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

• Nazli Yildiz Ikikardes , Jihyun Hwang and Daeyeoul Kim
From the journal Open Mathematics

## 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.

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

## 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 35. Similarly, using Lemmas 35, 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 35, a 1 ( a i ) = 0 . This completes the proof of Theorem 1.□

If