Introduction
Throughout this paper, for complex numbers a a a , b b b , and q q q with ∣ q ∣ < 1 |q| < 1 ∣ q ∣ < 1 , we use the standard notation
( a ; q ) ∞ = ∏ j ≥ 0 ( 1 − a q j ) , and ( a , b ; q ) ∞ = ( a ; q ) ∞ ( b ; q ) ∞ . (1) (a;q)_\infty= \prod_{j\ge0}(1-aq^j), \qquad\text{and} \qquad(a,b;q)_\infty= (a;q)_\infty(b;q)_\infty. \tag*{(1)} ( a ; q ) ∞ = j ≥ 0 ∏ ( 1 − a q j ) , and ( a , b ; q ) ∞ = ( a ; q ) ∞ ( b ; q ) ∞ . (1)
For convenience, we write f r = ( q r ; q r ) ∞ f_r=(q^r;q^r)_\infty f r = ( q r ; q r ) ∞ , where r r r is a positive integer.
A non-increasing finite sequence of positive integers β 1 ≥ β 2 ≥ ⋯ ≥ β s > 0 \beta_1 \ge\beta_2 \ge\cdots\ge\beta_s > 0 β 1 ≥ β 2 ≥ ⋯ ≥ β s > 0 is said to be a partition of a positive integer n n n if n = ∑ i = 1 s β i n=\sum_{i=1}^{s}\beta_i n = ∑ i = 1 s β i . The integers β i \beta_i β i are called the parts of the partition. If p ( n ) p(n) p ( n ) denotes the number of partitions of n n n , then its generating function [7 ] is given by
∑ n = 0 ∞ p ( n ) q n = 1 ( q ; q ) ∞ = 1 f 1 , p ( 0 ) = 1. \sum_{n=0}^{\infty} p(n)q^n=\frac{1}{(q;q)_{\infty}}=\frac{1}{f_1},\qquad p(0)=1. n = 0 ∑ ∞ p ( n ) q n = ( q ; q ) ∞ 1 = f 1 1 , p ( 0 ) = 1.
For example, p ( 7 ) = 15 p(7)=15 p ( 7 ) = 15 , and the corresponding partitions are 7 7 7 , 6 + 1 6+1 6 + 1 , 5 + 2 5+2 5 + 2 , 5 + 1 + 1 5+1+1 5 + 1 + 1 , 4 + 3 4+3 4 + 3 , 4 + 2 + 1 4+2+1 4 + 2 + 1 , 4 + 1 + 1 + 1 4+1+1+1 4 + 1 + 1 + 1 , 3 + 3 + 1 3+3+1 3 + 3 + 1 , 3 + 2 + 2 3+2+2 3 + 2 + 2 , 3 + 2 + 1 + 1 3+2+1+1 3 + 2 + 1 + 1 , 3 + 1 + 1 + 1 + 1 3+1+1+1+1 3 + 1 + 1 + 1 + 1 , 2 + 2 + 2 + 1 2+2+2+1 2 + 2 + 2 + 1 , 2 + 2 + 1 + 1 + 1 2+2+1+1+1 2 + 2 + 1 + 1 + 1 , 2 + 1 + 1 + 1 + 1 + 1 2+1+1+1+1+1 2 + 1 + 1 + 1 + 1 + 1 , and 1 + 1 + 1 + 1 + 1 + 1 + 1 1+1+1+1+1+1+1 1 + 1 + 1 + 1 + 1 + 1 + 1 . Among the most celebrated arithmetic properties of p ( n ) p(n) p ( n ) are the following congruences due to Ramanujan [11 ] :
p ( 5 n + 4 ) ≡ 0 ( m o d 5 ) , p ( 7 n + 5 ) ≡ 0 ( m o d 7 ) , p ( 11 n + 6 ) ≡ 0 ( m o d 11 ) . p(5n+4)\equiv0\pmod{5},\qquad p(7n+5)\equiv0\pmod{7},\qquad p(11n+6)\equiv0\pmod{11}. p ( 5 n + 4 ) ≡ 0 ( mod 5 ) , p ( 7 n + 5 ) ≡ 0 ( mod 7 ) , p ( 11 n + 6 ) ≡ 0 ( mod 11 ) .
valid for every nonnegative integer n n n . Ramanujan’s congruences have motivated extensive investigations of divisibility properties of several restricted partition functions.
Chan [5 ] studied the cubic partition function a ( n ) a(n) a ( n ) , in which even parts may appear in two distinct colors. The generating function of a ( n ) a(n) a ( n ) is given by
A ( q ) : = ∑ n ≥ 0 a ( n ) q n = 1 f 1 f 2 = ∏ m ≥ 1 1 ( 1 − q 2 m − 1 ) ( 1 − q 2 m ) 2 ; a ( 0 ) = 1. (2) A(q):=\sum_{n\geq0}a(n)q^n=\frac{1}{f_1f_2}=\prod_{m\geq1}\frac{1}{(1-q^{2m-1})(1-q^{2m})^2};\qquad a(0)=1. \tag*{(2)} A ( q ) := n ≥ 0 ∑ a ( n ) q n = f 1 f 2 1 = m ≥ 1 ∏ ( 1 − q 2 m − 1 ) ( 1 − q 2 m ) 2 1 ; a ( 0 ) = 1. (2)
For example, the cubic partitions of 3 are 3 3 3 , 2 r + 1 2_r+1 2 r + 1 , 2 g + 1 2_g+1 2 g + 1 , and 1 + 1 + 1 1+1+1 1 + 1 + 1 , where the subscripts r r r and g g g distinguish the two colors of the even part. Clearly, a ( 3 ) = 4 a(3)=4 a ( 3 ) = 4 . Chan [5 Theorem 1] proved the identity
∑ n ≥ 0 a ( 3 n + 2 ) q n = 3 f 3 3 f 6 3 f 1 4 f 2 4 . (3) \sum_{n\geq0}a(3n+2)q^n=3\frac{f_3^3f_6^3}{f_1^4f_2^4}. \tag*{(3)} n ≥ 0 ∑ a ( 3 n + 2 ) q n = 3 f 1 4 f 2 4 f 3 3 f 6 3 . (3)
which immediately gives the Ramanujan-type congruence a ( 3 n + 2 ) ≡ 0 ( m o d 3 ) a(3n+2)\equiv0\pmod{3} a ( 3 n + 2 ) ≡ 0 ( mod 3 ) . Chen and Lin [6 Theorems 1.3–1.4] proved that, for every n ≥ 0 n\geq0 n ≥ 0 ,
a ( 25 n + 22 ) ≡ 0 ( m o d 5 ) a(25n+22)\equiv0\pmod{5} a ( 25 n + 22 ) ≡ 0 ( mod 5 )
and
a ( 49 n + 15 ) ≡ 0 ( m o d 7 ) , a ( 49 n + 29 ) ≡ 0 ( m o d 7 ) , a ( 49 n + 36 ) ≡ 0 ( m o d 7 ) , a ( 49 n + 43 ) ≡ 0 ( m o d 7 ) . \begin{aligned}
a(49n+15)&\equiv0\pmod{7},\\
a(49n+29)&\equiv0\pmod{7},\\
a(49n+36)&\equiv0\pmod{7},\\
a(49n+43)&\equiv0\pmod{7}.
\end{aligned} a ( 49 n + 15 ) a ( 49 n + 29 ) a ( 49 n + 36 ) a ( 49 n + 43 ) ≡ 0 ( mod 7 ) , ≡ 0 ( mod 7 ) , ≡ 0 ( mod 7 ) , ≡ 0 ( mod 7 ) .
Xiong [14 Theorem 1.3] extended the modulo 5 result to an infinite family modulo powers of 5.
Recently, Andrews and Ghosh Dastidar [1 ] defined S O M E ( n ) SOME(n) S O M E ( n ) as the difference between the sum of all odd parts and the sum of all even parts in the partitions of a nonnegative integer n n n . They established several divisibility properties of S O M E ( n ) SOME(n) S O M E ( n ) . In particular, they proved [1 Theorem 4] that S O M E ( 5 n + 4 ) ≡ 0 ( m o d 5 ) SOME(5n+4) \equiv0 \pmod{5} S O M E ( 5 n + 4 ) ≡ 0 ( mod 5 ) , which, together with Ramanujan’s congruence p ( 5 n + 4 ) ≡ 0 ( m o d 5 ) p(5n+4) \equiv0 \pmod{5} p ( 5 n + 4 ) ≡ 0 ( mod 5 ) , implies that both the sum of all odd parts and the sum of all even parts in the partitions of 5 n + 4 5n+4 5 n + 4 are divisible by 5. They also conjectured that S O M E ( λ ) ≡ 0 ( m o d 5 α ) SOME(\lambda) \equiv0 \pmod{5^\alpha} S O M E ( λ ) ≡ 0 ( mod 5 α ) , where α \alpha α is a positive integer and λ \lambda λ is a nonnegative integer satisfying 24 λ ≡ 1 ( m o d 5 α ) 24\lambda\equiv1 \pmod{5^\alpha} 24 λ ≡ 1 ( mod 5 α ) , which was subsequently proved by Bardhan and Saikia [2 ] . Furthermore, Garvan and Sarma [8 ] studied sums of certain parts in overpartitions and partitions without repeated odd parts. Subsequently, Baruah and Gogoi [3 ] investigated analogous sums of overlined parts in overpartitions.
Motivated by the S O M E ( n ) SOME(n) S O M E ( n ) function and its properties, in this paper, we define the corresponding arithmetic functions C O S ( n ) COS(n) C O S ( n ) and C E S ( n ) CES(n) C E S ( n ) for cubic partitions and study their divisibility properties. The function C O S ( n ) COS(n) C O S ( n ) (resp. C E S ( n ) CES(n) C E S ( n ) ) denotes the sum of all odd (resp. even) parts in all cubic partitions of a positive integer n n n . Since the sum of all parts of each cubic partition of n n n is n n n , it follows that
C O S ( n ) + C E S ( n ) = n a ( n ) . (4) COS(n) + CES(n) = na(n). \tag*{(4)} C O S ( n ) + C E S ( n ) = na ( n ) . (4)
For example, C O S ( 3 ) = 8 COS(3)=8 C O S ( 3 ) = 8 and C E S ( 3 ) = 4 CES(3)=4 C E S ( 3 ) = 4 . For convenience, we set C O S ( 0 ) = C E S ( 0 ) = 0 COS(0)=CES(0)=0 C O S ( 0 ) = C E S ( 0 ) = 0 .
The paper is organized as follows. § 2 records the necessary q q q -identities and extraction formulas used in the subsequent proofs. In § 3 , we derive the generating functions and divisor-sum formulas for C O S ( n ) COS(n) C O S ( n ) and C E S ( n ) CES(n) C E S ( n ) and establish their congruences modulo 3, 4, 5, and 7. Finally, § 4 presents a conjecture for C O S ( n ) COS(n) C O S ( n ) and C E S ( n ) CES(n) C E S ( n ) modulo higher prime powers.
Preliminary results
Ramanujan’s general theta function is defined by [4 Eq. (1.2.1), p. 6]
f ( α , β ) = ∑ m ∈ Z α m ( m + 1 ) / 2 β m ( m − 1 ) / 2 , ∣ α β ∣ < 1. f(\alpha,\beta)=\sum_{m\in\mathbb{Z}}\alpha^{m(m+1)/2}\beta^{m(m-1)/2}, \qquad|\alpha\beta|<1. f ( α , β ) = m ∈ Z ∑ α m ( m + 1 ) /2 β m ( m − 1 ) /2 , ∣ α β ∣ < 1.
Three important special cases [4 Eqs. (1.2.2)–(1.2.4), pp. 6–7] are
ϕ ( q ) : = f ( q , q ) = ∑ m ∈ Z q m 2 = f 2 5 f 1 2 f 4 2 . (5) \phi(q):=f(q,q)=\sum_{m\in\mathbb{Z}}q^{m^2}=\frac{f_2^5}{f_1^2f_4^2}. \tag*{(5)} ϕ ( q ) := f ( q , q ) = m ∈ Z ∑ q m 2 = f 1 2 f 4 2 f 2 5 . (5)
ψ ( q ) : = f ( q , q 3 ) = ∑ m ≥ 0 q m ( m + 1 ) / 2 = f 2 2 f 1 . (6) \psi(q):=f(q,q^3)=\sum_{m\geq0}q^{m(m+1)/2}=\frac{f_2^2}{f_1}. \tag*{(6)} ψ ( q ) := f ( q , q 3 ) = m ≥ 0 ∑ q m ( m + 1 ) /2 = f 1 f 2 2 . (6)
and
f ( − q ) : = f ( − q , − q 2 ) = ∑ m ∈ Z ( − 1 ) m q m ( 3 m − 1 ) / 2 = f 1 . (7) f(-q) := f(-q,-q^2) = \sum_{m\in\mathbb{Z}} (-1)^m q^{m(3m-1)/2} = f_1. \tag*{(7)} f ( − q ) := f ( − q , − q 2 ) = m ∈ Z ∑ ( − 1 ) m q m ( 3 m − 1 ) /2 = f 1 . (7)
The product representations in (2.1), (2.2), and (2.3) follow from Jacobi’s triple product identity [4 Theorem 1.3.3] , given by
f ( α , β ) = ( − α ; α β ) ∞ ( − β ; α β ) ∞ ( α β ; α β ) ∞ , f(\alpha,\beta)=(-\alpha;\alpha\beta)_\infty(-\beta;\alpha\beta)_\infty(\alpha\beta;\alpha\beta)_\infty, f ( α , β ) = ( − α ; α β ) ∞ ( − β ; α β ) ∞ ( α β ; α β ) ∞ ,
which can be equivalently stated as
∑ m ∈ Z z m t m ( m − 1 ) / 2 = ( − z , − t / z ; t ) ∞ ( t ; t ) ∞ , 0 < ∣ t ∣ < 1 , z ≠ 0. (8) \sum_{m\in\mathbb{Z}} z^m t^{m(m-1)/2}=(-z,-t/z;t)_\infty(t;t)_\infty,\qquad0<|t|<1,\quad z\ne0. \tag*{(8)} m ∈ Z ∑ z m t m ( m − 1 ) /2 = ( − z , − t / z ; t ) ∞ ( t ; t ) ∞ , 0 < ∣ t ∣ < 1 , z = 0. (8)
Replacing q q q by − q -q − q in (2.1) and simplifying, we obtain
ϕ ( − q ) = ∑ m ∈ Z ( − 1 ) m q m 2 = f 1 2 f 2 . (9) \phi(-q)=\sum_{m\in\mathbb{Z}}(-1)^m q^{m^2}=\frac{f_1^2}{f_2}. \tag*{(9)} ϕ ( − q ) = m ∈ Z ∑ ( − 1 ) m q m 2 = f 2 f 1 2 . (9)
Replacing q q q by q 2 q^2 q 2 in (2.2), we obtain
ψ ( q 2 ) = f 4 2 f 2 . (10) \psi(q^2)=\frac{f_4^2}{f_2}. \tag*{(10)} ψ ( q 2 ) = f 2 f 4 2 . (10)
Also, the binomial theorem gives, for every prime p p p ,
f r p = ∏ m ≥ 1 ( 1 − q r m ) p ≡ ∏ m ≥ 1 ( 1 − q p r m ) = f p r ( m o d p ) . (11) f_r^p=\prod_{m\geq1}(1-q^{rm})^p\equiv\prod_{m\geq1}(1-q^{prm})=f_{pr}\pmod p. \tag*{(11)} f r p = m ≥ 1 ∏ ( 1 − q r m ) p ≡ m ≥ 1 ∏ ( 1 − q p r m ) = f p r ( mod p ) . (11)
The following identity due to Jacobi [4 Theorem 1.3.9] will also be used in the proofs:
f 1 3 = ∑ m ≥ 0 ( − 1 ) m ( 2 m + 1 ) q m ( m + 1 ) / 2 . (12) f_1^3=\sum_{m\geq0}(-1)^m(2m+1)q^{m(m+1)/2}. \tag*{(12)} f 1 3 = m ≥ 0 ∑ ( − 1 ) m ( 2 m + 1 ) q m ( m + 1 ) /2 . (12)
Next, we record some results related to Lambert series. Define
L ( q ) = ∑ m ≥ 1 m q m 1 − q m = ∑ n ≥ 1 σ ( n ) q n , (13) L(q)=\sum_{m\geq1}\frac{m q^m}{1-q^m}=\sum_{n\geq1}\sigma(n)q^n, \tag*{(13)} L ( q ) = m ≥ 1 ∑ 1 − q m m q m = n ≥ 1 ∑ σ ( n ) q n , (13)
where σ ( n ) \sigma(n) σ ( n ) denotes the sum of all positive divisors of the positive integer n n n . Throughout the paper, we will use the notation D : = q d d q D:=q\frac{d}{dq} D := q d q d . By logarithmic differentiation of the q q q -product, f r = ( q r ; q r ) ∞ f_r=(q^r;q^r)_\infty f r = ( q r ; q r ) ∞ , we obtain
D f r f r = − r ∑ m ≥ 1 m q r m 1 − q r m = − r L ( q r ) . (14) \frac{D f_r}{f_r}=-r\sum_{m\geq1}\frac{m q^{rm}}{1-q^{rm}}=-rL(q^r). \tag*{(14)} f r D f r = − r m ≥ 1 ∑ 1 − q r m m q r m = − r L ( q r ) . (14)
Throughout the rest of the paper, we will denote the coefficient of q n q^n q n in F ( q ) F(q) F ( q ) by [ q n ] F ( q ) [q^n]F(q) [ q n ] F ( q ) , where F ( q ) F(q) F ( q ) is a formal power series in the variable q q q with coefficients from the ring of integers Z \mathbb{Z} Z . We will use the standard notation Z [ [ q ] ] \mathbb{Z}[[q]] Z [[ q ]] to denote the ring of formal power series in the variable q q q with coefficients from Z \mathbb{Z} Z . All congruences between formal power series are understood coefficientwise. The operator D D D preserves these congruences: if F ( q ) − G ( q ) = m C ( q ) F(q)-G(q)=mC(q) F ( q ) − G ( q ) = m C ( q ) for a positive integer m m m and some C ( q ) ∈ Z [ [ q ] ] C(q)\in\mathbb{Z}[[q]] C ( q ) ∈ Z [[ q ]] , then D F ( q ) − D G ( q ) = m D C ( q ) ∈ m Z [ [ q ] ] DF(q)-DG(q)=mDC(q)\in m\mathbb{Z}[[q]] D F ( q ) − D G ( q ) = m D C ( q ) ∈ m Z [[ q ]] . Also, for any positive integer m m m and 0 ≤ r < m 0\le r<m 0 ≤ r < m , let the extraction operator E m , r \mathcal{E}_{m,r} E m , r be defined by
E m , r ( ∑ n ≥ 0 g ( n ) q n ) = ∑ n ≥ 0 g ( m n + r ) q n . (15) \mathcal{E}_{m,r}\left(\sum_{n\geq0}g(n)q^n\right)=\sum_{n\geq0}g(mn+r)q^n. \tag*{(15)} E m , r ( n ≥ 0 ∑ g ( n ) q n ) = n ≥ 0 ∑ g ( mn + r ) q n . (15)
Let B ( q ) = ∑ j ≥ 0 b ( j ) q j B(q)=\sum_{j\geq0}b(j)q^j B ( q ) = ∑ j ≥ 0 b ( j ) q j and G ( q ) = ∑ n ≥ 0 g ( n ) q n G(q)=\sum_{n\geq0}g(n)q^n G ( q ) = ∑ n ≥ 0 g ( n ) q n . Then for any prime p p p ,
E p , r ( B ( q p ) G ( q ) ) = ∑ n ≥ 0 ∑ j = 0 n b ( j ) g ( p ( n − j ) + r ) q n = B ( q ) E p , r ( G ( q ) ) (16) \mathcal{E}_{p,r}\left(B(q^p)G(q)\right)=\sum_{n\geq0}\sum_{j=0}^{n}b(j)g\bigl(p(n-j)+r\bigr)q^n=B(q)\mathcal{E}_{p,r}\bigl(G(q)\bigr) \tag*{(16)} E p , r ( B ( q p ) G ( q ) ) = n ≥ 0 ∑ j = 0 ∑ n b ( j ) g ( p ( n − j ) + r ) q n = B ( q ) E p , r ( G ( q ) ) (16)
and
E p , r ( ( D − r ) G ( q ) ) = ∑ n ≥ 0 p n g ( p n + r ) q n = p D ( E p , r ( G ( q ) ) ) . (17) \mathcal{E}_{p,r}\bigl((D-r)G(q)\bigr)=\sum_{n\geq0}pn\,g(pn+r)q^n=pD\left(\mathcal{E}_{p,r}\bigl(G(q)\bigr)\right). \tag*{(17)} E p , r ( ( D − r ) G ( q ) ) = n ≥ 0 ∑ p n g ( p n + r ) q n = p D ( E p , r ( G ( q ) ) ) . (17)
Further, if G ( q ) ∈ Z [ [ q ] ] G(q)\in\mathbb{Z}[[q]] G ( q ) ∈ Z [[ q ]] , then from (17) , we note that
E p , r ( ( D − r ) G ( q ) ) ≡ 0 ( m o d p ) (18) \mathcal{E}_{p,r}\bigl((D-r)G(q)\bigr)\equiv0\pmod p \tag*{(18)} E p , r ( ( D − r ) G ( q ) ) ≡ 0 ( mod p ) (18)
and for B ( q ) ∈ Z [ [ q ] ] B(q)\in\mathbb{Z}[[q]] B ( q ) ∈ Z [[ q ]] , from the binomial theorem and Fermat’s little theorem, we obtain
B ( q ) p ≡ ∑ j ≥ 0 b ( j ) p q p j ≡ ∑ j ≥ 0 b ( j ) q p j = B ( q p ) ( m o d p ) . (19) B(q)^p\equiv\sum_{j\geq0}b(j)^p q^{pj}\equiv\sum_{j\geq0}b(j)q^{pj}=B(q^p)\pmod p. \tag*{(19)} B ( q ) p ≡ j ≥ 0 ∑ b ( j ) p q p j ≡ j ≥ 0 ∑ b ( j ) q p j = B ( q p ) ( mod p ) . (19)
Moreover, it can be verified that
E p , s ( E p , r ( G ( q ) ) ) = ∑ n ≥ 0 g ( p 2 n + p s + r ) q n , 0 ≤ r , s < p . (20) \mathcal{E}_{p,s}\bigl(\mathcal{E}_{p,r}(G(q))\bigr)=\sum_{n\geq0}g(p^2n+ps+r)q^n,\qquad0\leq r,s<p. \tag*{(20)} E p , s ( E p , r ( G ( q )) ) = n ≥ 0 ∑ g ( p 2 n + p s + r ) q n , 0 ≤ r , s < p . (20)
Next, we record some lemmas involving the Lambert series L ( q ) L(q) L ( q ) and the three important Eisenstein series E 2 ( q ) E_2(q) E 2 ( q ) , E 4 ( q ) E_4(q) E 4 ( q ) , and E 6 ( q ) E_6(q) E 6 ( q ) [10 ] , which are defined by
E 2 ( q ) = 1 − 24 ∑ n ≥ 1 σ 1 ( n ) q n = 1 − 24 L ( q ) , (21) E_2(q)=1-24\sum_{n\geq1}\sigma_1(n)q^n=1-24L(q), \tag*{(21)} E 2 ( q ) = 1 − 24 n ≥ 1 ∑ σ 1 ( n ) q n = 1 − 24 L ( q ) , (21)
E 4 ( q ) = 1 + 240 ∑ n ≥ 1 σ 3 ( n ) q n , (22) E_4(q)=1+240\sum_{n\geq1}\sigma_3(n)q^n, \tag*{(22)} E 4 ( q ) = 1 + 240 n ≥ 1 ∑ σ 3 ( n ) q n , (22)
and
E 6 ( q ) = 1 − 504 ∑ n ≥ 1 σ 5 ( n ) q n (23) E_6(q)=1-504\sum_{n\geq1}\sigma_5(n)q^n \tag*{(23)} E 6 ( q ) = 1 − 504 n ≥ 1 ∑ σ 5 ( n ) q n (23)
where, for any positive integer k k k , σ k ( n ) = ∑ d ∣ n d k \sigma_k(n)=\sum_{d\mid n}d^k σ k ( n ) = ∑ d ∣ n d k . Note that σ 1 ( n ) = σ ( n ) \sigma_1(n)=\sigma(n) σ 1 ( n ) = σ ( n ) . The above three Eisenstein series satisfy the following differential equations due to Ramanujan [10 Eq. (30), § 7] :
D E 2 ( q ) = E 2 ( q ) 2 − E 4 ( q ) 12 (24) DE_2(q)=\frac{E_2(q)^2-E_4(q)}{12} \tag*{(24)} D E 2 ( q ) = 12 E 2 ( q ) 2 − E 4 ( q ) (24)
D E 4 ( q ) = E 2 ( q ) E 4 ( q ) − E 6 ( q ) 3 (25) DE_4(q)=\frac{E_2(q)E_4(q)-E_6(q)}{3} \tag*{(25)} D E 4 ( q ) = 3 E 2 ( q ) E 4 ( q ) − E 6 ( q ) (25)
and
D E 6 ( q ) = E 2 ( q ) E 6 ( q ) − E 4 ( q ) 2 2 . (26) DE_6(q)=\frac{E_2(q)E_6(q)-E_4(q)^2}{2}. \tag*{(26)} D E 6 ( q ) = 2 E 2 ( q ) E 6 ( q ) − E 4 ( q ) 2 . (26)
From (21) , it follows that
M : = M ( q ) : = 2 E 2 ( q 2 ) − E 2 ( q ) = 1 + 24 ( L ( q ) − 2 L ( q 2 ) ) . (27) M:=M(q):=2E_2(q^2)-E_2(q)=1+24\bigl(L(q)-2L(q^2)\bigr). \tag*{(27)} M := M ( q ) := 2 E 2 ( q 2 ) − E 2 ( q ) = 1 + 24 ( L ( q ) − 2 L ( q 2 ) ) . (27)
Lemma 2.1. We have
M = 2 ϕ ( q ) 4 − ϕ ( − q ) 4 = ϕ ( − q ) 4 + 32 q ψ ( q 2 ) 4 , (28) M=2\phi(q)^4-\phi(-q)^4=\phi(-q)^4+32q\psi(q^2)^4, \tag*{(28)} M = 2 ϕ ( q ) 4 − ϕ ( − q ) 4 = ϕ ( − q ) 4 + 32 q ψ ( q 2 ) 4 , (28) 5 M 2 = E 4 ( q ) + 4 E 4 ( q 2 ) , (29) 5M^2=E_4(q)+4E_4(q^2), \tag*{(29)} 5 M 2 = E 4 ( q ) + 4 E 4 ( q 2 ) , (29) 7 M 3 = 8 E 6 ( q 2 ) − E 6 ( q ) , (30) 7M^3=8E_6(q^2)-E_6(q), \tag*{(30)} 7 M 3 = 8 E 6 ( q 2 ) − E 6 ( q ) , (30) and
E 6 ( q 2 ) = M 3 − 72 q M f 2 16 f 1 8 . (31) E_6(q^2)=M^3-72qM\frac{f_2^{16}}{f_1^8}. \tag*{(31)} E 6 ( q 2 ) = M 3 − 72 q M f 1 8 f 2 16 . (31)
Proof. Jacobi’s four-square theorem [9 p. 188] states that the number of ordered representations of a positive integer n n n as a sum of four integer squares, with signs and zero coordinates included, is 8 ∑ d ∣ n , 4 ∣ d d 8\sum_{d\mid n,\,4\mid d}d 8 ∑ d ∣ n , 4 ∣ d d . Thus, by (13) , we obtain
ϕ ( q ) 4 = 1 + 8 ∑ n ≥ 1 ( ∑ d ∣ n 4 ∣ d d ) q n = 1 + 8 ∑ n ≥ 1 σ ( n ) q n − 8 ∑ n ≥ 1 ∑ d ∣ n 4 ∣ d d q n = 1 + 8 L ( q ) − 32 ∑ m ≥ 1 ( ∑ e ∣ m e ) q 4 m = 1 + 8 L ( q ) − 32 L ( q 4 ) . (32) \begin{aligned}
\phi(q)^4 &= 1+8\sum_{n\geq1}\left(\sum_{\substack{d\mid n\\4\mid d}}d\right)q^n\\
&=1+8\sum_{n\geq1}\sigma(n)q^n-8\sum_{n\geq1}\sum_{\substack{d\mid n\\4\mid d}}dq^n\\
&=1+8L(q)-32\sum_{m\geq1}\left(\sum_{e\mid m}e\right)q^{4m}\\
&=1+8L(q)-32L(q^4). \tag*{(32)}
\end{aligned} ϕ ( q ) 4 = 1 + 8 n ≥ 1 ∑ d ∣ n 4 ∣ d ∑ d q n = 1 + 8 n ≥ 1 ∑ σ ( n ) q n − 8 n ≥ 1 ∑ d ∣ n 4 ∣ d ∑ d q n = 1 + 8 L ( q ) − 32 m ≥ 1 ∑ e ∣ m ∑ e q 4 m = 1 + 8 L ( q ) − 32 L ( q 4 ) . (32)
Set n = 2 a u n=2^a u n = 2 a u , where a ≥ 0 a\geq0 a ≥ 0 and u u u is an odd integer. Then
σ ( 2 a u ) = ∑ j = 0 a ∑ d ∣ u 2 j d = ( 2 a + 1 − 1 ) σ ( u ) . (33) \sigma(2^a u)=\sum_{j=0}^{a}\sum_{d\mid u}2^j d=(2^{a+1}-1)\sigma(u). \tag*{(33)} σ ( 2 a u ) = j = 0 ∑ a d ∣ u ∑ 2 j d = ( 2 a + 1 − 1 ) σ ( u ) . (33)
From (33) , it follows that, for a = 1 a=1 a = 1 ,
σ ( 2 u ) = 3 σ ( u ) \sigma(2u)=3\sigma(u) σ ( 2 u ) = 3 σ ( u )
and for a ≥ 1 a\geq1 a ≥ 1 ,
3 σ ( 2 a u ) − 2 σ ( 2 a − 1 u ) = ( 3 ( 2 a + 1 − 1 ) − 2 ( 2 a − 1 ) ) σ ( u ) = σ ( 2 a + 1 u ) . 3\sigma(2^a u)-2\sigma(2^{a-1}u)=\left(3(2^{a+1}-1)-2(2^a-1)\right)\sigma(u)=\sigma(2^{a+1}u). 3 σ ( 2 a u ) − 2 σ ( 2 a − 1 u ) = ( 3 ( 2 a + 1 − 1 ) − 2 ( 2 a − 1 ) ) σ ( u ) = σ ( 2 a + 1 u ) .
Therefore, for every integer n ≥ 1 n\geq1 n ≥ 1 , we obtain
σ ( 2 n ) = 3 σ ( n ) − 2 σ ( n / 2 ) , (34) \sigma(2n)=3\sigma(n)-2\sigma(n/2), \tag*{(34)} σ ( 2 n ) = 3 σ ( n ) − 2 σ ( n /2 ) , (34)
where we used the convention that σ ( x ) = 0 \sigma(x)=0 σ ( x ) = 0 when x x x is not a positive integer. Thus, from (13) and (34) , we obtain
L ( − q ) = ∑ n ≥ 1 ( − 1 ) n σ ( n ) q n = − ∑ n ≥ 1 σ ( n ) q n + 2 ∑ n ≥ 1 n e v e n σ ( n ) q n = − L ( q ) + 2 ∑ m ≥ 1 σ ( 2 m ) q 2 m = − L ( q ) + 6 ∑ m ≥ 1 σ ( m ) q 2 m − 4 ∑ m ≥ 1 σ ( m / 2 ) q 2 m = − L ( q ) + 6 L ( q 2 ) − 4 ∑ r ≥ 1 σ ( r ) q 4 r = − L ( q ) + 6 L ( q 2 ) − 4 L ( q 4 ) . (35) \begin{aligned}
L(-q)&=\sum_{n\geq1}(-1)^n\sigma(n)q^n\\
&=-\sum_{n\geq1}\sigma(n)q^n+2\sum_{\substack{n\geq1\\n\ \mathrm{even}}}\sigma(n)q^n\\
&=-L(q)+2\sum_{m\geq1}\sigma(2m)q^{2m}\\
&=-L(q)+6\sum_{m\geq1}\sigma(m)q^{2m}-4\sum_{m\geq1}\sigma(m/2)q^{2m}\\
&=-L(q)+6L(q^2)-4\sum_{r\geq1}\sigma(r)q^{4r}\\
&=-L(q)+6L(q^2)-4L(q^4). \tag*{(35)}
\end{aligned} L ( − q ) = n ≥ 1 ∑ ( − 1 ) n σ ( n ) q n = − n ≥ 1 ∑ σ ( n ) q n + 2 n ≥ 1 n even ∑ σ ( n ) q n = − L ( q ) + 2 m ≥ 1 ∑ σ ( 2 m ) q 2 m = − L ( q ) + 6 m ≥ 1 ∑ σ ( m ) q 2 m − 4 m ≥ 1 ∑ σ ( m /2 ) q 2 m = − L ( q ) + 6 L ( q 2 ) − 4 r ≥ 1 ∑ σ ( r ) q 4 r = − L ( q ) + 6 L ( q 2 ) − 4 L ( q 4 ) . (35)
Replacing q q q by − q -q − q in (32) and then employing (35) , we obtain
ϕ ( − q ) 4 = 1 + 8 L ( − q ) − 32 L ( q 4 ) = 1 − 8 L ( q ) + 48 L ( q 2 ) − 64 L ( q 4 ) . (36) \phi(-q)^4 = 1 + 8L(-q) - 32L(q^4) = 1 - 8L(q) + 48L(q^2) - 64L(q^4). \tag*{(36)} ϕ ( − q ) 4 = 1 + 8 L ( − q ) − 32 L ( q 4 ) = 1 − 8 L ( q ) + 48 L ( q 2 ) − 64 L ( q 4 ) . (36)
From (27) , (32) , and (36) , we obtain
2 ϕ ( q ) 4 − ϕ ( − q ) 4 = 1 + 24 L ( q ) − 48 L ( q 2 ) = M . (37) 2\phi(q)^4 - \phi(-q)^4 = 1 + 24L(q) - 48L(q^2) = M. \tag*{(37)} 2 ϕ ( q ) 4 − ϕ ( − q ) 4 = 1 + 24 L ( q ) − 48 L ( q 2 ) = M . (37)
From [4 Eq. (3.6.8), p. 72] , we note that
ϕ ( q ) 4 − ϕ ( − q ) 4 = 16 q ψ ( q 2 ) 4 . (38) \phi(q)^4 - \phi(-q)^4 = 16q\psi(q^2)^4. \tag*{(38)} ϕ ( q ) 4 − ϕ ( − q ) 4 = 16 q ψ ( q 2 ) 4 . (38)
Employing (38) in (37) , we obtain
M = ϕ ( − q ) 4 + 2 ( ϕ ( q ) 4 − ϕ ( − q ) 4 ) = ϕ ( − q ) 4 + 32 q ψ ( q 2 ) 4 , M = \phi(-q)^4 + 2\bigl(\phi(q)^4 - \phi(-q)^4\bigr) = \phi(-q)^4 + 32q\psi(q^2)^4, M = ϕ ( − q ) 4 + 2 ( ϕ ( q ) 4 − ϕ ( − q ) 4 ) = ϕ ( − q ) 4 + 32 q ψ ( q 2 ) 4 ,
which is (28) .
To prove the remaining identities, set
z = ϕ ( q ) 2 and x = ϕ ( q ) 4 − ϕ ( − q ) 4 ϕ ( q ) 4 . (39) z = \phi(q)^2 \qquad\text{and}\qquad x = \frac{\phi(q)^4 - \phi(-q)^4}{\phi(q)^4}. \tag*{(39)} z = ϕ ( q ) 2 and x = ϕ ( q ) 4 ϕ ( q ) 4 − ϕ ( − q ) 4 . (39)
Using (28) and (39) , we obtain
M = z 2 ( 1 + x ) = ϕ ( q ) 4 + ϕ ( q ) 4 − ϕ ( − q ) 4 . (40) M = z^2(1+x) = \phi(q)^4 + \phi(q)^4 - \phi(-q)^4. \tag*{(40)} M = z 2 ( 1 + x ) = ϕ ( q ) 4 + ϕ ( q ) 4 − ϕ ( − q ) 4 . (40)
From [4 pp. 126–127, Theorems 5.4.11 and 5.4.12] , we note that
E 4 ( q ) = z 4 ( 1 + 14 x + x 2 ) , (41) E_4(q) = z^4(1 + 14x + x^2), \tag*{(41)} E 4 ( q ) = z 4 ( 1 + 14 x + x 2 ) , (41)
E 4 ( q 2 ) = z 4 ( 1 − x + x 2 ) , (42) E_4(q^2) = z^4(1 - x + x^2), \tag*{(42)} E 4 ( q 2 ) = z 4 ( 1 − x + x 2 ) , (42)
E 6 ( q ) = z 6 ( 1 + x ) ( 1 − 34 x + x 2 ) , (43) E_6(q) = z^6(1 + x)(1 - 34x + x^2), \tag*{(43)} E 6 ( q ) = z 6 ( 1 + x ) ( 1 − 34 x + x 2 ) , (43)
and
E 6 ( q 2 ) = z 6 ( 1 + x ) ( 1 − x / 2 ) ( 1 − 2 x ) . (44) E_6(q^2) = z^6(1 + x)(1 - x/2)(1 - 2x). \tag*{(44)} E 6 ( q 2 ) = z 6 ( 1 + x ) ( 1 − x /2 ) ( 1 − 2 x ) . (44)
Combining (40) , (41) , and (42) , we obtain
E 4 ( q ) + 4 E 4 ( q 2 ) = z 4 ( 1 + 14 x + x 2 + 4 − 4 x + 4 x 2 ) = 5 ( z 2 ( 1 + x ) ) 2 = 5 M 2 , E_4(q) + 4E_4(q^2) = z^4(1 + 14x + x^2 + 4 - 4x + 4x^2) = 5\bigl(z^2(1+x)\bigr)^2 = 5M^2, E 4 ( q ) + 4 E 4 ( q 2 ) = z 4 ( 1 + 14 x + x 2 + 4 − 4 x + 4 x 2 ) = 5 ( z 2 ( 1 + x ) ) 2 = 5 M 2 ,
which is (29) . Again, using (40) , (43) , and (44) , we obtain
8 E 6 ( q 2 ) − E 6 ( q ) = z 6 ( 1 + x ) ( 7 + 14 x + 7 x 2 ) = 7 ( z 2 ( 1 + x ) ) 3 = 7 M 3 , 8E_6(q^2) - E_6(q) = z^6(1+x)(7 + 14x + 7x^2) = 7\bigl(z^2(1+x)\bigr)^3 = 7M^3, 8 E 6 ( q 2 ) − E 6 ( q ) = z 6 ( 1 + x ) ( 7 + 14 x + 7 x 2 ) = 7 ( z 2 ( 1 + x ) ) 3 = 7 M 3 ,
which is (30) .
Finally, using (5) , (10) , (38) , (39) , (40) , and (44) , we obtain
M 3 − E 6 ( q 2 ) = z 6 ( 1 + x ) ( ( 1 + x ) 2 − ( 1 − x / 2 ) ( 1 − 2 x ) ) = 9 2 x z 6 ( 1 + x ) = 9 2 M ϕ ( q ) 4 ( ϕ ( q ) 4 − ϕ ( − q ) 4 ) = 72 q M ϕ ( q ) 4 ψ ( q 2 ) 4 = 72 q M f 2 20 f 1 8 f 4 8 ⋅ f 4 8 f 2 4 = 72 q M f 2 16 f 1 8 . \begin{aligned}
M^3-E_6(q^2)&=z^6(1+x)\left((1+x)^2-(1-x/2)(1-2x)\right)\\
&=\frac{9}{2}xz^6(1+x)\\
&=\frac{9}{2}M\phi(q)^4\left(\phi(q)^4-\phi(-q)^4\right)\\
&=72qM\phi(q)^4\psi(q^2)^4\\
&=72qM\frac{f_2^{20}}{f_1^8f_4^8}\cdot\frac{f_4^8}{f_2^4}\\
&=72qM\frac{f_2^{16}}{f_1^8}.
\end{aligned} M 3 − E 6 ( q 2 ) = z 6 ( 1 + x ) ( ( 1 + x ) 2 − ( 1 − x /2 ) ( 1 − 2 x ) ) = 2 9 x z 6 ( 1 + x ) = 2 9 M ϕ ( q ) 4 ( ϕ ( q ) 4 − ϕ ( − q ) 4 ) = 72 q M ϕ ( q ) 4 ψ ( q 2 ) 4 = 72 q M f 1 8 f 4 8 f 2 20 ⋅ f 2 4 f 4 8 = 72 q M f 1 8 f 2 16 .
This completes the proof of (31) .
Lemma 2.2. We have
L ( q ) ≡ L ( q 3 ) + q f 3 8 ( m o d 3 ) (45) L(q)\equiv L(q^3)+qf_3^8\pmod{3} \tag*{(45)} L ( q ) ≡ L ( q 3 ) + q f 3 8 ( mod 3 ) (45) and the following exact dissection holds:
ψ ( q ) = f 6 f 9 2 f 3 f 18 + q ψ ( q 9 ) . (46) \psi(q)=\frac{f_6f_9^2}{f_3f_{18}}+q\psi(q^9). \tag*{(46)} ψ ( q ) = f 3 f 18 f 6 f 9 2 + q ψ ( q 9 ) . (46) Proof. In the series on the right-hand side of Jacobi’s identity (12) , the exponent r ( r + 1 ) / 2 r(r+1)/2 r ( r + 1 ) /2 is congruent to 0 modulo 3 if r ≡ 0 r\equiv0 r ≡ 0 or 2 (mod 3) and is congruent to 1 modulo 3 if r ≡ 1 r\equiv1 r ≡ 1 (mod 3). Thus, putting r = 3 u + 1 r=3u+1 r = 3 u + 1 with u ≥ 0 u\ge0 u ≥ 0 , we obtain
∑ u ≥ 0 ( − 1 ) 3 u + 1 ( 6 u + 3 ) q ( 3 u + 1 ) ( 3 u + 2 ) / 2 = − 3 q ∑ u ≥ 0 ( − 1 ) u ( 2 u + 1 ) q 9 u ( u + 1 ) / 2 = − 3 q f 9 3 (47) \sum_{u\ge0}(-1)^{3u+1}(6u+3)q^{(3u+1)(3u+2)/2}
=-3q\sum_{u\ge0}(-1)^u(2u+1)q^{9u(u+1)/2}
=-3qf_9^3 \tag*{(47)} u ≥ 0 ∑ ( − 1 ) 3 u + 1 ( 6 u + 3 ) q ( 3 u + 1 ) ( 3 u + 2 ) /2 = − 3 q u ≥ 0 ∑ ( − 1 ) u ( 2 u + 1 ) q 9 u ( u + 1 ) /2 = − 3 q f 9 3 (47) Therefore, from (12) and (47) , we obtain
f 1 3 = B ( q 3 ) − 3 q f 9 3 , (48) f_1^3=B(q^3)-3qf_9^3, \tag*{(48)} f 1 3 = B ( q 3 ) − 3 q f 9 3 , (48) for some B ( q ) ∈ Z [ [ q ] ] B(q)\in\mathbb{Z}[[q]] B ( q ) ∈ Z [[ q ]] . Using (11) in (48) , we note that
B ( q 3 ) ≡ f 3 ( m o d 3 ) , B(q^3)\equiv f_3\pmod{3}, B ( q 3 ) ≡ f 3 ( mod 3 ) , which immediately implies
B ( q ) ≡ f 1 ( m o d 3 ) . B(q)\equiv f_1\pmod{3}. B ( q ) ≡ f 1 ( mod 3 ) . Applying the differential operator D D D to B ( q ) B(q) B ( q ) , replacing q q q by q 3 q^3 q 3 , and employing (14) , we obtain
D B ( q ) ∣ q ↦ q 3 ≡ D f 1 ∣ q ↦ q 3 = − L ( q 3 ) f 3 ( m o d 3 ) . (49) DB(q)\big|_{q\mapsto q^3}\equiv Df_1\big|_{q\mapsto q^3}=-L(q^3)f_3\pmod{3}. \tag*{(49)} D B ( q ) q ↦ q 3 ≡ D f 1 q ↦ q 3 = − L ( q 3 ) f 3 ( mod 3 ) . (49)
Differentiating (48) and simplifying by using (14) , we obtain
− 3 L ( q ) f 1 3 = 3 D B ( q ) ∣ q ↦ q 3 − 3 q f 9 3 − 3 q D ( f 9 3 ) . (50) -3L(q)f_1^3 = 3\,DB(q)\big|_{q\mapsto q^3} - 3qf_9^3 - 3qD(f_9^3). \tag*{(50)} − 3 L ( q ) f 1 3 = 3 D B ( q ) q ↦ q 3 − 3 q f 9 3 − 3 q D ( f 9 3 ) . (50)
Dividing the exact identity (50) by − 3 -3 − 3 , we obtain
L ( q ) f 1 3 = − D B ( q ) ∣ q ↦ q 3 + q f 9 3 + q D ( f 9 3 ) . L(q)f_1^3 = -DB(q)\big|_{q\mapsto q^3} + qf_9^3 + qD(f_9^3). L ( q ) f 1 3 = − D B ( q ) q ↦ q 3 + q f 9 3 + q D ( f 9 3 ) .
Since D ( f 9 3 ) = − 27 L ( q 9 ) f 9 3 D(f_9^3)=-27L(q^9)f_9^3 D ( f 9 3 ) = − 27 L ( q 9 ) f 9 3 , reduction modulo 3 3 3 , followed by (11) and (49) , gives
L ( q ) f 3 ≡ L ( q 3 ) f 3 + q f 9 3 ( m o d 3 ) . L(q)f_3 \equiv L(q^3)f_3 + qf_9^3 \pmod{3}. L ( q ) f 3 ≡ L ( q 3 ) f 3 + q f 9 3 ( mod 3 ) .
Since f 3 f_3 f 3 has constant term 1 1 1 , we obtain
L ( q ) ≡ L ( q 3 ) + q f 9 3 f 3 ≡ L ( q 3 ) + q f 3 9 f 3 ≡ L ( q 3 ) + q f 3 8 ( m o d 3 ) , L(q) \equiv L(q^3)+q\frac{f_9^3}{f_3}\equiv L(q^3)+q\frac{f_3^9}{f_3}\equiv L(q^3)+qf_3^8 \pmod{3}, L ( q ) ≡ L ( q 3 ) + q f 3 f 9 3 ≡ L ( q 3 ) + q f 3 f 3 9 ≡ L ( q 3 ) + q f 3 8 ( mod 3 ) ,
which is (45) .
Next, express (6) as
ψ ( q ) = ∑ u ≥ 0 q u ( 3 u + 1 ) / 2 + q ∑ u ≥ 0 q u ( 9 u + 1 ) / 2 + ∑ u ≥ 0 q ( u + 1 ) ( 3 u + 2 ) / 2 = ∑ t ∈ Z ( q 3 ) t ( 3 t + 1 ) / 2 + q ψ ( q 9 ) . (51) \psi(q)=\sum_{u\geq0}q^{u(3u+1)/2}+q\sum_{u\geq0}q^{u(9u+1)/2}+\sum_{u\geq0}q^{(u+1)(3u+2)/2}=\sum_{t\in\mathbb{Z}}(q^3)^{t(3t+1)/2}+q\psi(q^9). \tag*{(51)} ψ ( q ) = u ≥ 0 ∑ q u ( 3 u + 1 ) /2 + q u ≥ 0 ∑ q u ( 9 u + 1 ) /2 + u ≥ 0 ∑ q ( u + 1 ) ( 3 u + 2 ) /2 = t ∈ Z ∑ ( q 3 ) t ( 3 t + 1 ) /2 + q ψ ( q 9 ) . (51)
Setting t = q 3 t=q^3 t = q 3 and z = q 2 z=q^2 z = q 2 in (8) , we obtain
∑ n ∈ Z q n ( 3 n + 1 ) / 2 = ( − q 2 , − q ; q 3 ) ∞ f 3 = f 2 f 3 2 f 1 f 6 . (52) \sum_{n\in\mathbb{Z}}q^{n(3n+1)/2}=(-q^2,-q;q^3)_\infty f_3=\frac{f_2f_3^2}{f_1f_6}. \tag*{(52)} n ∈ Z ∑ q n ( 3 n + 1 ) /2 = ( − q 2 , − q ; q 3 ) ∞ f 3 = f 1 f 6 f 2 f 3 2 . (52)
Replacing q q q by q 3 q^3 q 3 in (52) , we obtain
∑ t ∈ Z ( q 3 ) t ( 3 t + 1 ) / 2 = f 6 f 9 2 f 3 f 18 . (53) \sum_{t\in\mathbb{Z}}(q^3)^{t(3t+1)/2}=\frac{f_6f_9^2}{f_3f_{18}}. \tag*{(53)} t ∈ Z ∑ ( q 3 ) t ( 3 t + 1 ) /2 = f 3 f 18 f 6 f 9 2 . (53)
Employing (53) in (51) , we complete the proof of (46) . □
Lemma 2.3. We have
E 5 , 2 ( f 1 2 ) = − f 5 2 , (54) \mathcal{E}_{5,2}(f_1^2)=-f_5^2, \tag*{(54)} E 5 , 2 ( f 1 2 ) = − f 5 2 , (54) E 5 , 2 ( f 1 f 4 2 f 2 ) = q f 5 f 20 2 f 10 , (55) \mathcal{E}_{5,2}\left(\frac{f_1f_4^2}{f_2}\right)=q\frac{f_5f_{20}^2}{f_{10}}, \tag*{(55)} E 5 , 2 ( f 2 f 1 f 4 2 ) = q f 10 f 5 f 20 2 , (55) E 7 , 1 ( f 2 2 ) = q f 14 2 , (56) \mathcal{E}_{7,1}(f_2^2)=qf_{14}^2, \tag*{(56)} E 7 , 1 ( f 2 2 ) = q f 14 2 , (56) and
E 7 , 1 ( ϕ ( q ) f 4 ) = 2 ϕ ( − q ) f 4 + q ϕ ( q 7 ) f 28 . (57) \mathcal{E}_{7,1}(\phi(q)f_4)=2\phi(-q)f_4+q\phi(q^7)f_{28}. \tag*{(57)} E 7 , 1 ( ϕ ( q ) f 4 ) = 2 ϕ ( − q ) f 4 + q ϕ ( q 7 ) f 28 . (57)
Proof. Set
T ( q ) = ( q 2 , q 3 ; q 5 ) ∞ ( q , q 4 ; q 5 ) ∞ . T(q)=\frac{(q^2,q^3;q^5)_\infty}{(q,q^4;q^5)_\infty}. T ( q ) = ( q , q 4 ; q 5 ) ∞ ( q 2 , q 3 ; q 5 ) ∞ .
From [12 Chapter 16] (see also [4 Theorem 7.4.1, p. 161] ), we note that
f 1 = f 25 ( T ( q 5 ) − q − q 2 T ( q 5 ) ) . (58) f_1=f_{25}\left(T(q^5)-q-\frac{q^2}{T(q^5)}\right). \tag*{(58)} f 1 = f 25 ( T ( q 5 ) − q − T ( q 5 ) q 2 ) . (58)
Squaring (58) , we obtain
f 1 2 = f 25 2 ( T ( q 5 ) 2 − 2 q T ( q 5 ) − q 2 + 2 q 3 T ( q 5 ) + q 4 T ( q 5 ) 2 ) . (59) f_1^2=f_{25}^2\left(T(q^5)^2-2qT(q^5)-q^2+\frac{2q^3}{T(q^5)}+\frac{q^4}{T(q^5)^2}\right). \tag*{(59)} f 1 2 = f 25 2 ( T ( q 5 ) 2 − 2 q T ( q 5 ) − q 2 + T ( q 5 ) 2 q 3 + T ( q 5 ) 2 q 4 ) . (59)
Extracting the terms involving q 5 n + 2 q^{5n+2} q 5 n + 2 from (59) , dividing by q 2 q^2 q 2 , and replacing q 5 q^5 q 5 by q q q , we arrive at (2.50).
Replacing q q q by q 4 q^4 q 4 in (7) , we obtain
f 4 = ∑ r ∈ Z ( − 1 ) r q 6 r 2 − 2 r . (60) f_4=\sum_{r\in\mathbb{Z}}(-1)^r q^{6r^2-2r}. \tag*{(60)} f 4 = r ∈ Z ∑ ( − 1 ) r q 6 r 2 − 2 r . (60)
Setting t = q 4 t=q^4 t = q 4 and z = − q z=-q z = − q in (8) , we obtain
∑ s ∈ Z ( − 1 ) s q 2 s 2 − s = ( q , q 3 ; q 4 ) ∞ f 4 = f 1 f 4 f 2 . (61) \sum_{s\in\mathbb{Z}}(-1)^s q^{2s^2-s}=(q,q^3;q^4)_\infty f_4=\frac{f_1f_4}{f_2}. \tag*{(61)} s ∈ Z ∑ ( − 1 ) s q 2 s 2 − s = ( q , q 3 ; q 4 ) ∞ f 4 = f 2 f 1 f 4 . (61)
Multiplying (60) and (61) , we obtain
f 1 f 4 2 f 2 = ∑ r , s ∈ Z ( − 1 ) r + s q 6 r 2 − 2 r + 2 s 2 − s . (62) \frac{f_1f_4^2}{f_2}=\sum_{r,s\in\mathbb{Z}}(-1)^{r+s}q^{6r^2-2r+2s^2-s}. \tag*{(62)} f 2 f 1 f 4 2 = r , s ∈ Z ∑ ( − 1 ) r + s q 6 r 2 − 2 r + 2 s 2 − s . (62)
Set
N 1 = 6 r 2 − 2 r + 2 s 2 − s . N_1=6r^2-2r+2s^2-s. N 1 = 6 r 2 − 2 r + 2 s 2 − s .
Then
24 N 1 + 7 = ( 12 r − 2 ) 2 + 3 ( 4 s − 1 ) 2 . (63) 24N_1+7=(12r-2)^2+3(4s-1)^2. \tag*{(63)} 24 N 1 + 7 = ( 12 r − 2 ) 2 + 3 ( 4 s − 1 ) 2 . (63)
Since − 3 ≡ 2 ( m o d 5 ) -3\equiv2\pmod{5} − 3 ≡ 2 ( mod 5 ) is a quadratic nonresidue modulo 5, (63) implies N 1 ≡ 2 ( m o d 5 ) N_1\equiv2\pmod{5} N 1 ≡ 2 ( mod 5 ) if and only if r = 1 − 5 u r=1-5u r = 1 − 5 u and s = 5 v − 1 s=5v-1 s = 5 v − 1 , where u u u and v v v are integers. Consequently, from (62) , we obtain
E 5 , 2 ( f 1 f 4 2 f 2 ) = ∑ u , v ∈ Z ( − 1 ) u + v q ( 7 + 25 ( 6 u 2 − 2 u + 2 v 2 − v ) − 2 ) / 5 = q ( ∑ u ∈ Z ( − 1 ) u q 5 ( 6 u 2 − 2 u ) ) ( ∑ v ∈ Z ( − 1 ) v q 5 ( 2 v 2 − v ) ) = q f 20 f 5 f 20 f 10 = q f 5 f 20 2 f 10 , \begin{aligned}
\mathcal{E}_{5,2}\left(\frac{f_1f_4^2}{f_2}\right)
= \sum_{u,v\in\mathbb{Z}} (-1)^{u+v}q^{(7+25(6u^2-2u+2v^2-v)-2)/5} \\
&= q\left(\sum_{u\in\mathbb{Z}}(-1)^u q^{5(6u^2-2u)}\right)\left(\sum_{v\in\mathbb{Z}}(-1)^v q^{5(2v^2-v)}\right) \\
&= qf_{20}\frac{f_5f_{20}}{f_{10}} \\
&= q\frac{f_5f_{20}^2}{f_{10}},
\end{aligned} E 5 , 2 ( f 2 f 1 f 4 2 ) = u , v ∈ Z ∑ ( − 1 ) u + v q ( 7 + 25 ( 6 u 2 − 2 u + 2 v 2 − v ) − 2 ) /5 = q ( u ∈ Z ∑ ( − 1 ) u q 5 ( 6 u 2 − 2 u ) ) ( v ∈ Z ∑ ( − 1 ) v q 5 ( 2 v 2 − v ) ) = q f 20 f 10 f 5 f 20 = q f 10 f 5 f 20 2 ,
which completes the proof of (55) .
Replacing q q q by q 2 q^2 q 2 in (7) and squaring, we obtain
f 2 2 = ∑ r , s ∈ Z ( − 1 ) r + s q 3 r 2 − r + 3 s 2 − s . (64) f_2^2=\sum_{r,s\in\mathbb{Z}}(-1)^{r+s}q^{3r^2-r+3s^2-s}. \tag*{(64)} f 2 2 = r , s ∈ Z ∑ ( − 1 ) r + s q 3 r 2 − r + 3 s 2 − s . (64)
Set
N 2 = 3 r 2 − r + 3 s 2 − s . N_2=3r^2-r+3s^2-s. N 2 = 3 r 2 − r + 3 s 2 − s .
Then
12 N 2 + 2 = ( 6 r − 1 ) 2 + ( 6 s − 1 ) 2 . (65) 12N_2+2=(6r-1)^2+(6s-1)^2. \tag*{(65)} 12 N 2 + 2 = ( 6 r − 1 ) 2 + ( 6 s − 1 ) 2 . (65)
Since − 1 -1 − 1 is a quadratic nonresidue modulo 7 7 7 , (65) gives N 2 ≡ 1 ( m o d 7 ) N_2\equiv1\pmod{7} N 2 ≡ 1 ( mod 7 ) if and only if r = 7 u − 1 r=7u-1 r = 7 u − 1 and s = 7 v − 1 s=7v-1 s = 7 v − 1 . Therefore,
E 7 , 1 ( f 2 2 ) = ∑ u , v ∈ Z ( − 1 ) u + v q ( 8 + 49 ( 3 u 2 − u + 3 v 2 − v ) − 1 ) / 7 = q ( ∑ u ∈ Z ( − 1 ) u q 7 ( 3 u 2 − u ) ) 2 = q f 14 2 , \mathcal{E}_{7,1}(f_2^2)=\sum_{u,v\in\mathbb{Z}}(-1)^{u+v}q^{(8+49(3u^2-u+3v^2-v)-1)/7}
=q\left(\sum_{u\in\mathbb{Z}}(-1)^u q^{7(3u^2-u)}\right)^2
=qf_{14}^2, E 7 , 1 ( f 2 2 ) = u , v ∈ Z ∑ ( − 1 ) u + v q ( 8 + 49 ( 3 u 2 − u + 3 v 2 − v ) − 1 ) /7 = q ( u ∈ Z ∑ ( − 1 ) u q 7 ( 3 u 2 − u ) ) 2 = q f 14 2 ,
which completes the proof of (56) .
From (5) and (60) , we obtain
ϕ ( q ) f 4 = ∑ r , s ∈ Z ( − 1 ) s q r 2 + 6 s 2 − 2 s . (66) \phi(q)f_4=\sum_{r,s\in\mathbb{Z}}(-1)^s q^{r^2+6s^2-2s}. \tag*{(66)} ϕ ( q ) f 4 = r , s ∈ Z ∑ ( − 1 ) s q r 2 + 6 s 2 − 2 s . (66)
The condition r 2 + 6 s 2 − 2 s ≡ 1 ( m o d 7 ) r^2+6s^2-2s\equiv1\pmod{7} r 2 + 6 s 2 − 2 s ≡ 1 ( mod 7 ) is equivalent to r ≡ ± ( s + 1 ) ( m o d 7 ) r\equiv\pm(s+1)\pmod{7} r ≡ ± ( s + 1 ) ( mod 7 ) . The class r ≡ − ( s + 1 ) ( m o d 7 ) r\equiv-(s+1)\pmod{7} r ≡ − ( s + 1 ) ( mod 7 ) is parametrized bijectively by
r = 6 t − v − 1 , s = t + v , t , v ∈ Z , r=6t-v-1,\qquad s=t+v,\qquad t,v\in\mathbb{Z}, r = 6 t − v − 1 , s = t + v , t , v ∈ Z ,
since t = ( r + s + 1 ) / 7 t=(r+s+1)/7 t = ( r + s + 1 ) /7 and v = s − t v=s-t v = s − t . Replacing r r r by − r -r − r parametrizes the other class. The two classes intersect when r ≡ 0 ( m o d 7 ) r\equiv0\pmod{7} r ≡ 0 ( mod 7 ) and s ≡ − 1 ( m o d 7 ) s\equiv-1\pmod{7} s ≡ − 1 ( mod 7 ) , so this intersection must be counted only once. For either parametrization, we have
r 2 + 6 s 2 − 2 s = 1 + 7 ( v 2 + 6 t 2 − 2 t ) and ( − 1 ) s = ( − 1 ) t + v . r^2+6s^2-2s=1+7(v^2+6t^2-2t)\quad\text{and}\quad(-1)^s=(-1)^{t+v}. r 2 + 6 s 2 − 2 s = 1 + 7 ( v 2 + 6 t 2 − 2 t ) and ( − 1 ) s = ( − 1 ) t + v .
So by (9) and (60) , we obtain
∑ t , v ∈ Z ( − 1 ) t + v q v 2 + 6 t 2 − 2 t = ( ∑ v ∈ Z ( − 1 ) v q v 2 ) ( ∑ t ∈ Z ( − 1 ) t q 6 t 2 − 2 t ) = ϕ ( − q ) f 4 . \sum_{t,v\in\mathbb{Z}}(-1)^{t+v}q^{v^2+6t^2-2t}=\left(\sum_{v\in\mathbb{Z}}(-1)^vq^{v^2}\right)\left(\sum_{t\in\mathbb{Z}}(-1)^tq^{6t^2-2t}\right)=\phi(-q)f_4. t , v ∈ Z ∑ ( − 1 ) t + v q v 2 + 6 t 2 − 2 t = ( v ∈ Z ∑ ( − 1 ) v q v 2 ) ( t ∈ Z ∑ ( − 1 ) t q 6 t 2 − 2 t ) = ϕ ( − q ) f 4 .
Also, when r ≡ 0 ( m o d 7 ) r\equiv0\pmod{7} r ≡ 0 ( mod 7 ) and s ≡ − 1 ( m o d 7 ) s\equiv-1\pmod{7} s ≡ − 1 ( mod 7 ) , writing r = 7 d r=7d r = 7 d and s = 7 u − 1 s=7u-1 s = 7 u − 1 for d , u ∈ Z d,u\in\mathbb{Z} d , u ∈ Z , we obtain
r 2 + 6 s 2 − 2 s = 8 + 49 ( d 2 + 6 u 2 − 2 u ) and ( − 1 ) s = − ( − 1 ) u . r^2+6s^2-2s=8+49(d^2+6u^2-2u)\quad\text{and}\quad(-1)^s=-(-1)^u. r 2 + 6 s 2 − 2 s = 8 + 49 ( d 2 + 6 u 2 − 2 u ) and ( − 1 ) s = − ( − 1 ) u .
So, when r ≡ 0 ( m o d 7 ) r\equiv0\pmod{7} r ≡ 0 ( mod 7 ) and s ≡ − 1 ( m o d 7 ) s\equiv-1\pmod{7} s ≡ − 1 ( mod 7 ) ,
− ∑ d , u ∈ Z ( − 1 ) u q 1 + 7 ( d 2 + 6 u 2 − 2 u ) = − q ϕ ( q 7 ) f 28 . -\sum_{d,u\in\mathbb{Z}}(-1)^u q^{1+7(d^2+6u^2-2u)}=-q\phi(q^7)f_{28}. − d , u ∈ Z ∑ ( − 1 ) u q 1 + 7 ( d 2 + 6 u 2 − 2 u ) = − q ϕ ( q 7 ) f 28 .
Therefore, we obtain
E 7 , 1 ( ϕ ( q ) f 4 ) = 2 ϕ ( − q ) f 4 − ( − q ϕ ( q 7 ) f 28 ) = 2 ϕ ( − q ) f 4 + q ϕ ( q 7 ) f 28 . \mathcal{E}_{7,1}(\phi(q)f_4)=2\phi(-q)f_4-\left(-q\phi(q^7)f_{28}\right)=2\phi(-q)f_4+q\phi(q^7)f_{28}. E 7 , 1 ( ϕ ( q ) f 4 ) = 2 ϕ ( − q ) f 4 − ( − q ϕ ( q 7 ) f 28 ) = 2 ϕ ( − q ) f 4 + q ϕ ( q 7 ) f 28 .
Hence, the proof of (57) is complete.
Lemma 2.4. We have
E 5 , 2 ( M A ( q ) ) ≡ M f 1 3 f 2 3 ( m o d 5 ) . (67) \mathcal{E}_{5,2}(MA(q))\equiv Mf_1^3f_2^3\pmod{5}. \tag*{(67)} E 5 , 2 ( M A ( q )) ≡ M f 1 3 f 2 3 ( mod 5 ) . (67) and
E 7 , 1 ( M A ( q ) ) ≡ 4 f 1 f 2 M 3 ( m o d 7 ) . (68) \mathcal{E}_{7,1}(MA(q))\equiv4f_1f_2M^3\pmod{7}. \tag*{(68)} E 7 , 1 ( M A ( q )) ≡ 4 f 1 f 2 M 3 ( mod 7 ) . (68)
Proof. Set
K = f 2 19 f 1 9 f 4 8 and J = f 1 7 f 2 5 . K=\frac{f_2^{19}}{f_1^9f_4^8}\quad\text{and}\quad J=\frac{f_1^7}{f_2^5}. K = f 1 9 f 4 8 f 2 19 and J = f 2 5 f 1 7 .
From (2) , (5) , (9) , and (28) , we obtain
M A ( q ) = 2 ϕ ( q ) 4 − ϕ ( − q ) 4 f 1 f 2 = 2 K − J . (69) MA(q)=\frac{2\phi(q)^4-\phi(-q)^4}{f_1f_2}=2K-J. \tag*{(69)} M A ( q ) = f 1 f 2 2 ϕ ( q ) 4 − ϕ ( − q ) 4 = 2 K − J . (69)
Using (11) , we obtain
K = f 2 20 f 1 10 f 4 10 f 1 f 4 2 f 2 ≡ f 10 4 f 5 2 f 20 2 f 1 f 4 2 f 2 ( m o d 5 ) , (70) K=\frac{f_2^{20}}{f_1^{10}f_4^{10}}\frac{f_1f_4^2}{f_2}\equiv\frac{f_{10}^4}{f_5^2f_{20}^2}\frac{f_1f_4^2}{f_2}\pmod{5}, \tag*{(70)} K = f 1 10 f 4 10 f 2 20 f 2 f 1 f 4 2 ≡ f 5 2 f 20 2 f 10 4 f 2 f 1 f 4 2 ( mod 5 ) , (70)
and
J = f 1 5 f 2 5 f 1 2 ≡ f 5 f 10 f 1 2 ( m o d 5 ) . (71) J=\frac{f_1^5}{f_2^5}f_1^2\equiv\frac{f_5}{f_{10}}f_1^2\pmod{5}. \tag*{(71)} J = f 2 5 f 1 5 f 1 2 ≡ f 10 f 5 f 1 2 ( mod 5 ) . (71)
Employing (2.66) and (2.67) in (2.65), and then simplifying by using (2.12), (2.50), and (2.51), we obtain
E 5 , 2 ( M A ( q ) ) = 2 f 2 4 f 1 2 f 4 2 ( q f 5 f 2 6 f 10 ) − f 1 f 2 ( − f 5 2 ) = 2 q f 2 4 f 5 f 2 10 f 1 2 f 4 2 f 10 + f 1 f 5 2 f 2 ( m o d 5 ) . (72) \mathcal{E}_{5,2}(MA(q))=2\frac{f_2^4}{f_1^2f_4^2}\left(q\frac{f_5f_2^6}{f_{10}}\right)-\frac{f_1}{f_2}(-f_5^2)=2q\frac{f_2^4f_5f_2^{10}}{f_1^2f_4^2f_{10}}+\frac{f_1f_5^2}{f_2}\pmod{5}. \tag*{(72)} E 5 , 2 ( M A ( q )) = 2 f 1 2 f 4 2 f 2 4 ( q f 10 f 5 f 2 6 ) − f 2 f 1 ( − f 5 2 ) = 2 q f 1 2 f 4 2 f 10 f 2 4 f 5 f 2 10 + f 2 f 1 f 5 2 ( mod 5 ) . (72)
Employing (2.5), (2.6), (2.7), and (2.24) in (2.68), we obtain
E 5 , 2 ( M A ( q ) ) ≡ 2 q f 2 4 f 1 2 f 4 2 f 1 5 f 4 10 f 2 5 + f 1 f 2 f 1 10 ( m o d 5 ) = 2 q f 1 3 f 4 8 f 2 + f 1 11 f 2 = f 1 3 f 2 3 ( 2 q f 4 8 f 2 4 + f 1 8 f 2 4 ) = f 1 3 f 2 3 ( 2 q ψ ( q 2 ) 4 + ϕ ( − q ) 4 ) ≡ M f 1 3 f 2 3 ( m o d 5 ) . \begin{aligned}
\mathcal{E}_{5,2}(MA(q))&\equiv2q\frac{f_2^4}{f_1^2f_4^2}\frac{f_1^5f_4^{10}}{f_2^5}+\frac{f_1}{f_2}f_1^{10}\pmod{5}\\
&=2q\frac{f_1^3f_4^8}{f_2}+\frac{f_1^{11}}{f_2}\\
&=f_1^3f_2^3\left(2q\frac{f_4^8}{f_2^4}+\frac{f_1^8}{f_2^4}\right)\\
&=f_1^3f_2^3\left(2q\psi(q^2)^4+\phi(-q)^4\right)\\
&\equiv Mf_1^3f_2^3\pmod{5}.
\end{aligned} E 5 , 2 ( M A ( q )) ≡ 2 q f 1 2 f 4 2 f 2 4 f 2 5 f 1 5 f 4 10 + f 2 f 1 f 1 10 ( mod 5 ) = 2 q f 2 f 1 3 f 4 8 + f 2 f 1 11 = f 1 3 f 2 3 ( 2 q f 2 4 f 4 8 + f 2 4 f 1 8 ) = f 1 3 f 2 3 ( 2 q ψ ( q 2 ) 4 + ϕ ( − q ) 4 ) ≡ M f 1 3 f 2 3 ( mod 5 ) .
So the proof of (2.63) is complete.
Similarly, from (2.7) and (2.1), we obtain
K = f 2 14 f 1 7 f 4 7 f 2 5 f 1 2 f 4 2 ≡ f 2 2 f 7 f 28 ϕ ( q ) f 4 ( m o d 7 ) . (73) K=\frac{f_2^{14}}{f_1^7f_4^7}\frac{f_2^5}{f_1^2f_4^2}\equiv\frac{f_2^2}{f_7f_{28}}\phi(q)f_4\pmod{7}. \tag*{(73)} K = f 1 7 f 4 7 f 2 14 f 1 2 f 4 2 f 2 5 ≡ f 7 f 28 f 2 2 ϕ ( q ) f 4 ( mod 7 ) . (73)
and
J = f 1 7 f 2 7 f 2 2 = f 7 f 14 f 2 2 ( m o d 7 ) . (74) J=\frac{f_1^7}{f_2^7}f_2^2=\frac{f_7}{f_{14}}f_2^2\pmod{7}. \tag*{(74)} J = f 2 7 f 1 7 f 2 2 = f 14 f 7 f 2 2 ( mod 7 ) . (74)
Employing (2.12) and (2.53) in (2.69), and then using (2.1), (2.5), (2.6), (2.7), and (2.15), we obtain
E 7 , 1 ( K ) ≡ f 2 2 f 1 f 4 ( 2 ϕ ( − q ) f 4 + q ϕ ( q 7 ) f 28 ) ( m o d 7 ) ≡ 2 f 1 f 2 + q f 2 2 f 4 6 f 1 2 f 4 ϕ ( q ) 7 ( m o d 7 ) = 2 f 1 f 2 + q f 2 37 f 1 15 f 4 8 = f 1 f 2 ( 2 + q f 4 8 f 2 4 f 2 40 f 1 16 f 4 16 ) = f 1 f 2 ( 2 + q ψ ( q 2 ) 4 ϕ ( q 8 ) ) . (75) \begin{aligned}
\mathcal{E}_{7,1}(K)&\equiv\frac{f_2^2}{f_1f_4}\left(2\phi(-q)f_4+q\phi(q^7)f_{28}\right)\pmod{7}\\
&\equiv2f_1f_2+q\frac{f_2^2f_4^6}{f_1^2f_4}\phi(q)^7\pmod{7}\\
&=2f_1f_2+q\frac{f_2^{37}}{f_1^{15}f_4^8}\\
&=f_1f_2\left(2+q\frac{f_4^8}{f_2^4}\frac{f_2^{40}}{f_1^{16}f_4^{16}}\right)\\
&=f_1f_2\left(2+q\psi(q^2)^4\phi(q^8)\right). \tag*{(75)}
\end{aligned} E 7 , 1 ( K ) ≡ f 1 f 4 f 2 2 ( 2 ϕ ( − q ) f 4 + q ϕ ( q 7 ) f 28 ) ( mod 7 ) ≡ 2 f 1 f 2 + q f 1 2 f 4 f 2 2 f 4 6 ϕ ( q ) 7 ( mod 7 ) = 2 f 1 f 2 + q f 1 15 f 4 8 f 2 37 = f 1 f 2 ( 2 + q f 2 4 f 4 8 f 1 16 f 4 16 f 2 40 ) = f 1 f 2 ( 2 + q ψ ( q 2 ) 4 ϕ ( q 8 ) ) . (75)
Similarly, employing (16) and (56) in (74) , and then using (11) , (5) , (9) , and (10) , we obtain
E 7 , 1 ( J ) ≡ f 1 f 2 q f 14 2 ( m o d 7 ) ≡ q f 1 f 2 13 ( m o d 7 ) = q f 1 f 2 ⋅ f 4 8 f 2 4 ⋅ f 2 20 f 1 8 f 4 8 ⋅ f 1 8 f 2 4 = q f 1 f 2 ψ ( q 2 ) 4 ϕ ( q ) 4 ϕ ( − q ) 4 . (76) \begin{aligned}
\mathcal{E}_{7,1}(J) &\equiv\frac{f_1}{f_2}qf_{14}^{2} \pmod{7}\\
&\equiv qf_1f_2^{13} \pmod{7}\\
&= qf_1f_2\cdot\frac{f_4^{8}}{f_2^{4}}\cdot\frac{f_2^{20}}{f_1^{8}f_4^{8}}\cdot\frac{f_1^{8}}{f_2^{4}}\\
&=qf_1f_2\psi(q^2)^4\phi(q)^4\phi(-q)^4. \tag*{(76)}
\end{aligned} E 7 , 1 ( J ) ≡ f 2 f 1 q f 14 2 ( mod 7 ) ≡ q f 1 f 2 13 ( mod 7 ) = q f 1 f 2 ⋅ f 2 4 f 4 8 ⋅ f 1 8 f 4 8 f 2 20 ⋅ f 2 4 f 1 8 = q f 1 f 2 ψ ( q 2 ) 4 ϕ ( q ) 4 ϕ ( − q ) 4 . (76)
Employing (75) and (76) in (69) , and then using (28) , (5) , and (10) , we obtain
E 7 , 1 ( M ⋅ A ( q ) ) ≡ f 1 f 2 ( 4 + 2 q ψ ( q 2 ) 4 ϕ ( q ) 8 − q ψ ( q 2 ) 4 ϕ ( q ) 4 ϕ ( − q ) 4 ) ( m o d 7 ) = f 1 f 2 ( 4 + q ψ ( q 2 ) 4 ϕ ( q ) 4 ( 2 ϕ ( q ) 4 − ϕ ( − q ) 4 ) ) = f 1 f 2 ( 4 + q M f 2 16 f 1 8 ) . (77) \begin{aligned}
\mathcal{E}_{7,1}(M\cdot A(q)) &\equiv f_1f_2\left(4+2q\psi(q^2)^4\phi(q)^8-q\psi(q^2)^4\phi(q)^4\phi(-q)^4\right)\pmod{7}\\
&=f_1f_2\left(4+q\psi(q^2)^4\phi(q)^4\left(2\phi(q)^4-\phi(-q)^4\right)\right)\\
&=f_1f_2\left(4+qM\frac{f_2^{16}}{f_1^8}\right). \tag*{(77)}
\end{aligned} E 7 , 1 ( M ⋅ A ( q )) ≡ f 1 f 2 ( 4 + 2 q ψ ( q 2 ) 4 ϕ ( q ) 8 − q ψ ( q 2 ) 4 ϕ ( q ) 4 ϕ ( − q ) 4 ) ( mod 7 ) = f 1 f 2 ( 4 + q ψ ( q 2 ) 4 ϕ ( q ) 4 ( 2 ϕ ( q ) 4 − ϕ ( − q ) 4 ) ) = f 1 f 2 ( 4 + q M f 1 8 f 2 16 ) . (77)
From (23) , it is easily seen that E 6 ( q 2 ) ≡ 1 ( m o d 7 ) E_6(q^2)\equiv1\pmod{7} E 6 ( q 2 ) ≡ 1 ( mod 7 ) . So from (31) , we obtain
q M f 2 16 f 1 8 ≡ 4 ( M 3 − 1 ) ( m o d 7 ) . (78) qM\frac{f_2^{16}}{f_1^8}\equiv4(M^3-1)\pmod{7}. \tag*{(78)} q M f 1 8 f 2 16 ≡ 4 ( M 3 − 1 ) ( mod 7 ) . (78)
Employing (78) in (77) , we obtain
E 7 , 1 ( M A ( q ) ) ≡ f 1 f 2 ( 4 + 4 ( M 3 − 1 ) ) = 4 f 1 f 2 M 3 ( m o d 7 ) , \mathcal{E}_{7,1}(MA(q))\equiv f_1f_2\left(4+4(M^3-1)\right)=4f_1f_2M^3\pmod{7}, E 7 , 1 ( M A ( q )) ≡ f 1 f 2 ( 4 + 4 ( M 3 − 1 ) ) = 4 f 1 f 2 M 3 ( mod 7 ) ,
which completes the proof of (68) .
Lemma 2.5. We have
E 5 , 2 ( A ( q ) ) ≡ 3 f 1 3 f 2 3 ( m o d 5 ) (79) \mathcal{E}_{5,2}(A(q))\equiv3f_1^3f_2^3\pmod{5} \tag*{(79)} E 5 , 2 ( A ( q )) ≡ 3 f 1 3 f 2 3 ( mod 5 ) (79) and
E 7 , 1 ( A ( q ) ) ≡ f 1 f 2 M 2 ( m o d 7 ) . (80) \mathcal{E}_{7,1}(A(q))\equiv f_1f_2M^2\pmod{7}. \tag*{(80)} E 7 , 1 ( A ( q )) ≡ f 1 f 2 M 2 ( mod 7 ) . (80)
Proof. First, we show that
( D − 2 ) ( M 2 A ( q ) ) ≡ ( M 5 + 2 M ) A ( q ) ( m o d 5 ) (81) (D-2)(M^2A(q))\equiv(M^5+2M)A(q)\pmod{5} \tag*{(81)} ( D − 2 ) ( M 2 A ( q )) ≡ ( M 5 + 2 M ) A ( q ) ( mod 5 ) (81)
and
( D − 1 ) ( M 3 A ( q ) ) ≡ ( M 7 + 5 M ) A ( q ) ( m o d 7 ) . (82) (D-1)(M^3A(q))\equiv(M^7+5M)A(q)\pmod{7}. \tag*{(82)} ( D − 1 ) ( M 3 A ( q )) ≡ ( M 7 + 5 M ) A ( q ) ( mod 7 ) . (82)
Set
P = E 2 ( q ) P=E_2(q) P = E 2 ( q ) , Q = E 2 ( q 2 ) Q=E_2(q^2) Q = E 2 ( q 2 ) , S = E 4 ( q ) S=E_4(q) S = E 4 ( q ) , R = E 4 ( q 2 ) R=E_4(q^2) R = E 4 ( q 2 ) .
From (1.2) and (2.10), we obtain
D A ( q ) A ( q ) − D f 1 f 1 − D f 2 f 2 = L ( q ) + 2 L ( q 2 ) . (83) \frac{DA(q)}{A(q)}-\frac{Df_1}{f_1}-\frac{Df_2}{f_2}=L(q)+2L(q^2). \tag*{(83)} A ( q ) D A ( q ) − f 1 D f 1 − f 2 D f 2 = L ( q ) + 2 L ( q 2 ) . (83)
Employing (2.17) in (2.79), we obtain
D A ( q ) A ( q ) = 1 − P 24 + 2 1 − Q 24 = 3 − P − 2 Q 24 . (84) \frac{DA(q)}{A(q)}=\frac{1-P}{24}+2\frac{1-Q}{24}=\frac{3-P-2Q}{24}. \tag*{(84)} A ( q ) D A ( q ) = 24 1 − P + 2 24 1 − Q = 24 3 − P − 2 Q . (84)
Differentiating (2.23) and using (2.20), we obtain
D M = 2 D ( E 2 ( q 2 ) ) − D E 2 ( q ) = 4 ( Q 2 − R ) − ( P 2 − S ) 12 = ( 2 Q − P ) ( 2 Q + P ) + S − 4 R 12 = M ( P + 2 Q ) + S − 4 R 12 . (85) \begin{aligned}
DM&=2D\left(E_2(q^2)\right)-DE_2(q)\\
&=\frac{4(Q^2-R)-(P^2-S)}{12}\\
&=\frac{(2Q-P)(2Q+P)+S-4R}{12}\\
&=\frac{M(P+2Q)+S-4R}{12}. \tag*{(85)}
\end{aligned} D M = 2 D ( E 2 ( q 2 ) ) − D E 2 ( q ) = 12 4 ( Q 2 − R ) − ( P 2 − S ) = 12 ( 2 Q − P ) ( 2 Q + P ) + S − 4 R = 12 M ( P + 2 Q ) + S − 4 R . (85)
From (2.18), it is easily seen that
S ≡ 1 ( m o d 5 ) and R ≡ 1 ( m o d 5 ) . (86) S\equiv1\pmod{5}\qquad\text{and}\qquad R\equiv1\pmod{5}. \tag*{(86)} S ≡ 1 ( mod 5 ) and R ≡ 1 ( mod 5 ) . (86)
So from (2.21), we obtain
0 ≡ 3 D S = P S − E 6 ( q ) ≡ P − E 6 ( q ) ( m o d 5 ) 0\equiv3DS=PS-E_6(q)\equiv P-E_6(q)\pmod{5} 0 ≡ 3 D S = P S − E 6 ( q ) ≡ P − E 6 ( q ) ( mod 5 )
and
0 ≡ 3 D R = 2 Q R − 2 E 6 ( q 2 ) ≡ 2 ( Q − E 6 ( q 2 ) ) ( m o d 5 ) , 0\equiv3DR=2QR-2E_6(q^2)\equiv2\left(Q-E_6(q^2)\right)\pmod{5}, 0 ≡ 3 D R = 2 QR − 2 E 6 ( q 2 ) ≡ 2 ( Q − E 6 ( q 2 ) ) ( mod 5 ) ,
which imply, respectively,
E 6 ( q ) ≡ P ( m o d 5 ) (87) E_6(q)\equiv P\pmod{5} \tag*{(87)} E 6 ( q ) ≡ P ( mod 5 ) (87)
and
E 6 ( q 2 ) ≡ Q ( m o d 5 ) . (88) E_6(q^2)\equiv Q\pmod{5}. \tag*{(88)} E 6 ( q 2 ) ≡ Q ( mod 5 ) . (88)
Employing (2.83) and (2.84) in (2.26), and then using (2.23), we obtain
2 M 3 ≡ 3 Q − P = Q + M ( m o d 5 ) , 2M^3\equiv3Q-P=Q+M\pmod{5}, 2 M 3 ≡ 3 Q − P = Q + M ( mod 5 ) ,
Q ≡ 2 M 3 − M ( m o d 5 ) , (89) Q\equiv2M^3-M\pmod{5}, \tag*{(89)} Q ≡ 2 M 3 − M ( mod 5 ) , (89)
and
P = 2 Q − M ≡ 4 M 3 − 3 M ( m o d 5 ) . (90) P = 2Q - M \equiv4M^{3} - 3M \pmod{5}. \tag*{(90)} P = 2 Q − M ≡ 4 M 3 − 3 M ( mod 5 ) . (90)
Using (89) and (90) , we obtain
P + 2 Q ≡ ( 4 M 3 − 3 M ) + 2 ( 2 M 3 − M ) = 8 M 3 − 5 M ≡ 3 M 3 ( m o d 5 ) . (91) P + 2Q \equiv(4M^{3} - 3M) + 2(2M^{3} - M) = 8M^{3} - 5M \equiv3M^{3} \pmod{5}. \tag*{(91)} P + 2 Q ≡ ( 4 M 3 − 3 M ) + 2 ( 2 M 3 − M ) = 8 M 3 − 5 M ≡ 3 M 3 ( mod 5 ) . (91)
Employing (91) in (2.80) and (2.81), and noting S ≡ R ≡ 1 ( m o d 5 ) S \equiv R \equiv1 \pmod{5} S ≡ R ≡ 1 ( mod 5 ) , we obtain
D A ( q ) A ( q ) ≡ 4 ( 3 − 3 M 3 ) ≡ 2 + 3 M 3 ( m o d 5 ) (92) \frac{D A(q)}{A(q)} \equiv4(3 - 3M^{3}) \equiv2 + 3M^{3} \pmod{5} \tag*{(92)} A ( q ) D A ( q ) ≡ 4 ( 3 − 3 M 3 ) ≡ 2 + 3 M 3 ( mod 5 ) (92)
and
D M ≡ 3 ( 3 M 4 + 1 − 4 ) ≡ 1 − M 4 ( m o d 5 ) , (93) D M \equiv3(3M^{4} + 1 - 4) \equiv1 - M^{4} \pmod{5}, \tag*{(93)} D M ≡ 3 ( 3 M 4 + 1 − 4 ) ≡ 1 − M 4 ( mod 5 ) , (93)
respectively. Combining (92) and (93) , we obtain
( D − 2 ) ( M 2 A ( q ) ) A ( q ) = 2 M D M + M 2 D A ( q ) A ( q ) − 2 M 2 ≡ M 5 + 2 M ( m o d 5 ) , \frac{(D - 2)(M^{2}A(q))}{A(q)} = 2M\,DM + M^{2}\frac{DA(q)}{A(q)} - 2M^{2} \equiv M^{5} + 2M \pmod{5}, A ( q ) ( D − 2 ) ( M 2 A ( q )) = 2 M D M + M 2 A ( q ) D A ( q ) − 2 M 2 ≡ M 5 + 2 M ( mod 5 ) ,
which is equivalent to (81) .
Again, from (23) , we note that
E 6 ( q ) ≡ 1 ( m o d 7 ) and E 6 ( q 2 ) ≡ 1 ( m o d 7 ) . E_{6}(q) \equiv1 \pmod{7} \qquad\text{and} \qquad E_{6}(q^{2}) \equiv1 \pmod{7}. E 6 ( q ) ≡ 1 ( mod 7 ) and E 6 ( q 2 ) ≡ 1 ( mod 7 ) .
Therefore, from (26) , we obtain
0 ≡ 2 D E 6 ( q ) = P E 6 ( q ) − S 2 ≡ P − S 2 ( m o d 7 ) 0 \equiv2D E_{6}(q) = P E_{6}(q) - S^{2} \equiv P - S^{2} \pmod{7} 0 ≡ 2 D E 6 ( q ) = P E 6 ( q ) − S 2 ≡ P − S 2 ( mod 7 )
and
0 ≡ D ( E 6 ( q 2 ) ) = Q E 6 ( q 2 ) − R 2 ≡ Q − R 2 ( m o d 7 ) , 0 \equiv D\left(E_{6}(q^{2})\right) = Q E_{6}(q^{2}) - R^{2} \equiv Q - R^{2} \pmod{7}, 0 ≡ D ( E 6 ( q 2 ) ) = Q E 6 ( q 2 ) − R 2 ≡ Q − R 2 ( mod 7 ) ,
which imply
P ≡ S 2 ( m o d 7 ) (94) P \equiv S^{2} \pmod{7} \tag*{(94)} P ≡ S 2 ( mod 7 ) (94)
and
Q ≡ R 2 ( m o d 7 ) , (95) Q \equiv R^{2} \pmod{7}, \tag*{(95)} Q ≡ R 2 ( mod 7 ) , (95)
respectively. Employing (94) and (95) in (27) , and then using (29) , we obtain
M ≡ 2 R 2 − S 2 = 2 R 2 − ( 5 M 2 − 4 R ) 2 ≡ 3 M 4 + 5 M 2 R ( m o d 7 ) . (96) M \equiv2R^{2} - S^{2} = 2R^{2} - (5M^{2} - 4R)^{2} \equiv3M^{4} + 5M^{2}R \pmod{7}. \tag*{(96)} M ≡ 2 R 2 − S 2 = 2 R 2 − ( 5 M 2 − 4 R ) 2 ≡ 3 M 4 + 5 M 2 R ( mod 7 ) . (96)
Since M M M has constant term 1, it is a unit in Z [ [ q ] ] \mathbb{Z}[[q]] Z [[ q ]] . Hence its inverse exists in Z [ [ q ] ] \mathbb{Z}[[q]] Z [[ q ]] and remains an inverse after reduction modulo 5 or 7. Multiplying (96) by 3 M − 2 3M^{-2} 3 M − 2 , and then using (29) , we obtain
R ≡ 3 M − 2 ( M − 3 M 4 ) ≡ 5 M 2 + 3 M − 1 ( m o d 7 ) (97) R \equiv3M^{-2}(M-3M^4) \equiv5M^2+3M^{-1}\pmod{7} \tag*{(97)} R ≡ 3 M − 2 ( M − 3 M 4 ) ≡ 5 M 2 + 3 M − 1 ( mod 7 ) (97)
and
S = 5 M 2 − 4 R ≡ 5 M 2 − 4 ( 5 M 2 + 3 M − 1 ) ≡ 6 M 2 + 2 M − 1 ( m o d 7 ) . (98) S=5M^2-4R\equiv5M^2-4(5M^2+3M^{-1})\equiv6M^2+2M^{-1}\pmod{7}. \tag*{(98)} S = 5 M 2 − 4 R ≡ 5 M 2 − 4 ( 5 M 2 + 3 M − 1 ) ≡ 6 M 2 + 2 M − 1 ( mod 7 ) . (98)
Employing (97) and (98) in (94) and (95) , we obtain
P ≡ ( 6 M 2 + 2 M − 1 ) 2 = 36 M 4 + 24 M + 4 M − 2 ≡ M 4 + 3 M + 4 M − 2 ( m o d 7 ) (99) P\equiv(6M^2+2M^{-1})^2=36M^4+24M+4M^{-2}\equiv M^4+3M+4M^{-2}\pmod{7} \tag*{(99)} P ≡ ( 6 M 2 + 2 M − 1 ) 2 = 36 M 4 + 24 M + 4 M − 2 ≡ M 4 + 3 M + 4 M − 2 ( mod 7 ) (99)
and
Q ≡ ( 5 M 2 + 3 M − 1 ) 2 = 25 M 4 + 30 M + 9 M − 2 ≡ 4 M 4 + 2 M + 2 M − 2 ( m o d 7 ) , (100) Q\equiv(5M^2+3M^{-1})^2=25M^4+30M+9M^{-2}\equiv4M^4+2M+2M^{-2}\pmod{7}, \tag*{(100)} Q ≡ ( 5 M 2 + 3 M − 1 ) 2 = 25 M 4 + 30 M + 9 M − 2 ≡ 4 M 4 + 2 M + 2 M − 2 ( mod 7 ) , (100)
respectively. From (97) and (98) , we obtain
S − 4 R ≡ − 14 M 2 − 10 M − 1 ≡ 4 M − 1 ( m o d 7 ) . (101) S-4R\equiv-14M^2-10M^{-1}\equiv4M^{-1}\pmod{7}. \tag*{(101)} S − 4 R ≡ − 14 M 2 − 10 M − 1 ≡ 4 M − 1 ( mod 7 ) . (101)
From (99) and (100) , we obtain
P + 2 Q ≡ 9 M 4 + 7 M + 8 M − 2 ≡ 2 M 4 + M − 2 ( m o d 7 ) . (102) P+2Q\equiv9M^4+7M+8M^{-2}\equiv2M^4+M^{-2}\pmod{7}. \tag*{(102)} P + 2 Q ≡ 9 M 4 + 7 M + 8 M − 2 ≡ 2 M 4 + M − 2 ( mod 7 ) . (102)
Employing (101) and (102) in (84) and (85) , we obtain
D A ( q ) A ( q ) ≡ 5 ( 3 − 2 M 4 − M − 2 ) ≡ 1 + 4 M 4 + 2 M − 2 ( m o d 7 ) (103) \frac{DA(q)}{A(q)}\equiv5(3-2M^4-M^{-2})\equiv1+4M^4+2M^{-2}\pmod{7} \tag*{(103)} A ( q ) D A ( q ) ≡ 5 ( 3 − 2 M 4 − M − 2 ) ≡ 1 + 4 M 4 + 2 M − 2 ( mod 7 ) (103)
and
D M ≡ 3 ( M ( 2 M 4 + M − 2 ) + 4 M − 1 ) = 6 M 5 + 15 M − 1 ≡ M − 1 − M 5 ( m o d 7 ) . (104) DM\equiv3\bigl(M(2M^4+M^{-2})+4M^{-1}\bigr)=6M^5+15M^{-1}\equiv M^{-1}-M^5\pmod{7}. \tag*{(104)} D M ≡ 3 ( M ( 2 M 4 + M − 2 ) + 4 M − 1 ) = 6 M 5 + 15 M − 1 ≡ M − 1 − M 5 ( mod 7 ) . (104)
respectively. Combining (103) and (104) , we obtain
( D − 1 ) ( M 3 A ( q ) ) A ( q ) = 3 M 2 D M + M 3 ( D A ( q ) A ( q ) ) − M 3 ≡ 3 M 2 ( M − 1 − M 5 ) + M 3 ( 1 + 4 M 4 + 2 M − 2 ) − M 3 = M 7 + 5 M ( m o d 7 ) , \begin{aligned}
\frac{(D-1)(M^3A(q))}{A(q)}
&=3M^2DM+M^3\left(\frac{DA(q)}{A(q)}\right)-M^3\\
&\equiv3M^2(M^{-1}-M^5)+M^3(1+4M^4+2M^{-2})-M^3\\
&=M^7+5M\pmod{7},
\end{aligned} A ( q ) ( D − 1 ) ( M 3 A ( q )) = 3 M 2 D M + M 3 ( A ( q ) D A ( q ) ) − M 3 ≡ 3 M 2 ( M − 1 − M 5 ) + M 3 ( 1 + 4 M 4 + 2 M − 2 ) − M 3 = M 7 + 5 M ( mod 7 ) ,
which is equivalent to (82) .
Now, employing (18) , (19) , (16) , and (67) in (81) , we obtain
0 ≡ E 5 , 2 ( ( D − 2 ) ( M 2 A ( q ) ) ) ≡ E 5 , 2 ( M 5 A ( q ) ) + 2 E 5 , 2 ( M A ( q ) ) ≡ E 5 , 2 ( M ( q 5 ) A ( q ) ) + 2 M f 1 3 f 2 3 = M E 5 , 2 ( A ( q ) ) + 2 M f 1 3 f 2 3 ( m o d 5 ) . (105) \begin{aligned}
0 &\equiv\mathcal{E}_{5,2}\left((D-2)(M^2A(q))\right)\\
&\equiv\mathcal{E}_{5,2}\left(M^5A(q)\right)+2\mathcal{E}_{5,2}\left(MA(q)\right)\\
&\equiv\mathcal{E}_{5,2}\left(M(q^5)A(q)\right)+2M f_1^3 f_2^3\\
&=M\mathcal{E}_{5,2}\left(A(q)\right)+2M f_1^3 f_2^3\pmod{5}. \tag*{(105)}
\end{aligned} 0 ≡ E 5 , 2 ( ( D − 2 ) ( M 2 A ( q )) ) ≡ E 5 , 2 ( M 5 A ( q ) ) + 2 E 5 , 2 ( M A ( q ) ) ≡ E 5 , 2 ( M ( q 5 ) A ( q ) ) + 2 M f 1 3 f 2 3 = M E 5 , 2 ( A ( q ) ) + 2 M f 1 3 f 2 3 ( mod 5 ) . (105)
Multiplying (105) by M − 1 M^{-1} M − 1 , we arrive at (79) .
Similarly, employing (18) , (19) , (16) , and (68) in (82) , we obtain
0 ≡ E 7 , 1 ( ( D − 1 ) ( M 3 A ( q ) ) ) ≡ E 7 , 1 ( M 7 A ( q ) ) + 5 E 7 , 1 ( M A ( q ) ) ≡ E 7 , 1 ( M ( q 7 ) A ( q ) ) + 20 f 1 f 2 M 3 = M E 7 , 1 ( A ( q ) ) + 20 f 1 f 2 M 3 ( m o d 7 ) . (106) \begin{aligned}
0 &\equiv\mathcal{E}_{7,1}\left((D-1)(M^3A(q))\right)\\
&\equiv\mathcal{E}_{7,1}\left(M^7A(q)\right)+5\mathcal{E}_{7,1}\left(MA(q)\right)\\
&\equiv\mathcal{E}_{7,1}\left(M(q^7)A(q)\right)+20f_1f_2M^3\\
&=M\mathcal{E}_{7,1}\left(A(q)\right)+20f_1f_2M^3\pmod{7}. \tag*{(106)}
\end{aligned} 0 ≡ E 7 , 1 ( ( D − 1 ) ( M 3 A ( q )) ) ≡ E 7 , 1 ( M 7 A ( q ) ) + 5 E 7 , 1 ( M A ( q ) ) ≡ E 7 , 1 ( M ( q 7 ) A ( q ) ) + 20 f 1 f 2 M 3 = M E 7 , 1 ( A ( q ) ) + 20 f 1 f 2 M 3 ( mod 7 ) . (106)
Multiplying (106) by M − 1 M^{-1} M − 1 , we complete the proof of (80) .
Lemma 2.6. For j = 2 , 3 j=2,3 j = 2 , 3 , let
H j ( q ) = f 1 f 2 M j . (107) H_j(q)=f_1f_2M^j. \tag*{(107)} H j ( q ) = f 1 f 2 M j . (107) Then
E 7 , 6 ( H j ( q ) ) = 7 2 j f 7 f 14 M ( q 7 ) j . (108) \mathcal{E}_{7,6}\left(H_j(q)\right)=7^{2j}f_7f_{14}M(q^7)^j. \tag*{(108)} E 7 , 6 ( H j ( q ) ) = 7 2 j f 7 f 14 M ( q 7 ) j . (108)
Proof. For r ∈ Z r\in\mathbb{Z} r ∈ Z , set
P r = r ( 3 r − 1 ) 2 . (109) P_r=\frac{r(3r-1)}{2}. \tag*{(109)} P r = 2 r ( 3 r − 1 ) . (109)
Ramanujan [13, p. 369]] recorded the following sequence of q q q -series, where k k k is a nonnegative integer:
V 2 k ( q ) = ∑ r ∈ Z ( − 1 ) r ( 6 r − 1 ) 2 k q r ( 3 r − 1 ) / 2 ∑ r ∈ Z ( − 1 ) r q r ( 3 r − 1 ) / 2 . (110) V_{2k}(q)=\frac{\sum_{r\in\mathbb{Z}}(-1)^r(6r-1)^{2k}q^{r(3r-1)/2}}{\sum_{r\in\mathbb{Z}}(-1)^rq^{r(3r-1)/2}}. \tag*{(110)} V 2 k ( q ) = ∑ r ∈ Z ( − 1 ) r q r ( 3 r − 1 ) /2 ∑ r ∈ Z ( − 1 ) r ( 6 r − 1 ) 2 k q r ( 3 r − 1 ) /2 . (110)
Using (7) in (110) , we obtain
∑ r ∈ Z ( − 1 ) r ( 6 r − 1 ) 2 k q P r = f 1 V 2 k ( q ) . (111) \sum_{r\in\mathbb{Z}}(-1)^r(6r-1)^{2k}q^{P_r}=f_1V_{2k}(q). \tag*{(111)} r ∈ Z ∑ ( − 1 ) r ( 6 r − 1 ) 2 k q P r = f 1 V 2 k ( q ) . (111)
Since ( 6 r − 1 ) 2 = 24 P r + 1 (6r-1)^2=24P_r+1 ( 6 r − 1 ) 2 = 24 P r + 1 , (111) gives
f 1 V 2 k + 2 ( q ) = ( 24 D + 1 ) ( f 1 V 2 k ( q ) ) , f_1V_{2k+2}(q)=(24D+1)\left(f_1V_{2k}(q)\right), f 1 V 2 k + 2 ( q ) = ( 24 D + 1 ) ( f 1 V 2 k ( q ) ) ,
V 2 k + 2 ( q ) = 24 D V 2 k ( q ) + E 2 ( q ) V 2 k ( q ) , V_{2k+2}(q)=24DV_{2k}(q)+E_2(q)V_{2k}(q), V 2 k + 2 ( q ) = 24 D V 2 k ( q ) + E 2 ( q ) V 2 k ( q ) ,
where we used (14) and (21) . Starting with V 0 ( q ) = 1 V_0(q)=1 V 0 ( q ) = 1 and employing (24) –(26) , we obtain the following identities recorded by Ramanujan [13 ] :
V 0 ( q ) = 1 , (112) V_0(q)=1, \tag*{(112)} V 0 ( q ) = 1 , (112)
V 2 ( q ) = E 2 ( q ) , (113) V_2(q)=E_2(q), \tag*{(113)} V 2 ( q ) = E 2 ( q ) , (113)
V 4 ( q ) = 3 E 2 ( q ) 2 − 2 E 4 ( q ) , (114) V_4(q)=3E_2(q)^2-2E_4(q), \tag*{(114)} V 4 ( q ) = 3 E 2 ( q ) 2 − 2 E 4 ( q ) , (114)
and
V 6 ( q ) = 15 E 2 ( q ) 3 − 30 E 2 ( q ) E 4 ( q ) + 16 E 6 ( q ) . (115) V_6(q)=15E_2(q)^3-30E_2(q)E_4(q)+16E_6(q). \tag*{(115)} V 6 ( q ) = 15 E 2 ( q ) 3 − 30 E 2 ( q ) E 4 ( q ) + 16 E 6 ( q ) . (115)
Now, for r , s ∈ Z r,s\in\mathbb{Z} r , s ∈ Z , set
x = 6 r − 1 , y = 6 s − 1 , and N = P r + 2 P s . (116) x=6r-1,\qquad y=6s-1,\qquad\text{and}\qquad N=P_r+2P_s. \tag*{(116)} x = 6 r − 1 , y = 6 s − 1 , and N = P r + 2 P s . (116)
Define
W 2 ( x , y ) = x 4 − 12 x 2 y 2 + 4 y 4 , (117) W_2(x,y)=x^4-12x^2y^2+4y^4, \tag*{(117)} W 2 ( x , y ) = x 4 − 12 x 2 y 2 + 4 y 4 , (117)
W 3 ( x , y ) = x 6 − 30 x 4 y 2 + 60 x 2 y 4 − 8 y 6 , (118) W_3(x,y)=x^6-30x^4y^2+60x^2y^4-8y^6, \tag*{(118)} W 3 ( x , y ) = x 6 − 30 x 4 y 2 + 60 x 2 y 4 − 8 y 6 , (118)
and set
S j ( q ) = ∑ r , s ∈ Z ( − 1 ) r + s W j ( x , y ) q N , j = 2 , 3. (119) S_j(q)=\sum_{r,s\in\mathbb{Z}}(-1)^{r+s}W_j(x,y)q^N,\qquad j=2,3. \tag*{(119)} S j ( q ) = r , s ∈ Z ∑ ( − 1 ) r + s W j ( x , y ) q N , j = 2 , 3. (119)
First we establish that
S 2 ( q ) = − 7 f 1 f 2 M 2 (120) S_2(q)=-7f_1f_2M^2 \tag*{(120)} S 2 ( q ) = − 7 f 1 f 2 M 2 (120)
and
S 3 ( q ) = 23 f 1 f 2 M 3 . (121) S_3(q)=23f_1f_2M^3. \tag*{(121)} S 3 ( q ) = 23 f 1 f 2 M 3 . (121)
Set P = E 2 ( q ) P=E_2(q) P = E 2 ( q ) and Q = E 2 ( q 2 ) Q=E_2(q^2) Q = E 2 ( q 2 ) . Employing (111) –(115) , and (117) in (119) , and then using (27) and (29) , we obtain
S 2 ( q ) f 1 f 2 = V 4 ( q ) − 12 V 2 ( q ) V 2 ( q 2 ) + 4 V 4 ( q 2 ) = 3 P 2 − 2 E 4 ( q ) − 12 P Q + 12 Q 2 − 8 E 4 ( q 2 ) = 3 ( P − 2 Q ) 2 − 2 ( E 4 ( q ) + 4 E 4 ( q 2 ) ) = 3 M 2 − 2 ( 5 M 2 ) = − 7 M 2 , \begin{aligned}
\frac{S_2(q)}{f_1f_2}
&=V_4(q)-12V_2(q)V_2(q^2)+4V_4(q^2)\\
&=3P^2-2E_4(q)-12PQ+12Q^2-8E_4(q^2)\\
&=3(P-2Q)^2-2\left(E_4(q)+4E_4(q^2)\right)\\
&=3M^2-2(5M^2)\\
&=-7M^2,
\end{aligned} f 1 f 2 S 2 ( q ) = V 4 ( q ) − 12 V 2 ( q ) V 2 ( q 2 ) + 4 V 4 ( q 2 ) = 3 P 2 − 2 E 4 ( q ) − 12 P Q + 12 Q 2 − 8 E 4 ( q 2 ) = 3 ( P − 2 Q ) 2 − 2 ( E 4 ( q ) + 4 E 4 ( q 2 ) ) = 3 M 2 − 2 ( 5 M 2 ) = − 7 M 2 ,
which is equivalent to (120) .
Similarly, employing (111) –(2.111}, and (118) in (119) , and then using (27) , (29) , and (30) , we obtain
S 3 ( q ) f 1 f 2 = V 6 ( q ) − 30 V 4 ( q ) Q + 60 P V 4 ( q 2 ) − 8 V 6 ( q 2 ) = 15 P 3 − 30 P E 4 ( q ) + 16 E 6 ( q ) − 90 P 2 Q + 60 Q E 4 ( q ) + 180 P Q 2 − 120 P E 4 ( q 2 ) − 120 Q 3 + 240 Q E 4 ( q 2 ) − 128 E 6 ( q 2 ) = 15 ( P − 2 Q ) 3 + 30 ( 2 Q − P ) E 4 ( q ) + 120 ( 2 Q − P ) E 4 ( q 2 ) + 16 ( E 6 ( q ) − 8 E 6 ( q 2 ) ) = − 15 M 3 + 30 M ( E 4 ( q ) + 4 E 4 ( q 2 ) ) + 16 ( E 6 ( q ) − 8 E 6 ( q 2 ) ) = − 15 M 3 + 30 M ( 5 M 2 ) + 16 ( − 7 M 3 ) = 23 M 3 , \begin{aligned}
\frac{S_3(q)}{f_1f_2}
&= V_6(q)-30V_4(q)Q+60PV_4(q^2)-8V_6(q^2)\\
&=15P^3-30PE_4(q)+16E_6(q)\\
&\quad-90P^2Q+60QE_4(q)+180PQ^2-120PE_4(q^2)\\
&\quad-120Q^3+240QE_4(q^2)-128E_6(q^2)\\
&=15(P-2Q)^3+30(2Q-P)E_4(q)\\
&\quad+120(2Q-P)E_4(q^2)+16\bigl(E_6(q)-8E_6(q^2)\bigr)\\
&=-15M^3+30M\bigl(E_4(q)+4E_4(q^2)\bigr)\\
&\quad+16\bigl(E_6(q)-8E_6(q^2)\bigr)\\
&=-15M^3+30M(5M^2)+16(-7M^3)\\
&=23M^3,
\end{aligned} f 1 f 2 S 3 ( q ) = V 6 ( q ) − 30 V 4 ( q ) Q + 60 P V 4 ( q 2 ) − 8 V 6 ( q 2 ) = 15 P 3 − 30 P E 4 ( q ) + 16 E 6 ( q ) − 90 P 2 Q + 60 Q E 4 ( q ) + 180 P Q 2 − 120 P E 4 ( q 2 ) − 120 Q 3 + 240 Q E 4 ( q 2 ) − 128 E 6 ( q 2 ) = 15 ( P − 2 Q ) 3 + 30 ( 2 Q − P ) E 4 ( q ) + 120 ( 2 Q − P ) E 4 ( q 2 ) + 16 ( E 6 ( q ) − 8 E 6 ( q 2 ) ) = − 15 M 3 + 30 M ( E 4 ( q ) + 4 E 4 ( q 2 ) ) + 16 ( E 6 ( q ) − 8 E 6 ( q 2 ) ) = − 15 M 3 + 30 M ( 5 M 2 ) + 16 ( − 7 M 3 ) = 23 M 3 ,
which is equivalent to (121) .
Now, from (109) and (116) , we obtain
x 2 + 2 y 2 = ( 24 P r + 1 ) + 2 ( 24 P s + 1 ) = 24 N + 3. (122) x^2+2y^2=(24P_r+1)+2(24P_s+1)=24N+3. \tag*{(122)} x 2 + 2 y 2 = ( 24 P r + 1 ) + 2 ( 24 P s + 1 ) = 24 N + 3. (122)
Since − 2 ≡ 5 ( m o d 7 ) -2\equiv5\pmod{7} − 2 ≡ 5 ( mod 7 ) is a quadratic nonresidue modulo 7, from (122) , we obtain
N ≡ 6 ( m o d 7 ) ⟺ { x ≡ 0 ( m o d 7 ) , y ≡ 0 ( m o d 7 ) , ⟺ { r = 7 u − 1 , s = 7 v − 1 , u , v ∈ Z . (123) N\equiv6\pmod{7}\Longleftrightarrow
\begin{cases}
x\equiv0\pmod{7},\\
y\equiv0\pmod{7},
\end{cases}
\Longleftrightarrow
\begin{cases}
r=7u-1,\\
s=7v-1,
\end{cases}
\qquad u,v\in\mathbb{Z}. \tag*{(123)} N ≡ 6 ( mod 7 ) ⟺ { x ≡ 0 ( mod 7 ) , y ≡ 0 ( mod 7 ) , ⟺ { r = 7 u − 1 , s = 7 v − 1 , u , v ∈ Z . (123)
Employing (123) in (109) and (116) , we obtain
P 7 u − 1 = 49 P u + 2 , (124) P_{7u-1}=49P_u+2, \tag*{(124)} P 7 u − 1 = 49 P u + 2 , (124)
N = 49 ( P u + 2 P v ) + 6 , (125) N=49(P_u+2P_v)+6, \tag*{(125)} N = 49 ( P u + 2 P v ) + 6 , (125)
( − 1 ) r + s = ( − 1 ) 7 u + 7 v − 2 = ( − 1 ) u + v , (126) (-1)^{r+s}=(-1)^{7u+7v-2}=(-1)^{u+v}, \tag*{(126)} ( − 1 ) r + s = ( − 1 ) 7 u + 7 v − 2 = ( − 1 ) u + v , (126)
and
x = 7 ( 6 u − 1 ) , y = 7 ( 6 v − 1 ) . (127) x=7(6u-1),\qquad y=7(6v-1). \tag*{(127)} x = 7 ( 6 u − 1 ) , y = 7 ( 6 v − 1 ) . (127)
Therefore, employing (123) , (124) , (125) , (126) , and (127) in (119) , we obtain
E 7 , 6 ( S j ( q ) ) = ∑ u , v ∈ Z ( − 1 ) u + v W j ( 7 ( 6 u − 1 ) , 7 ( 6 v − 1 ) ) q 7 ( P u + 2 P v ) = 7 2 j ∑ u , v ∈ Z ( − 1 ) u + v W j ( 6 u − 1 , 6 v − 1 ) q 7 ( P u + 2 P v ) = 7 2 j S j ( q 7 ) . (128) \begin{aligned}
\mathcal{E}_{7,6}(S_j(q))
&=\sum_{u,v\in\mathbb{Z}}(-1)^{u+v}W_j(7(6u-1),7(6v-1))q^{7(P_u+2P_v)}\\
&=7^{2j}\sum_{u,v\in\mathbb{Z}}(-1)^{u+v}W_j(6u-1,6v-1)q^{7(P_u+2P_v)}\\
&=7^{2j}S_j(q^7). \tag*{(128)}
\end{aligned} E 7 , 6 ( S j ( q )) = u , v ∈ Z ∑ ( − 1 ) u + v W j ( 7 ( 6 u − 1 ) , 7 ( 6 v − 1 )) q 7 ( P u + 2 P v ) = 7 2 j u , v ∈ Z ∑ ( − 1 ) u + v W j ( 6 u − 1 , 6 v − 1 ) q 7 ( P u + 2 P v ) = 7 2 j S j ( q 7 ) . (128)
where j = 2 , 3 j=2,3 j = 2 , 3 .
Employing (107) , (120) , and (121) in (128) , we obtain
− 7 E 7 , 6 ( H 2 ( q ) ) = − 7 ⋅ 7 4 H 2 ( q 7 ) (129) -7\mathcal{E}_{7,6}(H_2(q))=-7\cdot7^4H_2(q^7) \tag*{(129)} − 7 E 7 , 6 ( H 2 ( q )) = − 7 ⋅ 7 4 H 2 ( q 7 ) (129)
and
23 E 7 , 6 ( H 3 ( q ) ) = 23 ⋅ 7 6 H 3 ( q 7 ) . (130) 23\mathcal{E}_{7,6}(H_3(q))=23\cdot7^6H_3(q^7). \tag*{(130)} 23 E 7 , 6 ( H 3 ( q )) = 23 ⋅ 7 6 H 3 ( q 7 ) . (130)
Since (129) and (130) are exact identities in Z [ [ q ] ] \mathbb{Z}[[q]] Z [[ q ]] , we may cancel the nonzero integer factors − 7 -7 − 7 and 23 23 23 , respectively. Then, using (107) , we obtain
E 7 , 6 ( H j ( q ) ) = 7 2 j H j ( q 7 ) = 7 2 j f 7 f 14 M ( q 7 ) j , \mathcal{E}_{7,6}(H_j(q))=7^{2j}H_j(q^7)=7^{2j}f_7f_{14}M(q^7)^j, E 7 , 6 ( H j ( q )) = 7 2 j H j ( q 7 ) = 7 2 j f 7 f 14 M ( q 7 ) j ,
where j = 2 , 3 j=2,3 j = 2 , 3 . This completes the proof of (108) .
Generating functions and divisibility properties
In this section, we establish generating functions for C O S ( n ) COS(n) C O S ( n ) and C E S ( n ) CES(n) C E S ( n ) and prove their divisibility properties.
Theorem 3.1. We have
∑ n ≥ 0 C O S ( n ) q n = L ( q ) − 2 L ( q 2 ) f 1 f 2 = 2 ϕ ( q ) 4 − ϕ ( − q ) 4 − 1 24 f 1 f 2 = 1 24 ( 2 f 2 19 f 1 9 f 4 8 − f 1 7 f 2 5 − 1 f 1 f 2 ) (131) \sum_{n\geq0}COS(n)q^n=\frac{L(q)-2L(q^2)}{f_1f_2}=\frac{2\phi(q)^4-\phi(-q)^4-1}{24f_1f_2}=\frac{1}{24}\left(2\frac{f_2^{19}}{f_1^9f_4^8}-\frac{f_1^7}{f_2^5}-\frac{1}{f_1f_2}\right) \tag*{(131)} n ≥ 0 ∑ C O S ( n ) q n = f 1 f 2 L ( q ) − 2 L ( q 2 ) = 24 f 1 f 2 2 ϕ ( q ) 4 − ϕ ( − q ) 4 − 1 = 24 1 ( 2 f 1 9 f 4 8 f 2 19 − f 2 5 f 1 7 − f 1 f 2 1 ) (131) and
∑ n ≥ 0 C E S ( n ) q n = 4 L ( q 2 ) f 1 f 2 = 1 − E 2 ( q 2 ) 6 f 1 f 2 . (132) \sum_{n\geq0}CES(n)q^n=\frac{4L(q^2)}{f_1f_2}=\frac{1-E_2(q^2)}{6f_1f_2}. \tag*{(132)} n ≥ 0 ∑ C E S ( n ) q n = f 1 f 2 4 L ( q 2 ) = 6 f 1 f 2 1 − E 2 ( q 2 ) . (132)
Proof. From (2) , we obtain
∑ n ≥ 0 C O S ( n ) q n = ∂ ∂ z ∏ m ≥ 1 1 ( 1 − ( z q ) 2 m − 1 ) ( 1 − q 2 m ) 2 ∣ z = 1 = 1 f 1 f 2 ∑ m ≥ 1 ( 2 m − 1 ) q 2 m − 1 1 − q 2 m − 1 = L ( q ) − 2 L ( q 2 ) f 1 f 2 , (133) \begin{aligned}
\sum_{n\geq0}COS(n)q^n
&=\left.\frac{\partial}{\partial z}\prod_{m\geq1}\frac{1}{(1-(zq)^{2m-1})(1-q^{2m})^2}\right|_{z=1}\\
&=\frac{1}{f_1f_2}\sum_{m\geq1}\frac{(2m-1)q^{2m-1}}{1-q^{2m-1}}\\
&=\frac{L(q)-2L(q^2)}{f_1f_2}, \tag*{(133)}
\end{aligned} n ≥ 0 ∑ C O S ( n ) q n = ∂ z ∂ m ≥ 1 ∏ ( 1 − ( z q ) 2 m − 1 ) ( 1 − q 2 m ) 2 1 z = 1 = f 1 f 2 1 m ≥ 1 ∑ 1 − q 2 m − 1 ( 2 m − 1 ) q 2 m − 1 = f 1 f 2 L ( q ) − 2 L ( q 2 ) , (133)
where the exponent of z z z records the sum of the odd parts. Similarly, for two colors of each even part, we obtain
∑ n ≥ 0 C E S ( n ) q n = ∂ ∂ z ∏ m ≥ 1 1 ( 1 − q 2 m − 1 ) ( 1 − ( z q ) 2 m ) 2 ∣ z = 1 = 1 f 1 f 2 ∑ m ≥ 1 4 m q 2 m 1 − q 2 m = 4 L ( q 2 ) f 1 f 2 . (134) \sum_{n\ge0} CES(n)q^n=\left.\frac{\partial}{\partial z}\prod_{m\ge1}\frac{1}{(1-q^{2m-1})(1-(zq)^{2m})^{2}}\right|_{z=1}
=\frac{1}{f_1f_2}\sum_{m\ge1}\frac{4mq^{2m}}{1-q^{2m}}
=\frac{4L(q^2)}{f_1f_2}. \tag*{(134)} n ≥ 0 ∑ C E S ( n ) q n = ∂ z ∂ m ≥ 1 ∏ ( 1 − q 2 m − 1 ) ( 1 − ( z q ) 2 m ) 2 1 z = 1 = f 1 f 2 1 m ≥ 1 ∑ 1 − q 2 m 4 m q 2 m = f 1 f 2 4 L ( q 2 ) . (134)
where the exponent of z z z records the sum of the even parts. From (5) , (9) , (27) , and (28) , we obtain
L ( q ) − 2 L ( q 2 ) = M − 1 24 = 2 ϕ ( q ) 4 − ϕ ( − q ) 4 − 1 24 = 1 24 ( 2 f 2 20 f 1 8 f 4 8 − f 1 8 f 2 4 − 1 ) . (135) L(q)-2L(q^2)=\frac{M-1}{24}=\frac{2\phi(q)^4-\phi(-q)^4-1}{24}=\frac{1}{24}\left(2\frac{f_2^{20}}{f_1^8f_4^8}-\frac{f_1^8}{f_2^4}-1\right). \tag*{(135)} L ( q ) − 2 L ( q 2 ) = 24 M − 1 = 24 2 ϕ ( q ) 4 − ϕ ( − q ) 4 − 1 = 24 1 ( 2 f 1 8 f 4 8 f 2 20 − f 2 4 f 1 8 − 1 ) . (135)
Employing (135) in (133) and simplifying, we complete the proof of (131) .
Again, from (21) , we note that
4 L ( q 2 ) = ( 1 − E 2 ( q 2 ) ) / 6. (136) 4L(q^2)=(1-E_2(q^2))/6. \tag*{(136)} 4 L ( q 2 ) = ( 1 − E 2 ( q 2 )) /6. (136)
Employing (136) in (134) , we complete the proof of (132) . □
Theorem 3.2. For any integer n ≥ 0 n\ge0 n ≥ 0 , we have
C E S ( n ) = 4 ∑ j = 1 ⌊ n / 2 ⌋ σ ( j ) a ( n − 2 j ) (137) CES(n)=4\sum_{j=1}^{\lfloor n/2\rfloor}\sigma(j)a(n-2j) \tag*{(137)} C E S ( n ) = 4 j = 1 ∑ ⌊ n /2 ⌋ σ ( j ) a ( n − 2 j ) (137) and
C O S ( n ) = n a ( n ) − 4 ∑ j = 1 ⌊ n / 2 ⌋ σ ( j ) a ( n − 2 j ) . (138) COS(n)=na(n)-4\sum_{j=1}^{\lfloor n/2\rfloor}\sigma(j)a(n-2j). \tag*{(138)} C O S ( n ) = na ( n ) − 4 j = 1 ∑ ⌊ n /2 ⌋ σ ( j ) a ( n − 2 j ) . (138) Here ⌊ x ⌋ \lfloor x\rfloor ⌊ x ⌋ denotes the greatest integer not exceeding x x x , and the sums are interpreted as 0 0 0 when n = 0 n=0 n = 0 or n = 1 n=1 n = 1 .
Proof. Employing (2) and (13) in (134) , we obtain
∑ n ≥ 0 C E S ( n ) q n = 4 ( ∑ j ≥ 1 σ ( j ) q 2 j ) ( ∑ m ≥ 0 a ( m ) q m ) = 4 ∑ n ≥ 0 ( ∑ j = 1 ⌊ n / 2 ⌋ σ ( j ) a ( n − 2 j ) ) q n . (139) \sum_{n\ge0}CES(n)q^n=4\left(\sum_{j\ge1}\sigma(j)q^{2j}\right)\left(\sum_{m\ge0}a(m)q^m\right)=4\sum_{n\ge0}\left(\sum_{j=1}^{\lfloor n/2\rfloor}\sigma(j)a(n-2j)\right)q^n. \tag*{(139)} n ≥ 0 ∑ C E S ( n ) q n = 4 ( j ≥ 1 ∑ σ ( j ) q 2 j ) ( m ≥ 0 ∑ a ( m ) q m ) = 4 n ≥ 0 ∑ j = 1 ∑ ⌊ n /2 ⌋ σ ( j ) a ( n − 2 j ) q n . (139)
Equating the coefficients of q n q^n q n on both sides of (139) , we arrive at (137) . Employing (4) in (137) and simplifying, we complete the proof of (138) . □
The following corollary is an easy consequence of Theorem 3.2 .
Corollary 3.3. For every nonnegative integer n n n , we have
C E S ( n ) ≡ 0 ( m o d 4 ) CES(n) \equiv0 \pmod{4} C E S ( n ) ≡ 0 ( mod 4 ) and
C O S ( 4 n ) ≡ 0 ( m o d 4 ) . COS(4n) \equiv0 \pmod{4}. C O S ( 4 n ) ≡ 0 ( mod 4 ) .
Theorem 3.4. For any integer n ≥ 0 n \ge0 n ≥ 0 , we have
C E S ( 9 n + 1 ) ≡ 0 ( m o d 3 ) , (140) CES(9n+1) \equiv0 \pmod{3}, \tag*{(140)} C E S ( 9 n + 1 ) ≡ 0 ( mod 3 ) , (140) C E S ( 9 n + 8 ) ≡ 0 ( m o d 3 ) , (141) CES(9n+8) \equiv0 \pmod{3}, \tag*{(141)} C E S ( 9 n + 8 ) ≡ 0 ( mod 3 ) , (141) C O S ( 9 n + 8 ) ≡ 0 ( m o d 3 ) , (142) COS(9n+8) \equiv0 \pmod{3}, \tag*{(142)} C O S ( 9 n + 8 ) ≡ 0 ( mod 3 ) , (142) C O S ( 9 n + 7 ) ≡ 0 ( m o d 3 ) , (143) COS(9n+7) \equiv0 \pmod{3}, \tag*{(143)} C O S ( 9 n + 7 ) ≡ 0 ( mod 3 ) , (143) C O S ( 9 n ) ≡ 0 ( m o d 3 ) , (144) COS(9n) \equiv0 \pmod{3}, \tag*{(144)} C O S ( 9 n ) ≡ 0 ( mod 3 ) , (144) and
C E S ( 9 n ) ≡ 0 ( m o d 3 ) . (145) CES(9n) \equiv0 \pmod{3}. \tag*{(145)} C E S ( 9 n ) ≡ 0 ( mod 3 ) . (145)
Proof. Employing (6) , (11) , and (46) in (2) , we obtain
A ( q ) = f 2 2 / f 1 f 2 3 ≡ ψ ( q ) f 6 = f 9 2 f 3 f 18 + q ψ ( q 9 ) f 6 ( m o d 3 ) . (146) A(q)=\frac{f_2^2/f_1}{f_2^3}\equiv\frac{\psi(q)}{f_6}=\frac{f_9^2}{f_3f_{18}}+q\frac{\psi(q^9)}{f_6}\pmod{3}. \tag*{(146)} A ( q ) = f 2 3 f 2 2 / f 1 ≡ f 6 ψ ( q ) = f 3 f 18 f 9 2 + q f 6 ψ ( q 9 ) ( mod 3 ) . (146)
Employing (6) , (16) , and (11) in (146) , we obtain
E 3 , 2 ( A ( q ) ) ≡ 0 ( m o d 3 ) (147) \mathcal{E}_{3,2}(A(q))\equiv0\pmod{3} \tag*{(147)} E 3 , 2 ( A ( q )) ≡ 0 ( mod 3 ) (147)
and
E 3 , 1 ( A ( q ) ) ≡ ψ ( q 3 ) f 2 = f 6 2 f 3 f 2 = f 6 f 3 f 2 2 ( m o d 3 ) . (148) \mathcal{E}_{3,1}(A(q))\equiv\frac{\psi(q^3)}{f_2}=\frac{f_6^2}{f_3f_2}=\frac{f_6}{f_3}f_2^2\pmod{3}. \tag*{(148)} E 3 , 1 ( A ( q )) ≡ f 2 ψ ( q 3 ) = f 3 f 2 f 6 2 = f 3 f 6 f 2 2 ( mod 3 ) . (148)
Replacing q q q by q 2 q^2 q 2 in (45) , we obtain
L ( q 2 ) ≡ L ( q 6 ) + q 2 f 6 8 ( m o d 3 ) . (149) L(q^2)\equiv L(q^6)+q^2f_6^8\pmod{3}. \tag*{(149)} L ( q 2 ) ≡ L ( q 6 ) + q 2 f 6 8 ( mod 3 ) . (149)
Employing (146) and (149) in (132) and simplifying, we obtain
∑ n ≥ 0 C E S ( n ) q n ≡ f 9 2 L ( q 6 ) f 3 f 18 + q ψ ( q 9 ) L ( q 6 ) f 6 + q 2 f 6 8 f 9 2 f 3 f 18 + q 3 ψ ( q 9 ) f 6 7 ( m o d 3 ) . (150) \sum_{n\geq0}CES(n)q^n\equiv\frac{f_9^2L(q^6)}{f_3f_{18}}+q\frac{\psi(q^9)L(q^6)}{f_6}+q^2\frac{f_6^8f_9^2}{f_3f_{18}}+q^3\psi(q^9)f_6^7\pmod{3}. \tag*{(150)} n ≥ 0 ∑ C E S ( n ) q n ≡ f 3 f 18 f 9 2 L ( q 6 ) + q f 6 ψ ( q 9 ) L ( q 6 ) + q 2 f 3 f 18 f 6 8 f 9 2 + q 3 ψ ( q 9 ) f 6 7 ( mod 3 ) . (150)
Employing (6) , (11) , (16) , and (14) in (150) , we obtain
E 3 , 1 ( ∑ n ≥ 0 C E S ( n ) q n ) ≡ ψ ( q 3 ) L ( q 2 ) f 2 ≡ f 6 f 3 f 2 L ( q 2 ) ≡ − f 6 f 3 D ( f 2 2 ) ( m o d 3 ) . (151) \mathcal{E}_{3,1}\left(\sum_{n\geq0} CES(n)q^n\right) \equiv\frac{\psi(q^3)L(q^2)}{f_2} \equiv\frac{f_6}{f_3}f_2L(q^2) \equiv-\frac{f_6}{f_3}D(f_2^2) \pmod{3}. \tag*{(151)} E 3 , 1 ( n ≥ 0 ∑ C E S ( n ) q n ) ≡ f 2 ψ ( q 3 ) L ( q 2 ) ≡ f 3 f 6 f 2 L ( q 2 ) ≡ − f 3 f 6 D ( f 2 2 ) ( mod 3 ) . (151)
Applying E 3 , 2 \mathcal{E}_{3,2} E 3 , 2 to (150) , only the third summand contributes. Therefore, using f 2 3 ≡ f 6 ( m o d 3 ) f_2^3\equiv f_6\pmod{3} f 2 3 ≡ f 6 ( mod 3 ) , we obtain
E 3 , 2 ( ∑ n ≥ 0 C E S ( n ) q n ) ≡ f 2 8 f 3 2 f 1 f 6 ≡ f 3 2 f 1 f 2 f 2 9 f 6 ≡ f 2 3 f 6 2 f 1 f 2 ≡ f 3 2 f 6 2 A ( q ) ( m o d 3 ) . (152) \mathcal{E}_{3,2}\left(\sum_{n\geq0} CES(n)q^n\right) \equiv\frac{f_2^8f_3^2}{f_1f_6} \equiv\frac{f_3^2}{f_1f_2}\frac{f_2^9}{f_6} \equiv\frac{f_2^3f_6^2}{f_1f_2} \equiv f_3^2f_6^2A(q) \pmod{3}. \tag*{(152)} E 3 , 2 ( n ≥ 0 ∑ C E S ( n ) q n ) ≡ f 1 f 6 f 2 8 f 3 2 ≡ f 1 f 2 f 3 2 f 6 f 2 9 ≡ f 1 f 2 f 2 3 f 6 2 ≡ f 3 2 f 6 2 A ( q ) ( mod 3 ) . (152)
Employing (16) , (18) , and (20) in (151) and (152) , and then simplifying by using (147) , we obtain
∑ n ≥ 0 C E S ( 9 n + 1 ) q n ≡ − f 2 f 1 E 3 , 0 ( D ( f 2 2 ) ) ≡ 0 ( m o d 3 ) , (153) \sum_{n\geq0} CES(9n+1)q^n \equiv-\frac{f_2}{f_1}\mathcal{E}_{3,0}\left(D(f_2^2)\right) \equiv0 \pmod{3}, \tag*{(153)} n ≥ 0 ∑ C E S ( 9 n + 1 ) q n ≡ − f 1 f 2 E 3 , 0 ( D ( f 2 2 ) ) ≡ 0 ( mod 3 ) , (153)
and
∑ n ≥ 0 C E S ( 9 n + 8 ) q n ≡ f 1 2 f 2 2 E 3 , 2 ( A ( q ) ) ≡ 0 ( m o d 3 ) , (154) \sum_{n\geq0} CES(9n+8)q^n \equiv f_1^2f_2^2\mathcal{E}_{3,2}(A(q)) \equiv0 \pmod{3}, \tag*{(154)} n ≥ 0 ∑ C E S ( 9 n + 8 ) q n ≡ f 1 2 f 2 2 E 3 , 2 ( A ( q )) ≡ 0 ( mod 3 ) , (154)
respectively. Now, (140) and (141) follow immediately from (153) and (154) , respectively. Replacing n n n by 9 n + 8 9n+8 9 n + 8 in (4) and simplifying using (154) and (147) , we arrive at (142) .
Employing (18) , (148) , and (151) in (4) , we obtain
E 3 , 1 ( ∑ n ≥ 0 C O S ( n ) q n ) = E 3 , 1 ( D A ( q ) ) − E 3 , 1 ( ∑ n ≥ 0 C E S ( n ) q n ) ≡ f 6 f 3 f 2 2 + f 6 f 3 D ( f 2 2 ) ≡ f 6 f 3 ( 1 + D ) f 2 2 ( m o d 3 ) . (155) \begin{aligned}
\mathcal{E}_{3,1}\left(\sum_{n\geq0} COS(n)q^n\right)
&=\mathcal{E}_{3,1}(DA(q))-\mathcal{E}_{3,1}\left(\sum_{n\geq0} CES(n)q^n\right)\\
&\equiv\frac{f_6}{f_3}f_2^2+\frac{f_6}{f_3}D(f_2^2)\\
&\equiv\frac{f_6}{f_3}(1+D)f_2^2\pmod{3}. \tag*{(155)}
\end{aligned} E 3 , 1 ( n ≥ 0 ∑ C O S ( n ) q n ) = E 3 , 1 ( D A ( q )) − E 3 , 1 ( n ≥ 0 ∑ C E S ( n ) q n ) ≡ f 3 f 6 f 2 2 + f 3 f 6 D ( f 2 2 ) ≡ f 3 f 6 ( 1 + D ) f 2 2 ( mod 3 ) . (155)
From (17) , we note that
E 3 , 2 ( ( 1 + D ) f 2 2 ) = 3 ( D + 1 ) E 3 , 2 ( f 2 2 ) . \mathcal{E}_{3,2}\left((1+D)f_2^2\right)=3(D+1)\mathcal{E}_{3,2}(f_2^2). E 3 , 2 ( ( 1 + D ) f 2 2 ) = 3 ( D + 1 ) E 3 , 2 ( f 2 2 ) .
Employing (16) and (20) in (155) , we obtain
∑ n ≥ 0 C O S ( 9 n + 7 ) q n ≡ f 2 f 1 E 3 , 2 ( ( 1 + D ) f 2 2 ) ≡ 0 ( m o d 3 ) . (156) \sum_{n\geq0} COS(9n+7)q^n \equiv\frac{f_2}{f_1}\mathcal{E}_{3,2}\left((1+D)f_2^2\right) \equiv0 \pmod{3}. \tag*{(156)} n ≥ 0 ∑ C O S ( 9 n + 7 ) q n ≡ f 1 f 2 E 3 , 2 ( ( 1 + D ) f 2 2 ) ≡ 0 ( mod 3 ) . (156)
Now, (143) follows easily from (156) . Again, from (2) and (14) , we note that
D A ( q ) A ( q ) = − D f 1 f 1 − D f 2 f 2 = L ( q ) + 2 L ( q 2 ) . (157) \frac{DA(q)}{A(q)}=-\frac{Df_1}{f_1}-\frac{Df_2}{f_2}=L(q)+2L(q^2). \tag*{(157)} A ( q ) D A ( q ) = − f 1 D f 1 − f 2 D f 2 = L ( q ) + 2 L ( q 2 ) . (157)
Employing (157) and (9) in (133) , we obtain
∑ n ≥ 0 COS ( n ) q n = A ( q ) ( 2 L ( q ) − ( L ( q ) + 2 L ( q 2 ) ) ) = 2 A ( q ) L ( q ) − D A ( q ) = 2 ϕ ( − q ) L ( q ) f 1 3 − D A ( q ) , (158) \begin{aligned}
\sum_{n\geq0}\operatorname{COS}(n)q^n
&=A(q)\left(2L(q)-\left(L(q)+2L(q^2)\right)\right)\\
&=2A(q)L(q)-DA(q)\\
&=2\frac{\phi(-q)L(q)}{f_1^3}-DA(q), \tag*{(158)}
\end{aligned} n ≥ 0 ∑ COS ( n ) q n = A ( q ) ( 2 L ( q ) − ( L ( q ) + 2 L ( q 2 ) ) ) = 2 A ( q ) L ( q ) − D A ( q ) = 2 f 1 3 ϕ ( − q ) L ( q ) − D A ( q ) , (158)
Using (45) and (11) , we obtain
L ( q ) f 1 3 ≡ L ( q 3 ) + q f 3 8 f 3 = L ( q 3 ) f 3 + q f 3 7 ( m o d 3 ) . (159) \frac{L(q)}{f_1^3}\equiv\frac{L(q^3)+qf_3^8}{f_3}=\frac{L(q^3)}{f_3}+qf_3^7\pmod{3}. \tag*{(159)} f 1 3 L ( q ) ≡ f 3 L ( q 3 ) + q f 3 8 = f 3 L ( q 3 ) + q f 3 7 ( mod 3 ) . (159)
Also, from (9) , we note that
ϕ ( − q ) = ∑ n ∈ Z ( − 1 ) n q n 2 = ϕ ( − q 9 ) + ∑ n ∈ 3 Z ( − 1 ) n q n 2 . (160) \phi(-q)=\sum_{n\in\mathbb{Z}}(-1)^nq^{n^2}=\phi(-q^9)+\sum_{n\in3\mathbb{Z}}(-1)^nq^{n^2}. \tag*{(160)} ϕ ( − q ) = n ∈ Z ∑ ( − 1 ) n q n 2 = ϕ ( − q 9 ) + n ∈ 3 Z ∑ ( − 1 ) n q n 2 . (160)
Employing (14) , (16) , (18) , (159) , and (160) in (158) , we obtain
E 3 , 0 ( ∑ n ≥ 0 COS ( n ) q n ) ≡ 2 E 3 , 0 ( ϕ ( − q 9 ) L ( q 3 ) f 3 ) = 2 ϕ ( − q 3 ) L ( q ) f 1 = 2 ϕ ( − q 3 ) D ( 1 f 1 ) ( m o d 3 ) . (161) \begin{aligned}
\mathcal{E}_{3,0}\left(\sum_{n\geq0}\operatorname{COS}(n)q^n\right)
&\equiv2\mathcal{E}_{3,0}\left(\phi(-q^9)\frac{L(q^3)}{f_3}\right)\\
&=2\phi(-q^3)\frac{L(q)}{f_1}\\
&=2\phi(-q^3)D\left(\frac{1}{f_1}\right)\pmod{3}. \tag*{(161)}
\end{aligned} E 3 , 0 ( n ≥ 0 ∑ COS ( n ) q n ) ≡ 2 E 3 , 0 ( ϕ ( − q 9 ) f 3 L ( q 3 ) ) = 2 ϕ ( − q 3 ) f 1 L ( q ) = 2 ϕ ( − q 3 ) D ( f 1 1 ) ( mod 3 ) . (161)
Employing (16) , (18) , and (20) in (161) , we obtain
∑ n ≥ 0 COS ( 9 n ) q n ≡ 2 ϕ ( − q ) E 3 , 0 ( D ( 1 f 1 ) ) ≡ 0 ( m o d 3 ) . (162) \sum_{n\geq0}\operatorname{COS}(9n)q^n\equiv2\phi(-q)\mathcal{E}_{3,0}\left(D\left(\frac{1}{f_1}\right)\right)\equiv0\pmod{3}. \tag*{(162)} n ≥ 0 ∑ COS ( 9 n ) q n ≡ 2 ϕ ( − q ) E 3 , 0 ( D ( f 1 1 ) ) ≡ 0 ( mod 3 ) . (162)
Now, (144) follows immediately from (162) . Congruence (145) follows from (144) and (4) . □
Theorem 3.5. For any integer n ≥ 0 n \ge0 n ≥ 0 , we have
C O S ( 25 n + 7 ) ≡ 0 ( m o d 5 ) , (163) COS(25n+7) \equiv0 \pmod{5}, \tag*{(163)} C O S ( 25 n + 7 ) ≡ 0 ( mod 5 ) , (163) C O S ( 25 n + 22 ) ≡ 0 ( m o d 5 ) , (164) COS(25n+22) \equiv0 \pmod{5}, \tag*{(164)} C O S ( 25 n + 22 ) ≡ 0 ( mod 5 ) , (164) C E S ( 25 n + 12 ) ≡ 0 ( m o d 5 ) , (165) CES(25n+12) \equiv0 \pmod{5}, \tag*{(165)} C E S ( 25 n + 12 ) ≡ 0 ( mod 5 ) , (165) and
C E S ( 25 n + 22 ) ≡ 0 ( m o d 5 ) . (166) CES(25n+22) \equiv0 \pmod{5}. \tag*{(166)} C E S ( 25 n + 22 ) ≡ 0 ( mod 5 ) . (166)
Proof. Employing (2) and (27) in (131) , we obtain
∑ n ≥ 0 C O S ( n ) q n = M A ( q ) − A ( q ) 24 ≡ A ( q ) − M A ( q ) ( m o d 5 ) . (167) \sum_{n\geq0} COS(n)q^n = \frac{MA(q)-A(q)}{24} \equiv A(q)-MA(q) \pmod{5}. \tag*{(167)} n ≥ 0 ∑ C O S ( n ) q n = 24 M A ( q ) − A ( q ) ≡ A ( q ) − M A ( q ) ( mod 5 ) . (167)
Employing (15) , (67) , and (79) in (167) , we obtain
∑ n ≥ 0 C O S ( 5 n + 2 ) q n ≡ E 5 , 2 ( A ( q ) ) − E 5 , 2 ( M A ( q ) ) ≡ ( 3 − M ) H ( m o d 5 ) , (168) \sum_{n\geq0} COS(5n+2)q^n \equiv\mathcal{E}_{5,2}(A(q))-\mathcal{E}_{5,2}(MA(q)) \equiv(3-M)H \pmod{5}, \tag*{(168)} n ≥ 0 ∑ C O S ( 5 n + 2 ) q n ≡ E 5 , 2 ( A ( q )) − E 5 , 2 ( M A ( q )) ≡ ( 3 − M ) H ( mod 5 ) , (168)
where
H = f 1 3 f 2 3 . (169) H=f_1^3f_2^3. \tag*{(169)} H = f 1 3 f 2 3 . (169)
From (18) , (79) , and (169) , we obtain
E 5 , 2 ( D A ( q ) ) ≡ 2 E 5 , 2 ( A ( q ) ) ≡ 6 H ( m o d 5 ) . (170) \mathcal{E}_{5,2}(DA(q)) \equiv2\mathcal{E}_{5,2}(A(q)) \equiv6H \pmod{5}. \tag*{(170)} E 5 , 2 ( D A ( q )) ≡ 2 E 5 , 2 ( A ( q )) ≡ 6 H ( mod 5 ) . (170)
Employing (168) and (170) in (4) , we obtain
∑ n ≥ 0 C E S ( 5 n + 2 ) q n = E 5 , 2 ( D A ( q ) ) − ∑ n ≥ 0 C O S ( 5 n + 2 ) q n ≡ ( 3 + M ) H ( m o d 5 ) . (171) \begin{aligned}
\sum_{n\geq0} CES(5n+2)q^n
&= \mathcal{E}_{5,2}(DA(q))-\sum_{n\geq0} COS(5n+2)q^n \\
&\equiv(3+M)H \pmod{5}. \tag*{(171)}
\end{aligned} n ≥ 0 ∑ C E S ( 5 n + 2 ) q n = E 5 , 2 ( D A ( q )) − n ≥ 0 ∑ C O S ( 5 n + 2 ) q n ≡ ( 3 + M ) H ( mod 5 ) . (171)
Next, for any integers r , s ≥ 0 r,s \geq0 r , s ≥ 0 , set
T r = r ( r + 1 ) 2 T_r=\frac{r(r+1)}{2} T r = 2 r ( r + 1 )
and
x = 2 r + 1 , y = 2 s + 1 , m = T r + 2 T s . (172) x=2r+1,\qquad y=2s+1,\qquad m=T_r+2T_s. \tag*{(172)} x = 2 r + 1 , y = 2 s + 1 , m = T r + 2 T s . (172)
Replacing q q q by q 2 q^2 q 2 in (12) , we obtain
f 2 3 = ∑ m ≥ 0 ( − 1 ) m ( 2 m + 1 ) q m ( m + 1 ) . (173) f_2^3=\sum_{m\geq0}(-1)^m(2m+1)q^{m(m+1)}. \tag*{(173)} f 2 3 = m ≥ 0 ∑ ( − 1 ) m ( 2 m + 1 ) q m ( m + 1 ) . (173)
Simplifying (169) by using (12) , (173) , and (172) , we obtain
H = ∑ r , s ≥ 0 ( − 1 ) r + s x y q m . (174) H=\sum_{r,s\geq0}(-1)^{r+s}xyq^m. \tag*{(174)} H = r , s ≥ 0 ∑ ( − 1 ) r + s x y q m . (174)
From (14) and (21) , we obtain
( 8 D + 1 ) f 1 3 = ( 1 − 24 L ( q ) ) f 1 3 = E 2 ( q ) f 1 3 . (175) (8D+1)f_1^3=(1-24\mathcal{L}(q))f_1^3=E_2(q)f_1^3. \tag*{(175)} ( 8 D + 1 ) f 1 3 = ( 1 − 24 L ( q )) f 1 3 = E 2 ( q ) f 1 3 . (175)
and
( 4 D + 1 ) f 2 3 = ( 1 − 24 L ( q 2 ) ) f 2 3 = E 2 ( q 2 ) f 2 3 . (176) (4D+1)f_2^3=(1-24\mathcal{L}(q^2))f_2^3=E_2(q^2)f_2^3. \tag*{(176)} ( 4 D + 1 ) f 2 3 = ( 1 − 24 L ( q 2 )) f 2 3 = E 2 ( q 2 ) f 2 3 . (176)
Simplifying (175) and (176) by employing (12) and (172) , we obtain
E 2 ( q ) f 1 3 = ∑ r ≥ 0 ( − 1 ) r ( 2 r + 1 ) ( 8 T r + 1 ) q T r = ∑ r ≥ 0 ( − 1 ) r x 3 q T r . (177) E_2(q)f_1^3=\sum_{r\geq0}(-1)^r(2r+1)(8T_r+1)q^{T_r}=\sum_{r\geq0}(-1)^rx^3q^{T_r}. \tag*{(177)} E 2 ( q ) f 1 3 = r ≥ 0 ∑ ( − 1 ) r ( 2 r + 1 ) ( 8 T r + 1 ) q T r = r ≥ 0 ∑ ( − 1 ) r x 3 q T r . (177)
and
E 2 ( q 2 ) f 2 3 = ∑ s ≥ 0 ( − 1 ) s ( 2 s + 1 ) ( 8 T s + 1 ) q 2 T s = ∑ s ≥ 0 ( − 1 ) s y 3 q 2 T s , (178) E_2(q^2)f_2^3=\sum_{s\geq0}(-1)^s(2s+1)(8T_s+1)q^{2T_s}=\sum_{s\geq0}(-1)^sy^3q^{2T_s}, \tag*{(178)} E 2 ( q 2 ) f 2 3 = s ≥ 0 ∑ ( − 1 ) s ( 2 s + 1 ) ( 8 T s + 1 ) q 2 T s = s ≥ 0 ∑ ( − 1 ) s y 3 q 2 T s , (178)
respectively. Combining (12) , (27) , (177) and (178) , we obtain
M H = ( 2 E 2 ( q 2 ) − E 2 ( q ) ) f 1 3 f 2 3 = ∑ r , s ≥ 0 ( − 1 ) r + s x y ( 2 y 2 − x 2 ) q m . (179) MH=(2E_2(q^2)-E_2(q))f_1^3f_2^3=\sum_{r,s\geq0}(-1)^{r+s}xy(2y^2-x^2)q^m. \tag*{(179)} M H = ( 2 E 2 ( q 2 ) − E 2 ( q )) f 1 3 f 2 3 = r , s ≥ 0 ∑ ( − 1 ) r + s x y ( 2 y 2 − x 2 ) q m . (179)
Also, from (172) note that
x 2 + 2 y 2 = ( 8 T r + 1 ) + 2 ( 8 T s + 1 ) = 8 m + 3. (180) x^2+2y^2=(8T_r+1)+2(8T_s+1)=8m+3. \tag*{(180)} x 2 + 2 y 2 = ( 8 T r + 1 ) + 2 ( 8 T s + 1 ) = 8 m + 3. (180)
If x y ≡ 0 ( m o d 5 ) xy\equiv0\pmod{5} x y ≡ 0 ( mod 5 ) , then the corresponding summands in (174) and (179) vanish modulo 5 5 5 . Otherwise, each of x 2 x^2 x 2 and y 2 y^2 y 2 is congruent to 1 1 1 or 4 4 4 modulo 5 5 5 . Thus, by (180) , we obtain the following table for x y ≢ 0 ( m o d 5 ) xy\not\equiv0\pmod{5} x y ≡ 0 ( mod 5 ) , with all entries taken modulo 5 5 5 :
Table 1.
Thus, from the above table, we always have
x y ( 3 + x 2 − 2 y 2 ) ≡ 3 ( m − 1 ) ( m + 1 ) 3 x y ( m o d 5 ) . (181) xy(3+x^2-2y^2)\equiv3(m-1)(m+1)^3xy\pmod{5}. \tag*{(181)} x y ( 3 + x 2 − 2 y 2 ) ≡ 3 ( m − 1 ) ( m + 1 ) 3 x y ( mod 5 ) . (181)
Simplifying (174) and (179) and employing (181) , we obtain
( 3 − M ) H = ∑ r , s ≥ 0 ( − 1 ) r + s x y ( 3 + x 2 − 2 y 2 ) q m ≡ 3 ∑ r , s ≥ 0 ( − 1 ) r + s ( m − 1 ) ( m + 1 ) 3 x y q m = 3 ( D − 1 ) ( D + 1 ) 3 H ( m o d 5 ) . (182) \begin{aligned}
(3-M)H&=\sum_{r,s\geq0}(-1)^{r+s}xy(3+x^2-2y^2)q^m\\
&\equiv3\sum_{r,s\geq0}(-1)^{r+s}(m-1)(m+1)^3xyq^m\\
&=3(D-1)(D+1)^3H\pmod{5}. \tag*{(182)}
\end{aligned} ( 3 − M ) H = r , s ≥ 0 ∑ ( − 1 ) r + s x y ( 3 + x 2 − 2 y 2 ) q m ≡ 3 r , s ≥ 0 ∑ ( − 1 ) r + s ( m − 1 ) ( m + 1 ) 3 x y q m = 3 ( D − 1 ) ( D + 1 ) 3 H ( mod 5 ) . (182)
Employing (182) in (168) , we obtain
∑ n ≥ 0 C O S ( 5 n + 2 ) q n ≡ ( 3 − M ) H ≡ 3 ( D − 1 ) ( D + 1 ) 3 H ( m o d 5 ) . (183) \sum_{n\geq0}\mathit{COS}(5n+2)q^n\equiv(3-M)H\equiv3(D-1)(D+1)^3H\pmod{5}. \tag*{(183)} n ≥ 0 ∑ COS ( 5 n + 2 ) q n ≡ ( 3 − M ) H ≡ 3 ( D − 1 ) ( D + 1 ) 3 H ( mod 5 ) . (183)
Similarly, from (171) and (182) , we obtain
∑ n ≥ 0 C E S ( 5 n + 2 ) q n ≡ ( 3 + M ) H ≡ 3 ( 2 − ( D − 1 ) ( D + 1 ) 3 ) H ( m o d 5 ) . (184) \sum_{n\geq0}\mathit{CES}(5n+2)q^n\equiv(3+M)H\equiv3\left(2-(D-1)(D+1)^3\right)H\pmod{5}. \tag*{(184)} n ≥ 0 ∑ CES ( 5 n + 2 ) q n ≡ ( 3 + M ) H ≡ 3 ( 2 − ( D − 1 ) ( D + 1 ) 3 ) H ( mod 5 ) . (184)
Extracting the coefficients of q 5 n + 1 q^{5n+1} q 5 n + 1 and q 5 n + 2 q^{5n+2} q 5 n + 2 on both sides of (183) and (184) , respectively, we obtain
C O S ( 25 n + 7 ) ≡ 3 ( 5 n ) ( 5 n + 2 ) 3 [ q 5 n + 1 ] H ≡ 0 ( m o d 5 ) \mathit{COS}(25n+7)\equiv3(5n)(5n+2)^3[q^{5n+1}]H\equiv0\pmod{5} COS ( 25 n + 7 ) ≡ 3 ( 5 n ) ( 5 n + 2 ) 3 [ q 5 n + 1 ] H ≡ 0 ( mod 5 )
and
C E S ( 25 n + 12 ) ≡ 3 ( 2 − ( 5 n + 1 ) ( 5 n + 3 ) 3 ) [ q 5 n + 2 ] H ≡ 3 ( 2 − 27 ) [ q 5 n + 2 ] H ≡ 0 ( m o d 5 ) . \mathit{CES}(25n+12)\equiv3\left(2-(5n+1)(5n+3)^3\right)[q^{5n+2}]H\equiv3(2-27)[q^{5n+2}]H\equiv0\pmod{5}. CES ( 25 n + 12 ) ≡ 3 ( 2 − ( 5 n + 1 ) ( 5 n + 3 ) 3 ) [ q 5 n + 2 ] H ≡ 3 ( 2 − 27 ) [ q 5 n + 2 ] H ≡ 0 ( mod 5 ) .
So the proofs of (163) and (165) are complete.
Since − 2 ≡ 3 ( m o d 5 ) -2\equiv3\pmod{5} − 2 ≡ 3 ( mod 5 ) is a quadratic nonresidue modulo 5, from (180) , we obtain
m ≡ 4 ( m o d 5 ) if and only if { x ≡ 0 ( m o d 5 ) y ≡ 0 ( m o d 5 ) , if and only if { r = 5 u + 2 s = 5 v + 2 , (185) m\equiv4\pmod{5}\text{ if and only if }\left\{\begin{aligned}x&\equiv0\pmod{5}\\y&\equiv0\pmod{5},\end{aligned}\right.\text{ if and only if }\left\{\begin{aligned}r&=5u+2\\s&=5v+2,\end{aligned}\right. \tag*{(185)} m ≡ 4 ( mod 5 ) if and only if { x y ≡ 0 ( mod 5 ) ≡ 0 ( mod 5 ) , if and only if { r s = 5 u + 2 = 5 v + 2 , (185)
where u u u and v v v are non-negative integers.
From (172) and (185) , we obtain
T 5 u + 2 = ( 5 u + 2 ) ( 5 u + 3 ) 2 = 25 T u + 3 , (186) T_{5u+2}=\frac{(5u+2)(5u+3)}{2}=25T_u+3, \tag*{(186)} T 5 u + 2 = 2 ( 5 u + 2 ) ( 5 u + 3 ) = 25 T u + 3 , (186)
m = 25 ( T u + 2 T v ) + 9 , (187) m=25(T_u+2T_v)+9, \tag*{(187)} m = 25 ( T u + 2 T v ) + 9 , (187)
( − 1 ) r + s = ( − 1 ) 5 u + 5 v + 4 = ( − 1 ) u + v , (188) (-1)^{r+s}=(-1)^{5u+5v+4}=(-1)^{u+v}, \tag*{(188)} ( − 1 ) r + s = ( − 1 ) 5 u + 5 v + 4 = ( − 1 ) u + v , (188)
and
x y = ( 10 u + 5 ) ( 10 v + 5 ) = 25 ( 2 u + 1 ) ( 2 v + 1 ) . (189) xy=(10u+5)(10v+5)=25(2u+1)(2v+1). \tag*{(189)} x y = ( 10 u + 5 ) ( 10 v + 5 ) = 25 ( 2 u + 1 ) ( 2 v + 1 ) . (189)
Substituting (186) , (187) , (188) , and (189) in (174) and using (12) , we obtain
E 5 , 4 ( H ) = 25 ∑ u , v ≥ 0 ( − 1 ) u + v ( 2 u + 1 ) ( 2 v + 1 ) q ( 25 ( T u + 2 T v ) + 9 − 4 ) / 5 = 25 q ( ∑ u ≥ 0 ( − 1 ) u ( 2 u + 1 ) q 5 T u ) ( ∑ v ≥ 0 ( − 1 ) v ( 2 v + 1 ) q 10 T v ) = 25 q f 5 3 f 10 3 . (190) \begin{aligned}
\mathcal{E}_{5,4}(H)&=25\sum_{u,v\geq0}(-1)^{u+v}(2u+1)(2v+1)q^{(25(T_u+2T_v)+9-4)/5}\\
&=25q\left(\sum_{u\geq0}(-1)^u(2u+1)q^{5T_u}\right)\left(\sum_{v\geq0}(-1)^v(2v+1)q^{10T_v}\right)\\
&=25qf_5^3f_{10}^3. \tag*{(190)}
\end{aligned} E 5 , 4 ( H ) = 25 u , v ≥ 0 ∑ ( − 1 ) u + v ( 2 u + 1 ) ( 2 v + 1 ) q ( 25 ( T u + 2 T v ) + 9 − 4 ) /5 = 25 q ( u ≥ 0 ∑ ( − 1 ) u ( 2 u + 1 ) q 5 T u ) ( v ≥ 0 ∑ ( − 1 ) v ( 2 v + 1 ) q 10 T v ) = 25 q f 5 3 f 10 3 . (190)
Now, by (190) it is easily seen that
[ q 5 n + 4 ] H = 25 [ q n ] ( q f 5 3 f 10 3 ) ≡ 0 ( m o d 5 ) . [q^{5n+4}]H=25[q^n](qf_5^3f_{10}^3)\equiv0\pmod{5}. [ q 5 n + 4 ] H = 25 [ q n ] ( q f 5 3 f 10 3 ) ≡ 0 ( mod 5 ) .
Extracting the coefficients of q 5 n + 4 q^{5n+4} q 5 n + 4 on both sides of (183) and (184) , we obtain
C O S ( 25 n + 22 ) ≡ 3 ( 5 n + 3 ) ( 5 n + 5 ) 3 [ q 5 n + 4 ] H ≡ 0 ( m o d 5 ) \mathit{COS}(25n+22)\equiv3(5n+3)(5n+5)^3[q^{5n+4}]H\equiv0\pmod{5} COS ( 25 n + 22 ) ≡ 3 ( 5 n + 3 ) ( 5 n + 5 ) 3 [ q 5 n + 4 ] H ≡ 0 ( mod 5 )
and
C E S ( 25 n + 22 ) ≡ 3 ( 2 − ( 5 n + 3 ) ( 5 n + 5 ) 3 ) [ q 5 n + 4 ] H ≡ 6 [ q 5 n + 4 ] H ≡ 0 ( m o d 5 ) , \mathit{CES}(25n+22)\equiv3\left(2-(5n+3)(5n+5)^3\right)[q^{5n+4}]H\equiv6[q^{5n+4}]H\equiv0\pmod{5}, CES ( 25 n + 22 ) ≡ 3 ( 2 − ( 5 n + 3 ) ( 5 n + 5 ) 3 ) [ q 5 n + 4 ] H ≡ 6 [ q 5 n + 4 ] H ≡ 0 ( mod 5 ) ,
respectively. Hence, the proofs of (164) and (166) are complete. □
Theorem 3.6. For every nonnegative integer n n n , we have
C O S ( 49 n + 43 ) ≡ 0 ( m o d 7 ) (191) \mathit{COS}(49n+43)\equiv0\pmod{7} \tag*{(191)} COS ( 49 n + 43 ) ≡ 0 ( mod 7 ) (191) and
C E S ( 49 n + 43 ) ≡ 0 ( m o d 7 ) . (192) \mathit{CES}(49n+43)\equiv0\pmod{7}. \tag*{(192)} CES ( 49 n + 43 ) ≡ 0 ( mod 7 ) . (192)
Proof. Employing (2) and (27) in (131) , we obtain
∑ n ≥ 0 C O S ( n ) q n = M A ( q ) − A ( q ) 24 ≡ 5 ( M A ( q ) − A ( q ) ) ( m o d 7 ) . (193) \sum_{n\geq0}\mathit{COS}(n)q^n=\frac{MA(q)-A(q)}{24}\equiv5(MA(q)-A(q))\pmod{7}. \tag*{(193)} n ≥ 0 ∑ COS ( n ) q n = 24 M A ( q ) − A ( q ) ≡ 5 ( M A ( q ) − A ( q )) ( mod 7 ) . (193)
Employing (15) , (68) , and (80) in (193) , we obtain
∑ n ≥ 0 C O S ( 7 n + 1 ) q n ≡ 5 ( E 7 , 1 ( M A ( q ) ) − E 7 , 1 ( A ( q ) ) ) = 5 ( 4 f 1 f 2 M 3 − f 1 f 2 M 2 ) = f 1 f 2 M 2 ( 20 M − 5 ) ≡ f 1 f 2 M 2 ( 2 + 6 M ) ( m o d 7 ) . (194) \begin{aligned}
\sum_{n\geq0}\mathit{COS}(7n+1)q^n
&\equiv5\left(\mathcal{E}_{7,1}(MA(q))-\mathcal{E}_{7,1}(A(q))\right)\\
&=5(4f_1f_2M^3-f_1f_2M^2)\\
&=f_1f_2M^2(20M-5)\\
&\equiv f_1f_2M^2(2+6M)\pmod{7}. \tag*{(194)}
\end{aligned} n ≥ 0 ∑ COS ( 7 n + 1 ) q n ≡ 5 ( E 7 , 1 ( M A ( q )) − E 7 , 1 ( A ( q )) ) = 5 ( 4 f 1 f 2 M 3 − f 1 f 2 M 2 ) = f 1 f 2 M 2 ( 20 M − 5 ) ≡ f 1 f 2 M 2 ( 2 + 6 M ) ( mod 7 ) . (194)
Also, from (18) and (80) , we obtain
E 7 , 1 ( D A ( q ) ) ≡ E 7 , 1 ( A ( q ) ) ≡ f 1 f 2 M 2 ( m o d 7 ) . (195) \mathcal{E}_{7,1}(DA(q)) \equiv\mathcal{E}_{7,1}(A(q)) \equiv f_1f_2M^2 \pmod{7}. \tag*{(195)} E 7 , 1 ( D A ( q )) ≡ E 7 , 1 ( A ( q )) ≡ f 1 f 2 M 2 ( mod 7 ) . (195)
Employing (194) and (195) in (4) , we obtain
∑ n ≥ 0 C E S ( 7 n + 1 ) q n = E 7 , 1 ( D A ( q ) ) − ∑ n ≥ 0 C O S ( 7 n + 1 ) q n ≡ f 1 f 2 M 2 − f 1 f 2 M 2 ( 2 + 6 M ) = f 1 f 2 M 2 ( − 1 − 6 M ) ≡ f 1 f 2 M 2 ( 6 + M ) ( m o d 7 ) . (196) \begin{aligned}
\sum_{n\ge0} \mathit{CES}(7n+1)q^n
&= \mathcal{E}_{7,1}(DA(q))-\sum_{n\ge0}\mathit{COS}(7n+1)q^n \\
&\equiv f_1f_2M^2-f_1f_2M^2(2+6M) \\
&= f_1f_2M^2(-1-6M) \\
&\equiv f_1f_2M^2(6+M)\pmod{7}. \tag*{(196)}
\end{aligned} n ≥ 0 ∑ CES ( 7 n + 1 ) q n = E 7 , 1 ( D A ( q )) − n ≥ 0 ∑ COS ( 7 n + 1 ) q n ≡ f 1 f 2 M 2 − f 1 f 2 M 2 ( 2 + 6 M ) = f 1 f 2 M 2 ( − 1 − 6 M ) ≡ f 1 f 2 M 2 ( 6 + M ) ( mod 7 ) . (196)
Again, from (107) and (108) , we obtain
E 7 , 6 ( f 1 f 2 M 2 ) = 7 4 f 7 f 14 M ( q 7 ) 2 ≡ 0 ( m o d 7 ) (197) \mathcal{E}_{7,6}(f_1f_2M^2)=7^4f_7f_{14}M(q^7)^2\equiv0\pmod{7} \tag*{(197)} E 7 , 6 ( f 1 f 2 M 2 ) = 7 4 f 7 f 14 M ( q 7 ) 2 ≡ 0 ( mod 7 ) (197)
and
E 7 , 6 ( f 1 f 2 M 3 ) = 7 6 f 7 f 14 M ( q 7 ) 3 ≡ 0 ( m o d 7 ) . (198) \mathcal{E}_{7,6}(f_1f_2M^3)=7^6f_7f_{14}M(q^7)^3\equiv0\pmod{7}. \tag*{(198)} E 7 , 6 ( f 1 f 2 M 3 ) = 7 6 f 7 f 14 M ( q 7 ) 3 ≡ 0 ( mod 7 ) . (198)
Employing (20) , (197) , and (198) in (194) and (196) , we obtain
∑ n ≥ 0 C O S ( 49 n + 43 ) q n ≡ 2 E 7 , 6 ( f 1 f 2 M 2 ) + 6 E 7 , 6 ( f 1 f 2 M 3 ) ≡ 0 ( m o d 7 ) (199) \sum_{n\ge0}\mathit{COS}(49n+43)q^n\equiv2\mathcal{E}_{7,6}(f_1f_2M^2)+6\mathcal{E}_{7,6}(f_1f_2M^3)\equiv0\pmod{7} \tag*{(199)} n ≥ 0 ∑ COS ( 49 n + 43 ) q n ≡ 2 E 7 , 6 ( f 1 f 2 M 2 ) + 6 E 7 , 6 ( f 1 f 2 M 3 ) ≡ 0 ( mod 7 ) (199)
and
∑ n ≥ 0 C E S ( 49 n + 43 ) q n ≡ 6 E 7 , 6 ( f 1 f 2 M 2 ) + E 7 , 6 ( f 1 f 2 M 3 ) ≡ 0 ( m o d 7 ) , (200) \sum_{n\ge0}\mathit{CES}(49n+43)q^n\equiv6\mathcal{E}_{7,6}(f_1f_2M^2)+\mathcal{E}_{7,6}(f_1f_2M^3)\equiv0\pmod{7}, \tag*{(200)} n ≥ 0 ∑ CES ( 49 n + 43 ) q n ≡ 6 E 7 , 6 ( f 1 f 2 M 2 ) + E 7 , 6 ( f 1 f 2 M 3 ) ≡ 0 ( mod 7 ) , (200)
Now (191) and (192) follow easily from (199) and (200) , respectively.
Concluding remarks
Having established the above congruences modulo 3, 5, and 7, it is natural to ask whether corresponding congruences hold modulo higher powers. Based on numerical evidence, we make the following conjecture.
Conjecture. Let p ∈ { 3 , 5 , 7 } p\in\{3,5,7\} p ∈ { 3 , 5 , 7 } and let α \alpha α be a positive integer. Then, for every nonnegative integer n n n ,
C O S ( p 2 α n + 7 p 2 α + 1 8 ) ≡ C E S ( p 2 α n + 7 p 2 α + 1 8 ) ≡ 0 ( m o d p α ) . (201) \mathit{COS}\left(p^{2\alpha}n+\frac{7p^{2\alpha}+1}{8}\right)\equiv\mathit{CES}\left(p^{2\alpha}n+\frac{7p^{2\alpha}+1}{8}\right)\equiv0\pmod{p^\alpha}. \tag*{(201)} COS ( p 2 α n + 8 7 p 2 α + 1 ) ≡ CES ( p 2 α n + 8 7 p 2 α + 1 ) ≡ 0 ( mod p α ) . (201)
Since p p p is odd, p 2 α ≡ 1 ( m o d 8 ) p^{2\alpha}\equiv1\pmod{8} p 2 α ≡ 1 ( mod 8 ) , so 7 p 2 α + 1 8 \frac{7p^{2\alpha}+1}{8} 8 7 p 2 α + 1 is an integer. We note that the case α = 1 \alpha=1 α = 1 gives precisely the congruences proved above for p = 3 p=3 p = 3 , 5, and 7.
Declarations
Funding: This research did not receive funding.
Author Contributions. Both authors contributed equally to this work.
Conflict of Interest. The authors declare that there is no conflict of interest regarding the publication of this paper.
Human and Animal Rights. The authors declare that there is no research involving human participants or animals in the context of this paper.
Data Availability Statement. Data sharing is not applicable to this paper as no datasets were generated or analyzed during the current study.