2 if n ≡ 1 ( mod 2 ) , 1 if n ≡ 0 ( mod 2 ) .

Proof

It is well-known that

(4.1) 2 n + 1 − 1 ≡ 3 ( mod 12 ) if n ≡ 1 ( mod 2 ) , 7 ( mod 12 ) if n ≡ 0 ( mod 2 ) .

By (2.3) and (2.10), we obtain

(4.2) a 1 ( 2 n ) = − 2 sin ( 2 n + 1 − 1 ) π 6 .

This completes the proof of Lemma 11 by (4.1) and (4.2).□

Lemma 12

If N ≡ 2 ( mod 12 ) is a natural number, then a 1 ( N ) = − 2 E 1 , 5 ( N ; 12 ) .

Proof

Possible ordered pairs of N ω ¯ , ω ¯ satisfying N ω ⋅ ω ≡ 2 ( mod 12 ) and ω ∣ N are ( 1 ¯ , 2 ¯ ) , ( 2 ¯ , 1 ¯ ) , ( 2 ¯ , 7 ¯ ) , ( 7 ¯ , 2 ¯ ) , ( 5 ¯ , 10 ¯ ) , ( 10 ¯ , 5 ¯ ) , ( 10 ¯ , 11 ¯ ) , and ( 11 ¯ , 10 ¯ ) (Table 3). By (2.10), only odd divisors of N are possible. That is, possible pairs of N ω ¯ , ω ¯ are ( 2 ¯ , 1 ¯ ) , ( 2 ¯ , 7 ¯ ) , ( 10 ¯ , 5 ¯ ) , and ( 10 ¯ , 11 ¯ ) (see † in Table 3). Therefore,

a 1 ( N ) = − 2 ∑ ω ∣ N sin 2 N ω − ω π 6 = − 2 ∑ ω ∣ N ( 2 ¯ , 1 ¯ ) sin 3 π 6 + ∑ ω ∣ N ( 2 ¯ , 7 ¯ ) sin 9 π 6 + ∑ ω ∣ N ( 10 ¯ , 5 ¯ ) sin 3 π 6 + ∑ ω ∣ N ( 10 ¯ , 11 ¯ ) sin 9 π 6 = − 2 ∑ ω ∣ N ω ≡ 1 ( 12 ) 1 + ∑ ω ∣ N ω ≡ 7 ( 12 ) ( − 1 ) + ∑ ω ∣ N ω ≡ 5 ( 12 ) 1 + ∑ ω ∣ N ω ≡ 11 ( 12 ) ( − 1 ) = − 2 E 1 , 5 ( N ; 12 ) . □

Lemma 13

If N ≡ 10 ( mod 12 ) is a natural number, then a 1 ( N ) = E 1 , 5 ( N ; 12 ) ≡ 0 (mod 2).

Proof

By (2.10), possible ordered pairs of N ω ¯ , ω ¯ satisfying N ω ⋅ ω ≡ 10 ( mod 12 ) , ω ≡ 1 ( mod 2 ) , and ω ∣ N are ( 10 ¯ , 1 ¯ ) , ( 2 ¯ , 5 ¯ ) , ( 2 ¯ , 11 ¯ ) , and ( 10 ¯ , 7 ¯ ) (see • in Table 3). Thus, we obtain

a 1 ( N ) = − 2 ∑ ω ∣ N ( 10 ¯ , 1 ¯ ) sin 7 π 6 + ∑ ω ∣ N ( 2 ¯ , 5 ¯ ) sin 11 π 6 + ∑ ω ∣ N ( 2 ¯ , 11 ¯ ) sin 5 π 6 + ∑ ω ∣ N ( 10 ¯ , 7 ¯ ) sin π 6 = − 2 ∑ ω ∣ N ω ≡ 1 ( 12 ) − 1 2 + ∑ ω ∣ N ω ≡ 5 ( 12 ) − 1 2 + ∑ ω ∣ N ω ≡ 7 ( 12 ) 1 2 + ∑ ω ∣ N ω ≡ 11 ( 12 ) 1 2 = E 1 , 5 ( N ; 12 ) .

Since N ≡ 10 ( mod 12 ) , there exists an integer L ≡ 5 ( mod 6 ) satisfying N = 2 L . If ω is an odd divisor of N , then ω ∣ L . If ω ≡ 1 (resp., 5)(mod 6), then L ω ≡ 5 (resp., 1)(mod 6). So, we obtain

E 1 , 5 ( N ; 12 ) = E 1 , 5 ( L ; 12 ) = # L ω , ω ∣ L ω ¯ , ω ¯ = ( 1 ¯ , 5 ¯ ) , ω ∣ L ⋅ 2 + # L ω , ω ∣ L ω ¯ , ω ¯ = ( 1 ¯ , 11 ¯ ) , ω ∣ L ⋅ 0 + # L ω , ω ∣ L ω ¯ , ω ¯ = ( 7 ¯ , 5 ¯ ) , ω ∣ L ⋅ 0 + # L ω , ω ∣ L ω ¯ , ω ¯ = ( 7 ¯ , 11 ¯ ) , ω ∣ L ⋅ ( − 2 ) ≡ 0 ( mod 2 ) . □

Lemma 14

If N ≡ 4 ( mod 12 ) is a natural number, then a 1 ( N ) = E 1 , 5 ( N ; 12 ) .

Proof

By (2.10), possible ordered pairs of N ω ¯ , ω ¯ satisfying N ω ⋅ ω ≡ 4 ( mod 12 ) , ω ≡ 1 ( mod 2 ) , and ω ∣ N are ( 4 ¯ , 1 ¯ ) , ( 4 ¯ , 7 ¯ ) , ( 8 ¯ , 5 ¯ ) , and ( 8 ¯ , 11 ¯ ) (see □ in Table 3). Thus, we obtain

a 1 ( N ) = − 2 ∑ ω ∣ N ( 4 ¯ , 1 ¯ ) sin 7 π 6 + ∑ ω ∣ N ( 4 ¯ , 7 ¯ ) sin π 6 + ∑ ω ∣ N ( 8 ¯ , 5 ¯ ) sin 11 π 6 + ∑ ω ∣ N ( 8 ¯ , 11 ¯ ) sin 5 π 6 = − 2 ∑ ω ∣ N ω ≡ 1 ( 12 ) − 1 2 + ∑ ω ∣ N ω ≡ 5 ( 12 ) − 1 2 + ∑ ω ∣ N ω ≡ 7 ( 12 ) 1 2 + ∑ ω ∣ N ω ≡ 11 ( 12 ) 1 2 = E 1 , 5 ( N ; 12 ) . □

Lemma 15

If N ≡ 8 ( mod 12 ) is a natural number, then a 1 ( N ) = − 2 E 1 , 5 ( N ; 12 ) .

Proof

By (2.10), possible pairs of N ω ¯ , ω ¯ satisfying N ω ⋅ ω ≡ 8 ( mod 12 ) , ω ≡ 1 ( mod 2 ) , and ω ∣ N are ( 4 ¯ , 5 ¯ ) , ( 4 ¯ , 11 ¯ ) , ( 8 ¯ , 1 ¯ ) , and ( 8 ¯ , 7 ¯ ) (see ∇ in Table 3). Thus, we obtain

a 1 ( N ) = − 2 ∑ ω ∣ N ( 4 ¯ , 5 ¯ ) sin 3 π 6 + ∑ ω ∣ N ( 4 ¯ , 11 ¯ ) sin 9 π 6 + ∑ ω ∣ N ( 8 ¯ , 1 ¯ ) sin 3 π 6 + ∑ ω ∣ N ( 8 ¯ , 7 ¯ ) sin 9 π 6 = − 2 ∑ ω ∣ N ω ≡ 1 ( 12 ) 1 + ∑ ω ∣ N ω ≡ 5 ( 12 ) 1 + ∑ ω ∣ N ω ≡ 7 ( 12 ) ( − 1 ) + ∑ ω ∣ N ω ≡ 11 ( 12 ) ( − 1 ) = − 2 E 1 , 5 ( N ; 12 ) . □

5 Coefficient of a 1 ( N ) with N ≡ 6 (mod 12)

In this section, it is assumed that N is a positive integer that satisfies N ≡ 6 (mod 12). Then there exist integers L and M ( = 2 L + 1 ) such that N = 12 L + 6 = 6 ( 2 L + 1 ) = 6 M . Now, we classify odd M into six cases and compute each a 1 ( N ) step by step. Then, by (2.10), possible pairs of N ω ¯ , ω ¯ satisfying N ω ⋅ ω ≡ 6 ( mod 12 ) , ω ≡ 1 ( mod 2 ) , and ω ∣ N are ( 2 ¯ , 3 ¯ ) , ( 2 ¯ , 9 ¯ ) , ( 6 ¯ , 1 ¯ ) , ( 6 ¯ , 3 ¯ ) , ( 6 ¯ , 5 ¯ ) , ( 6 ¯ , 7 ¯ ) , ( 6 ¯ , 9 ¯ ) , ( 6 ¯ , 11 ¯ ) , ( 10 ¯ , 3 ¯ ) , and ( 10 ¯ , 9 ¯ ) (see ♮ in Table 3).

Lemma 16

Let N ≡ 6 (mod 12) and N = 6 M with gcd ( 6 , M ) = 1 . Then a 1 ( N ) = 0 .

Proof

Let ω ∣ N and ω ′ ∣ M with ω ≡ 1 (mod 2). Then, by (2.10) and Lemma 2, possible ordered pairs of N ω ¯ , ω ¯ = (even, odd) = 6 M ω ′ , ω ′ or 2 M ω ′ , 3 ω ′ .

First, we consider the case M ≡ 1 ( mod 12 ) . Possible ordered pairs 6 M ω ′ ¯ , ω ′ ¯ are ( 6 ¯ , 1 ¯ ) , ( 6 ¯ , 5 ¯ ) , ( 6 ¯ , 7 ¯ ) , and ( 6 ¯ , 11 ¯ ) and possible ordered pairs 2 M ω ′ ¯ , 3 ω ′ ¯ are ( 2 ⋅ 1 ¯ , 3 ⋅ 1 ¯ ) , ( 2 ⋅ 5 ¯ , 3 ⋅ 5 ¯ ) , ( 2 ⋅ 7 ¯ , 3 ⋅ 7 ¯ ) , and ( 2 ⋅ 11 ¯ , 3 ⋅ 11 ¯ ) . Thus, we obtain

a 1 ( N ) = − 2 T 6 M ω ′ , ω ′ ( N ) + ∑ ω ∣ N ( 2 ⋅ 1 ¯ , 3 ⋅ 1 ¯ ) sin π 6 + ∑ ω ∣ N ( 2 ⋅ 5 ¯ , 3 ⋅ 5 ¯ ) sin 5 π 6 + ∑ ω ∣ N ( 2 ⋅ 7 ¯ , 3 ⋅ 7 ¯ ) sin 7 π 6 + ∑ ω ∣ N ( 2 ⋅ 11 ¯ , 3 ⋅ 11 ¯ ) sin 11 π 6 = − 2 T 6 M ω ′ , ω ′ ( N ) + ∑ ω ∣ N ω ≡ 1 ( 12 ) 1 2 + ∑ ω ∣ N ω ≡ 5 ( 12 ) 1 2 + ∑ ω ∣ N ω ≡ 7 ( 12 ) − 1 2 + ∑ ω ∣ N ω ≡ 11 ( 12 ) − 1 2 = 0 .

Here,

(5.1) T 6 M ω ′ , ω ′ ( N ) ≔ ∑ ω ∣ N ( 6 ¯ , 1 ¯ ) sin 11 π 6 + ∑ ω ∣ N ( 6 ¯ , 5 ¯ ) sin 7 π 6 + ∑ ω ∣ N ( 6 ¯ , 7 ¯ ) sin 5 π 6 + ∑ ω ∣ N ( 6 ¯ , 11 ¯ ) sin π 6 = ∑ ω ∣ N ω ≡ 1 ( 12 ) − 1 2 + ∑ ω ∣ N ω ≡ 5 ( 12 ) − 1 2 + ∑ ω ∣ N ω ≡ 7 ( 12 ) 1 2 + ∑ ω ∣ N ω ≡ 11 ( 12 ) 1 2 .

Second, we consider the case M ≡ 11 ( mod 12 ) . By (2.10) and Lemma 2, possible ordered pairs 6 M ω ′ ¯ , ω ′ ¯ are the same with the case M ≡ 1 ( mod 12 ) and possible ordered pairs 2 M ω ′ ¯ , 3 ω ′ ¯ are ( 2 ⋅ 1 ¯ , 3 ⋅ 11 ¯ ) , ( 2 ⋅ 5 ¯ , 3 ⋅ 7 ¯ ) , ( 2 ⋅ 7 ¯ , 3 ⋅ 5 ¯ ) , and ( 2 ⋅ 11 ¯ , 3 ⋅ 1 ¯ ) . Using (5.1), we obtain

a 1 ( N ) = − 2 T 6 M ω ′ , ω ′ ( N ) + ∑ ω ∣ N ( 2 ⋅ 11 ¯ , 3 ⋅ 1 ¯ ) sin 5 π 6 + ∑ ω ∣ N ( 2 ⋅ 7 ¯ , 3 ⋅ 5 ¯ ) sin π 6 + ∑ ω ∣ N ( 2 ⋅ 5 ¯ , 3 ⋅ 7 ¯ ) sin 11 π 6 + ∑ ω ∣ N ( 2 ⋅ 1 ¯ , 3 ⋅ 11 ¯ ) sin 7 π 6 = − 2 T 6 M ω ′ , ω ′ ( N ) + ∑ ω ∣ N ω ≡ 1 ( 12 ) 1 2 + ∑ ω ∣ N ω ≡ 5 ( 12 ) 1 2 + ∑ ω ∣ N ω ≡ 7 ( 12 ) − 1 2 + ∑ ω ∣ N ω ≡ 11 ( 12 ) − 1 2 = 0 .

Let M ≡ 5 ( mod 12 ) be a positive integer. In the same method as above, we obtain possible ordered pairs ( 2 M ω ′ ¯ , 3 ω ′ ¯ ) are ( 2 ⋅ 1 ¯ , 3 ⋅ 5 ¯ ) , ( 2 ⋅ 5 ¯ , 3 ⋅ 1 ¯ ) , ( 2 ⋅ 7 ¯ , 3 ⋅ 11 ¯ ) , ( 2 ⋅ 11 ¯ , 3 ⋅ 7 ¯ ) , and

a 1 ( N ) = − 2 T 6 M ω ′ , ω ′ ( N ) + ∑ ω ∣ N ( 2 ⋅ 5 ¯ , 3 ⋅ 1 ¯ ) sin 5 π 6 + ∑ ω ∣ N ( 2 ⋅ 1 ¯ , 3 ⋅ 5 ¯ ) sin π 6 + ∑ ω ∣ N ( 2 ⋅ 11 ¯ , 3 ⋅ 7 ¯ ) sin 11 π 6 + ∑ ω ∣ N ( 2 ⋅ 7 ¯ , 3 ⋅ 11 ¯ ) sin 7 π 6 = 0 .

Let M ≡ 7 ( mod 12 ) be a positive integer. Then possible ordered pairs 2 M ω ′ ¯ , 3 ω ′ ¯ are ( 2 ⋅ 1 ¯ , 3 ⋅ 7 ¯ ) , ( 2 ⋅ 5 ¯ , 3 ⋅ 11 ¯ ) , ( 2 ⋅ 7 ¯ , 3 ⋅ 1 ¯ ) , ( 2 ⋅ 11 ¯ , 3 ⋅ 5 ¯ ) , and

a 1 ( N ) = − 2 T 6 M ω ′ , ω ′ ( N ) + ∑ ω ∣ N ( 2 ⋅ 7 ¯ , 3 ⋅ 1 ¯ ) sin π 6 + ∑ ω ∣ N ( 2 ⋅ 11 ¯ , 3 ⋅ 5 ¯ ) sin 5 π 6 + ∑ ω ∣ N ( 2 ⋅ 1 ¯ , 3 ⋅ 7 ¯ ) sin 7 π 6 + ∑ ω ∣ N ( 2 ⋅ 5 ¯ , 3 ⋅ 11 ¯ ) sin 11 π 6 = 0 .

This completes the proof of Lemma 16.□

The case of N = 6 M with gcd ( 6 , M ) = 3 will be shown in Remark 3.

6 Coefficient of a 1 ( N ) with N ≡ 0 (mod 12)

In this section, it is assumed that N is a positive integer that satisfies N ≡ 0 (mod 12). Then there exists an integer M satisfying N = 12 M . Now, we classify positive integers M into 12 cases.

Lemma 17

If M ≡ ± 1 , ± 5 ( mod 12 ) is a natural number, then a 1 ( N ) = 0 .

Proof

Let ω be a divisor of M . It is easily checked that ω 2 ≡ 1 ( mod 12 ) and ω ≡ 1 or − 1 or 5 or − 5 ( mod 12 ) . Thus, we have

(6.1) 24 M ω − ω ≡ − ω ( mod 12 )

and

(6.2) 8 M ω − 3 ω ≡ 8 ω 2 M ω − 3 ω ≡ ( 8 M − 3 ) ω ≡ 5 ω ( mod 12 ) if M ≡ 1 ( mod 6 ) , ω ( mod 12 ) if M ≡ 5 ( mod 6 ) .

Now we have, by (6.1) and (6.2) and by sin A + sin B = 2 sin A + B 2 cos A − B 2 ,

(6.3) ∑ ω ∣ N sin 24 M ω − ω π 6 + sin 8 M ω − 3 ω π 6 = ∑ ω ∣ N sin − ω π 6 + sin ν ω π 6 = 0 .

Here, ν = 1 or 5. Furthermore, it is easy to see that ( N ω , ω ) = ( even , odd ) ∣ ω ≡ 1 ( mod 2 ) , ω ∣ N = 12 M ω , ω ∣ ω ≡ 1 ( mod 2 ) , ω ∣ M ∪ 4 M ω , 3 ω ∣ ω ≡ 1 ( mod 2 ) , ω ∣ M and 12 M ω , ω ∣ ω ≡ 1 ( mod 2 ) , ω ∣ M ∩ 4 M ω , 3 ω ∣ ω ≡ 1 ( mod 2 ) , ω ∣ M = ∅ .

By (2.9), (6.3), and Lemma 2,

a 1 ( N ) = − 2 ∑ ω ∣ N 2 N ω − ω ≡ 1 ( mod 2 ) sin 2 N ω − ω π 6 = − 2 ∑ ω ∣ N ω odd sin 2 N ω − ω π 6

= − 2 ∑ ω ∣ N sin 24 M ω − ω π 6 + sin 8 M ω − 3 ω π 6 = 0 . □

Lemma 18

Let N = 2 2 3 2 k − 1 M with gcd ( 6 , M ) = 1 and k ∈ N . Then a 1 ( N ) = 0 .

Proof

Let d ∣ N and ω ∣ M . It is the only ordered pair N d , d = ( even, odd ) that satisfy 2 2 ⋅ 3 t ⋅ M ω , 3 2 k − t − 1 ⋅ ω with 0 ≤ t ≤ 2 k − 1 . Thus, by Lemma 2,

(6.4) a 1 ( N ) = − 2 ∑ ω ∣ M ∑ t = 0 2 k − 1 sin 2 3 ⋅ 3 t ⋅ M ω − 3 2 k − t − 1 ⋅ ω π 6 = − 2 ∑ ω ∣ M sin 2 3 ⋅ M ω − 3 2 k − 1 ⋅ ω π 6 + ∑ t = 1 2 k − 2 sin ( − 3 2 k − t − 1 ⋅ ω ) π 6 + sin − ω π 6 .

Using ω 2 ≡ 1 (mod 12), 3 2 t ≡ − 3 (mod 12), 3 2 t − 1 ≡ 3 (mod 12),

2 3 M ω − 3 2 k − 1 ω ≡ ( 8 M − 3 ) ω ≡ 5 ω if M ≡ 1 , 7 ( mod 12 ) , ω if M ≡ 5 , 11 ( mod 12 ) ,

and sin ω π 2 + sin − ω π 2 = 0 , equation (6.4) becomes

a 1 ( N ) = − 2 ∑ ω ∣ M sin ( 8 M − 3 ) ω π 6 + ( k − 1 ) sin − ω π 2 + sin ω π 2 + sin − ω π 6 = − 2 ∑ ω ∣ M sin ( 8 M − 3 ) ω π 6 + sin − ω π 6 = − 4 ∑ ω ∣ M sin ( 2 M − 1 ) ω π 3 cos ( 4 M − 1 ) ω π 6 = 0 .

Here, if M ≡ 1 ( mod 6 ) , then cos ( 4 M − 1 ) ω π 6 = 0 and M ≡ 5 ( mod 6 ) , then sin ( 2 M − 1 ) ω π 3 = 0 . Therefore, the proof of Lemma 18 is completed.□

Lemma 19

Let N = 2 2 3 2 k M with gcd ( 6 , M ) = 1 and k ∈ N . Then a 1 ( N ) = 4 E 1 , 5 ( M ; 12 ) . In particular, if M ≡ 7 , 11 (mod 12), then a 1 ( N ) = 0 .

Proof

It is easy to see that

(6.5) 2 3 M ω − 3 2 k ω ≡ ( 8 M − 9 ) ω ≡ 11 ω if M ≡ 1 , 7 ( mod 12 ) , 7 ω if M ≡ 5 , 11 ( mod 12 ) .

In a similar way to Lemma 18, we obtain

(6.6) a 1 ( N ) = − 2 ∑ ω ∣ M ∑ t = 0 2 k sin 2 3 ⋅ 3 t ⋅ M ω − 3 2 k − t ⋅ ω π 6 = − 2 ∑ ω ∣ M sin 2 3 ⋅ M ω − 3 2 k ⋅ ω π 6 + ∑ t = 1 2 k − 1 sin ( − 3 2 k − t ⋅ ω ) π 6 + sin − ω π 6 = − 2 ∑ ω ∣ M sin ( 8 M − 9 ) ω π 6 + sin ( − 3 ω ) π 6 + sin − ω π 6 = 4 E 1 , 5 ( M ; 12 )

by (6.5) and definition of E 1 , 5 ( M ; 12 ) . By Lemmas 3 and 4, the proof of Lemma 19 is completed.□

Corollary 4

Let N = 12 M be an integer with M ≡ 3 ( mod 6 ) . Then

a 1 ( N ) = 0 if M ≡ 9 ( mod 12 ) , 4 E 1 , 5 ( M ′ ; 12 ) if M ≡ 3 ( mod 12 ) , M = 3 t M ′ , and gcd ( 6 , M ′ ) = 1 .

In particular, if M ′ ≡ 7 , 11 (mod 12), then a 1 ( N ) = 0 .

Proof

Let M ≡ 9 ( mod 12 ) . For k ∈ N , we have

N = 2 2 ⋅ 3 ⋅ 3 2 k − 2 ⋅ M ′ with M ′ ≡ 1 , 5 ( mod 12 ) or 2 2 ⋅ 3 ⋅ 3 2 k − 1 ⋅ M ′ with M ′ ≡ 7 , 11 ( mod 12 ) .

By Lemmas 18 and 19, a 1 ( N ) = 0 .

Let M ≡ 3 ( mod 12 ) . For k ∈ N , we have

N = 2 2 ⋅ 3 ⋅ 3 2 k − 2 ⋅ M ′ with M ′ ≡ 7 , 11 ( mod 12 ) or 2 2 ⋅ 3 ⋅ 3 2 k − 1 ⋅ M ′ with M ′ ≡ 1 , 5 ( mod 12 ) .

If N = 2 2 ⋅ 3 ⋅ 3 2 k − 2 ⋅ M ′ with M ′ ≡ 7 , 11 ( mod 12 ) , then a 1 ( N ) = 4 E 1 , 5 ( M ′ ; 12 ) = 0 by Lemma 18. Finally, if M ′ ≡ 1 , 5 ( mod 12 ) , then a 1 ( 2 2 ⋅ 3 ⋅ 3 2 k − 1 ⋅ M ′ ) = 4 E 1 , 5 ( M ′ ; 12 ) by Lemma 19. This completes the proof of Corollary 4.□

Lemma 20

Let N = 2 k ⋅ 3 ⋅ M with gcd ( 6 , M ) = 1 and k ( ≥ 2 ) ∈ N . Then a 1 ( N ) = 0 .

Proof

Let d ∣ N and ω ∣ M . It is the only ordered pair N d , d = ( even, odd ) that satisfy 2 k ⋅ 3 ⋅ M ω , ω and 2 k ⋅ M ω , 3 ω . By elementary observation, we obtain

(6.7) 2 k ≡ 2 ( mod 12 ) if k = 1 , 4 ( mod 12 ) if k ≡ 0 ( mod 2 ) , 8 ( mod 12 ) if k ≡ 1 ( mod 2 ) , and k ≠ 1 ,

(6.8) sin ( 4 M − 3 ) ω π 6 = sin ω π 6 if M ≡ 1 ( mod 6 ) , sin 5 ω π 6 if M ≡ 5 ( mod 6 ) ,

and

(6.9) sin ( 8 M − 3 ) ω π 6 = sin 5 ω π 6 if M ≡ 1 ( mod 6 ) , sin ω π 6 if M ≡ 5 ( mod 6 ) .

First, let N = 2 2 n + 1 ⋅ 3 ⋅ M with n ∈ N . By Lemma 2, (6.7), and (6.8), we obtain

(6.10) a 1 ( N ) = − 2 ∑ ω ∣ M sin 2 2 n + 2 ⋅ 3 M ω − ω π 6 + sin 2 2 n + 2 ⋅ M ω − 3 ω π 6 = − 2 ∑ ω ∣ M sin − ω π 6 + sin ( 4 M − 3 ) ω π 6 = 0 .

Next, let N = 2 2 n ⋅ 3 ⋅ M with n ∈ N . By Lemma 2, (6.7), and (6.9), we obtain

(6.11) a 1 ( N ) = − 2 ∑ ω ∣ M sin 2 2 n + 1 ⋅ 3 M ω − ω π 6 + sin 2 2 n + 1 ⋅ M ω − 3 ω π 6 = − 2 ∑ ω ∣ M sin − ω π 6 + sin ( 8 M − 3 ) ω π 6 = 0 .

□

The following corollary is obtained from Lemma 20.

Corollary 5

Let N = 12 M be an integer with M ≡ ± 2 , ± 4 ( mod 12 ) . Then a 1 ( N ) = 0 .

Lemma 21

Let N = 2 3 ⋅ 3 m ⋅ M ′ with gcd ( 6 , M ′ ) = 1 and m ( ≥ 2 ) ∈ N . Then

a 1 ( N ) = 0 if m ≡ 1 ( mod 2 ) , 4 E 1 , 5 ( M ′ ; 12 ) otherwise .

In particular, if M ′ ≡ 7 , 11 (mod 12), then a 1 ( N ) = 0 . That is, if N = 12 M = 2 3 ⋅ 3 m ⋅ M ′ with M ≡ 6 (mod 12), then a 1 ( N ) = 0 or 4 E 1 , 5 ( M ′ ; 12 ) .

Proof

Let d ∣ N and ω ∣ M ′ . First, we set N = 2 3 ⋅ 3 2 k + 1 ⋅ M ′ . It is the only ordered pair N d , d = ( even, odd ) that satisfies 2 3 ⋅ 3 t ⋅ M ′ ω , 3 2 k − t + 1 ⋅ ω with 0 ≤ t ≤ 2 k + 1 . Thus, by Lemma 2 and (6.8),

a 1 ( N ) = − 2 ∑ ω ∣ M ′ ∑ t = 0 2 k + 1 sin 2 4 ⋅ 3 t ⋅ M ′ ω − 3 2 k − t + 1 ⋅ ω π 6 = − 2 ∑ ω ∣ M ′ sin ( 4 M ′ − 3 ) ω π 6 + sin − ω π 6 = 0 .

In the same way as before, we prove the case N = 2 3 ⋅ 3 2 k ⋅ M ′ . It is the only ordered pair N d , d = ( even, odd ) that satisfies ( 2 3 ⋅ 3 t ⋅ M ′ ω , 3 2 k − t ⋅ ω ) with 0 ≤ t ≤ 2 k . Using

sin ( 4 M ′ − 9 ) ω π 6 = sin 7 ω π 6 if M ′ ≡ 1 ( mod 6 ) sin − ω π 6 if M ′ ≡ 5 ( mod 6 ) ,

we obtain

(6.12) a 1 ( N ) = − 2 ∑ ω ∣ M ′ ∑ t = 0 2 k sin 2 4 ⋅ 3 t ⋅ M ′ ω − 3 2 k − t ⋅ ω π 6 = − 2 ∑ ω ∣ M ′ sin 4 ⋅ M ′ ω − 3 2 k ⋅ ω π 6 + ∑ t = 1 2 k − 1 sin ( − 3 2 k − t ⋅ ω ) π 6 + sin − ω π 6 = − 2 ∑ ω ∣ M ′ sin ( 4 M ′ − 9 ) ω π 6 + sin ( − 3 ω ) π 6 + sin − ω π 6 = 4 E 1 , 5 ( M ′ ; 12 ) .

If M ′ ≡ 7 , 11 (mod 12), then E 1 , 5 ( M ′ ; 12 ) = 0 and a 1 ( N ) = 4 E 1 , 5 ( M ′ ; 12 ) = 0 .□

Lemma 22

Let N = 2 n ⋅ 3 2 m + 1 ⋅ M ′ with gcd ( 6 , M ′ ) = 1 and n ( ≥ 4 ) , m ( ≥ 1 ) ∈ N . Then a 1 ( N ) = 0 .

Proof

Let d ∣ N and ω ∣ M ′ . First, we let N = 2 2 l ⋅ 3 2 m + 1 ⋅ M ′ with n = 2 l . It is the only ordered pair N d , d = ( even, odd ) that satisfy 2 2 l ⋅ 3 t ⋅ M ′ ω , 3 2 m − t + 1 ⋅ ω with 0 ≤ t ≤ 2 m + 1 . It is clear that

(6.13) 2 2 u ≡ 4 ( mod 12 ) , 2 2 u + 1 ≡ 8 ( mod 12 ) , 3 2 u ≡ 9 ( mod 12 ) , and 3 2 u + 1 ≡ 3 ( mod 12 )

with u ∈ N . By using (6.11) and (6.13), we see directly that

a 1 ( N ) = − 2 ∑ ω ∣ M ′ ∑ t = 0 2 m + 1 sin 2 2 l + 1 ⋅ 3 t ⋅ M ′ ω − 3 2 m − t + 1 ⋅ ω π 6

= − 2 ∑ ω ∣ M ′ sin 2 2 l + 1 ⋅ M ′ ω − 3 2 m + 1 ⋅ ω π 6 + ∑ t = 1 2 m sin ( − 3 2 m − t + 1 ⋅ ω ) π 6 + sin − ω π 6 = − 2 ∑ ω ∣ M ′ sin ( 8 M ′ − 3 ) ω π 6 + sin − ω π 6 = 0 .

Second, let N = 2 2 l + 1 ⋅ 3 2 m + 1 ⋅ M ′ . Then, by (6.10), we have

(6.14) a 1 ( N ) = − 2 ∑ ω ∣ M ′ ∑ t = 0 2 m + 1 sin 2 2 l + 2 ⋅ 3 t ⋅ M ′ ω − 3 2 m − t + 1 ⋅ ω π 6 = − 2 ∑ ω ∣ M ′ sin 2 2 l + 2 ⋅ M ′ ω − 3 2 m + 1 ⋅ ω π 6 + ∑ t = 1 2 m sin ( − 3 2 m − t + 1 ⋅ ω ) π 6 + sin − ω π 6 = − 2 ∑ ω ∣ M ′ sin ( 4 M ′ − 3 ) ω π 6 + sin − ω π 6 = 0 .

□

Lemma 23

Let N = 2 n ⋅ 3 2 m ⋅ M ′ with g c d ( 6 , M ′ ) = 1 and n ( ≥ 4 ) , m ( ≥ 1 ) ∈ N . Then a 1 ( N ) = 4 E 1 , 5 ( M ′ ; 12 ) . In particular, if M ′ ≡ 7 , 11 (mod 12) then a 1 ( N ) = 0 .

Proof

Let d ∣ N and ω ∣ M ′ . First, we let N = 2 2 l ⋅ 3 2 m ⋅ M ′ with n = 2 l . It is the only ordered pair N d , d = ( even, odd ) that satisfy 2 2 l ⋅ 3 t ⋅ M ′ ω , 3 2 m − t ⋅ ω with 0 ≤ t ≤ 2 m . Then, by (6.6), we obtain

a 1 ( N ) = − 2 ∑ ω ∣ M ′ ∑ t = 0 2 m sin 2 2 l + 1 ⋅ 3 t ⋅ M ′ ω − 3 2 m − t ⋅ ω π 6 = − 2 ∑ ω ∣ M ′ sin 2 2 l + 1 ⋅ M ′ ω − 3 2 m ⋅ ω π 6 + ∑ t = 1 2 m − 1 sin ( − 3 2 m − t ⋅ ω ) π 6 + sin − ω π 6 = − 2 ∑ ω ∣ M ′ sin ( 8 M ′ − 9 ) ω π 6 + sin − 3 ω π 6 + sin − ω π 6 = 4 E 1 , 5 ( M ′ ; 12 ) .

Second, let N = 2 2 l + 1 ⋅ 3 2 m ⋅ M ′ . Then, by (6.12), we have

(6.15) a 1 ( N ) = − 2 ∑ ω ∣ M ′ ∑ t = 0 2 m sin 2 2 l + 2 ⋅ 3 t ⋅ M ′ ω − 3 2 m − t ⋅ ω π 6 = − 2 ∑ ω ∣ M ′ sin 2 2 l + 2 ⋅ M ′ ω − 3 2 m ⋅ ω π 6 + ∑ t = 1 2 m − 1 sin ( − 3 2 m − t ⋅ ω ) π 6 + sin − ω π 6 = − 2 ∑ ω ∣ M ′ sin ( 4 M ′ − 9 ) ω π 6 + sin − 3 ω π 6 + sin − ω π 6 = 4 E 1 , 5 ( M ′ ; 12 ) .

If M ′ ≡ 7 , 11 (mod 12), then E 1 , 5 ( M ′ ; 12 ) = 0 and a 1 ( N ) = 4 E 1 , 5 ( M ′ ; 12 ) = 0 .□

By Lemmas 22 and 23, we obtain the following corollary.

Corollary 6

Let N = 12 M be an integer with M ≡ 0 ( mod 12 ) , that is, N = 2 n ⋅ 3 m ⋅ M ′ with gcd ( 6 , M ′ ) = 1 and n ( ≥ 4 ) , m ( ≥ 2 ) ∈ N . Then

a 1 ( N ) = 0 if m ≡ 1 ( mod 2 ) or M ′ ≡ 7 , 11 ( mod 12 ) , 4 E 1 , 5 ( M ′ ; 12 ) if m ≡ 0 ( mod 2 ) and M ′ ≡ 1 , 5 ( mod 12 ) .

Remark 3

We can easily obtain Lemma 16 again using the method in Section 6. In other words, let N = 2 ⋅ 3 2 m + 1 ⋅ M ′ (resp., N = 2 ⋅ 3 2 m + 2 ⋅ M ′ ) and gcd ( 6 , M ′ ) = 1 . Set n = 0 in (6.14) (resp., (6.15)):

a 1 ( N ) = 0 if N = 2 ⋅ 3 2 m + 1 ⋅ M ′ or M ′ ≡ 7 , 11 ( mod 12 ) , 4 E 1 , 5 ( M ′ ; 12 ) if N = 2 ⋅ 3 2 m + 2 ⋅ M ′ and M ′ ≡ 1 , 5 ( mod 12 ) .

Here, m ∈ N 0 .

7 Coefficient of a 2 ( N )

We obtain four cases in (2.4):

(7.1) n cos ( k − n ) π 3 = n if ( k − n ) ≡ 0 ( mod 6 ) , 1 2 n if ( k − n ) ≡ ± 1 ( mod 6 ) , − 1 2 n if ( k − n ) ≡ ± 2 ( mod 6 ) , − n if ( k − n ) ≡ 3 ( mod 6 ) .

Here, k , n ∈ N .

By (2.4) and (7.1), we obtain

∑ n = 0 ∞ a 2 ( n ) q n = ∏ n ≥ 1 ( 1 − q n ) 4 ( 1 − q n + q 2 n ) 2 = 1 − 2 ∑ N ≥ 1 ∑ n k = N n , k ≥ 1 n cos ( k − n ) π 3 q N = 1 − 2 ∑ N ≥ 1 ∑ d ∣ N d ≡ N d ( mod 6 ) d − ∑ d ∣ N d − N d ≡ 3 ( mod 6 ) d + ∑ d ∣ N d − N d ≡ ± 1 ( mod 6 ) d 2 − ∑ d ∣ N d − N d ≡ ± 2 ( mod 6 ) d 2 q N = 1 − ∑ N ≥ 1 σ ˆ ( N ) q N .

It follows that a 2 ( 0 ) = 1 and a 2 ( N ) = − σ ˆ ( N ) for N ≥ 1 .

Lemma 24

If N is a positive integer with N ≡ 1 ( mod 6 ) , then σ ˆ ( N ) = 2 σ ( N ) .

Proof

Let d be a positive divisor of N with d ≡ 1 ( resp . , − 1 ) ( mod 6 ) . Then we obtain N d ≡ 1 ( resp . , − 1 ) ( mod 6 ) . Hence, by the definition of σ ˆ ( N ) , we obtain that

σ ˆ ( N ) = 2 ∑ d ∣ N d ≡ N d ( mod 6 ) d − ∑ d ∣ N d − N d ≡ 3 ( mod 6 ) d + ∑ d ∣ N d − N d ≡ ± 1 ( mod 6 ) d − ∑ d ∣ N d − N d ≡ ± 2 ( mod 6 ) d = 2 ∑ d ∣ N d ≡ N d ( mod 6 ) d = 2 ∑ d ∣ N d = 2 σ ( N ) .

□

Similar to Lemma 24’s method, calculating σ ˆ ( N ) from N ≡ 2 ( mod 6 ) to N ≡ 6 ( mod 6 ) gives the following theorem.

Theorem 8

Let N , M ′ ∈ N and t , s ∈ N 0 . If N = 2 t 3 s M ′ is an integer with gcd ( 6 , M ′ ) = 1 , then we have

a 2 ( N ) = − σ ˆ ( N ) = − 2 σ ( M ′ ) if N ≡ 1 ( mod 6 ) , − 3 σ ( M ′ ) if N ≡ 2 ( mod 6 ) , 4 σ ( M ′ ) if N ≡ 3 ( mod 6 ) , 6 σ ( M ′ ) if N ≡ 4 ( mod 6 ) , σ ( M ′ ) if N ≡ 5 ( mod 6 ) , − 12 σ ( M ′ ) if N ≡ 0 ( mod 6 ) .

Proof of Theorem 3

If N ≡ 1 , 3 ( mod 6 ) is a positive integer, then a 2 ( N ) is even by Theorem 8. Let N ≡ 5 ( mod 6 ) be a positive integer. It is well-known that σ ( N ) is odd if and only if N is a perfect square in [21, p. 28]. But N ≡ 5 ( mod 6 ) is not a perfect square. This completes the proof of Theorem 3.□

Proof of Theorem 4

We see from (1.1) and Theorem 7 that

(7.2) ∑ k = 1 N − 1 e ˆ ( k ) e ˆ ( N − k ) = ∑ k = 1 N − 1 a 1 ( k ) a 1 ( N − k ) = − 2 a 1 ( N ) + a 2 ( N )

with N ≥ 2 .

Using (7.2), lemmas, corollaries, and Theorem 8 in this article, the proof of Theorem 4 is completed.□

8 Theory of modular forms

To prove Theorems 5 and 6, we use the theory of modular forms in [22,23]. For N ∈ N , we define the level N congruence subgroup Γ 0 ( N ) by

Γ 0 ( N ) ≔ a b c d ∈ SL 2 ( Z ) ∣ c ≡ 0 ( mod N ) .

Let H = x{ τ = u + i v ∣ v > 0 } be the complex upper half plane, and let k ∈ Z . Suppose that g is a holomorphic function on H and ρ is a Dirichlet character modulo N . Then g is called a modular form of weight k and level N with Nebentypus character ρ if the following hold:

  1. For any τ ∈ H and any a b c d ∈ Γ 0 ( N ) , g satisfies

    g a τ + b c τ + d = ρ ( d ) ( c τ + d ) k g ( τ ) .

  2. If γ = A B C D ∈ SL 2 ( Z ) , then ( C τ + D ) − k g A τ + B C τ + D has a Fourier expansion of the form

    ( C τ + D ) − k g A τ + B C τ + D = ∑ n ≥ 0 a γ ( n ) e π i n τ / N .

We denote by M k ( N , ρ ) the space of modular forms of weight k and level N with Nebentypus character ρ . If a γ ( 0 ) = 0 for all γ ∈ SL 2 ( Z ) , then g is called a cusp form.

We denote by S k ( N , ρ ) the space of cusp forms of weight k and level N with Nebentypus character ρ . If ρ is trivial, then we simply write M k ( N , ρ ) = M k ( N ) and S k ( N , ρ ) = S k ( N ) .

Before we prove the main theorems, we verify a property for σ ¯ 2 ( N ) and σ ˜ 2 ( N ) .

Lemma 25

Let N be an odd positive integer. Then σ ¯ 2 ( N ) = σ ˜ 2 ( N ) if N ≡ 1 ( mod 4 ) and σ ¯ 2 ( N ) = − σ ˜ 2 ( N ) if N ≡ 3 ( mod 4 ) .

Proof

Let d be any positive divisor of N . If N ≡ 1 ( mod 4 ) , then d ≡ ± 1 ( mod 4 ) if and only if N d ≡ ± 1 ( mod 4 ) . If N ≡ 3 ( mod 4 ) , then d ≡ ± 1 ( mod 4 ) if and only if N d ≡ ∓ 1 ( mod 4 ) . By the definitions of σ ¯ 2 ( N ) and σ ˜ 2 ( N ) , we have the result.□

We denote by L ( s , ρ ) a Dirichlet L -function.

Proof of Theorem 5

Let ρ be the Dirichlet character modulo 24 with conductor 4. That is, ρ ( 1 ) = ρ ( 5 ) = ρ ( 13 ) = ρ ( 17 ) = 1 and ρ ( 7 ) = ρ ( 11 ) = ρ ( 19 ) = ρ ( 23 ) = − 1 . Then H ( q ) 3 is a modular form in the space M 3 ( 24 , ρ ) by [23, Theorem 1.64, 1.65]. Using the dimension formula [23, Theorem 1.34], we obtain that the dimension of M 3 ( 24 , ρ ) is 12 and the dimension of S 3 ( 24 , ρ ) is 4. Also we can check that T 1 , T 2 , T 3 , and T 4 are in S 3 ( 24 , ρ ) and are linearly independent. Now let

E ¯ 3 ( q ) ≔ 1 2 L ( − 2 , ρ ¯ ) + ∑ n ≥ 1 σ ¯ 2 ( n ) q n , E ˜ 3 ( q ) ≔ ∑ n ≥ 1 σ ˜ 2 ( n ) q n ,

where ρ ¯ is the Dirichlet character modulo 4 such that ρ ¯ ( 3 ) = − 1 . In fact, L ( − 2 , ρ ¯ ) = − 1 2 which can be calculated by [23, Proposition 1.51]. By arguments in [22, §4.5], the set

{ T 1 ( q ) , T 2 ( q ) , T 3 ( q ) , T 4 ( q ) , E ¯ 3 ( q ) , E ¯ 3 ( q 2 ) , E ¯ 3 ( q 3 ) , E ¯ 3 ( q 6 ) , E ˜ 3 ( q ) , E ˜ 3 ( q 2 ) , E ˜ 3 ( q 3 ) , E ˜ 3 ( q 6 ) }

forms a basis for M 3 ( 24 , ρ ) . Hence, the modular form H ( q ) 3 can be expressed as a linear sum of the elements of the above basis. We have

H ( q ) 3 = 1 7 ( E ¯ 3 ( q ) − 2 E ¯ 3 ( q 2 ) + 27 E ¯ 3 ( q 3 ) − 54 E ¯ 3 ( q 6 ) − 4 E ˜ 3 ( q ) + 32 E ˜ 3 ( q 2 ) + 108 E ˜ 3 ( q 3 ) − 864 E 3 ˜ ( q 6 ) − 18 T 1 ( q ) − 36 T 2 ( q ) − 18 T 3 ( q ) − 36 T 4 ( q ) )

and then we obtain, for N ∈ N ,

(8.1) a 3 ( N ) = 1 7 σ ¯ 2 ( N ) − 2 σ ¯ 2 N 2 + 27 σ ¯ 2 N 3 − 54 σ ¯ 2 N 6 − 4 σ ˜ 2 ( N ) + 32 σ ˜ 2 N 2 + 108 σ ˜ 2 N 3 − 864 σ ˜ 2 N 6 − t ( N ) .

Meanwhile, H ( q ) 3 can be expressed as follows:

H ( q ) 3 = 1 − ∑ N ≥ 1 e ˆ ( N ) q N = 1 − 3 ∑ N ≥ 1 e ˆ ( N ) q N + 3 ∑ N ≥ 2 ∑ k = 1 N − 1 e ˆ ( k ) e ˆ ( N − k ) q N − ∑ N ≥ 3 ∑ a 1 + a 2 + a 3 = N a 1 , a 2 , a 3 ≥ 1 e ˆ ( a 1 ) e ˆ ( a 2 ) e ˆ ( a 3 ) q N .

Then, for N ≥ 3 , we have

∑ a 1 + a 2 + a 3 = N a 1 , a 2 , a 3 ≥ 1 e ˆ ( a 1 ) e ˆ ( a 2 ) e ˆ ( a 3 ) = − 3 e ˆ ( N ) + 3 ∑ k = 1 N − 1 e ˆ ( k ) e ˆ ( N − k ) − a 3 ( N ) .

From Theorem, 2, 4, and (8.1), we obtain

∑ a 1 + a 2 + a 3 = N a 1 , a 2 , a 3 ≥ 1 e ˆ ( a 1 ) e ˆ ( a 2 ) e ˆ ( a 3 ) = 3 E 1 , 5 ( N ; 12 ) − 6 σ ( N ) − 1 7 ( σ ¯ 2 ( N ) − 4 σ ˜ 2 ( N ) ) + t ( N ) if N ≡ 1 ( mod 12 ) , 12 σ ( M ′ ) − 1 7 σ ¯ 2 ( N ) + 27 σ ¯ 2 N 3 − 4 σ ˜ 2 ( N ) + 108 σ ˜ 2 N 3 + t ( N ) if N ≡ 3 ( mod 12 ) , − 12 E 1 ( N ; 12 ) + 3 σ ( N ) − 1 7 ( σ ¯ 2 ( N ) − 4 σ ˜ 2 ( N ) ) + t ( N ) if N ≡ 5 ( mod 12 ) , − 6 σ ( N ) − 1 7 ( σ ¯ 2 ( N ) − 4 σ ˜ 2 ( N ) ) + t ( N ) if N ≡ 7 ( mod 12 ) , 12 E 1 , 5 ( N ; 12 ) + 12 σ ( M ′ ) − 1 7 σ ¯ 2 ( N ) + 27 σ ¯ 2 N 3 − 4 σ ˜ 2 ( N ) + 108 σ ˜ 2 N 3 + t ( N ) if N ≡ 9 ( mod 12 ) , 3 σ ( N ) − 1 7 ( σ ¯ 2 ( N ) − 4 σ ˜ 2 ( N ) ) + t ( N ) if N ≡ 11 ( mod 12 ) .

Finally, by applying Lemma 25, the proof is completed.□

Proof of Theorem 6

The proof is similar to that of Theorem 5. H ( q ) 4 is in the 13-dimensional space M 4 ( 18 ) , which is spanned by the set

{ S 1 ( q ) , S 2 ( q ) , S 3 ( q ) , S 4 ( q ) , S 5 ( q ) , E 4 ( q ) , E 4 ( q 2 ) , E 4 ( q 3 ) , E 4 ( q 6 ) , E 4 ( q 9 ) , E 4 ( q 18 ) , F 4 ( q ) , F 4 ( q 2 ) } ,

where

E 4 ( q ) ≔ 1 240 + ∑ n ≥ 1 σ 3 ( n ) q n , F 4 ( q ) ≔ ∑ n ≥ 1 n ≡ 1 ( 3 ) σ 3 ( n ) q n − ∑ n ≥ 1 n ≡ 2 ( 3 ) σ 3 ( n ) q n .

Then H ( q ) 4 can be expressed as

(8.2) H ( q ) 4 = 1 10 ( − E 4 ( q ) + 16 E 4 ( q 2 ) + 84 E 4 ( q 3 ) − 1344 E 4 ( q 6 ) − 243 E 4 ( q 9 ) + 3888 E 4 ( q 18 ) + 3 F 4 ( q ) + 48 F 4 ( q 2 ) − 42 S 1 ( q ) − 384 S 2 ( q ) − 1152 S 3 ( q ) − 1152 S 4 ( q ) ) .

Meanwhile,

H ( q ) 4 = 1 − ∑ N ≥ 1 e ˆ ( N ) q N 4 = 1 − 4 ∑ N ≥ 1 e ˆ ( N ) q N + 6 ∑ N ≥ 2 ∑ k = 1 N − 1 e ˆ ( k ) e ˆ ( N − k ) q N − 4 ∑ N ≥ 3 ∑ a 1 + a 2 + a 3 = N a 1 , a 2 , a 3 ≥ 1 e ˆ ( a 1 ) e ˆ ( a 2 ) e ˆ ( a 3 ) q N + ∑ N ≥ 4 ∑ a 1 + a 2 + a 3 + a 4 = N a 1 , a 2 , a 3 , a 4 ≥ 1 e ˆ ( a 1 ) e ˆ ( a 2 ) e ˆ ( a 3 ) e ˆ ( a 4 ) q N ,

so we obtain, for N ≥ 4 ,

∑ a 1 + a 2 + a 3 + a 4 = N a 1 , a 2 , a 3 , a 4 ≥ 1 e ˆ ( a 1 ) e ˆ ( a 2 ) e ˆ ( a 3 ) e ˆ ( a 4 ) = 4 e ˆ ( N ) − 6 ∑ k = 1 N − 1 e ˆ ( k ) e ˆ ( N − k ) + 4 ∑ a 1 + a 2 + a 3 = N a 1 , a 2 , a 3 ≥ 1 e ˆ ( a 1 ) e ˆ ( a 2 ) e ˆ ( a 3 ) + a 4 ( N ) .

From Theorems 2, 4, 5, and (8.2), we obtain

∑ a 1 + a 2 + a 3 + a 4 = N a 1 , a 2 , a 3 , a 4 ≥ 1 e ˆ ( a 1 ) e ˆ ( a 2 ) e ˆ ( a 3 ) e ˆ ( a 4 ) = 4 E 1 , 5 ( N ; 12 ) − 12 σ ( N ) + 12 7 σ ¯ 2 ( N ) + 1 5 σ 3 ( N ) + s ( N ) if N ≡ 1 ( mod 12 ) , 24 σ ( M ′ ) − 20 7 σ ¯ 2 ( N ) − 540 7 σ ¯ 2 N 3 − 1 10 σ 3 ( N ) + 42 5 σ 3 N 3 − 243 10 σ 3 N 9 + s ( N ) if N ≡ 3 ( mod 12 ) , − 16 E 1 ( N ; 12 ) + 6 σ ( N ) + 12 7 σ ¯ 2 ( N ) − 2 5 σ 3 ( N ) + s ( N ) if N ≡ 5 ( mod 12 ) , − 12 σ ( N ) − 20 7 σ ¯ 2 ( N ) + 1 5 σ 3 ( N ) + s ( N ) if N ≡ 7 ( mod 12 ) , 16 E 1 , 5 ( N ; 12 ) + 24 σ ( M ′ ) + 12 7 σ ¯ 2 ( N ) + 324 7 σ ¯ 2 N 3 − 1 10 σ 3 ( N ) + 42 5 σ 3 N 3 − 243 10 σ 3 N 9 + s ( N ) if N ≡ 9 ( mod 12 ) , 6 σ ( N ) − 20 7 σ ¯ 2 ( N ) − 2 5 σ 3 ( N ) + s ( N ) if N ≡ 11 ( mod 12 ) .

We note that if N ≡ 3 , 15 , 21 , 33 ( mod 36 ) , then N is not divisible by 9. In these cases, σ 3 ( N ) = σ 3 ( 3 ) σ 3 N 3 = 28 σ 3 N 3 . Hence, we have − 1 10 σ 3 ( N ) + 42 5 σ 3 N 3 − 243 10 σ 3 N 9 = 28 5 σ 3 N 3 . Then we have the desired result.□

Acknowledgements

The corresponding author was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIT) (No. 2021R1F1A1051093).

  1. Conflict of interest: The authors declare that there is no conflict of interest.

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Received: 2022-01-14
Accepted: 2022-03-23
Published Online: 2022-04-19

© 2022 Nazli Yildiz Ikikardes et al., published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/math-2022-0031
Keywords for this article
restricted divisor functions; eta quotient; convolution sums; q-series
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  96. N-Tuples of weighted noncommutative Orlicz space and some geometrical properties
  97. The dimension-free estimate for the truncated maximal operator
  98. A human error risk priority number calculation methodology using fuzzy and TOPSIS grey
  99. Compact mappings and s-mappings at subsets
  100. The structural properties of the Gompertz-two-parameter-Lindley distribution and associated inference
  101. A monotone iteration for a nonlinear Euler-Bernoulli beam equation with indefinite weight and Neumann boundary conditions
  102. Delta waves of the isentropic relativistic Euler system coupled with an advection equation for Chaplygin gas
  103. Multiplicity and minimality of periodic solutions to fourth-order super-quadratic difference systems
  104. On the reciprocal sum of the fourth power of Fibonacci numbers
  105. Averaging principle for two-time-scale stochastic differential equations with correlated noise
  106. Phragmén-Lindelöf alternative results and structural stability for Brinkman fluid in porous media in a semi-infinite cylinder
  107. Study on r-truncated degenerate Stirling numbers of the second kind
  108. On 7-valent symmetric graphs of order 2pq and 11-valent symmetric graphs of order 4pq
  109. Some new characterizations of finite p-nilpotent groups
  110. A Billingsley type theorem for Bowen topological entropy of nonautonomous dynamical systems
  111. F4 and PSp (8, ℂ)-Higgs pairs understood as fixed points of the moduli space of E6-Higgs bundles over a compact Riemann surface
  112. On modules related to McCoy modules
  113. On generalized extragradient implicit method for systems of variational inequalities with constraints of variational inclusion and fixed point problems
  114. Solvability for a nonlocal dispersal model governed by time and space integrals
  115. Finite groups whose maximal subgroups of even order are MSN-groups
  116. Symmetric results of a Hénon-type elliptic system with coupled linear part
  117. On the connection between Sp-almost periodic functions defined on time scales and ℝ
  118. On a class of Harada rings
  119. On regular subgroup functors of finite groups
  120. Fast iterative solutions of Riccati and Lyapunov equations
  121. Weak measure expansivity of C2 dynamics
  122. Admissible congruences on type B semigroups
  123. Generalized fractional Hermite-Hadamard type inclusions for co-ordinated convex interval-valued functions
  124. Inverse eigenvalue problems for rank one perturbations of the Sturm-Liouville operator
  125. Data transmission mechanism of vehicle networking based on fuzzy comprehensive evaluation
  126. Dual uniformities in function spaces over uniform continuity
  127. Review Article
  128. On Hahn-Banach theorem and some of its applications
  129. Rapid Communication
  130. Discussion of foundation of mathematics and quantum theory
  131. Special Issue on Boundary Value Problems and their Applications on Biosciences and Engineering (Part II)
  132. A study of minimax shrinkage estimators dominating the James-Stein estimator under the balanced loss function
  133. Representations by degenerate Daehee polynomials
  134. Multilevel MC method for weak approximation of stochastic differential equation with the exact coupling scheme
  135. Multiple periodic solutions for discrete boundary value problem involving the mean curvature operator
  136. Special Issue on Evolution Equations, Theory and Applications (Part II)
  137. Coupled measure of noncompactness and functional integral equations
  138. Existence results for neutral evolution equations with nonlocal conditions and delay via fractional operator
  139. Global weak solution of 3D-NSE with exponential damping
  140. Special Issue on Fractional Problems with Variable-Order or Variable Exponents (Part I)
  141. Ground state solutions of nonlinear Schrödinger equations involving the fractional p-Laplacian and potential wells
  142. A class of p1(x, ⋅) & p2(x, ⋅)-fractional Kirchhoff-type problem with variable s(x, ⋅)-order and without the Ambrosetti-Rabinowitz condition in ℝN
  143. Jensen-type inequalities for m-convex functions
  144. Special Issue on Problems, Methods and Applications of Nonlinear Analysis (Part III)
  145. The influence of the noise on the exact solutions of a Kuramoto-Sivashinsky equation
  146. Basic inequalities for statistical submanifolds in Golden-like statistical manifolds
  147. Global existence and blow up of the solution for nonlinear Klein-Gordon equation with variable coefficient nonlinear source term
  148. Hopf bifurcation and Turing instability in a diffusive predator-prey model with hunting cooperation
  149. Efficient fixed-point iteration for generalized nonexpansive mappings and its stability in Banach spaces
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