Introduction
Let N \mathbb{N} N N 0 \mathbb{N}_0 N 0 Z \mathbb{Z} Z R \mathbb{R} R C \mathbb{C} C ( a , b ) (a,b) ( a , b ) a , b ∈ Z a,b \in\mathbb{Z} a , b ∈ Z { x } \{x\} { x } x ∈ R x \in\mathbb{R} x ∈ R [ x ] [x] [ x ] x ∈ R x \in\mathbb{R} x ∈ R [ x ] = x − { x } [x] = x - \{x\} [ x ] = x − { x } δ l , k \delta_{l,k} δ l , k l , k ∈ N 0 l,k \in\mathbb{N}_0 l , k ∈ N 0 δ l , k = 1 \delta_{l,k} = 1 δ l , k = 1 l = k l = k l = k 0 0 0 δ Z ( x ) \delta_{\mathbb{Z}}(x) δ Z ( x ) Z \mathbb{Z} Z δ Z ( x ) = 1 \delta_{\mathbb{Z}}(x) = 1 δ Z ( x ) = 1 x ∈ Z x \in\mathbb{Z} x ∈ Z 0 0 0 B ‾ n ( x ) \overline{B}_n(x) B n ( x ) n n n n ∈ N 0 n \in\mathbb{N}_0 n ∈ N 0 x ∈ R x \in\mathbb{R} x ∈ R
B ‾ n ( x ) = B n ( { x } ) + 1 2 δ 1 , n δ Z ( x ) . \overline{B}_n(x) = B_n(\{x\}) + \frac{1}{2}\delta_{1,n}\delta_{\mathbb{Z}}(x). B n ( x ) = B n ({ x }) + 2 1 δ 1 , n δ Z ( x ) . where B n ( x ) B_n(x) B n ( x ) n n n
t e x t e t − 1 = ∑ n = 0 ∞ B n ( x ) t n n ! ( ∣ t ∣ < 2 π ) . \frac{t e^{xt}}{e^t - 1} = \sum_{n=0}^{\infty} B_n(x) \frac{t^n}{n!} \qquad(|t| < 2\pi). e t − 1 t e x t = n = 0 ∑ ∞ B n ( x ) n ! t n ( ∣ t ∣ < 2 π ) . For a ∈ Z a \in\mathbb{Z} a ∈ Z b ∈ N b \in\mathbb{N} b ∈ N s ( a , b ) s(a,b) s ( a , b )
s ( a , b ) = ∑ r = 1 b B ‾ 1 ( a r b ) B ‾ 1 ( r b ) . (1) s(a,b) = \sum_{r=1}^{b} \overline{B}_1\left(\frac{ar}{b}\right)\overline{B}_1\left(\frac{r}{b}\right). \tag*{(1)} s ( a , b ) = r = 1 ∑ b B 1 ( b a r ) B 1 ( b r ) . (1) In the 1880s, Dedekind [12 ] introduced (1) in connection with the transformation formulas of the Dedekind eta-function. From this, he deduced his reciprocity theorem: if a , b ∈ N a,b \in\mathbb{N} a , b ∈ N ( a , b ) = 1 (a,b) = 1 ( a , b ) = 1
s ( a , b ) + s ( b , a ) = − 1 4 + 1 12 ( a b + b a + 1 a b ) . (2) s(a,b) + s(b,a) = -\frac{1}{4} + \frac{1}{12}\left(\frac{a}{b} + \frac{b}{a} + \frac{1}{ab}\right). \tag*{(2)} s ( a , b ) + s ( b , a ) = − 4 1 + 12 1 ( b a + a b + ab 1 ) . (2) Since the early 20th century, many mathematicians have studied these sums and their generalizations, and have established the corresponding reciprocity formulas; see, e.g., [2 , 3 , 7 , 16 ] and the references therein.
This paper is concerned with Dedekind character sums introduced by Berndt [4 , 6 ] , together with their generalizations. In 1973, while studying the transformation formulas of a generalized Eisenstein series, Berndt [4 ] was led to define the sums s ( a , b ; χ , χ ‾ ) s(a,b;\chi,\overline{\chi}) s ( a , b ; χ , χ )
s ( a , b ; χ , χ ‾ ) = ∑ r = 1 b q χ ( r ) B ‾ 1 ( a r b , χ ‾ ) B ‾ 1 ( r b q ) , (3) s(a,b;\chi,\overline{\chi})=\sum_{r=1}^{bq}\chi(r)\overline{B}_{1}\left(\frac{ar}{b},\overline{\chi}\right)\overline{B}_{1}\left(\frac{r}{bq}\right), \tag*{(3)} s ( a , b ; χ , χ ) = r = 1 ∑ b q χ ( r ) B 1 ( b a r , χ ) B 1 ( b q r ) , (3) where a , b ∈ Z a,b\in\mathbb{Z} a , b ∈ Z b ≥ 1 b\geq1 b ≥ 1 χ \chi χ q q q q ∈ N q\in\mathbb{N} q ∈ N χ ‾ \overline{\chi} χ χ \chi χ B ‾ n ( x , χ ) \overline{B}_{n}(x,\chi) B n ( x , χ ) n n n n ∈ N 0 n\in\mathbb{N}_{0} n ∈ N 0 x ∈ R x\in\mathbb{R} x ∈ R
B ‾ n ( x , χ ) = q n − 1 ∑ r = 1 q χ ( r ) B ‾ n ( r + x q ) . \overline{B}_{n}(x,\chi)=q^{n-1}\sum_{r=1}^{q}\chi(r)\overline{B}_{n}\left(\frac{r+x}{q}\right). B n ( x , χ ) = q n − 1 r = 1 ∑ q χ ( r ) B n ( q r + x ) . Meanwhile, Berndt [4 Theorem 4] used his transformation formulas to derive the following reciprocity theorem. Let q ∈ N q\in\mathbb{N} q ∈ N q ≥ 2 q\geq2 q ≥ 2 χ \chi χ q q q a , b ∈ N a,b\in\mathbb{N} a , b ∈ N ( a , b ) = 1 (a,b)=1 ( a , b ) = 1 a ≡ 0 a\equiv0 a ≡ 0 b ≡ 0 ( m o d q ) b\equiv0\pmod q b ≡ 0 ( mod q )
s ( a , b ; χ , χ ‾ ) + s ( b , a ; χ ‾ , χ ) = B 1 ( χ ) B 1 ( χ ‾ ) , (4) s(a,b;\chi,\overline{\chi})+s(b,a;\overline{\chi},\chi)=B_{1}(\chi)B_{1}(\overline{\chi}), \tag*{(4)} s ( a , b ; χ , χ ) + s ( b , a ; χ , χ ) = B 1 ( χ ) B 1 ( χ ) , (4) where B n ( χ ) = B ‾ n ( 0 , χ ) B_{n}(\chi)=\overline{B}_{n}(0,\chi) B n ( χ ) = B n ( 0 , χ ) [6 ] introduced the sums s 1 ( a , b ; χ ) s_{1}(a,b;\chi) s 1 ( a , b ; χ ) s 2 ( a , b ; χ ) s_{2}(a,b;\chi) s 2 ( a , b ; χ ) χ \chi χ q q q q ∈ N q\in\mathbb{N} q ∈ N a , b ∈ Z a,b\in\mathbb{Z} a , b ∈ Z b ≥ 1 b\geq1 b ≥ 1
s 1 ( a , b ; χ ) = ∑ r = 1 b q χ ( r ) B ‾ 1 ( a r b ) B ‾ 1 ( r b q ) , (5) s_{1}(a,b;\chi)=\sum_{r=1}^{bq}\chi(r)\overline{B}_{1}\left(\frac{ar}{b}\right)\overline{B}_{1}\left(\frac{r}{bq}\right), \tag*{(5)} s 1 ( a , b ; χ ) = r = 1 ∑ b q χ ( r ) B 1 ( b a r ) B 1 ( b q r ) , (5) and
s 2 ( a , b ; χ ) = ∑ r = 1 b q B ‾ 1 ( a r b , χ ) B ‾ 1 ( r b q ) . (6) s_{2}(a,b;\chi)=\sum_{r=1}^{bq}\overline{B}_{1}\left(\frac{ar}{b},\chi\right)\overline{B}_{1}\left(\frac{r}{bq}\right). \tag*{(6)} s 2 ( a , b ; χ ) = r = 1 ∑ b q B 1 ( b a r , χ ) B 1 ( b q r ) . (6) Moreover, by applying his transformation formulas for another generalized Eisenstein series, Berndt [6 Theorem 7] showed that for a , b , q ∈ N a,b,q\in\mathbb{N} a , b , q ∈ N ( a , b ) = 1 (a,b)=1 ( a , b ) = 1 q ≥ 2 q\geq2 q ≥ 2 χ \chi χ q q q ( b , q ) = 1 (b,q)=1 ( b , q ) = 1
s 1 ( a , b ; χ ) + s 2 ( b , a ; χ ) = a 2 b B 2 ( χ ) ; (7) s_{1}(a,b;\chi)+s_{2}(b,a;\chi)=\frac{a}{2b}B_{2}(\chi); \tag*{(7)} s 1 ( a , b ; χ ) + s 2 ( b , a ; χ ) = 2 b a B 2 ( χ ) ; (7) and if b ≡ 0 ( m o d q ) b\equiv0\pmod q b ≡ 0 ( mod q )
s 1 ( a , b ; χ ) + s 2 ( b , a ; χ ) = a 2 b B 2 ( χ ) + χ ‾ ( a ) 2 a b B 2 ( χ ) . (8) s_{1}(a,b;\chi)+s_{2}(b,a;\chi)=\frac{a}{2b}B_{2}(\chi)+\frac{\overline{\chi}(a)}{2ab}B_{2}(\chi). \tag*{(8)} s 1 ( a , b ; χ ) + s 2 ( b , a ; χ ) = 2 b a B 2 ( χ ) + 2 ab χ ( a ) B 2 ( χ ) . (8) In 1977, Berndt [7 ] provided an alternative proof of (4) , (7) and (8) by applying the (periodic) Poisson summation formula. In 2007, Cenkci, Can and Kurt [9 ] studied some generalizations of (3) and found the corresponding reciprocity formulas. In 2017, Dağlı and Can [11 ] explored similar generalizations by applying Berndt’s character analogue of the Euler–Maclaurin summation formula [5 ] . Furthermore, in 2020, Can [8 ] investigated the sums s m , n ( a , b , c ; χ 1 , χ 2 ; x , y , z ) s_{m,n}(a,b,c;\chi_{1},\chi_{2};x,y,z) s m , n ( a , b , c ; χ 1 , χ 2 ; x , y , z ) s m , n ∗ ( a , b , c ; χ 1 , χ 2 ; x , y , z ) s^{*}_{m,n}(a,b,c;\chi_{1},\chi_{2};x,y,z) s m , n ∗ ( a , b , c ; χ 1 , χ 2 ; x , y , z ) χ 1 \chi_{1} χ 1 χ 2 \chi_{2} χ 2 q q q q ∈ N q\in\mathbb{N} q ∈ N m , n ∈ N 0 m,n\in\mathbb{N}_{0} m , n ∈ N 0 a , b , c ∈ Z a,b,c\in\mathbb{Z} a , b , c ∈ Z c ≥ 1 c\geq1 c ≥ 1 x , y , z ∈ R x,y,z\in\mathbb{R} x , y , z ∈ R
s m , n ( a , b , c ; χ 1 , χ 2 ; x , y , z ) = ∑ r = 1 c q χ 1 ( r ) B ‾ m ( a ( r + z ) c + x , χ 2 ) B ‾ n ( b ( r + z ) c q + y ) , (9) \begin{aligned}
s_{m,n}(a,b,c;\chi_{1},\chi_{2};x,y,z)
&=\sum_{r=1}^{cq}\chi_{1}(r)\overline{B}_{m}\left(\frac{a(r+z)}{c}+x,\chi_{2}\right)\overline{B}_{n}\left(\frac{b(r+z)}{cq}+y\right), \tag*{(9)}
\end{aligned} s m , n ( a , b , c ; χ 1 , χ 2 ; x , y , z ) = r = 1 ∑ c q χ 1 ( r ) B m ( c a ( r + z ) + x , χ 2 ) B n ( c q b ( r + z ) + y ) , (9) and
s m , n ∗ ( a , b , c ; χ 1 , χ 2 ; x , y , z ) = ∑ r = 1 c B ‾ m ( a q ( r + z ) c + x , χ 1 ) B ‾ n ( b q ( r + z ) c + y , χ 2 ) . (10) s^{*}_{m,n}(a,b,c;\chi_1,\chi_2;x,y,z)=\sum_{r=1}^{c}\overline{B}_m\left(\frac{aq(r+z)}{c}+x,\chi_1\right)\overline{B}_n\left(\frac{bq(r+z)}{c}+y,\chi_2\right). \tag*{(10)} s m , n ∗ ( a , b , c ; χ 1 , χ 2 ; x , y , z ) = r = 1 ∑ c B m ( c a q ( r + z ) + x , χ 1 ) B n ( c b q ( r + z ) + y , χ 2 ) . (10) In particular, Can [8 Theorem 2.2] established a reciprocity formula relating the sums (1.9) and (1.10), which holds for m , n , a , b , c , q ∈ N m,n,a,b,c,q \in\mathbb{N} m , n , a , b , c , q ∈ N ( a , b ) = ( b , c ) = ( c , a ) = 1 (a,b)=(b,c)=(c,a)=1 ( a , b ) = ( b , c ) = ( c , a ) = 1 q ≥ 2 q\ge2 q ≥ 2 χ 1 \chi_1 χ 1 χ 2 \chi_2 χ 2 q q q
Naturally, one may ask whether the hypothesis ( a , b ) = ( b , c ) = ( c , a ) = 1 (a,b)=(b,c)=(c,a)=1 ( a , b ) = ( b , c ) = ( c , a ) = 1 [8 Theorem 2.2] can be removed. One may also ask whether Berndt’s reciprocity formulas (1.7) and (1.8) can be generalized in the same way as (1.4).
To address these two questions, we study the aforementioned sums from a different perspective. Specifically, we apply the Fourier series technique to establish the following formula for the product of the generalized Bernoulli function and the Bernoulli function, which plays a key role in the sequel.
Theorem 1.1. Let m , n , q ∈ N m,n,q\in\mathbb{N} m , n , q ∈ N a , b ∈ Z a,b\in\mathbb{Z} a , b ∈ Z a ≠ 0 a\ne0 a = 0 b ≠ 0 b\ne0 b = 0 χ \chi χ q q q x , y , z ∈ R x,y,z\in\mathbb{R} x , y , z ∈ R
B ‾ m ( a x + y , χ ) B ‾ n ( b x q + z ) = n q m − 1 b n − 1 χ ( − 1 ) sgn ( b ) ∑ j = 1 m ( m j ) ( − 1 ) j a m − j m + n − j ∑ r = 1 q χ ‾ ( r ) × ∑ l = 1 ∣ b ∣ B ‾ j ( a ( l + z ) b − y − r q ) B ‾ m + n − j ( x q + l + z b ) + m ( a q ) m − 1 χ ( − 1 ) sgn ( a ) ∑ j = 1 n ( n j ) ( − 1 ) j b n − j m + n − j ∑ r = 1 q χ ‾ ( r ) × ∑ l = 1 ∣ a ∣ B ‾ j ( b l a + b y − b r a q − z ) B ‾ m + n − j ( x q + l a + y − r a q ) + m b n B ‾ m + n ( a x + y , χ ‾ ) ( a q ) n ( m + n ) + n q m − 1 a m χ ( 0 ) G ( 1 , χ ‾ ) B ‾ m + n ( b x q + z ) b m ( m + n ) + ( − 1 ) n − 1 m ! n ! ( a , b ) m + n χ ( b ( a , b ) ) B ‾ m + n ( b y − a q z ( a , b ) , χ ‾ ) ( a q ) n b m ( m + n ) ! − δ 1 , m δ 1 , n sgn ( a b ) δ Z ( a x + y ) δ Z ( b x q + z ) χ ‾ ( − a x − y ) 4 . (11) \begin{aligned}
\overline{B}_m(ax+y,\chi)\overline{B}_n\left(\frac{bx}{q}+z\right)
={}&nq^{m-1}b^{n-1}\chi(-1)\operatorname{sgn}(b)\sum_{j=1}^{m}\binom{m}{j}\frac{(-1)^j a^{m-j}}{m+n-j}\sum_{r=1}^{q}\overline{\chi}(r)\\
&\quad\times\sum_{l=1}^{|b|}\overline{B}_j\left(\frac{a(l+z)}{b}-\frac{y-r}{q}\right)\overline{B}_{m+n-j}\left(\frac{x}{q}+\frac{l+z}{b}\right)\\
&+m(aq)^{m-1}\chi(-1)\operatorname{sgn}(a)\sum_{j=1}^{n}\binom{n}{j}\frac{(-1)^j b^{n-j}}{m+n-j}\sum_{r=1}^{q}\overline{\chi}(r)\\
&\quad\times\sum_{l=1}^{|a|}\overline{B}_j\left(\frac{bl}{a}+\frac{by-br}{aq}-z\right)\overline{B}_{m+n-j}\left(\frac{x}{q}+\frac{l}{a}+\frac{y-r}{aq}\right)\\
&+\frac{mb^n\overline{B}_{m+n}(ax+y,\overline{\chi})}{(aq)^n(m+n)}
+\frac{nq^{m-1}a^m\chi(0)G(1,\overline{\chi})\overline{B}_{m+n}\left(\frac{bx}{q}+z\right)}{b^m(m+n)}\\
&+\frac{(-1)^{n-1}m!n!(a,b)^{m+n}\chi\left(\frac{b}{(a,b)}\right)\overline{B}_{m+n}\left(\frac{by-aqz}{(a,b)},\overline{\chi}\right)}{(aq)^n b^m(m+n)!}\\
&-\delta_{1,m}\delta_{1,n}\operatorname{sgn}(ab)\frac{\delta_{\mathbb{Z}}(ax+y)\delta_{\mathbb{Z}}\left(\frac{bx}{q}+z\right)\overline{\chi}(-ax-y)}{4}. \tag*{(11)}
\end{aligned} B m ( a x + y , χ ) B n ( q b x + z ) = n q m − 1 b n − 1 χ ( − 1 ) sgn ( b ) j = 1 ∑ m ( j m ) m + n − j ( − 1 ) j a m − j r = 1 ∑ q χ ( r ) × l = 1 ∑ ∣ b ∣ B j ( b a ( l + z ) − q y − r ) B m + n − j ( q x + b l + z ) + m ( a q ) m − 1 χ ( − 1 ) sgn ( a ) j = 1 ∑ n ( j n ) m + n − j ( − 1 ) j b n − j r = 1 ∑ q χ ( r ) × l = 1 ∑ ∣ a ∣ B j ( a b l + a q b y − b r − z ) B m + n − j ( q x + a l + a q y − r ) + ( a q ) n ( m + n ) m b n B m + n ( a x + y , χ ) + b m ( m + n ) n q m − 1 a m χ ( 0 ) G ( 1 , χ ) B m + n ( q b x + z ) + ( a q ) n b m ( m + n )! ( − 1 ) n − 1 m ! n ! ( a , b ) m + n χ ( ( a , b ) b ) B m + n ( ( a , b ) b y − a q z , χ ) − δ 1 , m δ 1 , n sgn ( ab ) 4 δ Z ( a x + y ) δ Z ( q b x + z ) χ ( − a x − y ) . (11) Clearly, the case q = 1 q=1 q = 1 [14 Theorem 1.1] . Moreover, it can be used to establish a reciprocity formula for the generalized Dedekind–Rademacher sums introduced by Hall, Wilson and Zagier [13 ] ; see [14 Theorem 1.2] for details.
As another application of Theorem 1.1 , we are able to remove the hypothesis ( a , b ) = ( b , c ) = ( c , a ) = 1 (a,b)=(b,c)=(c,a)=1 ( a , b ) = ( b , c ) = ( c , a ) = 1 [8 Theorem 2.2] . Our result is stated as follows.
Theorem 1.2. Let m , n , a , b , c , q ∈ N m,n,a,b,c,q \in\mathbb{N} m , n , a , b , c , q ∈ N q ≥ 2 q \ge2 q ≥ 2 χ 1 \chi_1 χ 1 χ 2 \chi_2 χ 2 q q q x , y , z ∈ R x,y,z \in\mathbb{R} x , y , z ∈ R
s m , n ( a , b , c ; χ 1 , χ 2 ‾ ; x , y , z ) = n ( b c q ) n − 1 ∑ j = 1 n − 1 ( m j ) ( a c ) m − j m + n − j s j , m + n − j ∗ ( − a , c , b ; χ 2 ‾ , χ 1 ; x , z , y ) + m ( a c ) m − 1 ∑ j = 0 n ( n j ) ( − 1 ) j ( b c q ) n − j m + n − j s m + n − j , j ( c , b , a ; χ 2 ‾ , χ 1 ; z , − y , x ) − δ 1 , m δ 1 , n χ 2 ( − 1 ) 4 ∑ k = 1 c q χ 1 ( k ) δ Z ( a ( k + z ) c + x ) δ Z ( b ( k + z ) c q + y ) − × χ 2 ‾ ( a ( k + z ) c + x ) . (12) \begin{aligned}
s_{m,n}(a,b,c;\chi_1,\overline{\chi_2};x,y,z)
={}&n\left(\frac{b}{cq}\right)^{n-1}\sum_{j=1}^{n-1}\binom{m}{j}\frac{\left(\frac{a}{c}\right)^{m-j}}{m+n-j}s^*_{j,m+n-j}(-a,c,b;\overline{\chi_2},\chi_1;x,z,y)\\
&+m\left(\frac{a}{c}\right)^{m-1}\sum_{j=0}^{n}\binom{n}{j}\frac{(-1)^j\left(\frac{b}{cq}\right)^{n-j}}{m+n-j}s_{m+n-j,j}(c,b,a;\overline{\chi_2},\chi_1;z,-y,x)\\
&-\frac{\delta_{1,m}\delta_{1,n}\chi_2(-1)}{4}\sum_{k=1}^{cq}\chi_1(k)\delta_{\mathbb{Z}}\left(\frac{a(k+z)}{c}+x\right)\delta_{\mathbb{Z}}\left(\frac{b(k+z)}{cq}+y\right)\\
&\mathrel{\phantom{-}}\times\overline{\chi_2}\left(\frac{a(k+z)}{c}+x\right). \tag*{(12)}
\end{aligned} s m , n ( a , b , c ; χ 1 , χ 2 ; x , y , z ) = n ( c q b ) n − 1 j = 1 ∑ n − 1 ( j m ) m + n − j ( c a ) m − j s j , m + n − j ∗ ( − a , c , b ; χ 2 , χ 1 ; x , z , y ) + m ( c a ) m − 1 j = 0 ∑ n ( j n ) m + n − j ( − 1 ) j ( c q b ) n − j s m + n − j , j ( c , b , a ; χ 2 , χ 1 ; z , − y , x ) − 4 δ 1 , m δ 1 , n χ 2 ( − 1 ) k = 1 ∑ c q χ 1 ( k ) δ Z ( c a ( k + z ) + x ) δ Z ( c q b ( k + z ) + y ) − × χ 2 ( c a ( k + z ) + x ) . (12) In order to generalize Berndt’s reciprocity formulas (7) and (8) in the same spirit as Theorem 1.2 , we introduce the following three sums:
s m , n ( 1 ) ( a , b , c ; χ ; x , y , z ) , s m , n ( 2 ) ( a , b , c ; χ ; x , y , z ) , s m , n ( 3 ) ( a , b , c ; χ ; x , y , z ) , s^{(1)}_{m,n}(a,b,c;\chi;x,y,z),\quad s^{(2)}_{m,n}(a,b,c;\chi;x,y,z),\quad s^{(3)}_{m,n}(a,b,c;\chi;x,y,z), s m , n ( 1 ) ( a , b , c ; χ ; x , y , z ) , s m , n ( 2 ) ( a , b , c ; χ ; x , y , z ) , s m , n ( 3 ) ( a , b , c ; χ ; x , y , z ) , defined for a Dirichlet character χ \chi χ q q q q ∈ N q \in\mathbb{N} q ∈ N m , n ∈ N 0 m,n \in\mathbb{N}_0 m , n ∈ N 0 a , b , c ∈ Z a,b,c \in\mathbb{Z} a , b , c ∈ Z c ≥ 1 c \ge1 c ≥ 1 x , y , z ∈ R x,y,z \in\mathbb{R} x , y , z ∈ R
s m , n ( 1 ) ( a , b , c ; χ ; x , y , z ) = ∑ r = 1 c q χ ( r ) B ‾ m ( a ( r + z ) c + x ) B ‾ n ( b ( r + z ) c q + y ) , (13) s^{(1)}_{m,n}(a,b,c;\chi;x,y,z)=\sum_{r=1}^{cq}\chi(r)\overline{B}_m\left(\frac{a(r+z)}{c}+x\right)\overline{B}_n\left(\frac{b(r+z)}{cq}+y\right), \tag*{(13)} s m , n ( 1 ) ( a , b , c ; χ ; x , y , z ) = r = 1 ∑ c q χ ( r ) B m ( c a ( r + z ) + x ) B n ( c q b ( r + z ) + y ) , (13) s m , n ( 2 ) ( a , b , c ; χ ; x , y , z ) = ∑ r = 1 c q B ‾ m ( a ( r + z ) c + x , χ ) B ‾ n ( b ( r + z ) c q + y ) , (14) s^{(2)}_{m,n}(a,b,c;\chi;x,y,z)=\sum_{r=1}^{cq}\overline{B}_m\left(\frac{a(r+z)}{c}+x,\chi\right)\overline{B}_n\left(\frac{b(r+z)}{cq}+y\right), \tag*{(14)} s m , n ( 2 ) ( a , b , c ; χ ; x , y , z ) = r = 1 ∑ c q B m ( c a ( r + z ) + x , χ ) B n ( c q b ( r + z ) + y ) , (14) and
s m , n ( 3 ) ( a , b , c ; χ ; x , y , z ) = ∑ r = 1 c B ‾ m ( a q ( r + z ) c + x , χ ) B ‾ n ( b q ( r + z ) c + y ) . (15) s^{(3)}_{m,n}(a,b,c;\chi;x,y,z)=\sum_{r=1}^{c}\overline{B}_m\left(\frac{aq(r+z)}{c}+x,\chi\right)\overline{B}_n\left(\frac{bq(r+z)}{c}+y\right). \tag*{(15)} s m , n ( 3 ) ( a , b , c ; χ ; x , y , z ) = r = 1 ∑ c B m ( c a q ( r + z ) + x , χ ) B n ( c b q ( r + z ) + y ) . (15) Applying Theorem 1.1 , we obtain a reciprocity formula relating the sums (13) , (14) and (15) . This formula recovers (7) and (8) , as well as another reciprocity formulas of Berndt in [7 Corollary 8.5] , as special cases. The precise statement reads as follows.
Theorem 1.3. Let m , n , a , b , c , q ∈ N m,n,a,b,c,q \in\mathbb{N} m , n , a , b , c , q ∈ N q ≥ 2 q \ge2 q ≥ 2 χ \chi χ non-principal primitive Dirichlet character modulo q q q x , y , z ∈ R x,y,z \in\mathbb{R} x , y , z ∈ R
s m , n ( 2 ) ( a , b , c ; χ ; x , y , z ) = n ( b c q ) n − 1 ∑ j = 1 m ( m j ) ( a c ) m − j m + n − j s j , m + n − j ( 3 ) ( − a , c , b ; χ ; x , z , y ) + m ( a c ) m − 1 ∑ j = 0 n ( n j ) ( − 1 ) j ( b c q ) n − j m + n − j s m + n − j , j ( 1 ) ( c , b , a ; χ ; z , − y , x ) + ( − 1 ) n − 1 m ! n ! ( a , b ) m + n c χ ( b ( a , b ) ) ‾ B ‾ m + n ( b x − a q y ( a , b ) , χ ) a n q n − 1 b m ( m + n ) ! − δ 1 , m δ 1 , n χ ( − 1 ) 4 ∑ k = 1 c q δ Z ( a ( k + z ) c + x ) δ Z ( b ( k + z ) c q + y ) × χ ( a ( k + z ) c + x ) . (16) \begin{aligned}
s^{(2)}_{m,n}(a,b,c;\chi;x,y,z)
={}&n\left(\frac{b}{cq}\right)^{n-1}\sum_{j=1}^{m}\binom{m}{j}\frac{\left(\frac{a}{c}\right)^{m-j}}{m+n-j}s^{(3)}_{j,m+n-j}(-a,c,b;\chi;x,z,y)\\
&+m\left(\frac{a}{c}\right)^{m-1}\sum_{j=0}^{n}\binom{n}{j}\frac{(-1)^j\left(\frac{b}{cq}\right)^{n-j}}{m+n-j}s^{(1)}_{m+n-j,j}(c,b,a;\chi;z,-y,x)\\
&+\frac{(-1)^{n-1}m!n!(a,b)^{m+n}c\overline{\chi\left(\frac{b}{(a,b)}\right)}\overline{B}_{m+n}\left(\frac{bx-aqy}{(a,b)},\chi\right)}{a^nq^{n-1}b^m(m+n)!}\\
&-\frac{\delta_{1,m}\delta_{1,n}\chi(-1)}{4}\sum_{k=1}^{cq}\delta_{\mathbb{Z}}\left(\frac{a(k+z)}{c}+x\right)\delta_{\mathbb{Z}}\left(\frac{b(k+z)}{cq}+y\right)\\
&\qquad\times\chi\left(\frac{a(k+z)}{c}+x\right). \tag*{(16)}
\end{aligned} s m , n ( 2 ) ( a , b , c ; χ ; x , y , z ) = n ( c q b ) n − 1 j = 1 ∑ m ( j m ) m + n − j ( c a ) m − j s j , m + n − j ( 3 ) ( − a , c , b ; χ ; x , z , y ) + m ( c a ) m − 1 j = 0 ∑ n ( j n ) m + n − j ( − 1 ) j ( c q b ) n − j s m + n − j , j ( 1 ) ( c , b , a ; χ ; z , − y , x ) + a n q n − 1 b m ( m + n )! ( − 1 ) n − 1 m ! n ! ( a , b ) m + n c χ ( ( a , b ) b ) B m + n ( ( a , b ) b x − a q y , χ ) − 4 δ 1 , m δ 1 , n χ ( − 1 ) k = 1 ∑ c q δ Z ( c a ( k + z ) + x ) δ Z ( c q b ( k + z ) + y ) × χ ( c a ( k + z ) + x ) . (16) Finally, as a natural complement to Theorem 1.2 , we introduce two more sums s ^ m , n ( a , b , c ; χ 1 , χ 2 ; x , y , z ) \widehat{s}_{m,n}(a,b,c;\chi_1,\chi_2;x,y,z) s m , n ( a , b , c ; χ 1 , χ 2 ; x , y , z ) s ~ m , n ( a , b , c ; χ 1 , χ 2 ; x , y , z ) \widetilde{s}_{m,n}(a,b,c;\chi_1,\chi_2;x,y,z) s m , n ( a , b , c ; χ 1 , χ 2 ; x , y , z ) χ 1 \chi_1 χ 1 q 1 q_1 q 1 χ 2 \chi_2 χ 2 q 2 q_2 q 2 q 1 , q 2 ∈ N q_1,q_2 \in\mathbb{N} q 1 , q 2 ∈ N m , n ∈ N 0 m,n \in\mathbb{N}_0 m , n ∈ N 0 a , b , c ∈ Z a,b,c \in\mathbb{Z} a , b , c ∈ Z c ≥ 1 c \ge1 c ≥ 1 x , y , z ∈ R x,y,z \in\mathbb{R} x , y , z ∈ R
s ^ m , n ( a , b , c ; χ 1 , χ 2 ; x , y , z ) = ∑ r = 1 c q 1 q 2 χ 1 ( r ) B ‾ m ( a ( r + z ) c + x , χ 2 ) B ‾ n ( b ( r + z ) c q 1 q 2 + y ) , (17) \widehat{s}_{m,n}(a,b,c;\chi_1,\chi_2;x,y,z)
=\sum_{r=1}^{cq_1q_2}\chi_1(r)\overline{B}_m\left(\frac{a(r+z)}{c}+x,\chi_2\right)\overline{B}_n\left(\frac{b(r+z)}{cq_1q_2}+y\right), \tag*{(17)} s m , n ( a , b , c ; χ 1 , χ 2 ; x , y , z ) = r = 1 ∑ c q 1 q 2 χ 1 ( r ) B m ( c a ( r + z ) + x , χ 2 ) B n ( c q 1 q 2 b ( r + z ) + y ) , (17) and
s ~ m , n ( a , b , c ; χ 1 , χ 2 ; x , y , z ) = ∑ r = 1 c B ‾ m ( a q 1 q 2 ( r + z ) c + x , χ 1 ) B ‾ n ( b q 1 q 2 ( r + z ) c + y , χ 2 ) . (18) \widetilde{s}_{m,n}(a,b,c;\chi_1,\chi_2;x,y,z)
=\sum_{r=1}^{c}\overline{B}_m\left(\frac{aq_1q_2(r+z)}{c}+x,\chi_1\right)\overline{B}_n\left(\frac{bq_1q_2(r+z)}{c}+y,\chi_2\right). \tag*{(18)} s m , n ( a , b , c ; χ 1 , χ 2 ; x , y , z ) = r = 1 ∑ c B m ( c a q 1 q 2 ( r + z ) + x , χ 1 ) B n ( c b q 1 q 2 ( r + z ) + y , χ 2 ) . (18) Just as the sums (9) and (10) do, the sums (17) and (18) satisfy a reciprocity formula of the same type, stated in the following theorem.
Theorem 1.4. Let m , n , a , b , c , q 1 , q 2 ∈ N m,n,a,b,c,q_1,q_2 \in\mathbb{N} m , n , a , b , c , q 1 , q 2 ∈ N q 1 ≥ 2 q_1 \ge2 q 1 ≥ 2 q 2 ≥ 2 q_2 \ge2 q 2 ≥ 2 χ 1 \chi_1 χ 1 χ 2 \chi_2 χ 2 non-principal primitive Dirichlet characters modulo q 1 q_1 q 1 q 2 q_2 q 2 x , y , z ∈ R x,y,z \in\mathbb{R} x , y , z ∈ R
s ^ m , n ( a , b , c ; χ 1 , χ ‾ 2 ; x , y , z ) = n ( b c q 1 q 2 ) n − 1 ∑ j = 1 m ( m j ) ( a c ) m − j m + n − j s ~ j , m + n − j ( − a , c , b ; χ ‾ 2 , χ 1 ; x , z , y ) + m ( a c ) m − 1 ∑ j = 0 n ( n j ) ( − 1 ) j ( b c q 1 q 2 ) n − j m + n − j s ^ m + n − j , j ( c , b , a ; χ ‾ 2 , χ 1 ; z , − y , x ) − δ 1 , m δ 1 , n χ 2 ( − 1 ) 4 ∑ k = 1 c q 1 q 2 χ 1 ( k ) δ Z ( a ( k + z ) c + x ) δ Z ( b ( k + z ) c q 1 q 2 + y ) × χ ‾ 2 ( a ( k + z ) c + x ) . (19) \begin{aligned}
\widehat{s}_{m,n}(a,b,c;\chi_1,\overline{\chi}_2;x,y,z)
={}&n\left(\frac{b}{cq_1q_2}\right)^{n-1}\sum_{j=1}^{m}\binom{m}{j}\frac{\left(\frac{a}{c}\right)^{m-j}}{m+n-j}\widetilde{s}_{j,m+n-j}(-a,c,b;\overline{\chi}_2,\chi_1;x,z,y)\\
&+m\left(\frac{a}{c}\right)^{m-1}\sum_{j=0}^{n}\binom{n}{j}\frac{(-1)^j\left(\frac{b}{cq_1q_2}\right)^{n-j}}{m+n-j}\widehat{s}_{m+n-j,j}(c,b,a;\overline{\chi}_2,\chi_1;z,-y,x)\\
&-\frac{\delta_{1,m}\delta_{1,n}\chi_2(-1)}{4}\sum_{k=1}^{cq_1q_2}\chi_1(k)\delta_{\mathbb{Z}}\left(\frac{a(k+z)}{c}+x\right)\delta_{\mathbb{Z}}\left(\frac{b(k+z)}{cq_1q_2}+y\right)\\
&\qquad\times\overline{\chi}_2\left(\frac{a(k+z)}{c}+x\right). \tag*{(19)}
\end{aligned} s m , n ( a , b , c ; χ 1 , χ 2 ; x , y , z ) = n ( c q 1 q 2 b ) n − 1 j = 1 ∑ m ( j m ) m + n − j ( c a ) m − j s j , m + n − j ( − a , c , b ; χ 2 , χ 1 ; x , z , y ) + m ( c a ) m − 1 j = 0 ∑ n ( j n ) m + n − j ( − 1 ) j ( c q 1 q 2 b ) n − j s m + n − j , j ( c , b , a ; χ 2 , χ 1 ; z , − y , x ) − 4 δ 1 , m δ 1 , n χ 2 ( − 1 ) k = 1 ∑ c q 1 q 2 χ 1 ( k ) δ Z ( c a ( k + z ) + x ) δ Z ( c q 1 q 2 b ( k + z ) + y ) × χ 2 ( c a ( k + z ) + x ) . (19) This paper is organized as follows. In Section 2, we give some preliminary results. Section 3 is devoted to the detailed proofs of Theorems 1.1, 1.2, 1.3 and 1.4. In Section 4, we discuss some special cases of Theorems 1.2, 1.3 and 1.4. We show that Berndt’s reciprocity formulas (1.4), (1.7) and (1.8), as well as the reciprocity formulas in [7 , 9 , 11 ] , hold as special cases.
Some preliminary results
We first recall the following definitions of the generalized Apostol-Bernoulli polynomials and the generalized Apostol-Bernoulli functions, given in [15 Definitions 2.1 and 2.7] .
Definition 2.1. Let n ∈ N 0 n \in\mathbb{N}_0 n ∈ N 0 q ∈ N q \in\mathbb{N} q ∈ N λ ∈ C ∖ { 0 } \lambda\in\mathbb{C} \setminus\{0\} λ ∈ C ∖ { 0 } χ \chi χ q q q β n , χ ( x , λ ) \beta_{n,\chi}(x,\lambda) β n , χ ( x , λ )
∑ r = 1 q χ ( r ) λ r t e ( r + x ) t λ q e q t − 1 = ∑ n = 0 ∞ β n , χ ( x , λ ) t n n ! ( ∣ t + log λ ∣ < 2 π / q ) . (20) \sum_{r=1}^{q} \frac{\chi(r)\lambda^{r}t e^{(r+x)t}}{\lambda^{q}e^{qt}-1} = \sum_{n=0}^{\infty} \beta_{n,\chi}(x,\lambda)\frac{t^{n}}{n!} \qquad(|t+\log\lambda|<2\pi/q). \tag*{(20)} r = 1 ∑ q λ q e q t − 1 χ ( r ) λ r t e ( r + x ) t = n = 0 ∑ ∞ β n , χ ( x , λ ) n ! t n ( ∣ t + log λ ∣ < 2 π / q ) . (20) where λ z \lambda^{z} λ z λ z = e z log λ \lambda^{z}=e^{z\log\lambda} λ z = e z l o g λ z ∈ C z \in\mathbb{C} z ∈ C
log λ = log ∣ λ ∣ + i arg λ ( − π < arg λ ≤ π ) . \log\lambda= \log|\lambda| + i\arg\lambda\qquad(-\pi<\arg\lambda\le\pi). log λ = log ∣ λ ∣ + i arg λ ( − π < arg λ ≤ π ) . In particular, β n , χ ( λ ) = β n , χ ( 0 , λ ) \beta_{n,\chi}(\lambda)=\beta_{n,\chi}(0,\lambda) β n , χ ( λ ) = β n , χ ( 0 , λ )
Definition 2.2. Let n ∈ N 0 n \in\mathbb{N}_0 n ∈ N 0 q ∈ N q \in\mathbb{N} q ∈ N x ∈ R x \in\mathbb{R} x ∈ R λ ∈ C ∖ { 0 } \lambda\in\mathbb{C} \setminus\{0\} λ ∈ C ∖ { 0 } χ \chi χ q q q β ‾ n , χ ( x , λ ) \overline{\beta}_{n,\chi}(x,\lambda) β n , χ ( x , λ )
β ‾ n , χ ( x , λ ) = β n , χ ( { x } χ , λ ) + 1 2 δ 1 , n δ Z ( x ) λ − x χ ( − x ) . (21) \overline{\beta}_{n,\chi}(x,\lambda)=\beta_{n,\chi}(\{x\}_{\chi},\lambda)+\frac{1}{2}\delta_{1,n}\delta_{\mathbb{Z}}(x)\lambda^{-x}\chi(-x). \tag*{(21)} β n , χ ( x , λ ) = β n , χ ({ x } χ , λ ) + 2 1 δ 1 , n δ Z ( x ) λ − x χ ( − x ) . (21) where
β n , χ ( { x } χ , λ ) = q n − 1 ∑ k = 1 q χ ( k ) λ k − q ⌈ x + k q ⌉ β n ( { x + k q } , λ q ) , \beta_{n,\chi}(\{x\}_{\chi},\lambda)=q^{n-1}\sum_{k=1}^{q}\chi(k)\lambda^{k-q\left\lceil\frac{x+k}{q}\right\rceil}\beta_n\left(\left\{\frac{x+k}{q}\right\},\lambda^q\right), β n , χ ({ x } χ , λ ) = q n − 1 k = 1 ∑ q χ ( k ) λ k − q ⌈ q x + k ⌉ β n ( { q x + k } , λ q ) , and β n ( x , λ ) \beta_n(x,\lambda) β n ( x , λ ) q = 1 q=1 q = 1 λ β n ( x , λ ) = β n , χ ( x − 1 , λ ) \lambda\beta_n(x,\lambda)=\beta_{n,\chi}(x-1,\lambda) λ β n ( x , λ ) = β n , χ ( x − 1 , λ )
It should be noted that Definition 2.1 includes information on the Bernoulli, Euler, Apostol-Bernoulli, Apostol-Euler, generalized Bernoulli, and generalized Euler polynomials, while Definition 2.2 includes information on the corresponding functions. It is also shown in [15 ] that the generalized Apostol-Bernoulli functions have the following Fourier series.
Lemma 2.3. Let n , q ∈ N n,q \in\mathbb{N} n , q ∈ N x ∈ R x \in\mathbb{R} x ∈ R λ ∈ C ∖ { 0 } \lambda\in\mathbb{C} \setminus\{0\} λ ∈ C ∖ { 0 } χ \chi χ q q q
β ‾ n , χ ( x , λ ) = − q n − 1 n ! λ x ( 2 π i ) n ∑ k = − ∞ + ∞ G ( k , χ ) e 2 π i k x q ( k − q log λ 2 π i ) n . (22) \overline{\beta}_{n,\chi}(x,\lambda)=-\frac{q^{n-1}n!}{\lambda^{x}(2\pi i)^{n}}\sum_{k=-\infty}^{+\infty}\frac{G(k,\chi)e^{\frac{2\pi ikx}{q}}}{\left(k-\frac{q\log\lambda}{2\pi i}\right)^{n}}. \tag*{(22)} β n , χ ( x , λ ) = − λ x ( 2 π i ) n q n − 1 n ! k = − ∞ ∑ + ∞ ( k − 2 π i q l o g λ ) n G ( k , χ ) e q 2 π ik x . (22) where the dash denotes throughout that undefined terms are excluded from the sum, and G ( k , χ ) G(k,\chi) G ( k , χ )
G ( k , χ ) = ∑ r = 1 q χ ( r ) e 2 π i r k q ( k ∈ Z ) . G(k,\chi)=\sum_{r=1}^{q}\chi(r)e^{\frac{2\pi irk}{q}} \qquad(k\in\mathbb{Z}). G ( k , χ ) = r = 1 ∑ q χ ( r ) e q 2 π i r k ( k ∈ Z ) . Proof. See [15 Theorem 3.1] for details. □
Clearly, for a primitive character χ \chi χ q q q q ∈ N q \in\mathbb{N} q ∈ N [1 Theorem 8.15] ):
G ( k , χ ) = χ ‾ ( k ) G ( 1 , χ ) ( k ∈ Z ) . (23) G(k,\chi)=\overline{\chi}(k)G(1,\chi)\qquad(k\in\mathbb{Z}). \tag*{(23)} G ( k , χ ) = χ ( k ) G ( 1 , χ ) ( k ∈ Z ) . (23) Hence, taking λ = 1 \lambda=1 λ = 1 (23) , we obtain that, for a primitive Dirichlet character χ \chi χ q q q q ∈ N q\in\mathbb{N} q ∈ N n ∈ N n\in\mathbb{N} n ∈ N x ∈ R x\in\mathbb{R} x ∈ R
B ‾ n ( x , χ ‾ ) = − q n − 1 n ! G ( 1 , χ ‾ ) ( 2 π i ) n ∑ k = − ∞ + ∞ χ ( k ) e 2 π i k x q k n . (24) \overline{B}_n(x,\overline{\chi})=-\frac{q^{n-1}n!G(1,\overline{\chi})}{(2\pi i)^n}\sum_{k=-\infty}^{+\infty}\frac{\chi(k)e^{\frac{2\pi ikx}{q}}}{k^n}. \tag*{(24)} B n ( x , χ ) = − ( 2 π i ) n q n − 1 n ! G ( 1 , χ ) k = − ∞ ∑ + ∞ k n χ ( k ) e q 2 π ik x . (24) In particular, the case q = 1 q=1 q = 1 (24) gives the Fourier series of the Bernoulli functions:
B ‾ n ( x ) = − n ! ( 2 π i ) n ∑ k = − ∞ + ∞ e 2 π i k x k n ( n ∈ N , x ∈ R ) . (25) \overline{B}_n(x)=-\frac{n!}{(2\pi i)^n}\sum_{k=-\infty}^{+\infty}\frac{e^{2\pi ikx}}{k^n}\qquad(n\in\mathbb{N},x\in\mathbb{R}). \tag*{(25)} B n ( x ) = − ( 2 π i ) n n ! k = − ∞ ∑ + ∞ k n e 2 π ik x ( n ∈ N , x ∈ R ) . (25) It is worth noting that (25) implies
B ‾ n ( − x ) = ( − 1 ) n B ‾ n ( x ) ( n ∈ N , x ∈ R ) . (26) \overline{B}_n(-x)=(-1)^n\overline{B}_n(x)\qquad(n\in\mathbb{N},x\in\mathbb{R}). \tag*{(26)} B n ( − x ) = ( − 1 ) n B n ( x ) ( n ∈ N , x ∈ R ) . (26) Moreover, for a Dirichlet character χ \chi χ q q q q ∈ N q\in\mathbb{N} q ∈ N n ∈ N n\in\mathbb{N} n ∈ N x ∈ R x\in\mathbb{R} x ∈ R
q n − 1 ∑ r = 1 q χ ‾ ( r ) B ‾ n ( x + r q ) = B ‾ n ( x , χ ‾ ) . (27) q^{n-1}\sum_{r=1}^{q}\overline{\chi}(r)\overline{B}_n\left(\frac{x+r}{q}\right)=\overline{B}_n(x,\overline{\chi}). \tag*{(27)} q n − 1 r = 1 ∑ q χ ( r ) B n ( q x + r ) = B n ( x , χ ) . (27) The next result appears in [14 Lemma 2.7] . Here we provide an alternative proof.
Lemma 2.4. Let j ∈ N j\in\mathbb{N} j ∈ N b , r ∈ Z b,r\in\mathbb{Z} b , r ∈ Z b ≠ 0 b\ne0 b = 0 x ∈ R x\in\mathbb{R} x ∈ R
∑ d = − ∞ + ∞ e 2 π i d x ( d + r b ) j = ( − 1 ) j − 1 ( 2 π i ) j sgn ( b ) j ! ∣ b ∣ 1 − j ∑ l = 1 ∣ b ∣ e 2 π i ( l − x ) r b B ‾ j ( l − x b ) . (28) \sum_{d=-\infty}^{+\infty}\frac{e^{2\pi idx}}{\left(d+\frac{r}{b}\right)^j}=\frac{(-1)^{j-1}(2\pi i)^j\operatorname{sgn}(b)}{j!|b|^{1-j}}\sum_{l=1}^{|b|}e^{\frac{2\pi i(l-x)r}{b}}\overline{B}_j\left(\frac{l-x}{b}\right). \tag*{(28)} d = − ∞ ∑ + ∞ ( d + b r ) j e 2 π i d x = j ! ∣ b ∣ 1 − j ( − 1 ) j − 1 ( 2 π i ) j sgn ( b ) l = 1 ∑ ∣ b ∣ e b 2 π i ( l − x ) r B j ( b l − x ) . (28) Proof. We only prove the case b > 0 b>0 b > 0 b < 0 b<0 b < 0 r = b q + s r=bq+s r = b q + s q , s ∈ Z q,s\in\mathbb{Z} q , s ∈ Z 0 ≤ s ≤ b − 1 0\le s\le b-1 0 ≤ s ≤ b − 1
∑ d = − ∞ + ∞ e 2 π i d x ( d + r b ) j = e − 2 π i q x ∑ d = − ∞ + ∞ e 2 π i d x ( d + s b ) j . \sum_{d=-\infty}^{+\infty}\frac{e^{2\pi idx}}{\left(d+\frac{r}{b}\right)^j}=e^{-2\pi iqx}\sum_{d=-\infty}^{+\infty}\frac{e^{2\pi idx}}{\left(d+\frac{s}{b}\right)^j}. d = − ∞ ∑ + ∞ ( d + b r ) j e 2 π i d x = e − 2 π i q x d = − ∞ ∑ + ∞ ( d + b s ) j e 2 π i d x . Thus, it suffices to prove the case 0 ≤ r ≤ b − 1 0\le r\le b-1 0 ≤ r ≤ b − 1 (28) . It is easy to see that
arg e 2 π i r b = { 2 π r b , if 0 ≤ r ≤ b / 2 , 2 π r b − 2 π , if b / 2 < r ≤ b − 1. (29) \arg e^{\frac{2\pi ir}{b}}=
\begin{cases}
\frac{2\pi r}{b}, & \text{if }0\le r\le b/2,\\
\frac{2\pi r}{b}-2\pi, & \text{if }b/2<r\le b-1.
\end{cases} \tag*{(29)} arg e b 2 π i r = { b 2 π r , b 2 π r − 2 π , if 0 ≤ r ≤ b /2 , if b /2 < r ≤ b − 1. (29) Setting q = 1 q=1 q = 1 λ = e 2 π i r b \lambda=e^{\frac{2\pi ir}{b}} λ = e b 2 π i r (29) that
B ‾ j ( x , e 2 π i r b ) = − j ! e − 2 π i r x b ( 2 π i ) j ∑ d = − ∞ + ∞ e 2 π i d x ( d − r b ) j = − ( − 1 ) j j ! e − 2 π i r x b ( 2 π i ) j ∑ d = − ∞ + ∞ e − 2 π i d x ( d + r b ) j . (30) \begin{aligned}
\overline{B}_j\left(x,e^{\frac{2\pi ir}{b}}\right)
=-\frac{j!e^{-\frac{2\pi irx}{b}}}{(2\pi i)^j}\sum_{d=-\infty}^{+\infty}\frac{e^{2\pi idx}}{\left(d-\frac{r}{b}\right)^j} \\
&=-\frac{(-1)^jj!e^{-\frac{2\pi irx}{b}}}{(2\pi i)^j}\sum_{d=-\infty}^{+\infty}\frac{e^{-2\pi idx}}{\left(d+\frac{r}{b}\right)^j}. \tag*{(30)}
\end{aligned} B j ( x , e b 2 π i r ) = − ( 2 π i ) j j ! e − b 2 π i r x d = − ∞ ∑ + ∞ ( d − b r ) j e 2 π i d x = − ( 2 π i ) j ( − 1 ) j j ! e − b 2 π i r x d = − ∞ ∑ + ∞ ( d + b r ) j e − 2 π i d x . (30) where β ‾ j ( x , λ ) \overline{\beta}_j(x,\lambda) β j ( x , λ ) q = 1 q=1 q = 1 β ‾ j ( x , λ ) = β ‾ j , χ ( x , λ ) \overline{\beta}_j(x,\lambda)=\overline{\beta}_{j,\chi}(x,\lambda) β j ( x , λ ) = β j , χ ( x , λ ) [15 Corollary 3.9] that, for a , b , j ∈ N a,b,j\in\mathbb{N} a , b , j ∈ N λ , μ ∈ C ∖ { 0 } \lambda,\mu\in\mathbb{C}\setminus\{0\} λ , μ ∈ C ∖ { 0 } λ a = μ b \lambda^a=\mu^b λ a = μ b
a j − 1 ∑ l = 0 a − 1 λ l β ‾ j ( b x + b l a , μ ) = b j − 1 ∑ l = 0 b − 1 μ l β ‾ j ( a x + a l b , λ ) . (31) a^{j-1}\sum_{l=0}^{a-1}\lambda^l\overline{\beta}_j\left(bx+\frac{bl}{a},\mu\right)=b^{j-1}\sum_{l=0}^{b-1}\mu^l\overline{\beta}_j\left(ax+\frac{al}{b},\lambda\right). \tag*{(31)} a j − 1 l = 0 ∑ a − 1 λ l β j ( b x + a b l , μ ) = b j − 1 l = 0 ∑ b − 1 μ l β j ( a x + b a l , λ ) . (31) Taking a = λ = 1 a=\lambda=1 a = λ = 1 μ = e 2 π i r b \mu=e^{\frac{2\pi ir}{b}} μ = e b 2 π i r x x x x / b x/b x / b (31) , we obtain
β ‾ j ( x , e 2 π i r b ) = b j − 1 ∑ l = 0 b − 1 e 2 π i r l b B ‾ j ( x + l b ) = b j − 1 ∑ l = 1 b e 2 π i r l b B ‾ j ( x + l b ) . (32) \begin{aligned}
\overline{\beta}_j\left(x,e^{\frac{2\pi ir}{b}}\right)=b^{j-1}\sum_{l=0}^{b-1}e^{\frac{2\pi irl}{b}}\overline{B}_j\left(\frac{x+l}{b}\right) \\
&=b^{j-1}\sum_{l=1}^{b}e^{\frac{2\pi irl}{b}}\overline{B}_j\left(\frac{x+l}{b}\right). \tag*{(32)}
\end{aligned} β j ( x , e b 2 π i r ) = b j − 1 l = 0 ∑ b − 1 e b 2 π i r l B j ( b x + l ) = b j − 1 l = 1 ∑ b e b 2 π i r l B j ( b x + l ) . (32) Therefore, combining (2.11) and (2.13), we conclude that the case 0 ≤ r ≤ b − 1 0\le r\le b-1 0 ≤ r ≤ b − 1
We remark that taking r = 0 r=0 r = 0 b b b a a a [17 ] : for a , j ∈ N a,j\in\mathbb{N} a , j ∈ N x ∈ R x\in\mathbb{R} x ∈ R
a j − 1 ∑ l = 0 a − 1 B ‾ j ( x + l a ) = B ‾ j ( x ) . (33) a^{j-1}\sum_{l=0}^{a-1}\overline{B}_j\left(\frac{x+l}{a}\right)=\overline{B}_j(x). \tag*{(33)} a j − 1 l = 0 ∑ a − 1 B j ( a x + l ) = B j ( x ) . (33) It follows from Lemma 2.4 that the following result holds.
Lemma 2.5. Let j , q ∈ N j,q\in\mathbb{N} j , q ∈ N y , z ∈ R y,z\in\mathbb{R} y , z ∈ R a , b , a ˉ , b ˉ , k ∈ Z a,b,\bar{a},\bar{b},k\in\mathbb{Z} a , b , a ˉ , b ˉ , k ∈ Z a ≠ 0 a\ne0 a = 0 b ≠ 0 b\ne0 b = 0 a a ˉ + b b ˉ = ( a , b ) a\bar{a}+b\bar{b}=(a,b) a a ˉ + b b ˉ = ( a , b ) ( a , b ) ∣ k (a,b)\mid k ( a , b ) ∣ k χ \chi χ q q q
∑ d = − ∞ + ∞ χ ( k a ˉ ( a , b ) + b d ( a , b ) ) e 2 π i d ( b y − a q z ) ( a , b ) q ( d + k a ˉ b ˉ ) j = ( − 1 ) j − 1 ( 2 π i ) j sgn ( b ) ( a , b ) 1 − j G ( 1 , χ ) q j ! b 1 − j ∑ r = 1 q χ ‾ ( r ) \sum_{d=-\infty}^{+\infty}\frac{\chi\left(\frac{k\bar{a}}{(a,b)}+\frac{bd}{(a,b)}\right)e^{\frac{2\pi i d(by-aqz)}{(a,b)q}}}{\left(d+\frac{k\bar{a}}{\bar{b}}\right)^j}
=
\frac{(-1)^{j-1}(2\pi i)^j\operatorname{sgn}(b)(a,b)^{1-j}G(1,\chi)}{qj!b^{1-j}}
\sum_{r=1}^{q}\overline{\chi}(r) d = − ∞ ∑ + ∞ ( d + b ˉ k a ˉ ) j χ ( ( a , b ) k a ˉ + ( a , b ) b d ) e ( a , b ) q 2 π i d ( b y − a q z ) = q j ! b 1 − j ( − 1 ) j − 1 ( 2 π i ) j sgn ( b ) ( a , b ) 1 − j G ( 1 , χ ) r = 1 ∑ q χ ( r ) × ∑ l = 1 ∣ b ˉ ( a , b ) ∣ e 2 π i ( l − b y − a q z ( a , b ) q ) k a ˉ b B ‾ j ( ( a , b ) l − b y − b r − a q z q b ) , (34) \times\sum_{l=1}^{\left|\frac{\bar{b}}{(a,b)}\right|}e^{\frac{2\pi i\left(l-\frac{by-aqz}{(a,b)q}\right)k\bar{a}}{b}}\overline{B}_j\left(\frac{(a,b)l-\frac{by-br-aqz}{q}}{b}\right), \tag*{(34)} × l = 1 ∑ ∣ ( a , b ) b ˉ ∣ e b 2 π i ( l − ( a , b ) q b y − a q z ) k a ˉ B j ( b ( a , b ) l − q b y − b r − a q z ) , (34) and
∑ d = − ∞ + ∞ χ ( k a ˉ ( a , b ) + b d ( a , b ) ) e 2 π i d ( b y − a q z ) ( a , b ) q ( d − k b ˉ a ) j = ( − 1 ) j − 1 ( 2 π i ) j sgn ( a ) ( a , b ) 1 − j G ( 1 , χ ) q j ! a 1 − j ∑ r = 1 q χ ‾ ( r ) e − 2 π i k b ˉ r a q \sum_{d=-\infty}^{+\infty}\frac{\chi\left(\frac{k\bar{a}}{(a,b)}+\frac{bd}{(a,b)}\right)e^{\frac{2\pi i d(by-aqz)}{(a,b)q}}}{\left(d-\frac{k\bar{b}}{a}\right)^j}
=
\frac{(-1)^{j-1}(2\pi i)^j\operatorname{sgn}(a)(a,b)^{1-j}G(1,\chi)}{qj!a^{1-j}}
\sum_{r=1}^{q}\overline{\chi}(r)e^{-\frac{2\pi i k\bar{b}r}{aq}} d = − ∞ ∑ + ∞ ( d − a k b ˉ ) j χ ( ( a , b ) k a ˉ + ( a , b ) b d ) e ( a , b ) q 2 π i d ( b y − a q z ) = q j ! a 1 − j ( − 1 ) j − 1 ( 2 π i ) j sgn ( a ) ( a , b ) 1 − j G ( 1 , χ ) r = 1 ∑ q χ ( r ) e − a q 2 π ik b ˉ r × ∑ l = 1 ∣ a ˉ ( a , b ) ∣ e − 2 π i ( l − b y − a q z ( a , b ) q ) k b ˉ a B ‾ j ( ( a , b ) l − b y − b r − a q z q a ) . (35) \times\sum_{l=1}^{\left|\frac{\bar{a}}{(a,b)}\right|}e^{-\frac{2\pi i\left(l-\frac{by-aqz}{(a,b)q}\right)k\bar{b}}{a}}\overline{B}_j\left(\frac{(a,b)l-\frac{by-br-aqz}{q}}{a}\right). \tag*{(35)} × l = 1 ∑ ∣ ( a , b ) a ˉ ∣ e − a 2 π i ( l − ( a , b ) q b y − a q z ) k b ˉ B j ( a ( a , b ) l − q b y − b r − a q z ) . (35) Proof. It is well known that the finite Fourier expansion of a primitive Dirichlet character χ \chi χ q q q q ∈ N q \in\mathbb{N} q ∈ N [1 Theorem 8.20] ):
χ ( m ) = G ( 1 , χ ) q ∑ r = 1 q χ ‾ ( r ) e − 2 π i m r q ( m ∈ N ) . (36) \chi(m)=\frac{G(1,\chi)}{q}\sum_{r=1}^{q}\overline{\chi}(r)e^{-\frac{2\pi imr}{q}}\quad(m\in\mathbb{N}). \tag*{(36)} χ ( m ) = q G ( 1 , χ ) r = 1 ∑ q χ ( r ) e − q 2 π im r ( m ∈ N ) . (36) From Lemma 2.4 and (36) , it follows that
∑ d = − ∞ + ∞ χ ( k a ‾ ( a , b ) + b d ( a , b ) ) e 2 π i d ( b y − a q z ) ( a , b ) q ( d + k a ‾ b ) j = G ( 1 , χ ) q ∑ r = 1 q χ ‾ ( r ) e − 2 π i k a r ( a , b ) q ∑ d = − ∞ + ∞ e 2 π i d ( b y − b r − a q z ) ( a , b ) q ( d + k a ‾ b ) j = ( − 1 ) j − 1 ( 2 π i ) j sgn ( b ) ( a , b ) 1 − j G ( 1 , χ ) q j ! b 1 − j ∑ r = 1 q χ ‾ ( r ) × ∑ l = 1 ∣ b ( a , b ) ∣ e 2 π i ( l − b y − a q z ( a , b ) q ) k a ‾ b B ‾ j ( ( a , b ) l − b y − b r − a q z q b ) , \begin{aligned}
\sum_{d=-\infty}^{+\infty}\frac{\chi\left(\frac{k\overline{a}}{(a,b)}+\frac{bd}{(a,b)}\right)e^{\frac{2\pi id(by-aqz)}{(a,b)q}}}{\left(d+\frac{k\overline{a}}{b}\right)^j}
&=\frac{G(1,\chi)}{q}\sum_{r=1}^{q}\overline{\chi}(r)e^{-\frac{2\pi ikar}{(a,b)q}}\sum_{d=-\infty}^{+\infty}\frac{e^{\frac{2\pi id(by-br-aqz)}{(a,b)q}}}{\left(d+\frac{k\overline{a}}{b}\right)^j}\\
&=\frac{(-1)^{j-1}(2\pi i)^j\operatorname{sgn}(b)(a,b)^{1-j}G(1,\chi)}{qj!b^{1-j}}\sum_{r=1}^{q}\overline{\chi}(r)\\
&\quad\times\sum_{l=1}^{\left|\frac{b}{(a,b)}\right|}e^{\frac{2\pi i\left(l-\frac{by-aqz}{(a,b)q}\right)k\overline{a}}{b}}\overline{B}_j\left(\frac{(a,b)l-\frac{by-br-aqz}{q}}{b}\right),
\end{aligned} d = − ∞ ∑ + ∞ ( d + b k a ) j χ ( ( a , b ) k a + ( a , b ) b d ) e ( a , b ) q 2 π i d ( b y − a q z ) = q G ( 1 , χ ) r = 1 ∑ q χ ( r ) e − ( a , b ) q 2 π ik a r d = − ∞ ∑ + ∞ ( d + b k a ) j e ( a , b ) q 2 π i d ( b y − b r − a q z ) = q j ! b 1 − j ( − 1 ) j − 1 ( 2 π i ) j sgn ( b ) ( a , b ) 1 − j G ( 1 , χ ) r = 1 ∑ q χ ( r ) × l = 1 ∑ ∣ ( a , b ) b ∣ e b 2 π i ( l − ( a , b ) q b y − a q z ) k a B j ( b ( a , b ) l − q b y − b r − a q z ) , and
∑ d = − ∞ + ∞ χ ( k a ‾ ( a , b ) + b d ( a , b ) ) e 2 π i d ( b y − a q z ) ( a , b ) q ( d − k b ‾ a ) j = G ( 1 , χ ) q ∑ r = 1 q χ ‾ ( r ) e − 2 π i k a ‾ r ( a , b ) q ∑ d = − ∞ + ∞ e 2 π i d ( b y − b r − a q z ) ( a , b ) q ( d − k b ‾ a ) j = ( − 1 ) j − 1 ( 2 π i ) j sgn ( a ) ( a , b ) 1 − j G ( 1 , χ ) q j ! a 1 − j ∑ r = 1 q χ ‾ ( r ) e − 2 π i k a ‾ r a q × ∑ l = 1 ∣ a ( a , b ) ∣ e − 2 π i ( l − b y − a q z ( a , b ) q ) k b ‾ a B ‾ j ( ( a , b ) l − b y − b r − a q z q a ) , \begin{aligned}
\sum_{d=-\infty}^{+\infty}\frac{\chi\left(\frac{k\overline{a}}{(a,b)}+\frac{bd}{(a,b)}\right)e^{\frac{2\pi id(by-aqz)}{(a,b)q}}}{\left(d-\frac{k\overline{b}}{a}\right)^j}
&=\frac{G(1,\chi)}{q}\sum_{r=1}^{q}\overline{\chi}(r)e^{-\frac{2\pi ik\overline{a}r}{(a,b)q}}\sum_{d=-\infty}^{+\infty}\frac{e^{\frac{2\pi id(by-br-aqz)}{(a,b)q}}}{\left(d-\frac{k\overline{b}}{a}\right)^j}\\
&=\frac{(-1)^{j-1}(2\pi i)^j\operatorname{sgn}(a)(a,b)^{1-j}G(1,\chi)}{qj!a^{1-j}}\sum_{r=1}^{q}\overline{\chi}(r)e^{-\frac{2\pi ik\overline{a}r}{aq}}\\
&\quad\times\sum_{l=1}^{\left|\frac{a}{(a,b)}\right|}e^{-\frac{2\pi i\left(l-\frac{by-aqz}{(a,b)q}\right)k\overline{b}}{a}}\overline{B}_j\left(\frac{(a,b)l-\frac{by-br-aqz}{q}}{a}\right),
\end{aligned} d = − ∞ ∑ + ∞ ( d − a k b ) j χ ( ( a , b ) k a + ( a , b ) b d ) e ( a , b ) q 2 π i d ( b y − a q z ) = q G ( 1 , χ ) r = 1 ∑ q χ ( r ) e − ( a , b ) q 2 π ik a r d = − ∞ ∑ + ∞ ( d − a k b ) j e ( a , b ) q 2 π i d ( b y − b r − a q z ) = q j ! a 1 − j ( − 1 ) j − 1 ( 2 π i ) j sgn ( a ) ( a , b ) 1 − j G ( 1 , χ ) r = 1 ∑ q χ ( r ) e − a q 2 π ik a r × l = 1 ∑ ∣ ( a , b ) a ∣ e − a 2 π i ( l − ( a , b ) q b y − a q z ) k b B j ( a ( a , b ) l − q b y − b r − a q z ) , as desired. □
Lemma 2.6. Let m , n ∈ N m,n\in\mathbb{N} m , n ∈ N d , x , y ∈ R d,x,y\in\mathbb{R} d , x , y ∈ R
1 ( d − x ) m ( d − y ) n = ∑ j = 1 m ( m + n − j − 1 n − 1 ) ( − 1 ) m − j ( x − y ) m + n − j ( d − x ) j + ∑ j = 1 n ( m + n − j − 1 m − 1 ) ( − 1 ) n − j ( y − x ) m + n − j ( d − y ) j . (37) \frac{1}{(d-x)^m(d-y)^n}
=\sum_{j=1}^{m}\binom{m+n-j-1}{n-1}\frac{(-1)^{m-j}}{(x-y)^{m+n-j}(d-x)^j}
+\sum_{j=1}^{n}\binom{m+n-j-1}{m-1}\frac{(-1)^{n-j}}{(y-x)^{m+n-j}(d-y)^j}. \tag*{(37)} ( d − x ) m ( d − y ) n 1 = j = 1 ∑ m ( n − 1 m + n − j − 1 ) ( x − y ) m + n − j ( d − x ) j ( − 1 ) m − j + j = 1 ∑ n ( m − 1 m + n − j − 1 ) ( y − x ) m + n − j ( d − y ) j ( − 1 ) n − j . (37) Proof. See [14 Lemma 2.2] for details. □
Lemma 2.7. Let m , n ∈ N m,n\in\mathbb{N} m , n ∈ N
∑ j = 1 m ( m + n − j − 1 n − 1 ) = ( m + n − 1 n ) . (38) \sum_{j=1}^{m}\binom{m+n-j-1}{n-1}=\binom{m+n-1}{n}. \tag*{(38)} j = 1 ∑ m ( n − 1 m + n − j − 1 ) = ( n m + n − 1 ) . (38) Proof. See [14 Lemma 2.8] for details. □
Proof of Theorems
The proof of Theorem 1.1. Let
f ( x ) = B ‾ m ( a x + y , χ ‾ ) B ‾ n ( b x q + z ) . f(x)=\overline{B}_m(ax+y,\overline{\chi})\overline{B}_n\left(\frac{bx}{q}+z\right). f ( x ) = B m ( a x + y , χ ) B n ( q b x + z ) . Since f ( x ) f(x) f ( x ) f ( x ) f(x) f ( x )
f ( x + ) + f ( x − ) 2 = ∑ k = − ∞ + ∞ c k ( f ) e 2 π i k x q , (39) \frac{f(x^+)+f(x^-)}{2}=\sum_{k=-\infty}^{+\infty}c_k(f)e^{\frac{2\pi ikx}{q}}, \tag*{(39)} 2 f ( x + ) + f ( x − ) = k = − ∞ ∑ + ∞ c k ( f ) e q 2 π ik x , (39) where the Fourier coefficients c k ( f ) c_k(f) c k ( f )
c k ( f ) = 1 q ∫ 0 q B ‾ m ( a x + y , χ ‾ ) B ‾ n ( b x q + z ) e − 2 π i k x q d x . (40) c_k(f)=\frac{1}{q}\int_0^q\overline{B}_m(ax+y,\overline{\chi})\overline{B}_n\left(\frac{bx}{q}+z\right)e^{-\frac{2\pi ikx}{q}}\,dx. \tag*{(40)} c k ( f ) = q 1 ∫ 0 q B m ( a x + y , χ ) B n ( q b x + z ) e − q 2 π ik x d x . (40) Using Parseval’s formula from [10 Proposition 3.1.10] or [18 p. 81] , together with (2.5) and (2.6), we may rewrite (3.2) as
c k ( f ) = q m − 1 m ! n ! G ( 1 , χ ‾ ) ( 2 π i ) m + n ∑ l = − ∞ + ∞ ∑ j = − ∞ a l + b j = k + ∞ χ ( l ) e 2 π i ( l y q + j z ) l m j n . (41) c_k(f)=\frac{q^{m-1}m!n!G(1,\overline{\chi})}{(2\pi i)^{m+n}}\sum_{l=-\infty}^{+\infty}\sum_{\substack{j=-\infty\\ al+bj=k}}^{+\infty}\frac{\chi(l)e^{2\pi i\left(\frac{ly}{q}+jz\right)}}{l^m j^n}. \tag*{(41)} c k ( f ) = ( 2 π i ) m + n q m − 1 m ! n ! G ( 1 , χ ) l = − ∞ ∑ + ∞ j = − ∞ a l + bj = k ∑ + ∞ l m j n χ ( l ) e 2 π i ( q l y + j z ) . (41) Since the Diophantine equation a x + b y = k ax+by=k a x + b y = k ( a , b ) ∣ k (a,b)\mid k ( a , b ) ∣ k
x = k a ‾ ( a , b ) + b d ( a , b ) , y = k b ‾ ( a , b ) − a d ( a , b ) , x=\frac{k\overline{a}}{(a,b)}+\frac{bd}{(a,b)},\qquad y=\frac{k\overline{b}}{(a,b)}-\frac{ad}{(a,b)}, x = ( a , b ) k a + ( a , b ) b d , y = ( a , b ) k b − ( a , b ) a d , where a ‾ , b ‾ , d ∈ Z \overline{a},\overline{b},d\in\mathbb{Z} a , b , d ∈ Z a a ‾ + b b ‾ = ( a , b ) a\overline{a}+b\overline{b}=(a,b) a a + b b = ( a , b ) ( a , b ) ∤ k (a,b)\nmid k ( a , b ) ∤ k
c k ( f ) = 0 ; (42) c_k(f)=0; \tag*{(42)} c k ( f ) = 0 ; (42) and if ( a , b ) ∣ k (a,b)\mid k ( a , b ) ∣ k
c k ( f ) = q m − 1 m ! n ! ( a , b ) m + n G ( 1 , χ ‾ ) e 2 π i k ( a y ‾ + b z ‾ ) ( a , b ) q ( 2 π i ) m + n × ∑ d = − ∞ + ∞ χ ( k a ‾ ( a , b ) + b d ( a , b ) ) e 2 π i d ( b y ‾ − a q z ‾ ) ( a , b ) q ( k a ‾ + b d ) m ( k b ‾ − a d ) n = ( − 1 ) n q m − 1 m ! n ! ( a , b ) m + n G ( 1 , χ ‾ ) e 2 π i k ( a y ‾ + b z ‾ ) ( a , b ) q a n b m ( 2 π i ) m + n × ∑ d = − ∞ + ∞ χ ( k a ‾ ( a , b ) + b d ( a , b ) ) e 2 π i d ( b y ‾ − a q z ‾ ) ( a , b ) q ( d + k a ‾ a ) m ( d − k b ‾ b ) n . (43) \begin{aligned}
c_k(f)
&=\frac{q^{m-1}m!n!(a,b)^{m+n}G(1,\overline{\chi})e^{\frac{2\pi ik(a\overline{y}+b\overline{z})}{(a,b)q}}}{(2\pi i)^{m+n}}\\
&\quad\times\sum_{d=-\infty}^{+\infty}
\frac{\chi\left(\frac{k\overline{a}}{(a,b)}+\frac{bd}{(a,b)}\right)e^{\frac{2\pi id(b\overline{y}-a q\overline{z})}{(a,b)q}}}{(k\overline{a}+bd)^m(k\overline{b}-ad)^n}\\
&=\frac{(-1)^nq^{m-1}m!n!(a,b)^{m+n}G(1,\overline{\chi})e^{\frac{2\pi ik(a\overline{y}+b\overline{z})}{(a,b)q}}}{a^n b^m(2\pi i)^{m+n}}\\
&\quad\times\sum_{d=-\infty}^{+\infty}
\frac{\chi\left(\frac{k\overline{a}}{(a,b)}+\frac{bd}{(a,b)}\right)e^{\frac{2\pi id(b\overline{y}-a q\overline{z})}{(a,b)q}}}{\left(d+\frac{k\overline{a}}{a}\right)^m\left(d-\frac{k\overline{b}}{b}\right)^n}. \tag*{(43)}
\end{aligned} c k ( f ) = ( 2 π i ) m + n q m − 1 m ! n ! ( a , b ) m + n G ( 1 , χ ) e ( a , b ) q 2 π ik ( a y + b z ) × d = − ∞ ∑ + ∞ ( k a + b d ) m ( k b − a d ) n χ ( ( a , b ) k a + ( a , b ) b d ) e ( a , b ) q 2 π i d ( b y − a q z ) = a n b m ( 2 π i ) m + n ( − 1 ) n q m − 1 m ! n ! ( a , b ) m + n G ( 1 , χ ) e ( a , b ) q 2 π ik ( a y + b z ) × d = − ∞ ∑ + ∞ ( d + a k a ) m ( d − b k b ) n χ ( ( a , b ) k a + ( a , b ) b d ) e ( a , b ) q 2 π i d ( b y − a q z ) . (43) In particular, we see from (2.5) and (3.5) that
c 0 ( f ) = ( − 1 ) n − 1 m ! n ! ( a , b ) m + n χ ( b ( a , b ) ) B ‾ m + n ( b y ‾ − a q z ‾ ( a , b ) , χ ‾ ) ( a q ) n b m ( m + n ) ! . (44) c_0(f)=\frac{(-1)^{n-1}m!n!(a,b)^{m+n}\chi\left(\frac{b}{(a,b)}\right)\overline{B}_{m+n}\left(\frac{b\overline{y}-a q\overline{z}}{(a,b)},\overline{\chi}\right)}{(aq)^n b^m(m+n)!}. \tag*{(44)} c 0 ( f ) = ( a q ) n b m ( m + n )! ( − 1 ) n − 1 m ! n ! ( a , b ) m + n χ ( ( a , b ) b ) B m + n ( ( a , b ) b y − a q z , χ ) . (44) We next consider the case k ≠ 0 k \ne0 k = 0 (43) . Note that there is an implicit restriction
( a ( a , b ) , b ‾ ) = ( b ( a , b ) , a ‾ ) = 1. \left(\frac{a}{(a,b)},\overline{b}\right)=\left(\frac{b}{(a,b)},\overline{a}\right)=1. ( ( a , b ) a , b ) = ( ( a , b ) b , a ) = 1. Hence, by taking x = − k a ‾ / b x=-k\overline{a}/b x = − k a / b y = k b ‾ / a y=k\overline{b}/a y = k b / a (43) that if ( a , b ) ∣ k (a,b)\mid k ( a , b ) ∣ k
c k ( f ) = ( − 1 ) n q m − 1 m ! n ! ( a , b ) m + n G ( 1 , χ ‾ ) e 2 π i k ( a ‾ y + b ‾ x ) ( a , b ) q a n b m ( 2 π i ) m + n × ( ∑ j = 1 m ( m + n − j − 1 n − 1 ) ( − 1 ) m − j ( − a b k ( a , b ) ) m + n − j × ( ∑ d = − ∞ + ∞ χ ( k a ‾ ( a , b ) + b d ( a , b ) ) e 2 π i d ( b y − a q z ) ( a , b ) q ( d + k a ‾ b ) j − ϵ ( a , k ) χ ( k a ) e 2 π i k ( b y − a q z ) a ( a , b ) q ( − a b k ( a , b ) ) j ) + ∑ j = 1 n ( m + n − j − 1 m − 1 ) ( − 1 ) n − j ( a b k ( a , b ) ) m + n − j × ( ∑ d = − ∞ + ∞ χ ( k a ‾ ( a , b ) + b d ( a , b ) ) e 2 π i d ( b y − a q z ) ( a , b ) q ( d − k b ‾ a ) j − ϵ ( b , k ) χ ( 0 ) e − 2 π i k ( b y − a q z ) b ( a , b ) q ( − a b k ( a , b ) ) j ) ) , (45) \begin{aligned}
c_k(f)={}&\frac{(-1)^{n}q^{m-1}m!n!(a,b)^{m+n}G(1,\overline{\chi})e^{\frac{2\pi i k(\overline{a}y+\overline{b}x)}{(a,b)q}}}{a^n b^m(2\pi i)^{m+n}}\\
&\times\left(\sum_{j=1}^{m}\binom{m+n-j-1}{n-1}(-1)^{m-j}\left(-\frac{ab}{k(a,b)}\right)^{m+n-j}\right.\\
&\qquad\times\left(\sum_{d=-\infty}^{+\infty}\frac{\chi\left(\frac{k\overline{a}}{(a,b)}+\frac{bd}{(a,b)}\right)e^{\frac{2\pi i d(by-aqz)}{(a,b)q}}}{\left(d+\frac{k\overline{a}}{b}\right)^j}\right.\\
&\qquad\qquad\left.-\epsilon(a,k)\chi\left(\frac{k}{a}\right)e^{\frac{2\pi i k(by-aqz)}{a(a,b)q}}\left(-\frac{ab}{k(a,b)}\right)^j\right)\\
&\quad+\sum_{j=1}^{n}\binom{m+n-j-1}{m-1}(-1)^{n-j}\left(\frac{ab}{k(a,b)}\right)^{m+n-j}\\
&\qquad\times\left(\sum_{d=-\infty}^{+\infty}\frac{\chi\left(\frac{k\overline{a}}{(a,b)}+\frac{bd}{(a,b)}\right)e^{\frac{2\pi i d(by-aqz)}{(a,b)q}}}{\left(d-\frac{k\overline{b}}{a}\right)^j}\right.\\
&\qquad\qquad\left.\left.-\epsilon(b,k)\chi(0)e^{-\frac{2\pi i k(by-aqz)}{b(a,b)q}}\left(-\frac{ab}{k(a,b)}\right)^j\right)\right), \tag*{(45)}
\end{aligned} c k ( f ) = a n b m ( 2 π i ) m + n ( − 1 ) n q m − 1 m ! n ! ( a , b ) m + n G ( 1 , χ ) e ( a , b ) q 2 π ik ( a y + b x ) × ( j = 1 ∑ m ( n − 1 m + n − j − 1 ) ( − 1 ) m − j ( − k ( a , b ) ab ) m + n − j × d = − ∞ ∑ + ∞ ( d + b k a ) j χ ( ( a , b ) k a + ( a , b ) b d ) e ( a , b ) q 2 π i d ( b y − a q z ) − ϵ ( a , k ) χ ( a k ) e a ( a , b ) q 2 π ik ( b y − a q z ) ( − k ( a , b ) ab ) j ) + j = 1 ∑ n ( m − 1 m + n − j − 1 ) ( − 1 ) n − j ( k ( a , b ) ab ) m + n − j × d = − ∞ ∑ + ∞ ( d − a k b ) j χ ( ( a , b ) k a + ( a , b ) b d ) e ( a , b ) q 2 π i d ( b y − a q z ) − ϵ ( b , k ) χ ( 0 ) e − b ( a , b ) q 2 π ik ( b y − a q z ) ( − k ( a , b ) ab ) j ) ) , (45) where ϵ ( a , k ) = 1 \epsilon(a,k)=1 ϵ ( a , k ) = 1 a ∣ k a\mid k a ∣ k (23) and the geometric sum in [1 Theorem 8.1] , we see that
G ( 1 , χ ) G ( 1 , χ ‾ ) = G ( 1 , χ ) ∑ r = 1 q χ ‾ ( r ) e 2 π i r q = ∑ r = 1 q G ( r , χ ) e 2 π i r q = ∑ j = 1 q χ ( j ) ∑ r = 1 q e 2 π i ( j + 1 ) r q = χ ( − 1 ) q . (46) \begin{aligned}
G(1,\chi)G(1,\overline{\chi})
&=G(1,\chi)\sum_{r=1}^{q}\overline{\chi}(r)e^{\frac{2\pi i r}{q}}\\
&=\sum_{r=1}^{q}G(r,\chi)e^{\frac{2\pi i r}{q}}\\
&=\sum_{j=1}^{q}\chi(j)\sum_{r=1}^{q}e^{\frac{2\pi i(j+1)r}{q}}\\
&=\chi(-1)q. \tag*{(46)}
\end{aligned} G ( 1 , χ ) G ( 1 , χ ) = G ( 1 , χ ) r = 1 ∑ q χ ( r ) e q 2 π i r = r = 1 ∑ q G ( r , χ ) e q 2 π i r = j = 1 ∑ q χ ( j ) r = 1 ∑ q e q 2 π i ( j + 1 ) r = χ ( − 1 ) q . (46) Using Lemma 2.5 and (3.8), we find that (3.7) can be rewritten as
c k ( f ) = − q m − 1 b n − 1 m ! n ! ( a , b ) χ ( − 1 ) sgn ( b ) ∑ j = 1 m ( m + n − j − 1 n − 1 ) ( − 1 ) j a m − j j ! ( 2 π i k ) m + n − j × ∑ r = 1 q χ ‾ ( r ) ∑ l = 1 ∣ b ( a , b ) ∣ e 2 π i k ( z + a l ) b B ‾ j ( ( a , b ) l − b y − b r − a q z q b ) − ϵ ( a , k ) q m − 1 a m b n m ! n ! χ ( k a ) G ( 1 , χ ‾ ) e 2 π i k y a q ( 2 π i k ) m + n ∑ j = 1 m ( m + n − j − 1 n − 1 ) − ( a q ) m − 1 m ! n ! ( a , b ) χ ( − 1 ) sgn ( a ) ∑ j = 1 n ( m + n − j − 1 m − 1 ) b n − j j ! ( 2 π i k ) m + n − j × ∑ r = 1 q χ ‾ ( r ) e − 2 π i k r a q ∑ l = 1 ∣ a ( a , b ) ∣ e 2 π i k ( y q − b ‾ l ) a B ‾ j ( ( a , b ) l − b y − b r − a q z q a ) − ϵ ( b , k ) q m − 1 a m b n m ! n ! χ ( 0 ) G ( 1 , χ ‾ ) e 2 π i k z b ( 2 π i k ) m + n ∑ j = 1 n ( m + n − j − 1 m − 1 ) . \begin{aligned}
c_k(f)={}&-q^{m-1}b^{n-1}m!n!(a,b)\chi(-1)\operatorname{sgn}(b)\sum_{j=1}^{m}\binom{m+n-j-1}{n-1}\frac{(-1)^j a^{m-j}}{j!(2\pi ik)^{m+n-j}}\\
&\times\sum_{r=1}^{q}\overline{\chi}(r)\sum_{l=1}^{\left\lvert\frac{b}{(a,b)}\right\rvert}e^{\frac{2\pi ik(z+al)}{b}}\overline{B}_j\left(\frac{(a,b)l-\frac{by-br-aqz}{q}}{b}\right)\\
&-\epsilon(a,k)\frac{q^{m-1}a^m b^n m!n!\chi\left(\frac{k}{a}\right)G(1,\overline{\chi})e^{\frac{2\pi iky}{aq}}}{(2\pi ik)^{m+n}}\sum_{j=1}^{m}\binom{m+n-j-1}{n-1}\\
&-(aq)^{m-1}m!n!(a,b)\chi(-1)\operatorname{sgn}(a)\sum_{j=1}^{n}\binom{m+n-j-1}{m-1}\frac{b^{n-j}}{j!(2\pi ik)^{m+n-j}}\\
&\times\sum_{r=1}^{q}\overline{\chi}(r)e^{-\frac{2\pi ikr}{aq}}\sum_{l=1}^{\left\lvert\frac{a}{(a,b)}\right\rvert}e^{\frac{2\pi ik\left(\frac{y}{q}-\overline{b}l\right)}{a}}\overline{B}_j\left(\frac{(a,b)l-\frac{by-br-aqz}{q}}{a}\right)\\
&-\epsilon(b,k)\frac{q^{m-1}a^m b^n m!n!\chi(0)G(1,\overline{\chi})e^{\frac{2\pi ikz}{b}}}{(2\pi ik)^{m+n}}\sum_{j=1}^{n}\binom{m+n-j-1}{m-1}.
\end{aligned} c k ( f ) = − q m − 1 b n − 1 m ! n ! ( a , b ) χ ( − 1 ) sgn ( b ) j = 1 ∑ m ( n − 1 m + n − j − 1 ) j ! ( 2 π ik ) m + n − j ( − 1 ) j a m − j × r = 1 ∑ q χ ( r ) l = 1 ∑ ∣ ( a , b ) b ∣ e b 2 π ik ( z + a l ) B j ( b ( a , b ) l − q b y − b r − a q z ) − ϵ ( a , k ) ( 2 π ik ) m + n q m − 1 a m b n m ! n ! χ ( a k ) G ( 1 , χ ) e a q 2 π ik y j = 1 ∑ m ( n − 1 m + n − j − 1 ) − ( a q ) m − 1 m ! n ! ( a , b ) χ ( − 1 ) sgn ( a ) j = 1 ∑ n ( m − 1 m + n − j − 1 ) j ! ( 2 π ik ) m + n − j b n − j × r = 1 ∑ q χ ( r ) e − a q 2 π ik r l = 1 ∑ ∣ ( a , b ) a ∣ e a 2 π ik ( q y − b l ) B j ( a ( a , b ) l − q b y − b r − a q z ) − ϵ ( b , k ) ( 2 π ik ) m + n q m − 1 a m b n m ! n ! χ ( 0 ) G ( 1 , χ ) e b 2 π ik z j = 1 ∑ n ( m − 1 m + n − j − 1 ) . This, together with Lemma 2.7, yields that if ( a , b ) ∣ k (a,b)\mid k ( a , b ) ∣ k
c k ( f ) = − n q m − 1 b n − 1 ( a , b ) χ ( − 1 ) sgn ( b ) ∑ j = 1 m ( m j ) ( − 1 ) j a m − j ( m + n − j − 1 ) ! ( 2 π i k ) m + n − j × ∑ r = 1 q χ ‾ ( r ) ∑ l = 1 ∣ b ( a , b ) ∣ e 2 π i k ( z + a l ) b B ‾ j ( ( a , b ) l − b y − b r − a q z q b ) − m ( a q ) m − 1 ( a , b ) χ ( − 1 ) sgn ( a ) ∑ j = 1 n ( n j ) b n − j ( m + n − j − 1 ) ! ( 2 π i k ) m + n − j × ∑ r = 1 q χ ‾ ( r ) e − 2 π i k r a q ∑ l = 1 ∣ a ( a , b ) ∣ e 2 π i k ( y q − b ‾ l ) a B ‾ j ( ( a , b ) l − b y − b r − a q z q a ) − ϵ ( a , k ) e 2 π i k y a q m ( m + n − 1 ) ! q m − 1 a m b n χ ( k a ) G ( 1 , χ ‾ ) ( 2 π i k ) m + n − ϵ ( b , k ) e 2 π i k z b n ( m + n − 1 ) ! q m − 1 a m b n χ ( 0 ) G ( 1 , χ ‾ ) ( 2 π i k ) m + n . (47) \begin{aligned}
c_k(f)={}&-nq^{m-1}b^{n-1}(a,b)\chi(-1)\operatorname{sgn}(b)\sum_{j=1}^{m}\binom{m}{j}\frac{(-1)^j a^{m-j}(m+n-j-1)!}{(2\pi ik)^{m+n-j}}\\
&\times\sum_{r=1}^{q}\overline{\chi}(r)\sum_{l=1}^{\left\lvert\frac{b}{(a,b)}\right\rvert}e^{\frac{2\pi ik(z+al)}{b}}\overline{B}_j\left(\frac{(a,b)l-\frac{by-br-aqz}{q}}{b}\right)\\
&-m(aq)^{m-1}(a,b)\chi(-1)\operatorname{sgn}(a)\sum_{j=1}^{n}\binom{n}{j}\frac{b^{n-j}(m+n-j-1)!}{(2\pi ik)^{m+n-j}}\\
&\times\sum_{r=1}^{q}\overline{\chi}(r)e^{-\frac{2\pi ikr}{aq}}\sum_{l=1}^{\left\lvert\frac{a}{(a,b)}\right\rvert}e^{\frac{2\pi ik\left(\frac{y}{q}-\overline{b}l\right)}{a}}\overline{B}_j\left(\frac{(a,b)l-\frac{by-br-aqz}{q}}{a}\right)\\
&-\epsilon(a,k)e^{\frac{2\pi iky}{aq}}\frac{m(m+n-1)!q^{m-1}a^m b^n\chi\left(\frac{k}{a}\right)G(1,\overline{\chi})}{(2\pi ik)^{m+n}}\\
&-\epsilon(b,k)e^{\frac{2\pi ikz}{b}}\frac{n(m+n-1)!q^{m-1}a^m b^n\chi(0)G(1,\overline{\chi})}{(2\pi ik)^{m+n}}. \tag*{(47)}
\end{aligned} c k ( f ) = − n q m − 1 b n − 1 ( a , b ) χ ( − 1 ) sgn ( b ) j = 1 ∑ m ( j m ) ( 2 π ik ) m + n − j ( − 1 ) j a m − j ( m + n − j − 1 )! × r = 1 ∑ q χ ( r ) l = 1 ∑ ∣ ( a , b ) b ∣ e b 2 π ik ( z + a l ) B j ( b ( a , b ) l − q b y − b r − a q z ) − m ( a q ) m − 1 ( a , b ) χ ( − 1 ) sgn ( a ) j = 1 ∑ n ( j n ) ( 2 π ik ) m + n − j b n − j ( m + n − j − 1 )! × r = 1 ∑ q χ ( r ) e − a q 2 π ik r l = 1 ∑ ∣ ( a , b ) a ∣ e a 2 π ik ( q y − b l ) B j ( a ( a , b ) l − q b y − b r − a q z ) − ϵ ( a , k ) e a q 2 π ik y ( 2 π ik ) m + n m ( m + n − 1 )! q m − 1 a m b n χ ( a k ) G ( 1 , χ ) − ϵ ( b , k ) e b 2 π ik z ( 2 π ik ) m + n n ( m + n − 1 )! q m − 1 a m b n χ ( 0 ) G ( 1 , χ ) . (47) Since ( a ( a , b ) , b ( a , b ) ) = 1 \left(\frac{a}{(a,b)},\frac{b}{(a,b)}\right)=1 ( ( a , b ) a , ( a , b ) b ) = 1 a l ( a , b ) \frac{al}{(a,b)} ( a , b ) a l ∣ b ( a , b ) ∣ \left\lvert\frac{b}{(a,b)}\right\rvert ( a , b ) b l l l − b l ( a , b ) \frac{-bl}{(a,b)} ( a , b ) − b l ∣ a ( a , b ) ∣ \left\lvert\frac{a}{(a,b)}\right\rvert ( a , b ) a l l l ( a , b ) ∣ k (a,b)\mid k ( a , b ) ∣ k
c k ( f ) = − n q m − 1 b n − 1 ( a , b ) χ ( − 1 ) sgn ( b ) ∑ j = 1 m ( m j ) ( − 1 ) j a m − j ( m + n − j − 1 ) ! ( 2 π i k ) m + n − j × ∑ r = 1 q χ ‾ ( r ) ∑ l = 1 ∣ b ( a , b ) ∣ e 2 π i k ( l + z ) b B ‾ j ( a ( l + z ) b − y − r q ) − m ( a q ) m − 1 ( a , b ) χ ( − 1 ) sgn ( a ) ∑ j = 1 n ( n j ) ( − 1 ) j b n − j ( m + n − j − 1 ) ! ( 2 π i k ) m + n − j × ∑ r = 1 q χ ‾ ( r ) e − 2 π i k r a q ∑ l = 1 ∣ a ( a , b ) ∣ e 2 π i k ( l + y q ) a B ‾ j ( b l + b y − b r − a q z q a ) − ϵ ( a , k ) e 2 π i k x a q m ( m + n − 1 ) ! q m − 1 a m b n χ ( k a ) G ( 1 , χ ‾ ) ( 2 π i k ) m + n − ϵ ( b , k ) e 2 π i k x b n ( m + n − 1 ) ! q m − 1 a m b n χ ( 0 ) G ( 1 , χ ‾ ) ( 2 π i k ) m + n . (48) \begin{aligned}
c_k(f)={}&-nq^{m-1}b^{n-1}(a,b)\chi(-1)\operatorname{sgn}(b)\sum_{j=1}^{m}\binom{m}{j}\frac{(-1)^j a^{m-j}(m+n-j-1)!}{(2\pi i k)^{m+n-j}}\\
&\times\sum_{r=1}^{q}\overline{\chi}(r)\sum_{l=1}^{\left|\frac{b}{(a,b)}\right|}e^{\frac{2\pi i k(l+z)}{b}}\overline{B}_j\left(\frac{a(l+z)}{b}-\frac{y-r}{q}\right)\\
&-m(aq)^{m-1}(a,b)\chi(-1)\operatorname{sgn}(a)\sum_{j=1}^{n}\binom{n}{j}\frac{(-1)^j b^{n-j}(m+n-j-1)!}{(2\pi i k)^{m+n-j}}\\
&\times\sum_{r=1}^{q}\overline{\chi}(r)e^{-\frac{2\pi i k r}{aq}}\sum_{l=1}^{\left|\frac{a}{(a,b)}\right|}e^{\frac{2\pi i k(l+\frac{y}{q})}{a}}\overline{B}_j\left(\frac{bl+\frac{by-br-aqz}{q}}{a}\right)\\
&-\epsilon(a,k)e^{\frac{2\pi i kx}{aq}}\frac{m(m+n-1)!q^{m-1}a^mb^n\chi\left(\frac{k}{a}\right)G(1,\overline{\chi})}{(2\pi i k)^{m+n}}\\
&-\epsilon(b,k)e^{\frac{2\pi i kx}{b}}\frac{n(m+n-1)!q^{m-1}a^mb^n\chi(0)G(1,\overline{\chi})}{(2\pi i k)^{m+n}}. \tag*{(48)}
\end{aligned} c k ( f ) = − n q m − 1 b n − 1 ( a , b ) χ ( − 1 ) sgn ( b ) j = 1 ∑ m ( j m ) ( 2 π ik ) m + n − j ( − 1 ) j a m − j ( m + n − j − 1 )! × r = 1 ∑ q χ ( r ) l = 1 ∑ ∣ ( a , b ) b ∣ e b 2 π ik ( l + z ) B j ( b a ( l + z ) − q y − r ) − m ( a q ) m − 1 ( a , b ) χ ( − 1 ) sgn ( a ) j = 1 ∑ n ( j n ) ( 2 π ik ) m + n − j ( − 1 ) j b n − j ( m + n − j − 1 )! × r = 1 ∑ q χ ( r ) e − a q 2 π ik r l = 1 ∑ ∣ ( a , b ) a ∣ e a 2 π ik ( l + q y ) B j ( a b l + q b y − b r − a q z ) − ϵ ( a , k ) e a q 2 π ik x ( 2 π ik ) m + n m ( m + n − 1 )! q m − 1 a m b n χ ( a k ) G ( 1 , χ ) − ϵ ( b , k ) e b 2 π ik x ( 2 π ik ) m + n n ( m + n − 1 )! q m − 1 a m b n χ ( 0 ) G ( 1 , χ ) . (48) Inserting (3.4), (3.6) and (3.10) into (3.1), we arrive at
B ‾ m ( a x + y , χ ‾ ) B ‾ n ( b x q + z ) + δ 1 , m δ 1 , n sgn ( a b ) δ Z ( a x + y ) δ Z ( b x q + z ) χ ‾ ( − a x − y ) 4 = ∑ k = − ∞ ( a , b ) ∣ k + ∞ ′ ( − n q m − 1 b n − 1 ( a , b ) χ ( − 1 ) sgn ( b ) ∑ j = 1 m ( m j ) ( − 1 ) j a m − j ( 2 π i k ) m + n − j × ( m + n − j − 1 ) ! ∑ r = 1 q χ ‾ ( r ) ∑ l = 1 ∣ b ( a , b ) ∣ e 2 π i k ( l + z ) b B ‾ j ( a ( l + z ) b − y − r q ) ) − m ( a q ) m − 1 ( a , b ) χ ( − 1 ) sgn ( a ) ∑ j = 1 n ( n j ) ( − 1 ) j b n − j ( m + n − j − 1 ) ! ( 2 π i k ) m + n − j × ∑ r = 1 q χ ‾ ( r ) e − 2 π i k r a q ∑ l = 1 ∣ a ( a , b ) ∣ e 2 π i k ( l + y q ) a B ‾ j ( b l + b y − b r q a − z ) e 2 π i k x q − ∑ k = − ∞ a ∣ k + ∞ ′ ( m ( m + n − 1 ) ! q m − 1 a m b n χ ( k a ) G ( 1 , χ ‾ ) e 2 π i k y a q ( 2 π i k ) m + n ) e 2 π i k x q − ∑ k = − ∞ b ∣ k + ∞ ′ ( n ( m + n − 1 ) ! q m − 1 a m b n χ ( 0 ) G ( 1 , χ ‾ ) e 2 π i k z b ( 2 π i k ) m + n ) e 2 π i k x q + ( − 1 ) n − 1 m ! n ! ( a , b ) m + n χ ( b ( a , b ) ) B ‾ m + n ( b y − a q z ( a , b ) , χ ‾ ) ( a q ) n b m ( m + n ) ! . (49) \begin{aligned}
\overline{B}_m(ax+y,\overline{\chi})\overline{B}_n\left(\frac{bx}{q}+z\right)
&+\delta_{1,m}\delta_{1,n}\operatorname{sgn}(ab)\frac{\delta_{\mathbb{Z}}(ax+y)\delta_{\mathbb{Z}}\left(\frac{bx}{q}+z\right)\overline{\chi}(-ax-y)}{4}\\
={}&\sum_{\substack{k=-\infty\\(a,b)\mid k}}^{+\infty\prime}\left(-nq^{m-1}b^{n-1}(a,b)\chi(-1)\operatorname{sgn}(b)\sum_{j=1}^{m}\binom{m}{j}\frac{(-1)^j a^{m-j}}{(2\pi i k)^{m+n-j}}\right.\\
&\qquad\left.\times(m+n-j-1)!\sum_{r=1}^{q}\overline{\chi}(r)\sum_{l=1}^{\left|\frac{b}{(a,b)}\right|}e^{\frac{2\pi i k(l+z)}{b}}\overline{B}_j\left(\frac{a(l+z)}{b}-\frac{y-r}{q}\right)\right)\\
&-m(aq)^{m-1}(a,b)\chi(-1)\operatorname{sgn}(a)\sum_{j=1}^{n}\binom{n}{j}\frac{(-1)^j b^{n-j}(m+n-j-1)!}{(2\pi i k)^{m+n-j}}\\
&\qquad\times\sum_{r=1}^{q}\overline{\chi}(r)e^{-\frac{2\pi i k r}{aq}}\sum_{l=1}^{\left|\frac{a}{(a,b)}\right|}e^{\frac{2\pi i k(l+\frac{y}{q})}{a}}\overline{B}_j\left(\frac{bl+\frac{by-br}{q}}{a}-z\right)e^{\frac{2\pi i kx}{q}}\\
&-\sum_{\substack{k=-\infty\\a\mid k}}^{+\infty\prime}\left(\frac{m(m+n-1)!q^{m-1}a^mb^n\chi\left(\frac{k}{a}\right)G(1,\overline{\chi})e^{\frac{2\pi i ky}{aq}}}{(2\pi i k)^{m+n}}\right)e^{\frac{2\pi i kx}{q}}\\
&-\sum_{\substack{k=-\infty\\b\mid k}}^{+\infty\prime}\left(\frac{n(m+n-1)!q^{m-1}a^mb^n\chi(0)G(1,\overline{\chi})e^{\frac{2\pi i kz}{b}}}{(2\pi i k)^{m+n}}\right)e^{\frac{2\pi i kx}{q}}\\
&+\frac{(-1)^{n-1}m!n!(a,b)^{m+n}\chi\left(\frac{b}{(a,b)}\right)\overline{B}_{m+n}\left(\frac{by-aqz}{(a,b)},\overline{\chi}\right)}{(aq)^n b^m(m+n)!}. \tag*{(49)}
\end{aligned} B m ( a x + y , χ ) B n ( q b x + z ) = + δ 1 , m δ 1 , n sgn ( ab ) 4 δ Z ( a x + y ) δ Z ( q b x + z ) χ ( − a x − y ) k = − ∞ ( a , b ) ∣ k ∑ + ∞′ ( − n q m − 1 b n − 1 ( a , b ) χ ( − 1 ) sgn ( b ) j = 1 ∑ m ( j m ) ( 2 π ik ) m + n − j ( − 1 ) j a m − j × ( m + n − j − 1 )! r = 1 ∑ q χ ( r ) l = 1 ∑ ∣ ( a , b ) b ∣ e b 2 π ik ( l + z ) B j ( b a ( l + z ) − q y − r ) − m ( a q ) m − 1 ( a , b ) χ ( − 1 ) sgn ( a ) j = 1 ∑ n ( j n ) ( 2 π ik ) m + n − j ( − 1 ) j b n − j ( m + n − j − 1 )! × r = 1 ∑ q χ ( r ) e − a q 2 π ik r l = 1 ∑ ∣ ( a , b ) a ∣ e a 2 π ik ( l + q y ) B j ( a b l + q b y − b r − z ) e q 2 π ik x − k = − ∞ a ∣ k ∑ + ∞′ ( ( 2 π ik ) m + n m ( m + n − 1 )! q m − 1 a m b n χ ( a k ) G ( 1 , χ ) e a q 2 π ik y ) e q 2 π ik x − k = − ∞ b ∣ k ∑ + ∞′ ( ( 2 π ik ) m + n n ( m + n − 1 )! q m − 1 a m b n χ ( 0 ) G ( 1 , χ ) e b 2 π ik z ) e q 2 π ik x + ( a q ) n b m ( m + n )! ( − 1 ) n − 1 m ! n ! ( a , b ) m + n χ ( ( a , b ) b ) B m + n ( ( a , b ) b y − a q z , χ ) . (49) Thus, from (24) , (25) and (49) , we obtain
B ‾ m ( a x + y , x ‾ ) B ‾ n ( b x q + z ) + δ 1 , m δ 1 , n sgn ( a b ) δ Z ( a x + y ) δ Z ( b x q + z ) χ ‾ ( − a x − y ) 4 = n q m − 1 b n − 1 χ ( − 1 ) sgn ( b ) ∑ j = 1 m ( m j ) ( − 1 ) j a m − j ( a , b ) m + n − j − 1 ( m + n − j ) × ∑ r = 1 q χ ‾ ( r ) ∑ l = 1 ⌊ ∣ b ( a , b ) ∣ ⌋ B ‾ j ( a ( l + z ) b − y − r q ) × B ‾ m + n − j ( ( a , b ) x q + ( a , b ) ( l + z ) b ) + m ( a q ) m − 1 χ ( − 1 ) sgn ( a ) ∑ j = 1 n ( n j ) ( − 1 ) j b n − j ( a , b ) m + n − j − 1 ( m + n − j ) × ∑ r = 1 q χ ‾ ( r ) ∑ l = 1 ⌊ ∣ a ( a , b ) ∣ ⌋ B ‾ j ( b l + b y − b r q a − z ) × B ‾ m + n − j ( ( a , b ) x q + ( a , b ) ( l + y q ) a − ( a , b ) r a q ) + m b n B ‾ m + n ( a x + y , x ‾ ) ( a q ) n ( m + n ) + n q m − 1 a m χ ( 0 ) G ( 1 , χ ‾ ) B ‾ m + n ( b x q + z ) b m ( m + n ) + ( − 1 ) n − 1 m ! n ! ( a , b ) m + n χ ( b ( a , b ) ) B ‾ m + n ( b y − a q z ( a , b ) , x ‾ ) ( a q ) n b m ( m + n ) ! . (50) \begin{aligned}
\overline{B}_m(ax+y,\overline{x})\overline{B}_n\left(\frac{bx}{q}+z\right)
&+\delta_{1,m}\delta_{1,n}\operatorname{sgn}(ab)
\frac{\delta_{\mathbb{Z}}(ax+y)\delta_{\mathbb{Z}}\left(\frac{bx}{q}+z\right)\overline{\chi}(-ax-y)}{4}\\
&=nq^{m-1}b^{n-1}\chi(-1)\operatorname{sgn}(b)
\sum_{j=1}^{m}\binom{m}{j}
\frac{(-1)^j a^{m-j}}{(a,b)^{m+n-j-1}(m+n-j)}\\
&\quad\times\sum_{r=1}^{q}\overline{\chi}(r)
\sum_{l=1}^{\left\lfloor\left|\frac{b}{(a,b)}\right|\right\rfloor}
\overline{B}_j\left(\frac{a(l+z)}{b}-\frac{y-r}{q}\right)\\
&\quad\times\overline{B}_{m+n-j}\left(\frac{(a,b)x}{q}+\frac{(a,b)(l+z)}{b}\right)\\
&\quad+m(aq)^{m-1}\chi(-1)\operatorname{sgn}(a)
\sum_{j=1}^{n}\binom{n}{j}
\frac{(-1)^j b^{n-j}}{(a,b)^{m+n-j-1}(m+n-j)}\\
&\quad\times\sum_{r=1}^{q}\overline{\chi}(r)
\sum_{l=1}^{\left\lfloor\left|\frac{a}{(a,b)}\right|\right\rfloor}
\overline{B}_j\left(\frac{bl+\frac{by-br}{q}}{a}-z\right)\\
&\quad\times\overline{B}_{m+n-j}\left(\frac{(a,b)x}{q}+\frac{(a,b)\left(l+\frac{y}{q}\right)}{a}-\frac{(a,b)r}{aq}\right)\\
&\quad+\frac{mb^n\overline{B}_{m+n}(ax+y,\overline{x})}{(aq)^n(m+n)}
+\frac{nq^{m-1}a^m\chi(0)G(1,\overline{\chi})\overline{B}_{m+n}\left(\frac{bx}{q}+z\right)}{b^m(m+n)}\\
&\quad+\frac{(-1)^{n-1}m!n!(a,b)^{m+n}\chi\left(\frac{b}{(a,b)}\right)\overline{B}_{m+n}\left(\frac{by-aqz}{(a,b)},\overline{x}\right)}{(aq)^n b^m(m+n)!}. \tag*{(50)}
\end{aligned} B m ( a x + y , x ) B n ( q b x + z ) + δ 1 , m δ 1 , n sgn ( ab ) 4 δ Z ( a x + y ) δ Z ( q b x + z ) χ ( − a x − y ) = n q m − 1 b n − 1 χ ( − 1 ) sgn ( b ) j = 1 ∑ m ( j m ) ( a , b ) m + n − j − 1 ( m + n − j ) ( − 1 ) j a m − j × r = 1 ∑ q χ ( r ) l = 1 ∑ ⌊ ∣ ( a , b ) b ∣ ⌋ B j ( b a ( l + z ) − q y − r ) × B m + n − j ( q ( a , b ) x + b ( a , b ) ( l + z ) ) + m ( a q ) m − 1 χ ( − 1 ) sgn ( a ) j = 1 ∑ n ( j n ) ( a , b ) m + n − j − 1 ( m + n − j ) ( − 1 ) j b n − j × r = 1 ∑ q χ ( r ) l = 1 ∑ ⌊ ∣ ( a , b ) a ∣ ⌋ B j ( a b l + q b y − b r − z ) × B m + n − j q ( a , b ) x + a ( a , b ) ( l + q y ) − a q ( a , b ) r + ( a q ) n ( m + n ) m b n B m + n ( a x + y , x ) + b m ( m + n ) n q m − 1 a m χ ( 0 ) G ( 1 , χ ) B m + n ( q b x + z ) + ( a q ) n b m ( m + n )! ( − 1 ) n − 1 m ! n ! ( a , b ) m + n χ ( ( a , b ) b ) B m + n ( ( a , b ) b y − a q z , x ) . (50) It can be shown from (33) , the property of residue systems, and the division algorithm that
1 ( a , b ) m + n − j − 1 ∑ l = 1 ⌊ ∣ b ( a , b ) ∣ ⌋ B ‾ j ( a ( l + z ) b − y − r q ) × B ‾ m + n − j ( ( a , b ) x q + ( a , b ) ( l + z ) b ) = ∑ k = 0 ( a , b ) − 1 ∑ l = 1 ⌊ ∣ b ( a , b ) ∣ ⌋ B ‾ j ( a ( l + z + k ⌊ b ( a , b ) ⌋ ) b − y − r q ) × B ‾ m + n − j ( x q + l + z + k ⌊ b ( a , b ) ⌋ b ) = ∑ l = 1 ∣ b ∣ B ‾ j ( a ( l + z ) b − y − r q ) B ‾ m + n − j ( x q + l + z b ) , (51) \begin{aligned}
\frac{1}{(a,b)^{m+n-j-1}}
\sum_{l=1}^{\left\lfloor\left|\frac{b}{(a,b)}\right|\right\rfloor}
\overline{B}_j\left(\frac{a(l+z)}{b}-\frac{y-r}{q}\right)
&\times\overline{B}_{m+n-j}\left(\frac{(a,b)x}{q}+\frac{(a,b)(l+z)}{b}\right)\\
&=\sum_{k=0}^{(a,b)-1}\sum_{l=1}^{\left\lfloor\left|\frac{b}{(a,b)}\right|\right\rfloor}
\overline{B}_j\left(\frac{a\left(l+z+k\left\lfloor\frac{b}{(a,b)}\right\rfloor\right)}{b}-\frac{y-r}{q}\right)\\
&\quad\times\overline{B}_{m+n-j}\left(\frac{x}{q}+\frac{l+z+k\left\lfloor\frac{b}{(a,b)}\right\rfloor}{b}\right)\\
&=\sum_{l=1}^{|b|}\overline{B}_j\left(\frac{a(l+z)}{b}-\frac{y-r}{q}\right)\overline{B}_{m+n-j}\left(\frac{x}{q}+\frac{l+z}{b}\right), \tag*{(51)}
\end{aligned} ( a , b ) m + n − j − 1 1 l = 1 ∑ ⌊ ∣ ( a , b ) b ∣ ⌋ B j ( b a ( l + z ) − q y − r ) × B m + n − j ( q ( a , b ) x + b ( a , b ) ( l + z ) ) = k = 0 ∑ ( a , b ) − 1 l = 1 ∑ ⌊ ∣ ( a , b ) b ∣ ⌋ B j b a ( l + z + k ⌊ ( a , b ) b ⌋ ) − q y − r × B m + n − j q x + b l + z + k ⌊ ( a , b ) b ⌋ = l = 1 ∑ ∣ b ∣ B j ( b a ( l + z ) − q y − r ) B m + n − j ( q x + b l + z ) , (51) and
1 ( a , b ) m + n − j − 1 ∑ l = 1 ∣ a ( a , b ) ∣ B ‾ j ( b l a + b y − b r q − z ) × B ‾ m + n − j ( ( a , b ) x q + ( a , b ) ( l + y q ) a − ( a , b ) r a q ) = ∑ k = 0 ( a , b ) − 1 ∑ l = 1 ∣ a ( a , b ) ∣ B ‾ j ( b ( l + k ⌊ a ( a , b ) ⌋ ) a + b y − b r q − z ) × B ‾ m + n − j ( x q + l + k ⌊ a ( a , b ) ⌋ + y q a − r a q ) = ∑ l = 1 ∣ a ∣ B ‾ j ( b l a + b y − b r a q − z ) B ‾ m + n − j ( x q + l a + y − r a q ) . (52) \begin{aligned}
&\frac{1}{(a,b)^{m+n-j-1}}\sum_{l=1}^{\left\lvert\frac{a}{(a,b)}\right\rvert}\overline{B}_j\left(\frac{bl}{a}+\frac{by-br}{q}-z\right)\\
&\quad\times\overline{B}_{m+n-j}\left(\frac{(a,b)x}{q}+\frac{(a,b)\left(l+\frac{y}{q}\right)}{a}-\frac{(a,b)r}{aq}\right)\\
&=\sum_{k=0}^{(a,b)-1}\sum_{l=1}^{\left\lvert\frac{a}{(a,b)}\right\rvert}\overline{B}_j\left(\frac{b\left(l+k\left\lfloor\frac{a}{(a,b)}\right\rfloor\right)}{a}+\frac{by-br}{q}-z\right)\\
&\quad\times\overline{B}_{m+n-j}\left(\frac{x}{q}+\frac{l+k\left\lfloor\frac{a}{(a,b)}\right\rfloor+\frac{y}{q}}{a}-\frac{r}{aq}\right)\\
&=\sum_{l=1}^{|a|}\overline{B}_j\left(\frac{bl}{a}+\frac{by-br}{aq}-z\right)\overline{B}_{m+n-j}\left(\frac{x}{q}+\frac{l}{a}+\frac{y-r}{aq}\right). \tag*{(52)}
\end{aligned} ( a , b ) m + n − j − 1 1 l = 1 ∑ ∣ ( a , b ) a ∣ B j ( a b l + q b y − b r − z ) × B m + n − j q ( a , b ) x + a ( a , b ) ( l + q y ) − a q ( a , b ) r = k = 0 ∑ ( a , b ) − 1 l = 1 ∑ ∣ ( a , b ) a ∣ B j a b ( l + k ⌊ ( a , b ) a ⌋ ) + q b y − b r − z × B m + n − j q x + a l + k ⌊ ( a , b ) a ⌋ + q y − a q r = l = 1 ∑ ∣ a ∣ B j ( a b l + a q b y − b r − z ) B m + n − j ( q x + a l + a q y − r ) . (52) Therefore, applying (51) and (52) to the right-hand side of (50) , we obtain (1.11), which completes the proof of Theorem 1.1.
The proof of Theorem 1.2. It is easily seen from (26) , (27) and the property of residue systems that, for j ∈ N j\in\mathbb{N} j ∈ N
χ 2 ( − 1 ) ∑ r = 1 q χ ‾ 2 ( r ) B ‾ j ( a ( l + z ) b − y − r q ) = ( − 1 ) j ∑ r = 1 q χ ‾ 2 ( − r ) B ‾ j ( − a ( l + z ) b + y − r q ) = ( − 1 ) j ∑ r = 1 q χ ‾ 2 ( r ) B ‾ j ( − a ( l + z ) b + y + r q ) = ( − 1 ) j q j − 1 B ‾ j ( − a q ( l + z ) b + y , χ ‾ 2 ) . (53) \begin{aligned}
\chi_2(-1)\sum_{r=1}^{q}\overline{\chi}_2(r)\overline{B}_j\left(\frac{a(l+z)}{b}-\frac{y-r}{q}\right)
&=(-1)^j\sum_{r=1}^{q}\overline{\chi}_2(-r)\overline{B}_j\left(-\frac{a(l+z)}{b}+\frac{y-r}{q}\right)\\
&=(-1)^j\sum_{r=1}^{q}\overline{\chi}_2(r)\overline{B}_j\left(-\frac{a(l+z)}{b}+\frac{y+r}{q}\right)\\
&=\frac{(-1)^j}{q^{j-1}}\overline{B}_j\left(-\frac{aq(l+z)}{b}+y,\overline{\chi}_2\right). \tag*{(53)}
\end{aligned} χ 2 ( − 1 ) r = 1 ∑ q χ 2 ( r ) B j ( b a ( l + z ) − q y − r ) = ( − 1 ) j r = 1 ∑ q χ 2 ( − r ) B j ( − b a ( l + z ) + q y − r ) = ( − 1 ) j r = 1 ∑ q χ 2 ( r ) B j ( − b a ( l + z ) + q y + r ) = q j − 1 ( − 1 ) j B j ( − b a q ( l + z ) + y , χ 2 ) . (53) Similarly, it follows from (27) , (33) and the property of residue systems that
χ 2 ( − 1 ) ∑ r = 1 q χ ‾ 2 ( r ) ∑ l = 1 a B ‾ m + n ( k + x c q + l a + y − r a q ) = 1 a m + n − 1 ∑ r = 1 q χ ‾ 2 ( − r ) B ‾ m + n ( a ( k + x ) c q + y − r q ) = 1 a m + n − 1 ∑ r = 1 q χ ‾ 2 ( r ) B ‾ m + n ( a ( k + x ) c q + y + r q ) = 1 ( a q ) m + n − 1 B ‾ m + n ( a ( k + x ) c + y , χ ‾ 2 ) . (54) \begin{aligned}
\chi_2(-1)\sum_{r=1}^{q}\overline{\chi}_2(r)\sum_{l=1}^{a}\overline{B}_{m+n}\left(\frac{k+x}{cq}+\frac{l}{a}+\frac{y-r}{aq}\right)
&=\frac{1}{a^{m+n-1}}\sum_{r=1}^{q}\overline{\chi}_2(-r)\overline{B}_{m+n}\left(\frac{a(k+x)}{cq}+\frac{y-r}{q}\right)\\
&=\frac{1}{a^{m+n-1}}\sum_{r=1}^{q}\overline{\chi}_2(r)\overline{B}_{m+n}\left(\frac{a(k+x)}{cq}+\frac{y+r}{q}\right)\\
&=\frac{1}{(aq)^{m+n-1}}\overline{B}_{m+n}\left(\frac{a(k+x)}{c}+y,\overline{\chi}_2\right). \tag*{(54)}
\end{aligned} χ 2 ( − 1 ) r = 1 ∑ q χ 2 ( r ) l = 1 ∑ a B m + n ( c q k + x + a l + a q y − r ) = a m + n − 1 1 r = 1 ∑ q χ 2 ( − r ) B m + n ( c q a ( k + x ) + q y − r ) = a m + n − 1 1 r = 1 ∑ q χ 2 ( r ) B m + n ( c q a ( k + x ) + q y + r ) = ( a q ) m + n − 1 1 B m + n ( c a ( k + x ) + y , χ 2 ) . (54) Hence, setting χ = χ 2 \chi=\chi_2 χ = χ 2 x x x ( k + x ) / c (k+x)/c ( k + x ) / c ( 1.11 ) (1.11) ( 1.11 ) k = 1 , … , c q k=1,\ldots,cq k = 1 , … , c q χ 1 ( k ) \chi_1(k) χ 1 ( k ) ( 3.15 ) (3.15) ( 3.15 ) ( 3.16 ) (3.16) ( 3.16 ) [1 Theorem 6.10] that
∑ k = 1 c q χ 1 ( k ) B ‾ m ( a ( k + x ) c + y , χ 2 ) B ‾ n ( b ( k + x ) c q + z ) = n q m − 1 b n − 1 ∑ j = 1 m ( m j ) a m − j q j − 1 ( m + n − j ) ∑ l = 1 b B ‾ j ( − a q ( l + z ) b + y , χ 2 ) × ∑ k = 1 c q χ 1 ( k ) B ‾ m + n − j ( k + x c q + l + z b ) + m ( a q ) m − 1 χ 2 ( − 1 ) ∑ j = 0 n ( n j ) ( − 1 ) j b n − j m + n − j ∑ k = 1 c q χ 1 ( k ) ∑ r = 1 q χ ‾ 2 ( r ) × ∑ l = 1 a B ‾ j ( b l a + b y − b r a q − z ) B ‾ m + n − j ( k + x c q + l a + y − r a q ) − δ 1 , m δ 1 , n χ 2 ( − 1 ) 4 ∑ k = 1 c q χ 1 ( k ) δ Z ( a ( k + x ) c + y ) δ Z ( b ( k + x ) c q + z ) × χ 2 ( a ( k + x ) c + y ) . (55) \begin{aligned}
\sum_{k=1}^{cq}\chi_1(k)\overline{B}_m\left(\frac{a(k+x)}{c}+y,\chi_2\right)\overline{B}_n\left(\frac{b(k+x)}{cq}+z\right)
={}&nq^{m-1}b^{n-1}\sum_{j=1}^{m}\binom{m}{j}\frac{a^{m-j}}{q^{j-1}(m+n-j)}\sum_{l=1}^{b}\overline{B}_j\left(\frac{-aq(l+z)}{b}+y,\chi_2\right)\\
&\times\sum_{k=1}^{cq}\chi_1(k)\overline{B}_{m+n-j}\left(\frac{k+x}{cq}+\frac{l+z}{b}\right)\\
&+m(aq)^{m-1}\chi_2(-1)\sum_{j=0}^{n}\binom{n}{j}\frac{(-1)^jb^{n-j}}{m+n-j}\sum_{k=1}^{cq}\chi_1(k)\sum_{r=1}^{q}\overline{\chi}_2(r)\\
&\times\sum_{l=1}^{a}\overline{B}_j\left(\frac{bl}{a}+\frac{by-br}{aq}-z\right)\overline{B}_{m+n-j}\left(\frac{k+x}{cq}+\frac{l}{a}+\frac{y-r}{aq}\right)\\
&-\frac{\delta_{1,m}\delta_{1,n}\chi_2(-1)}{4}\sum_{k=1}^{cq}\chi_1(k)\delta_{\mathbb{Z}}\left(\frac{a(k+x)}{c}+y\right)\delta_{\mathbb{Z}}\left(\frac{b(k+x)}{cq}+z\right)\\
&\times\chi_2\left(\frac{a(k+x)}{c}+y\right). \tag*{(55)}
\end{aligned} k = 1 ∑ c q χ 1 ( k ) B m ( c a ( k + x ) + y , χ 2 ) B n ( c q b ( k + x ) + z ) = n q m − 1 b n − 1 j = 1 ∑ m ( j m ) q j − 1 ( m + n − j ) a m − j l = 1 ∑ b B j ( b − a q ( l + z ) + y , χ 2 ) × k = 1 ∑ c q χ 1 ( k ) B m + n − j ( c q k + x + b l + z ) + m ( a q ) m − 1 χ 2 ( − 1 ) j = 0 ∑ n ( j n ) m + n − j ( − 1 ) j b n − j k = 1 ∑ c q χ 1 ( k ) r = 1 ∑ q χ 2 ( r ) × l = 1 ∑ a B j ( a b l + a q b y − b r − z ) B m + n − j ( c q k + x + a l + a q y − r ) − 4 δ 1 , m δ 1 , n χ 2 ( − 1 ) k = 1 ∑ c q χ 1 ( k ) δ Z ( c a ( k + x ) + y ) δ Z ( c q b ( k + x ) + z ) × χ 2 ( c a ( k + x ) + y ) . (55) Note that from ( 2.8 ) (2.8) ( 2.8 ) ( 2.14 ) (2.14) ( 2.14 )
∑ k = 1 c q χ 1 ( k ) B ‾ m + n − j ( k + x c q + l + z b ) = ∑ l = 0 c − 1 ∑ s = 1 q χ 1 ( q l + s ) B ‾ m + n − j ( q l + s + x c q + l + z b ) = 1 c m + n − j − 1 ∑ s = 1 q χ 1 ( s ) B ‾ m + n − j ( s + x q + c ( l + z ) b ) = 1 ( c q ) m + n − j − 1 B ‾ m + n − j ( c q ( l + z ) b + x , χ 1 ) . (56) \begin{aligned}
\sum_{k=1}^{cq}\chi_1(k)\overline{B}_{m+n-j}\left(\frac{k+x}{cq}+\frac{l+z}{b}\right)
={}&\sum_{l=0}^{c-1}\sum_{s=1}^{q}\chi_1(ql+s)\overline{B}_{m+n-j}\left(\frac{ql+s+x}{cq}+\frac{l+z}{b}\right)\\
={}&\frac{1}{c^{m+n-j-1}}\sum_{s=1}^{q}\chi_1(s)\overline{B}_{m+n-j}\left(\frac{s+x}{q}+\frac{c(l+z)}{b}\right)\\
={}&\frac{1}{(cq)^{m+n-j-1}}\overline{B}_{m+n-j}\left(\frac{cq(l+z)}{b}+x,\chi_1\right). \tag*{(56)}
\end{aligned} k = 1 ∑ c q χ 1 ( k ) B m + n − j ( c q k + x + b l + z ) = = = l = 0 ∑ c − 1 s = 1 ∑ q χ 1 ( q l + s ) B m + n − j ( c q q l + s + x + b l + z ) c m + n − j − 1 1 s = 1 ∑ q χ 1 ( s ) B m + n − j ( q s + x + b c ( l + z ) ) ( c q ) m + n − j − 1 1 B m + n − j ( b c q ( l + z ) + x , χ 1 ) . (56) and
∑ k = 1 c q χ 1 ( k ) ∑ r = 1 q χ 2 ‾ ( r ) ∑ l = 1 a B ‾ j ( b l a + b y − b r a q − z ) B ‾ m + n − j ( k + x c q + l a + y − r a q ) = ∑ t = 0 c − 1 ∑ s = 1 q χ 1 ( q t + s ) ∑ r = 1 q χ 2 ‾ ( r ) ∑ l = 1 a B ‾ j ( b l a + b y − b r a q − z ) × B ‾ m + n − j ( t c + s + x c q + l a + y − r a q ) = 1 c m + n − j − 1 ∑ s = 1 q χ 1 ( s ) ∑ r = 1 q χ 2 ‾ ( r ) ∑ l = 1 a B ‾ j ( b l a + b y − b r a q − z ) × B ‾ m + n − j ( s + x q + c l a + c ( y − r ) a q ) = χ 2 ( − 1 ) ( c q ) m + n − j − 1 ∑ r = 1 q ∑ l = 1 a χ 2 ‾ ( q l − r ) B ‾ j ( b l a + b y − b r a q − z ) × B ‾ m + n − j ( x + c q l a + c ( y − r ) a , χ 1 ) = χ 2 ( − 1 ) ( c q ) m + n − j − 1 ∑ k = 1 a q χ 2 ‾ ( k ) B ‾ j ( b ( k + y ) a q − z ) B ‾ m + n − j ( c ( k + y ) a + x , χ 1 ) . \begin{aligned}
&\sum_{k=1}^{cq}\chi_1(k)\sum_{r=1}^{q}\overline{\chi_2}(r)\sum_{l=1}^{a}\overline{B}_j\left(\frac{bl}{a}+\frac{by-br}{aq}-z\right)\overline{B}_{m+n-j}\left(\frac{k+x}{cq}+\frac{l}{a}+\frac{y-r}{aq}\right)\\
={}&\sum_{t=0}^{c-1}\sum_{s=1}^{q}\chi_1(qt+s)\sum_{r=1}^{q}\overline{\chi_2}(r)\sum_{l=1}^{a}\overline{B}_j\left(\frac{bl}{a}+\frac{by-br}{aq}-z\right)\\
&\qquad\times\overline{B}_{m+n-j}\left(\frac{t}{c}+\frac{s+x}{cq}+\frac{l}{a}+\frac{y-r}{aq}\right)\\
={}&\frac{1}{c^{m+n-j-1}}\sum_{s=1}^{q}\chi_1(s)\sum_{r=1}^{q}\overline{\chi_2}(r)\sum_{l=1}^{a}\overline{B}_j\left(\frac{bl}{a}+\frac{by-br}{aq}-z\right)\\
&\qquad\times\overline{B}_{m+n-j}\left(\frac{s+x}{q}+\frac{cl}{a}+\frac{c(y-r)}{aq}\right)\\
={}&\frac{\chi_2(-1)}{(cq)^{m+n-j-1}}\sum_{r=1}^{q}\sum_{l=1}^{a}\overline{\chi_2}(ql-r)\overline{B}_j\left(\frac{bl}{a}+\frac{by-br}{aq}-z\right)\\
&\qquad\times\overline{B}_{m+n-j}\left(x+\frac{cql}{a}+\frac{c(y-r)}{a},\chi_1\right)\\
={}&\frac{\chi_2(-1)}{(cq)^{m+n-j-1}}\sum_{k=1}^{aq}\overline{\chi_2}(k)\overline{B}_j\left(\frac{b(k+y)}{aq}-z\right)\overline{B}_{m+n-j}\left(\frac{c(k+y)}{a}+x,\chi_1\right).
\end{aligned} = = = = k = 1 ∑ c q χ 1 ( k ) r = 1 ∑ q χ 2 ( r ) l = 1 ∑ a B j ( a b l + a q b y − b r − z ) B m + n − j ( c q k + x + a l + a q y − r ) t = 0 ∑ c − 1 s = 1 ∑ q χ 1 ( q t + s ) r = 1 ∑ q χ 2 ( r ) l = 1 ∑ a B j ( a b l + a q b y − b r − z ) × B m + n − j ( c t + c q s + x + a l + a q y − r ) c m + n − j − 1 1 s = 1 ∑ q χ 1 ( s ) r = 1 ∑ q χ 2 ( r ) l = 1 ∑ a B j ( a b l + a q b y − b r − z ) × B m + n − j ( q s + x + a c l + a q c ( y − r ) ) ( c q ) m + n − j − 1 χ 2 ( − 1 ) r = 1 ∑ q l = 1 ∑ a χ 2 ( q l − r ) B j ( a b l + a q b y − b r − z ) × B m + n − j ( x + a c q l + a c ( y − r ) , χ 1 ) ( c q ) m + n − j − 1 χ 2 ( − 1 ) k = 1 ∑ a q χ 2 ( k ) B j ( a q b ( k + y ) − z ) B m + n − j ( a c ( k + y ) + x , χ 1 ) . Thus, inserting the above two identities into (3.17) and then replacing x x x z z z y y y x x x z z z y y y
The proof of Theorem 1.3. Replacing x x x ( k + x ) / c (k+x)/c ( k + x ) / c k = 1 , … , c q k=1,\ldots,cq k = 1 , … , c q
∑ k = 1 c q B ‾ m ( a ( k + x ) c + y , χ ‾ ) B ‾ n ( b ( k + x ) c q + z ) = n ( b c q ) m − 1 ∑ j = 1 n − 1 ( m j ) ( a c ) m − j m + n − j ∑ l = 1 b B ‾ j ( − a q ( l + z ) b + y , χ ‾ ) × B ‾ m + n − j ( c q ( l + z ) b + x ) + m ( a c ) m − 1 χ ( − 1 ) ∑ j = 0 n ( n j ) ( − 1 ) j ( b c q ) n − j m + n − j ∑ r = 1 q χ ‾ ( r ) × ∑ l = 1 a B ‾ j ( b ( q l − r + y ) a q − z ) B ‾ m + n − j ( c ( q l − r + y ) a + x ) + ( − 1 ) n − 1 m ! n ! ( a , b ) m + n χ ( b ( a , b ) ) B ‾ m + n ( b y − a q z ( a , b ) , χ ‾ ) a n q n − 1 b m ( m + n ) ! − δ 1 , m δ 1 , n χ ( − 1 ) 4 ∑ k = 1 c q δ Z ( a ( k + x ) c + y ) δ Z ( b ( k + x ) c q + z ) × χ ‾ ( a ( k + x ) c + y ) . (57) \begin{aligned}
&\sum_{k=1}^{cq}\overline{B}_m\left(\frac{a(k+x)}{c}+y,\overline{\chi}\right)\overline{B}_n\left(\frac{b(k+x)}{cq}+z\right)\\
={}&n\left(\frac{b}{cq}\right)^{m-1}\sum_{j=1}^{n-1}\binom{m}{j}\frac{\left(\frac{a}{c}\right)^{m-j}}{m+n-j}\sum_{l=1}^{b}\overline{B}_j\left(-\frac{aq(l+z)}{b}+y,\overline{\chi}\right)\\
&\qquad\times\overline{B}_{m+n-j}\left(\frac{cq(l+z)}{b}+x\right)\\
&\quad+m\left(\frac{a}{c}\right)^{m-1}\chi(-1)\sum_{j=0}^{n}\binom{n}{j}\frac{(-1)^j\left(\frac{b}{cq}\right)^{n-j}}{m+n-j}\sum_{r=1}^{q}\overline{\chi}(r)\\
&\qquad\times\sum_{l=1}^{a}\overline{B}_j\left(\frac{b(ql-r+y)}{aq}-z\right)\overline{B}_{m+n-j}\left(\frac{c(ql-r+y)}{a}+x\right)\\
&\quad+\frac{(-1)^{n-1}m!n!(a,b)^{m+n}\chi\left(\frac{b}{(a,b)}\right)\overline{B}_{m+n}\left(\frac{by-aqz}{(a,b)},\overline{\chi}\right)}{a^nq^n-1^bm(m+n)!}\\
&\quad-\frac{\delta_{1,m}\delta_{1,n}\chi(-1)}{4}\sum_{k=1}^{cq}\delta_{\mathbb Z}\left(\frac{a(k+x)}{c}+y\right)\delta_{\mathbb Z}\left(\frac{b(k+x)}{cq}+z\right)\\
&\qquad\times\overline{\chi}\left(\frac{a(k+x)}{c}+y\right). \tag*{(57)}
\end{aligned} = k = 1 ∑ c q B m ( c a ( k + x ) + y , χ ) B n ( c q b ( k + x ) + z ) n ( c q b ) m − 1 j = 1 ∑ n − 1 ( j m ) m + n − j ( c a ) m − j l = 1 ∑ b B j ( − b a q ( l + z ) + y , χ ) × B m + n − j ( b c q ( l + z ) + x ) + m ( c a ) m − 1 χ ( − 1 ) j = 0 ∑ n ( j n ) m + n − j ( − 1 ) j ( c q b ) n − j r = 1 ∑ q χ ( r ) × l = 1 ∑ a B j ( a q b ( q l − r + y ) − z ) B m + n − j ( a c ( q l − r + y ) + x ) + a n q n − 1 b m ( m + n )! ( − 1 ) n − 1 m ! n ! ( a , b ) m + n χ ( ( a , b ) b ) B m + n ( ( a , b ) b y − a q z , χ ) − 4 δ 1 , m δ 1 , n χ ( − 1 ) k = 1 ∑ c q δ Z ( c a ( k + x ) + y ) δ Z ( c q b ( k + x ) + z ) × χ ( c a ( k + x ) + y ) . (57) Observe that
χ ( − 1 ) ∑ r = 1 q χ ‾ ( r ) ∑ l = 1 a B ‾ j ( b ( q l − r + y ) a q − z ) B ‾ m + n − j ( c ( q l − r + y ) a + x ) = ∑ l = 1 a q χ ‾ ( l ) B ‾ j ( b ( l + y ) a q − z ) B ‾ m + n − j ( c ( l + y ) a + x ) . (58) \chi(-1)\sum_{r=1}^{q}\overline{\chi}(r)\sum_{l=1}^{a}\overline{B}_{j}\left(\frac{b(ql-r+y)}{aq}-z\right)\overline{B}_{m+n-j}\left(\frac{c(ql-r+y)}{a}+x\right)
=
\sum_{l=1}^{aq}\overline{\chi}(l)\overline{B}_{j}\left(\frac{b(l+y)}{aq}-z\right)\overline{B}_{m+n-j}\left(\frac{c(l+y)}{a}+x\right). \tag*{(58)} χ ( − 1 ) r = 1 ∑ q χ ( r ) l = 1 ∑ a B j ( a q b ( q l − r + y ) − z ) B m + n − j ( a c ( q l − r + y ) + x ) = l = 1 ∑ a q χ ( l ) B j ( a q b ( l + y ) − z ) B m + n − j ( a c ( l + y ) + x ) . (58) Now, inserting (3.20) into (3.19) and then replacing x x x z z z y y y x x x z z z y y y χ ‾ \overline{\chi} χ χ \chi χ
The proof of Theorem 1.4. Replacing q q q q 2 q_{2} q 2 j ∈ N j\in\mathbb{N} j ∈ N
χ 2 ( − 1 ) ∑ r = 1 q 2 χ ‾ 2 ( r ) B ‾ j ( a ( l + z ) b − y − r q 2 ) = ( − 1 ) j q 2 j − 1 B ‾ j ( − a q 2 ( l + z ) b + y , χ ‾ 2 ) . (59) \chi_{2}(-1)\sum_{r=1}^{q_{2}}\overline{\chi}_{2}(r)\overline{B}_{j}\left(\frac{a(l+z)}{b}-\frac{y-r}{q_{2}}\right)
=
\frac{(-1)^{j}}{q_{2}^{j-1}}\overline{B}_{j}\left(-\frac{aq_{2}(l+z)}{b}+y,\overline{\chi}_{2}\right). \tag*{(59)} χ 2 ( − 1 ) r = 1 ∑ q 2 χ 2 ( r ) B j ( b a ( l + z ) − q 2 y − r ) = q 2 j − 1 ( − 1 ) j B j ( − b a q 2 ( l + z ) + y , χ 2 ) . (59) Similarly, replacing q q q q 2 q_{2} q 2 c c c c q 1 cq_{1} c q 1
χ 2 ( − 1 ) ∑ r = 1 q 2 χ ‾ 2 ( r ) ∑ l = 1 a B ‾ m + n ( k + x c q 1 q 2 + l a + y − r a q 2 ) = 1 ( a q 2 ) m + n − 1 B ‾ m + n ( a ( k + x ) c q 1 + y , χ ‾ 2 ) . (60) \chi_{2}(-1)\sum_{r=1}^{q_{2}}\overline{\chi}_{2}(r)\sum_{l=1}^{a}\overline{B}_{m+n}\left(\frac{k+x}{cq_{1}q_{2}}+\frac{l}{a}+\frac{y-r}{aq_{2}}\right)
=
\frac{1}{(aq_{2})^{m+n-1}}\overline{B}_{m+n}\left(\frac{a(k+x)}{cq_{1}}+y,\overline{\chi}_{2}\right). \tag*{(60)} χ 2 ( − 1 ) r = 1 ∑ q 2 χ 2 ( r ) l = 1 ∑ a B m + n ( c q 1 q 2 k + x + a l + a q 2 y − r ) = ( a q 2 ) m + n − 1 1 B m + n ( c q 1 a ( k + x ) + y , χ 2 ) . (60) Hence, setting χ = χ 2 \chi=\chi_{2} χ = χ 2 x x x ( k + x ) / c q 1 (k+x)/cq_{1} ( k + x ) / c q 1 k = 1 , … , c q 1 q 2 k=1,\ldots,cq_{1}q_{2} k = 1 , … , c q 1 q 2 χ 1 ( k ) \chi_{1}(k) χ 1 ( k )
∑ k = 1 c q 1 q 2 χ 1 ( k ) B ‾ m ( a ( k + x ) c q 1 + y , χ ‾ 2 ) B ‾ n ( b ( k + x ) c q 1 q 2 + z ) = n q 2 m − 1 b n − 1 ∑ j = 1 m ( m j ) a m − j q 2 j − 1 ( m + n − j ) ∑ l = 1 b B ‾ j ( − a q 2 ( l + z ) b + y , χ ‾ 2 ) × ∑ k = 1 c q 1 q 2 χ 1 ( k ) B ‾ m + n − j ( k + x c q 1 q 2 + l + z b ) + m ( a q 2 ) m − 1 χ 2 ( − 1 ) ∑ j = 0 n ( n j ) ( − 1 ) j b n − j m + n − j ∑ k = 1 c q 1 q 2 χ 1 ( k ) ∑ r = 1 q 2 χ ‾ 2 ( r ) × ∑ l = 1 a B ‾ j ( b l a + b y − b r a q 2 − z ) B ‾ m + n − j ( k + x c q 1 q 2 + l a + y − r a q 2 ) − δ 1 , m δ 1 , n χ 2 ( − 1 ) 4 ∑ k = 1 c q 1 q 2 χ 1 ( k ) δ Z ( a ( k + x ) c q 1 + y ) δ Z ( b ( k + x ) c q 1 q 2 + z ) × χ ‾ 2 ( a ( k + x ) c q 1 + y ) . (61) \begin{aligned}
\sum_{k=1}^{cq_{1}q_{2}}\chi_{1}(k)\overline{B}_{m}\left(\frac{a(k+x)}{cq_{1}}+y,\overline{\chi}_{2}\right)\overline{B}_{n}\left(\frac{b(k+x)}{cq_{1}q_{2}}+z\right)
={}&nq_{2}^{m-1}b^{n-1}\sum_{j=1}^{m}\binom{m}{j}\frac{a^{m-j}}{q_{2}^{j-1}(m+n-j)}
\sum_{l=1}^{b}\overline{B}_{j}\left(-\frac{aq_{2}(l+z)}{b}+y,\overline{\chi}_{2}\right)\\
&\times\sum_{k=1}^{cq_{1}q_{2}}\chi_{1}(k)\overline{B}_{m+n-j}\left(\frac{k+x}{cq_{1}q_{2}}+\frac{l+z}{b}\right)\\
&+m(aq_{2})^{m-1}\chi_{2}(-1)\sum_{j=0}^{n}\binom{n}{j}\frac{(-1)^{j}b^{n-j}}{m+n-j}
\sum_{k=1}^{cq_{1}q_{2}}\chi_{1}(k)\sum_{r=1}^{q_{2}}\overline{\chi}_{2}(r)\\
&\times\sum_{l=1}^{a}\overline{B}_{j}\left(\frac{bl}{a}+\frac{by-br}{aq_{2}}-z\right)\overline{B}_{m+n-j}\left(\frac{k+x}{cq_{1}q_{2}}+\frac{l}{a}+\frac{y-r}{aq_{2}}\right)\\
&-\frac{\delta_{1,m}\delta_{1,n}\chi_{2}(-1)}{4}\sum_{k=1}^{cq_{1}q_{2}}\chi_{1}(k)\delta_{Z}\left(\frac{a(k+x)}{cq_{1}}+y\right)\delta_{Z}\left(\frac{b(k+x)}{cq_{1}q_{2}}+z\right)\\
&\times\overline{\chi}_{2}\left(\frac{a(k+x)}{cq_{1}}+y\right). \tag*{(61)}
\end{aligned} k = 1 ∑ c q 1 q 2 χ 1 ( k ) B m ( c q 1 a ( k + x ) + y , χ 2 ) B n ( c q 1 q 2 b ( k + x ) + z ) = n q 2 m − 1 b n − 1 j = 1 ∑ m ( j m ) q 2 j − 1 ( m + n − j ) a m − j l = 1 ∑ b B j ( − b a q 2 ( l + z ) + y , χ 2 ) × k = 1 ∑ c q 1 q 2 χ 1 ( k ) B m + n − j ( c q 1 q 2 k + x + b l + z ) + m ( a q 2 ) m − 1 χ 2 ( − 1 ) j = 0 ∑ n ( j n ) m + n − j ( − 1 ) j b n − j k = 1 ∑ c q 1 q 2 χ 1 ( k ) r = 1 ∑ q 2 χ 2 ( r ) × l = 1 ∑ a B j ( a b l + a q 2 b y − b r − z ) B m + n − j ( c q 1 q 2 k + x + a l + a q 2 y − r ) − 4 δ 1 , m δ 1 , n χ 2 ( − 1 ) k = 1 ∑ c q 1 q 2 χ 1 ( k ) δ Z ( c q 1 a ( k + x ) + y ) δ Z ( c q 1 q 2 b ( k + x ) + z ) × χ 2 ( c q 1 a ( k + x ) + y ) . (61) Note that replacing q q q q 1 q_1 q 1 c c c c q 2 cq_2 c q 2 (56) yields
∑ k = 1 c q 1 q 2 χ 1 ( k ) B ‾ m + n − j ( k + x c q 1 q 2 + l + z b ) = 1 ( c q 1 q 2 ) m + n − j − 1 B ‾ m + n − j ( c q 1 q 2 ( l + z ) b + x , χ 1 ) ; (62) \begin{aligned}
\sum_{k=1}^{cq_1q_2} \chi_1(k)\overline{B}_{m+n-j}\left(\frac{k+x}{cq_1q_2}+\frac{l+z}{b}\right)
&=\frac{1}{(cq_1q_2)^{m+n-j-1}}\overline{B}_{m+n-j}\left(\frac{cq_1q_2(l+z)}{b}+x,\chi_1\right); \tag*{(62)}
\end{aligned} k = 1 ∑ c q 1 q 2 χ 1 ( k ) B m + n − j ( c q 1 q 2 k + x + b l + z ) = ( c q 1 q 2 ) m + n − j − 1 1 B m + n − j ( b c q 1 q 2 ( l + z ) + x , χ 1 ) ; (62) and using (27) , (33) and the division algorithm, we obtain
∑ k = 1 c q 1 q 2 χ 1 ( k ) ∑ r = 1 q 2 χ ‾ 2 ( r ) ∑ l = 1 a B ‾ j ( b l a + b y − b r a q 2 − z ) B ‾ m + n − j ( k + x c q 1 q 2 + l a + y − r a q 2 ) = ∑ t = 0 c q 2 − 1 ∑ s = 1 q 1 χ 1 ( q 1 t + s ) ∑ r = 1 q 2 χ ‾ 2 ( r ) ∑ l = 1 a B ‾ j ( b l a + b y − b r a q 2 − z ) B ‾ m + n − j ( t c q 2 + s + x c q 1 q 2 + l a + y − r a q 2 ) = 1 ( c q 2 ) m + n − j − 1 ∑ s = 1 q 1 χ 1 ( s ) ∑ r = 1 q 2 χ ‾ 2 ( r ) ∑ l = 1 a B ‾ j ( b l a + b y − b r a q 2 − z ) B ‾ m + n − j ( s + x q 1 + c q 2 l a + c ( y − r ) a ) = χ 2 ( − 1 ) ( c q 1 q 2 ) m + n − j − 1 ∑ r = 1 q 2 ∑ l = 1 a χ ‾ 2 ( q 2 l − r ) B ‾ j ( b l a + b y − b r a q 2 − z ) B ‾ m + n − j ( x + c q 1 q 2 l a + c q 1 ( y − r ) a , χ 1 ) = χ 2 ( − 1 ) ( c q 1 q 2 ) m + n − j − 1 ∑ k = 1 a q 2 χ ‾ 2 ( k ) B ‾ j ( b ( k + y ) a q 2 − z ) B ‾ m + n − j ( c q 1 ( k + y ) a + x , χ 1 ) . (63) \begin{aligned}
&\sum_{k=1}^{cq_1q_2}\chi_1(k)\sum_{r=1}^{q_2}\overline{\chi}_2(r)\sum_{l=1}^{a}\overline{B}_j\left(\frac{bl}{a}+\frac{by-br}{aq_2}-z\right)\overline{B}_{m+n-j}\left(\frac{k+x}{cq_1q_2}+\frac{l}{a}+\frac{y-r}{aq_2}\right)\\
&=\sum_{t=0}^{cq_2-1}\sum_{s=1}^{q_1}\chi_1(q_1t+s)\sum_{r=1}^{q_2}\overline{\chi}_2(r)\sum_{l=1}^{a}\overline{B}_j\left(\frac{bl}{a}+\frac{by-br}{aq_2}-z\right)\overline{B}_{m+n-j}\left(\frac{t}{cq_2}+\frac{s+x}{cq_1q_2}+\frac{l}{a}+\frac{y-r}{aq_2}\right)\\
&=\frac{1}{(cq_2)^{m+n-j-1}}\sum_{s=1}^{q_1}\chi_1(s)\sum_{r=1}^{q_2}\overline{\chi}_2(r)\sum_{l=1}^{a}\overline{B}_j\left(\frac{bl}{a}+\frac{by-br}{aq_2}-z\right)\overline{B}_{m+n-j}\left(\frac{s+x}{q_1}+\frac{cq_2l}{a}+\frac{c(y-r)}{a}\right)\\
&=\frac{\chi_2(-1)}{(cq_1q_2)^{m+n-j-1}}\sum_{r=1}^{q_2}\sum_{l=1}^{a}\overline{\chi}_2(q_2l-r)\overline{B}_j\left(\frac{bl}{a}+\frac{by-br}{aq_2}-z\right)\overline{B}_{m+n-j}\left(x+\frac{cq_1q_2l}{a}+\frac{cq_1(y-r)}{a},\chi_1\right)\\
&=\frac{\chi_2(-1)}{(cq_1q_2)^{m+n-j-1}}\sum_{k=1}^{aq_2}\overline{\chi}_2(k)\overline{B}_j\left(\frac{b(k+y)}{aq_2}-z\right)\overline{B}_{m+n-j}\left(\frac{cq_1(k+y)}{a}+x,\chi_1\right). \tag*{(63)}
\end{aligned} k = 1 ∑ c q 1 q 2 χ 1 ( k ) r = 1 ∑ q 2 χ 2 ( r ) l = 1 ∑ a B j ( a b l + a q 2 b y − b r − z ) B m + n − j ( c q 1 q 2 k + x + a l + a q 2 y − r ) = t = 0 ∑ c q 2 − 1 s = 1 ∑ q 1 χ 1 ( q 1 t + s ) r = 1 ∑ q 2 χ 2 ( r ) l = 1 ∑ a B j ( a b l + a q 2 b y − b r − z ) B m + n − j ( c q 2 t + c q 1 q 2 s + x + a l + a q 2 y − r ) = ( c q 2 ) m + n − j − 1 1 s = 1 ∑ q 1 χ 1 ( s ) r = 1 ∑ q 2 χ 2 ( r ) l = 1 ∑ a B j ( a b l + a q 2 b y − b r − z ) B m + n − j ( q 1 s + x + a c q 2 l + a c ( y − r ) ) = ( c q 1 q 2 ) m + n − j − 1 χ 2 ( − 1 ) r = 1 ∑ q 2 l = 1 ∑ a χ 2 ( q 2 l − r ) B j ( a b l + a q 2 b y − b r − z ) B m + n − j ( x + a c q 1 q 2 l + a c q 1 ( y − r ) , χ 1 ) = ( c q 1 q 2 ) m + n − j − 1 χ 2 ( − 1 ) k = 1 ∑ a q 2 χ 2 ( k ) B j ( a q 2 b ( k + y ) − z ) B m + n − j ( a c q 1 ( k + y ) + x , χ 1 ) . (63) Therefore, inserting (62) and (63) into (61) and then replacing a a a a q 1 aq_1 a q 1 x x x z z z y y y x x x z z z y y y (19) . This concludes the proof of Theorem 1.4.
Some special cases
We begin by discussing a special case of Theorem 1.2. We define the sums s n ( a , b ; χ 1 , χ ‾ 2 ; x , y ) s_n(a,b;\chi_1,\overline{\chi}_2;x,y) s n ( a , b ; χ 1 , χ 2 ; x , y )
s n ( a , b ; χ 1 , χ ‾ 2 ; x , y ) = ∑ r = 1 b q χ 1 ( r ) B ‾ n ( a ( r + y ) b + x , χ ‾ 2 ) B ‾ 1 ( r + y b q ) . (64) s_n(a,b;\chi_1,\overline{\chi}_2;x,y)=\sum_{r=1}^{bq}\chi_1(r)\overline{B}_n\left(\frac{a(r+y)}{b}+x,\overline{\chi}_2\right)\overline{B}_1\left(\frac{r+y}{bq}\right). \tag*{(64)} s n ( a , b ; χ 1 , χ 2 ; x , y ) = r = 1 ∑ b q χ 1 ( r ) B n ( b a ( r + y ) + x , χ 2 ) B 1 ( b q r + y ) . (64) where n , q ∈ N n,q\in\mathbb{N} n , q ∈ N a , b ∈ Z a,b\in\mathbb{Z} a , b ∈ Z b ≥ 1 b\geq1 b ≥ 1 χ 1 \chi_1 χ 1 χ 2 \chi_2 χ 2 q q q x , y ∈ R x,y\in\mathbb{R} x , y ∈ R (64) is given in the following corollary.
Corollary 4.1. Let a , b , n , q ∈ N a,b,n,q \in\mathbb{N} a , b , n , q ∈ N q ≥ 2 q \ge2 q ≥ 2 χ 1 \chi_1 χ 1 χ 2 \chi_2 χ 2 q q q x , y ∈ R x,y \in\mathbb{R} x , y ∈ R
a b n s n ( a , b ; χ 1 , χ ‾ 2 ; x , y ) + b a n s n ( b , a ; χ ‾ 2 , χ 1 ; y , x ) = 1 n + 1 ∑ j = 1 n ( n + 1 j ) a j b n + 1 − j B ‾ j ( y , χ 1 ) B ‾ n + 1 − j ( x , χ ‾ 2 ) + n q n − 1 ( a , b ) n + 1 n + 1 ∑ k = 1 q ∑ j = 1 q χ 1 ( k ) χ ‾ 2 ( j ) B ‾ n + 1 ( a ( k + y ) + b ( j + x ) q ( a , b ) ) − δ 1 , n χ 2 ( − 1 ) a b 4 ∑ k = 1 b q χ 1 ( k ) δ Z ( a ( k + y ) b + x ) δ Z ( k + y b q ) × χ ‾ 2 ( a ( k + y ) b + x ) . (65) \begin{aligned}
&ab^n s_n(a,b;\chi_1,\overline{\chi}_2;x,y)+ba^n s_n(b,a;\overline{\chi}_2,\chi_1;y,x)\\
&=\frac{1}{n+1}\sum_{j=1}^{n}\binom{n+1}{j}a^j b^{n+1-j}\overline{B}_j(y,\chi_1)\overline{B}_{n+1-j}(x,\overline{\chi}_2)\\
&\quad+\frac{nq^{n-1}(a,b)^{n+1}}{n+1}\sum_{k=1}^{q}\sum_{j=1}^{q}\chi_1(k)\overline{\chi}_2(j)\overline{B}_{n+1}\left(\frac{a(k+y)+b(j+x)}{q(a,b)}\right)\\
&\quad-\frac{\delta_{1,n}\chi_2(-1)ab}{4}\sum_{k=1}^{bq}\chi_1(k)\delta_{\mathbb{Z}}\left(\frac{a(k+y)}{b}+x\right)\delta_{\mathbb{Z}}\left(\frac{k+y}{bq}\right)\\
&\qquad\times\overline{\chi}_2\left(\frac{a(k+y)}{b}+x\right). \tag*{(65)}
\end{aligned} a b n s n ( a , b ; χ 1 , χ 2 ; x , y ) + b a n s n ( b , a ; χ 2 , χ 1 ; y , x ) = n + 1 1 j = 1 ∑ n ( j n + 1 ) a j b n + 1 − j B j ( y , χ 1 ) B n + 1 − j ( x , χ 2 ) + n + 1 n q n − 1 ( a , b ) n + 1 k = 1 ∑ q j = 1 ∑ q χ 1 ( k ) χ 2 ( j ) B n + 1 ( q ( a , b ) a ( k + y ) + b ( j + x ) ) − 4 δ 1 , n χ 2 ( − 1 ) ab k = 1 ∑ b q χ 1 ( k ) δ Z ( b a ( k + y ) + x ) δ Z ( b q k + y ) × χ 2 ( b a ( k + y ) + x ) . (65) Proof. Taking n = b = 1 n=b=1 n = b = 1 y = 0 y=0 y = 0 Theorem 1.2 and then replacing m m m n n n c c c b b b z z z y y y
s n ( a , b ; χ 1 , χ ‾ 2 ; x , y ) + ( a b ) n − 1 s n ( b , a ; χ ‾ 2 , χ 1 ; y , x ) = 1 n + 1 ∑ j = 1 n ( n + 1 j ) ( a b ) n − j B ‾ j ( x , χ ‾ 2 ) B ‾ n + 1 − j ( y , χ 1 ) + n a n − 1 b n q ( n + 1 ) ∑ r = 1 a q χ ‾ 2 ( r ) B ‾ n + 1 ( b ( r + x ) a + y , χ 1 ) − δ 1 , n χ 2 ( − 1 ) 4 ∑ k = 1 b q χ 1 ( k ) δ Z ( a ( k + y ) b + x ) δ Z ( k + y b q ) × χ ‾ 2 ( a ( k + y ) b + x ) . (66) \begin{aligned}
&s_n(a,b;\chi_1,\overline{\chi}_2;x,y)+\left(\frac{a}{b}\right)^{n-1}s_n(b,a;\overline{\chi}_2,\chi_1;y,x)\\
&=\frac{1}{n+1}\sum_{j=1}^{n}\binom{n+1}{j}\left(\frac{a}{b}\right)^{n-j}\overline{B}_j(x,\overline{\chi}_2)\overline{B}_{n+1-j}(y,\chi_1)\\
&\quad+\frac{na^{n-1}}{b^nq(n+1)}\sum_{r=1}^{aq}\overline{\chi}_2(r)\overline{B}_{n+1}\left(\frac{b(r+x)}{a}+y,\chi_1\right)\\
&\quad-\frac{\delta_{1,n}\chi_2(-1)}{4}\sum_{k=1}^{bq}\chi_1(k)\delta_{\mathbb{Z}}\left(\frac{a(k+y)}{b}+x\right)\delta_{\mathbb{Z}}\left(\frac{k+y}{bq}\right)\\
&\qquad\times\overline{\chi}_2\left(\frac{a(k+y)}{b}+x\right). \tag*{(66)}
\end{aligned} s n ( a , b ; χ 1 , χ 2 ; x , y ) + ( b a ) n − 1 s n ( b , a ; χ 2 , χ 1 ; y , x ) = n + 1 1 j = 1 ∑ n ( j n + 1 ) ( b a ) n − j B j ( x , χ 2 ) B n + 1 − j ( y , χ 1 ) + b n q ( n + 1 ) n a n − 1 r = 1 ∑ a q χ 2 ( r ) B n + 1 ( a b ( r + x ) + y , χ 1 ) − 4 δ 1 , n χ 2 ( − 1 ) k = 1 ∑ b q χ 1 ( k ) δ Z ( b a ( k + y ) + x ) δ Z ( b q k + y ) × χ 2 ( b a ( k + y ) + x ) . (66) From (33) , the property of residue systems, and the division algorithm, it is easily seen that, for a , j ∈ N a,j \in\mathbb{N} a , j ∈ N b ∈ Z b \in\mathbb{Z} b ∈ Z b ≠ 0 b \ne0 b = 0 x ∈ R x \in\mathbb{R} x ∈ R
∑ l = 0 a − 1 B ‾ j ( x + b l a ) = ∑ r = 0 a ( a , b ) − 1 ∑ j = 0 ( a , b ) ( a , b ) − 1 B ‾ j ( x ( a , b ) + b ( a , b ) ( a ( a , b ) r + j ) a ( a , b ) ) = ( a , b ) ∑ j = 0 a ( a , b ) − 1 B ‾ j ( x ( a , b ) + b j ( a , b ) a ( a , b ) ) = ( a , b ) ∑ j = 0 a ( a , b ) − 1 B ‾ j ( x ( a , b ) + j a ( a , b ) ) = ( a , b ) j a − 1 − j B ‾ j ( x ( a , b ) ) . (67) \begin{aligned}
\sum_{l=0}^{a-1}\overline{B}_j\left(\frac{x+bl}{a}\right)
&=\sum_{r=0}^{\frac{a}{(a,b)}-1}\sum_{j=0}^{\frac{(a,b)}{(a,b)}-1}\overline{B}_j\left(\frac{\frac{x}{(a,b)}+\frac{b}{(a,b)}\left(\frac{a}{(a,b)}r+j\right)}{\frac{a}{(a,b)}}\right)\\
&=(a,b)\sum_{j=0}^{\frac{a}{(a,b)}-1}\overline{B}_j\left(\frac{\frac{x}{(a,b)}+\frac{bj}{(a,b)}}{\frac{a}{(a,b)}}\right)\\
&=(a,b)\sum_{j=0}^{\frac{a}{(a,b)}-1}\overline{B}_j\left(\frac{\frac{x}{(a,b)}+j}{\frac{a}{(a,b)}}\right)\\
&=(a,b)j^{a-1-j}\overline{B}_j\left(\frac{x}{(a,b)}\right). \tag*{(67)}
\end{aligned} l = 0 ∑ a − 1 B j ( a x + b l ) = r = 0 ∑ ( a , b ) a − 1 j = 0 ∑ ( a , b ) ( a , b ) − 1 B j ( a , b ) a ( a , b ) x + ( a , b ) b ( ( a , b ) a r + j ) = ( a , b ) j = 0 ∑ ( a , b ) a − 1 B j ( ( a , b ) a ( a , b ) x + ( a , b ) bj ) = ( a , b ) j = 0 ∑ ( a , b ) a − 1 B j ( ( a , b ) a ( a , b ) x + j ) = ( a , b ) j a − 1 − j B j ( ( a , b ) x ) . (67) It follows from (2.8) and (4.4) that
∑ r = 1 a q χ ‾ 2 ( r ) B ‾ n + 1 ( b ( r + x ) a + y , χ 1 ) = q n ∑ r = 1 a q χ ‾ 2 ( r ) ∑ k = 1 q χ 1 ( k ) B ‾ n + 1 ( b ( r + x ) a q + k + y q ) = q n ∑ l = 0 q − 1 ∑ j = 1 q χ ‾ 2 ( q l + j ) ∑ k = 1 q χ 1 ( k ) B ‾ n + 1 ( b ( j + x ) a q + b l a + k + y q ) = q n ( a , b ) n + 1 a n ∑ j = 1 q χ ‾ 2 ( j ) ∑ k = 1 q χ 1 ( k ) B ‾ n + 1 ( b ( j + x ) + a ( k + y ) q ( a , b ) ) . \begin{aligned}
\sum_{r=1}^{aq}\overline{\chi}_2(r)\overline{B}_{n+1}\left(\frac{b(r+x)}{a}+y,\chi_1\right)
&=q^n\sum_{r=1}^{aq}\overline{\chi}_2(r)\sum_{k=1}^{q}\chi_1(k)\overline{B}_{n+1}\left(\frac{b(r+x)}{aq}+\frac{k+y}{q}\right)\\
&=q^n\sum_{l=0}^{q-1}\sum_{j=1}^{q}\overline{\chi}_2(ql+j)\sum_{k=1}^{q}\chi_1(k)\overline{B}_{n+1}\left(\frac{b(j+x)}{aq}+\frac{bl}{a}+\frac{k+y}{q}\right)\\
&=\frac{q^n(a,b)^{n+1}}{a^n}\sum_{j=1}^{q}\overline{\chi}_2(j)\sum_{k=1}^{q}\chi_1(k)\overline{B}_{n+1}\left(\frac{b(j+x)+a(k+y)}{q(a,b)}\right).
\end{aligned} r = 1 ∑ a q χ 2 ( r ) B n + 1 ( a b ( r + x ) + y , χ 1 ) = q n r = 1 ∑ a q χ 2 ( r ) k = 1 ∑ q χ 1 ( k ) B n + 1 ( a q b ( r + x ) + q k + y ) = q n l = 0 ∑ q − 1 j = 1 ∑ q χ 2 ( q l + j ) k = 1 ∑ q χ 1 ( k ) B n + 1 ( a q b ( j + x ) + a b l + q k + y ) = a n q n ( a , b ) n + 1 j = 1 ∑ q χ 2 ( j ) k = 1 ∑ q χ 1 ( k ) B n + 1 ( q ( a , b ) b ( j + x ) + a ( k + y ) ) . This, together with (4.3), yields
s n ( a , b ; χ 1 , χ ‾ 2 ; x , y ) + ( a b ) n − 1 s n ( b , a ; χ ‾ 2 , χ 1 ; y , x ) = 1 n + 1 ∑ j = 1 n ( n + 1 j ) ( a b ) n − j B ‾ j ( x , χ ‾ 2 ) B ‾ n + 1 − j ( y , χ 1 ) + n q n − 1 ( a , b ) n + 1 a b n ( n + 1 ) ∑ k = 1 q ∑ j = 1 q χ 1 ( k ) χ ‾ 2 ( j ) B ‾ n + 1 ( a ( k + y ) + b ( j + x ) q ( a , b ) ) − δ 1 , n χ 2 ( − 1 ) 4 ∑ k = 1 b q χ 1 ( k ) δ Z ( a ( k + y ) b + x ) δ Z ( k + y b q ) × χ ‾ 2 ( a ( k + y ) b + x ) . (68) \begin{aligned}
s_n(a,b;\chi_1,\overline{\chi}_2;x,y)+\left(\frac{a}{b}\right)^{n-1}s_n(b,a;\overline{\chi}_2,\chi_1;y,x)
&=\frac{1}{n+1}\sum_{j=1}^{n}\binom{n+1}{j}\left(\frac{a}{b}\right)^{n-j}\overline{B}_j(x,\overline{\chi}_2)\overline{B}_{n+1-j}(y,\chi_1)\\
&\quad+\frac{nq^{n-1}(a,b)^{n+1}}{ab^n(n+1)}\sum_{k=1}^{q}\sum_{j=1}^{q}\chi_1(k)\overline{\chi}_2(j)\overline{B}_{n+1}\left(\frac{a(k+y)+b(j+x)}{q(a,b)}\right)\\
&\quad-\frac{\delta_{1,n}\chi_2(-1)}{4}\sum_{k=1}^{bq}\chi_1(k)\delta_{\mathbb Z}\left(\frac{a(k+y)}{b}+x\right)\delta_{\mathbb Z}\left(\frac{k+y}{bq}\right)\\
&\qquad\qquad\qquad\qquad\qquad\qquad\times\overline{\chi}_2\left(\frac{a(k+y)}{b}+x\right). \tag*{(68)}
\end{aligned} s n ( a , b ; χ 1 , χ 2 ; x , y ) + ( b a ) n − 1 s n ( b , a ; χ 2 , χ 1 ; y , x ) = n + 1 1 j = 1 ∑ n ( j n + 1 ) ( b a ) n − j B j ( x , χ 2 ) B n + 1 − j ( y , χ 1 ) + a b n ( n + 1 ) n q n − 1 ( a , b ) n + 1 k = 1 ∑ q j = 1 ∑ q χ 1 ( k ) χ 2 ( j ) B n + 1 ( q ( a , b ) a ( k + y ) + b ( j + x ) ) − 4 δ 1 , n χ 2 ( − 1 ) k = 1 ∑ b q χ 1 ( k ) δ Z ( b a ( k + y ) + x ) δ Z ( b q k + y ) × χ 2 ( b a ( k + y ) + x ) . (68) Now, multiplying both sides of (4.5) by a b n ab^n a b n
It is clear that, taking n = 1 n=1 n = 1 χ 1 = χ 2 = χ \chi_1=\chi_2=\chi χ 1 = χ 2 = χ ( a , b ) = 1 (a,b)=1 ( a , b ) = 1 a ≡ 0 a\equiv0 a ≡ 0 b ≡ 0 ( m o d q ) b\equiv0\pmod q b ≡ 0 ( mod q ) x = y = 0 x=y=0 x = y = 0
It is also easy to see that, taking χ 1 = χ 2 = χ \chi_1=\chi_2=\chi χ 1 = χ 2 = χ ( a , b ) = 1 (a,b)=1 ( a , b ) = 1 [9 Theorem 5.4] . In addition, Corollary 4.1 also implies that, for a , b , n , q ∈ N a,b,n,q\in\mathbb{N} a , b , n , q ∈ N q ≥ 2 q\ge2 q ≥ 2 χ 1 \chi_1 χ 1 χ 2 \chi_2 χ 2 q q q ( − 1 ) n + 1 χ 1 ( − 1 ) χ 2 ( − 1 ) = 1 (-1)^{n+1}\chi_1(-1)\chi_2(-1)=1 ( − 1 ) n + 1 χ 1 ( − 1 ) χ 2 ( − 1 ) = 1
a b n s n ( a , b ; χ 1 , χ ‾ 2 ; 0 , 0 ) + b a n s n ( b , a ; χ ‾ 2 , χ 1 ; 0 , 0 ) = 1 n + 1 ∑ j = 1 n ( n + 1 j ) a j b n + 1 − j B j ( χ 1 ) B n + 1 − j ( χ ‾ 2 ) + n q n − 1 ( a , b ) n + 1 n + 1 ∑ k = 1 q ∑ j = 1 q χ 1 ( k ) χ ‾ 2 ( j ) B ‾ n + 1 ( a k + b j q ( a , b ) ) . (69) \begin{aligned}
ab^n s_n(a,b;\chi_1,\overline{\chi}_2;0,0)+ba^n s_n(b,a;\overline{\chi}_2,\chi_1;0,0)
&=\frac{1}{n+1}\sum_{j=1}^{n}\binom{n+1}{j}a^j b^{n+1-j}B_j(\chi_1)B_{n+1-j}(\overline{\chi}_2)\\
&\quad+\frac{nq^{n-1}(a,b)^{n+1}}{n+1}\sum_{k=1}^{q}\sum_{j=1}^{q}\chi_1(k)\overline{\chi}_2(j)\overline{B}_{n+1}\left(\frac{ak+bj}{q(a,b)}\right). \tag*{(69)}
\end{aligned} a b n s n ( a , b ; χ 1 , χ 2 ; 0 , 0 ) + b a n s n ( b , a ; χ 2 , χ 1 ; 0 , 0 ) = n + 1 1 j = 1 ∑ n ( j n + 1 ) a j b n + 1 − j B j ( χ 1 ) B n + 1 − j ( χ 2 ) + n + 1 n q n − 1 ( a , b ) n + 1 k = 1 ∑ q j = 1 ∑ q χ 1 ( k ) χ 2 ( j ) B n + 1 ( q ( a , b ) ak + bj ) . (69) This means that Dağlı and Can’s reciprocity formula [11 Theorem 1] also holds in the case n = 1 n=1 n = 1
Similarly, we define the sums s n ( 1 ) ( a , b ; χ ; x , y ) s_n^{(1)}(a,b;\chi;x,y) s n ( 1 ) ( a , b ; χ ; x , y ) s n ( 2 ) ( a , b ; χ ; x , y ) s_n^{(2)}(a,b;\chi;x,y) s n ( 2 ) ( a , b ; χ ; x , y )
s n ( 1 ) ( a , b ; χ ; x , y ) = ∑ r = 1 b q χ ( r ) B ‾ n ( a ( r + y ) b + x ) B ‾ 1 ( r + y b q ) , (70) s_n^{(1)}(a,b;\chi;x,y)=\sum_{r=1}^{bq}\chi(r)\overline{B}_n\left(\frac{a(r+y)}{b}+x\right)\overline{B}_1\left(\frac{r+y}{bq}\right), \tag*{(70)} s n ( 1 ) ( a , b ; χ ; x , y ) = r = 1 ∑ b q χ ( r ) B n ( b a ( r + y ) + x ) B 1 ( b q r + y ) , (70) and
s n ( 2 ) ( a , b ; χ ; x , y ) = ∑ r = 1 b q B ‾ n ( a ( r + y ) b + x ) B ‾ 1 ( r + y b q ) , (71) s_n^{(2)}(a,b;\chi;x,y)=\sum_{r=1}^{bq}\overline{B}_n\left(\frac{a(r+y)}{b}+x\right)\overline{B}_1\left(\frac{r+y}{bq}\right), \tag*{(71)} s n ( 2 ) ( a , b ; χ ; x , y ) = r = 1 ∑ b q B n ( b a ( r + y ) + x ) B 1 ( b q r + y ) , (71) where n , q ∈ N n,q\in\mathbb{N} n , q ∈ N a , b ∈ Z a,b\in\mathbb{Z} a , b ∈ Z b ≥ 1 b\geq1 b ≥ 1 χ \chi χ q q q x , y ∈ R x,y\in\mathbb{R} x , y ∈ R
As a consequence of Theorem 1.3 , we have the reciprocity formula relating the sums (4.7) and (4.8) as follows.
Corollary 4.2. Let a , b , n , q ∈ N a,b,n,q\in\mathbb{N} a , b , n , q ∈ N q ≥ 2 q\geq2 q ≥ 2 χ \chi χ q q q x , y ∈ R x,y\in\mathbb{R} x , y ∈ R
a b n s n ( 1 ) ( a , b ; χ ; x , y ) + b a n s n ( 2 ) ( b , a ; χ ; y , x ) = 1 n + 1 ∑ j = 1 n + 1 ( n + 1 j ) a j b n + 1 − j B ‾ j ( y , χ ) B ‾ n + 1 − j ( x ) + n ( b , a q ) n + 1 q ( n + 1 ) ∑ k = 1 q χ ( k ) B ‾ n + 1 ( a ( k + y ) + b x ( b , a q ) ) − δ 1 , n χ ( − 1 ) a b 4 ∑ k = 1 a q δ Z ( b ( k + x ) a + y ) δ Z ( k + x a q ) × χ ( b ( k + x ) a + y ) . (72) \begin{aligned}
ab^n s_n^{(1)}(a,b;\chi;x,y)+ba^n s_n^{(2)}(b,a;\chi;y,x)
={}&\frac{1}{n+1}\sum_{j=1}^{n+1}\binom{n+1}{j}a^j b^{n+1-j}\overline{B}_j(y,\chi)\overline{B}_{n+1-j}(x)\\
&+\frac{n(b,aq)^{n+1}}{q(n+1)}\sum_{k=1}^{q}\chi(k)\overline{B}_{n+1}\left(\frac{a(k+y)+bx}{(b,aq)}\right)\\
&-\frac{\delta_{1,n}\chi(-1)ab}{4}\sum_{k=1}^{aq}\delta_{\mathbb{Z}}\left(\frac{b(k+x)}{a}+y\right)\delta_{\mathbb{Z}}\left(\frac{k+x}{aq}\right)\\
&\qquad\times\chi\left(\frac{b(k+x)}{a}+y\right). \tag*{(72)}
\end{aligned} a b n s n ( 1 ) ( a , b ; χ ; x , y ) + b a n s n ( 2 ) ( b , a ; χ ; y , x ) = n + 1 1 j = 1 ∑ n + 1 ( j n + 1 ) a j b n + 1 − j B j ( y , χ ) B n + 1 − j ( x ) + q ( n + 1 ) n ( b , a q ) n + 1 k = 1 ∑ q χ ( k ) B n + 1 ( ( b , a q ) a ( k + y ) + b x ) − 4 δ 1 , n χ ( − 1 ) ab k = 1 ∑ a q δ Z ( a b ( k + x ) + y ) δ Z ( a q k + x ) × χ ( a b ( k + x ) + y ) . (72) Proof. Taking n = b = 1 n=b=1 n = b = 1 y = 0 y=0 y = 0 Theorem 1.3 and then replacing m m m n n n c c c b b b z z z y y y
s n ( 2 ) ( a , b ; χ ; x , y ) + ( a b ) n − 1 s n ( 1 ) ( b , a ; χ ; y , x ) = 1 n + 1 ∑ j = 1 n + 1 ( n + 1 j ) ( a b ) n − j B ‾ j ( x , χ ) B ‾ n + 1 − j ( y ) + n a n − 1 b n q ( n + 1 ) ∑ r = 1 a q χ ( r ) B ‾ n + 1 ( b ( r + x ) a + y ) − δ 1 , n χ ( − 1 ) 4 ∑ k = 1 b q δ Z ( a ( k + y ) b + x ) δ Z ( k + y b q ) × χ ( a ( k + y ) b + x ) . (73) \begin{aligned}
s_n^{(2)}(a,b;\chi;x,y)+\left(\frac{a}{b}\right)^{n-1}s_n^{(1)}(b,a;\chi;y,x)
={}&\frac{1}{n+1}\sum_{j=1}^{n+1}\binom{n+1}{j}\left(\frac{a}{b}\right)^{n-j}\overline{B}_j(x,\chi)\overline{B}_{n+1-j}(y)\\
&+\frac{na^{n-1}}{b^nq(n+1)}\sum_{r=1}^{aq}\chi(r)\overline{B}_{n+1}\left(\frac{b(r+x)}{a}+y\right)\\
&-\frac{\delta_{1,n}\chi(-1)}{4}\sum_{k=1}^{bq}\delta_{\mathbb{Z}}\left(\frac{a(k+y)}{b}+x\right)\delta_{\mathbb{Z}}\left(\frac{k+y}{bq}\right)\\
&\qquad\times\chi\left(\frac{a(k+y)}{b}+x\right). \tag*{(73)}
\end{aligned} s n ( 2 ) ( a , b ; χ ; x , y ) + ( b a ) n − 1 s n ( 1 ) ( b , a ; χ ; y , x ) = n + 1 1 j = 1 ∑ n + 1 ( j n + 1 ) ( b a ) n − j B j ( x , χ ) B n + 1 − j ( y ) + b n q ( n + 1 ) n a n − 1 r = 1 ∑ a q χ ( r ) B n + 1 ( a b ( r + x ) + y ) − 4 δ 1 , n χ ( − 1 ) k = 1 ∑ b q δ Z ( b a ( k + y ) + x ) δ Z ( b q k + y ) × χ ( b a ( k + y ) + x ) . (73) Note that from (67) and the division algorithm we have
∑ r = 1 a q χ ( r ) B ‾ n + 1 ( b ( r + x ) a + y ) = ∑ l = 0 a − 1 ∑ k = 1 q χ ( q l + k ) B ‾ n + 1 ( b ( q l + k + x ) a + y ) = ( a , b q ) n + 1 a n ∑ k = 1 q χ ( k ) B ‾ n + 1 ( b ( k + x ) + a y ( a , b q ) ) . (74) \begin{aligned}
\sum_{r=1}^{aq}\chi(r)\overline{B}_{n+1}\left(\frac{b(r+x)}{a}+y\right)
&=\sum_{l=0}^{a-1}\sum_{k=1}^{q}\chi(ql+k)\overline{B}_{n+1}\left(\frac{b(ql+k+x)}{a}+y\right)\\
&=\frac{(a,bq)^{n+1}}{a^n}\sum_{k=1}^{q}\chi(k)\overline{B}_{n+1}\left(\frac{b(k+x)+ay}{(a,bq)}\right). \tag*{(74)}
\end{aligned} r = 1 ∑ a q χ ( r ) B n + 1 ( a b ( r + x ) + y ) = l = 0 ∑ a − 1 k = 1 ∑ q χ ( q l + k ) B n + 1 ( a b ( q l + k + x ) + y ) = a n ( a , b q ) n + 1 k = 1 ∑ q χ ( k ) B n + 1 ( ( a , b q ) b ( k + x ) + a y ) . (74) Applying (74) to the right-hand side of (73) , we obtain
s n ( 2 ) ( a , b ; χ ; x , y ) + ( a b ) n − 1 s n ( 1 ) ( b , a ; χ ; y , x ) = 1 n + 1 ∑ j = 1 n + 1 ( n + 1 j ) ( a b ) n − j B ‾ j ( x , χ ) B ‾ n + 1 − j ( y ) + n ( a , b q ) n + 1 a b n q ( n + 1 ) ∑ k = 1 q χ ( k ) B ‾ n + 1 ( b ( k + x ) + a y ( a , b q ) ) − δ 1 , n χ ( − 1 ) 4 ∑ k = 1 b q δ Z ( a ( k + y ) b + x ) δ Z ( k + y b q ) χ ( a ( k + y ) b + x ) . \begin{aligned}
s_n^{(2)}(a,b;\chi;x,y)+\left(\frac{a}{b}\right)^{n-1}s_n^{(1)}(b,a;\chi;y,x)
&=\frac{1}{n+1}\sum_{j=1}^{n+1}\binom{n+1}{j}\left(\frac{a}{b}\right)^{n-j}\overline{B}_j(x,\chi)\overline{B}_{n+1-j}(y)\\
&\quad+\frac{n(a,bq)^{n+1}}{ab^nq(n+1)}\sum_{k=1}^{q}\chi(k)\overline{B}_{n+1}\left(\frac{b(k+x)+ay}{(a,bq)}\right)\\
&\quad-\frac{\delta_{1,n}\chi(-1)}{4}\sum_{k=1}^{bq}\delta_{\mathbb{Z}}\left(\frac{a(k+y)}{b}+x\right)\delta_{\mathbb{Z}}\left(\frac{k+y}{bq}\right)\chi\left(\frac{a(k+y)}{b}+x\right).
\end{aligned} s n ( 2 ) ( a , b ; χ ; x , y ) + ( b a ) n − 1 s n ( 1 ) ( b , a ; χ ; y , x ) = n + 1 1 j = 1 ∑ n + 1 ( j n + 1 ) ( b a ) n − j B j ( x , χ ) B n + 1 − j ( y ) + a b n q ( n + 1 ) n ( a , b q ) n + 1 k = 1 ∑ q χ ( k ) B n + 1 ( ( a , b q ) b ( k + x ) + a y ) − 4 δ 1 , n χ ( − 1 ) k = 1 ∑ b q δ Z ( b a ( k + y ) + x ) δ Z ( b q k + y ) χ ( b a ( k + y ) + x ) . Therefore, multiplying both sides of the above identity by a b n ab^n a b n a a a b b b x x x y y y
It is obvious that, taking x = y = 0 x=y=0 x = y = 0 a , b , n , q ∈ N a,b,n,q\in\mathbb{N} a , b , n , q ∈ N q ≥ 2 q\ge2 q ≥ 2 χ \chi χ q q q ( − 1 ) n + 1 χ ( − 1 ) = 1 (-1)^{n+1}\chi(-1)=1 ( − 1 ) n + 1 χ ( − 1 ) = 1
a b n s n ( 1 ) ( a , b ; χ ; 0 , 0 ) + b a n s n ( 2 ) ( b , a ; χ ; 0 , 0 ) = 1 n + 1 ∑ j = 1 n + 1 ( n + 1 j ) a j b n + 1 − j B j ( χ ) B ‾ n + 1 − j ( 0 ) + n ( b , a q ) n + 1 q ( n + 1 ) ∑ k = 1 q χ ( k ) B ‾ n + 1 ( a k ( b , a q ) ) . (75) \begin{aligned}
ab^n s_n^{(1)}(a,b;\chi;0,0)+ba^n s_n^{(2)}(b,a;\chi;0,0)
&=\frac{1}{n+1}\sum_{j=1}^{n+1}\binom{n+1}{j}a^jb^{n+1-j}B_j(\chi)\overline{B}_{n+1-j}(0)\\
&\quad+\frac{n(b,aq)^{n+1}}{q(n+1)}\sum_{k=1}^{q}\chi(k)\overline{B}_{n+1}\left(\frac{ak}{(b,aq)}\right). \tag*{(75)}
\end{aligned} a b n s n ( 1 ) ( a , b ; χ ; 0 , 0 ) + b a n s n ( 2 ) ( b , a ; χ ; 0 , 0 ) = n + 1 1 j = 1 ∑ n + 1 ( j n + 1 ) a j b n + 1 − j B j ( χ ) B n + 1 − j ( 0 ) + q ( n + 1 ) n ( b , a q ) n + 1 k = 1 ∑ q χ ( k ) B n + 1 ( ( b , a q ) ak ) . (75) In particular, the case n = 1 n=1 n = 1 ( a , b ) = 1 (a,b)=1 ( a , b ) = 1 (75) gives Berndt’s reciprocity formulas (7) and (8) .
As another consequence of Theorem 1.3, we have the following three-term reciprocity formula, which removes the hypotheses ( b , q ) = 1 (b,q)=1 ( b , q ) = 1 ( a , b ) = ( b , c ) = ( c , a ) = 1 (a,b)=(b,c)=(c,a)=1 ( a , b ) = ( b , c ) = ( c , a ) = 1 [7 Corollary 8.5] .
Corollary 4.3. Let a , b , c , q ∈ N a,b,c,q \in\mathbb{N} a , b , c , q ∈ N q ≥ 2 q \ge2 q ≥ 2 χ \chi χ q q q
s 1 , 1 ( 1 ) ( c , b , a ; χ ; 0 , 0 , 0 ) + s 1 , 1 ( 2 ) ( a , b , c ; χ ; 0 , 0 , 0 ) + s 1 , 1 ( 3 ) ( a , c , b ; χ ; 0 , 0 , 0 ) = b ( a , c q ) 2 2 a c q ∑ j = 1 q χ ( j ) B ‾ 2 ( c j ( a , c q ) ) + ( a , b ) 2 χ ‾ ( a ( a , b ) ) B 2 ( χ ) 2 a b − 1 4 ∑ k = 1 c q δ Z ( a k c ) δ Z ( b k c q ) χ ( a k c ) . (76) \begin{aligned}
&s^{(1)}_{1,1}(c,b,a;\chi;0,0,0)+s^{(2)}_{1,1}(a,b,c;\chi;0,0,0)+s^{(3)}_{1,1}(a,c,b;\chi;0,0,0)\\
&=\frac{b(a,cq)^2}{2acq}\sum_{j=1}^{q}\chi(j)\overline{B}_2\left(\frac{cj}{(a,cq)}\right)+\frac{(a,b)^2\overline{\chi}\left(\frac{a}{(a,b)}\right)B_2(\chi)}{2ab}\\
&\quad-\frac{1}{4}\sum_{k=1}^{cq}\delta_{\mathbb{Z}}\left(\frac{ak}{c}\right)\delta_{\mathbb{Z}}\left(\frac{bk}{cq}\right)\chi\left(\frac{ak}{c}\right). \tag*{(76)}
\end{aligned} s 1 , 1 ( 1 ) ( c , b , a ; χ ; 0 , 0 , 0 ) + s 1 , 1 ( 2 ) ( a , b , c ; χ ; 0 , 0 , 0 ) + s 1 , 1 ( 3 ) ( a , c , b ; χ ; 0 , 0 , 0 ) = 2 a c q b ( a , c q ) 2 j = 1 ∑ q χ ( j ) B 2 ( ( a , c q ) c j ) + 2 ab ( a , b ) 2 χ ( ( a , b ) a ) B 2 ( χ ) − 4 1 k = 1 ∑ c q δ Z ( c ak ) δ Z ( c q bk ) χ ( c ak ) . (76) Proof. Taking m = n = 1 m=n=1 m = n = 1 x = y = z = 0 x=y=z=0 x = y = z = 0 Theorem 1.3 and using (26) , we have
s 1 , 1 ( 1 ) ( c , b , a ; χ ; 0 , 0 , 0 ) + s 1 , 1 ( 2 ) ( a , b , c ; χ ; 0 , 0 , 0 ) + s 1 , 1 ( 3 ) ( a , c , b ; χ ; 0 , 0 , 0 ) = b 2 c q ∑ r = 1 c q χ ( r ) B ‾ 2 ( c r a ) + ( a , b ) 2 χ ‾ ( a ( a , b ) ) B 2 ( χ ) 2 a b − 1 4 ∑ k = 1 c q δ Z ( a k c ) δ Z ( b k c q ) χ ( a k c ) . (77) \begin{aligned}
&s^{(1)}_{1,1}(c,b,a;\chi;0,0,0)+s^{(2)}_{1,1}(a,b,c;\chi;0,0,0)+s^{(3)}_{1,1}(a,c,b;\chi;0,0,0)\\
&=\frac{b}{2cq}\sum_{r=1}^{cq}\chi(r)\overline{B}_2\left(\frac{cr}{a}\right)+\frac{(a,b)^2\overline{\chi}\left(\frac{a}{(a,b)}\right)B_2(\chi)}{2ab}\\
&\quad-\frac{1}{4}\sum_{k=1}^{cq}\delta_{\mathbb{Z}}\left(\frac{ak}{c}\right)\delta_{\mathbb{Z}}\left(\frac{bk}{cq}\right)\chi\left(\frac{ak}{c}\right). \tag*{(77)}
\end{aligned} s 1 , 1 ( 1 ) ( c , b , a ; χ ; 0 , 0 , 0 ) + s 1 , 1 ( 2 ) ( a , b , c ; χ ; 0 , 0 , 0 ) + s 1 , 1 ( 3 ) ( a , c , b ; χ ; 0 , 0 , 0 ) = 2 c q b r = 1 ∑ c q χ ( r ) B 2 ( a cr ) + 2 ab ( a , b ) 2 χ ( ( a , b ) a ) B 2 ( χ ) − 4 1 k = 1 ∑ c q δ Z ( c ak ) δ Z ( c q bk ) χ ( c ak ) . (77) It is easily seen from (67) that
∑ r = 1 a q χ ( r ) B ‾ 2 ( c r a ) = ∑ l = 0 a − 1 ∑ j = 1 q χ ( q l + j ) B ‾ 2 ( c ( q l + j ) a ) = ( a , c q ) 2 a ∑ j = 1 q χ ( j ) B ‾ 2 ( c j ( a , c q ) ) . (78) \begin{aligned}
\sum_{r=1}^{aq}\chi(r)\overline{B}_2\left(\frac{cr}{a}\right)=\sum_{l=0}^{a-1}\sum_{j=1}^{q}\chi(ql+j)\overline{B}_2\left(\frac{c(ql+j)}{a}\right) \\
&=\frac{(a,cq)^2}{a}\sum_{j=1}^{q}\chi(j)\overline{B}_2\left(\frac{cj}{(a,cq)}\right). \tag*{(78)}
\end{aligned} r = 1 ∑ a q χ ( r ) B 2 ( a cr ) = l = 0 ∑ a − 1 j = 1 ∑ q χ ( q l + j ) B 2 ( a c ( q l + j ) ) = a ( a , c q ) 2 j = 1 ∑ q χ ( j ) B 2 ( ( a , c q ) c j ) . (78) Thus, the desired result follows immediately by applying (78) to the right-hand side of (77) . □
We next consider a special case of Theorem 1.4 . Like the sums (64) , the sums s ^ n ( a , b ; χ 1 , χ 2 ; x , y ) \widehat{s}_n(a,b;\chi_1,\chi_2;x,y) s n ( a , b ; χ 1 , χ 2 ; x , y )
s ^ n ( a , b ; χ 1 , χ 2 ; x , y ) = ∑ r = 1 c q 1 q 2 χ 1 ( r ) B ‾ n ( a ( r + y ) c + x , χ 2 ) B ‾ 1 ( r + y c q 1 q 2 ) . (79) \widehat{s}_n(a,b;\chi_1,\chi_2;x,y)=\sum_{r=1}^{cq_1q_2}\chi_1(r)\overline{B}_n\left(\frac{a(r+y)}{c}+x,\chi_2\right)\overline{B}_1\left(\frac{r+y}{cq_1q_2}\right). \tag*{(79)} s n ( a , b ; χ 1 , χ 2 ; x , y ) = r = 1 ∑ c q 1 q 2 χ 1 ( r ) B n ( c a ( r + y ) + x , χ 2 ) B 1 ( c q 1 q 2 r + y ) . (79) where n , q 1 , q 2 ∈ N n,q_1,q_2\in\mathbb{N} n , q 1 , q 2 ∈ N a , b ∈ Z a,b\in\mathbb{Z} a , b ∈ Z b ≥ 1 b\ge1 b ≥ 1 χ 1 \chi_1 χ 1 χ 2 \chi_2 χ 2 q 1 q_1 q 1 q 2 q_2 q 2 x , y ∈ R x,y\in\mathbb{R} x , y ∈ R (79) .
Corollary 4.4. Let a , b , n , q 1 , q 2 ∈ N a,b,n,q_1,q_2 \in\mathbb{N} a , b , n , q 1 , q 2 ∈ N q 1 ≥ 2 q_1 \ge2 q 1 ≥ 2 q 2 ≥ 2 q_2 \ge2 q 2 ≥ 2 χ 1 \chi_1 χ 1 χ 2 \chi_2 χ 2 q 1 q_1 q 1 q 2 q_2 q 2 x , y ∈ R x,y \in\mathbb{R} x , y ∈ R
a b n s ^ n ( a , b ; χ 1 , χ 2 ‾ ; x , y ) + b a n s ^ n ( b , a ; χ 2 ‾ , χ 1 ; y , x ) = 1 n + 1 ∑ j = 1 n ( n + 1 j ) a j b n + 1 − j B ‾ j ( y , χ 1 ) B ‾ n + 1 − j ( x , χ 2 ‾ ) + n ( a q 1 , b q 2 ) n + 1 q 1 q 2 ( n + 1 ) ∑ k = 1 q 1 ∑ j = 1 q 2 χ 1 ( k ) χ 2 ‾ ( j ) B ‾ n + 1 ( a ( k + y ) + b ( j + x ) ( a q 1 , b q 2 ) ) − δ 1 , n χ 2 ( − 1 ) a b 4 ∑ k = 1 b q 1 q 2 χ 1 ( k ) δ Z ( a ( k + y ) b + x ) δ Z ( k + y b q 1 q 2 ) × χ 2 ‾ ( a ( k + y ) b + x ) . (80) \begin{aligned}
ab^n\widehat{s}_n(a,b;\chi_1,\overline{\chi_2};x,y)+ba^n\widehat{s}_n(b,a;\overline{\chi_2},\chi_1;y,x)
&=\frac{1}{n+1}\sum_{j=1}^{n}\binom{n+1}{j}a^jb^{n+1-j}\overline{B}_j(y,\chi_1)\overline{B}_{n+1-j}(x,\overline{\chi_2})\\
&+\frac{n(aq_1,bq_2)^{n+1}}{q_1q_2(n+1)}
\sum_{k=1}^{q_1}\sum_{j=1}^{q_2}\chi_1(k)\overline{\chi_2}(j)\overline{B}_{n+1}\left(\frac{a(k+y)+b(j+x)}{(aq_1,bq_2)}\right)\\
&-\frac{\delta_{1,n}\chi_2(-1)ab}{4}\sum_{k=1}^{bq_1q_2}\chi_1(k)\delta_{\mathbb{Z}}\left(\frac{a(k+y)}{b}+x\right)\delta_{\mathbb{Z}}\left(\frac{k+y}{bq_1q_2}\right)\\
&\quad\times\overline{\chi_2}\left(\frac{a(k+y)}{b}+x\right). \tag*{(80)}
\end{aligned} a b n s n ( a , b ; χ 1 , χ 2 ; x , y ) + b a n s n ( b , a ; χ 2 , χ 1 ; y , x ) = n + 1 1 j = 1 ∑ n ( j n + 1 ) a j b n + 1 − j B j ( y , χ 1 ) B n + 1 − j ( x , χ 2 ) + q 1 q 2 ( n + 1 ) n ( a q 1 , b q 2 ) n + 1 k = 1 ∑ q 1 j = 1 ∑ q 2 χ 1 ( k ) χ 2 ( j ) B n + 1 ( ( a q 1 , b q 2 ) a ( k + y ) + b ( j + x ) ) − 4 δ 1 , n χ 2 ( − 1 ) ab k = 1 ∑ b q 1 q 2 χ 1 ( k ) δ Z ( b a ( k + y ) + x ) δ Z ( b q 1 q 2 k + y ) × χ 2 ( b a ( k + y ) + x ) . (80) Proof. Taking n = b = 1 n=b=1 n = b = 1 y = 0 y=0 y = 0 m m m n n n c c c b b b z z z y y y
s ^ n ( a , b ; χ 1 , χ 2 ‾ ; x , y ) + ( a b ) n − 1 s ^ n ( b , a ; χ 2 ‾ , χ 1 ; y , x ) = 1 n + 1 ∑ j = 1 n ( n + 1 j ) ( a b ) n − j B ‾ j ( x , χ 2 ‾ ) B ‾ n + 1 − j ( y , χ 1 ) + n a n − 1 b n q 1 q 2 ( n + 1 ) ∑ r = 1 a q 1 q 2 χ 2 ‾ ( r ) B ‾ n + 1 ( b ( r + x ) a + y , χ 1 ) − δ 1 , n χ 2 ( − 1 ) 4 ∑ k = 1 b q 1 q 2 χ 1 ( k ) δ Z ( a ( k + y ) b + x ) δ Z ( k + y b q 1 q 2 ) × χ 2 ‾ ( a ( k + y ) b + x ) . (81) \begin{aligned}
\widehat{s}_n(a,b;\chi_1,\overline{\chi_2};x,y)+\left(\frac{a}{b}\right)^{n-1}\widehat{s}_n(b,a;\overline{\chi_2},\chi_1;y,x)
&=\frac{1}{n+1}\sum_{j=1}^{n}\binom{n+1}{j}\left(\frac{a}{b}\right)^{n-j}\overline{B}_j(x,\overline{\chi_2})\overline{B}_{n+1-j}(y,\chi_1)\\
&+\frac{na^{n-1}}{b^nq_1q_2(n+1)}\sum_{r=1}^{aq_1q_2}\overline{\chi_2}(r)\overline{B}_{n+1}\left(\frac{b(r+x)}{a}+y,\chi_1\right)\\
&-\frac{\delta_{1,n}\chi_2(-1)}{4}\sum_{k=1}^{bq_1q_2}\chi_1(k)\delta_{\mathbb{Z}}\left(\frac{a(k+y)}{b}+x\right)\delta_{\mathbb{Z}}\left(\frac{k+y}{bq_1q_2}\right)\\
&\quad\times\overline{\chi_2}\left(\frac{a(k+y)}{b}+x\right). \tag*{(81)}
\end{aligned} s n ( a , b ; χ 1 , χ 2 ; x , y ) + ( b a ) n − 1 s n ( b , a ; χ 2 , χ 1 ; y , x ) = n + 1 1 j = 1 ∑ n ( j n + 1 ) ( b a ) n − j B j ( x , χ 2 ) B n + 1 − j ( y , χ 1 ) + b n q 1 q 2 ( n + 1 ) n a n − 1 r = 1 ∑ a q 1 q 2 χ 2 ( r ) B n + 1 ( a b ( r + x ) + y , χ 1 ) − 4 δ 1 , n χ 2 ( − 1 ) k = 1 ∑ b q 1 q 2 χ 1 ( k ) δ Z ( b a ( k + y ) + x ) δ Z ( b q 1 q 2 k + y ) × χ 2 ( b a ( k + y ) + x ) . (81) From (27) and (67) , we obtain
∑ r = 1 a q 1 q 2 χ 2 ‾ ( r ) B ‾ n + 1 ( b ( r + x ) a + y , χ 1 ) = q 1 n ∑ r = 1 a q 1 q 2 χ 2 ‾ ( r ) ∑ k = 1 q 1 χ 1 ( k ) B ‾ n + 1 ( b ( r + x ) a q 1 + k + y q 1 ) = q 1 n ∑ ℓ = 0 a q 1 − 1 ∑ j = 1 q 2 χ 2 ‾ ( q 2 ℓ + j ) ∑ k = 1 q 1 χ 1 ( k ) B ‾ n + 1 ( b ( j + x ) a q 1 + b q 2 ℓ a q 1 + k + y q 1 ) = ( a q 1 , b q 2 ) n + 1 a n ∑ j = 1 q 2 χ 2 ‾ ( j ) ∑ k = 1 q 1 χ 1 ( k ) B ‾ n + 1 ( b ( j + x ) + a ( k + y ) ( a q 1 , b q 2 ) ) . \begin{aligned}
\sum_{r=1}^{aq_1q_2}\overline{\chi_2}(r)\overline{B}_{n+1}\left(\frac{b(r+x)}{a}+y,\chi_1\right)
&=q_1^n\sum_{r=1}^{aq_1q_2}\overline{\chi_2}(r)\sum_{k=1}^{q_1}\chi_1(k)\overline{B}_{n+1}\left(\frac{b(r+x)}{aq_1}+\frac{k+y}{q_1}\right)\\
&=q_1^n\sum_{\ell=0}^{aq_1-1}\sum_{j=1}^{q_2}\overline{\chi_2}(q_2\ell+j)\sum_{k=1}^{q_1}\chi_1(k)\overline{B}_{n+1}\left(\frac{b(j+x)}{aq_1}+\frac{bq_2\ell}{aq_1}+\frac{k+y}{q_1}\right)\\
&=\frac{(aq_1,bq_2)^{n+1}}{a^n}\sum_{j=1}^{q_2}\overline{\chi_2}(j)\sum_{k=1}^{q_1}\chi_1(k)\overline{B}_{n+1}\left(\frac{b(j+x)+a(k+y)}{(aq_1,bq_2)}\right).
\end{aligned} r = 1 ∑ a q 1 q 2 χ 2 ( r ) B n + 1 ( a b ( r + x ) + y , χ 1 ) = q 1 n r = 1 ∑ a q 1 q 2 χ 2 ( r ) k = 1 ∑ q 1 χ 1 ( k ) B n + 1 ( a q 1 b ( r + x ) + q 1 k + y ) = q 1 n ℓ = 0 ∑ a q 1 − 1 j = 1 ∑ q 2 χ 2 ( q 2 ℓ + j ) k = 1 ∑ q 1 χ 1 ( k ) B n + 1 ( a q 1 b ( j + x ) + a q 1 b q 2 ℓ + q 1 k + y ) = a n ( a q 1 , b q 2 ) n + 1 j = 1 ∑ q 2 χ 2 ( j ) k = 1 ∑ q 1 χ 1 ( k ) B n + 1 ( ( a q 1 , b q 2 ) b ( j + x ) + a ( k + y ) ) . This, together with (81) , yields
s ^ n ( a , b ; χ 1 , χ 2 ‾ ; x , y ) + ( a b ) n − 1 s ^ n ( b , a ; χ 2 ‾ , χ 1 ; y , x ) = 1 n + 1 ∑ j = 1 n ( n + 1 j ) ( a b ) n − j B ‾ j ( x , χ 2 ‾ ) B ‾ n + 1 − j ( y , χ 1 ) + n ( a q 1 , b q 2 ) n + 1 a b n q 1 q 2 ( n + 1 ) ∑ k = 1 q 1 ∑ j = 1 q 2 χ 1 ( k ) χ 2 ‾ ( j ) B ‾ n + 1 ( a ( k + y ) + b ( j + x ) ( a q 1 , b q 2 ) ) − δ 1 , n χ 2 ( − 1 ) 4 ∑ k = 1 b q 1 q 2 χ 1 ( k ) δ Z ( a ( k + y ) b + x ) δ Z ( k + y b q 1 q 2 ) χ 2 ‾ ( a ( k + y ) b + x ) . (82) \widehat{s}_n(a,b;\chi_1,\overline{\chi_2};x,y)+\left(\frac{a}{b}\right)^{n-1}\widehat{s}_n(b,a;\overline{\chi_2},\chi_1;y,x)
= \frac{1}{n+1}\sum_{j=1}^{n}\binom{n+1}{j}\left(\frac{a}{b}\right)^{n-j}\overline{B}_j(x,\overline{\chi_2})\overline{B}_{n+1-j}(y,\chi_1)
+ \frac{n(aq_1,bq_2)^{n+1}}{ab^nq_1q_2(n+1)}\sum_{k=1}^{q_1}\sum_{j=1}^{q_2}\chi_1(k)\overline{\chi_2}(j)\overline{B}_{n+1}\left(\frac{a(k+y)+b(j+x)}{(aq_1,bq_2)}\right)
- \frac{\delta_{1,n}\chi_2(-1)}{4}\sum_{k=1}^{bq_1q_2}\chi_1(k)\delta_{\mathbb{Z}}\left(\frac{a(k+y)}{b}+x\right)\delta_{\mathbb{Z}}\left(\frac{k+y}{bq_1q_2}\right)\overline{\chi_2}\left(\frac{a(k+y)}{b}+x\right). \tag*{(82)} s n ( a , b ; χ 1 , χ 2 ; x , y ) + ( b a ) n − 1 s n ( b , a ; χ 2 , χ 1 ; y , x ) = n + 1 1 j = 1 ∑ n ( j n + 1 ) ( b a ) n − j B j ( x , χ 2 ) B n + 1 − j ( y , χ 1 ) + a b n q 1 q 2 ( n + 1 ) n ( a q 1 , b q 2 ) n + 1 k = 1 ∑ q 1 j = 1 ∑ q 2 χ 1 ( k ) χ 2 ( j ) B n + 1 ( ( a q 1 , b q 2 ) a ( k + y ) + b ( j + x ) ) − 4 δ 1 , n χ 2 ( − 1 ) k = 1 ∑ b q 1 q 2 χ 1 ( k ) δ Z ( b a ( k + y ) + x ) δ Z ( b q 1 q 2 k + y ) χ 2 ( b a ( k + y ) + x ) . (82) Thus, multiplying both sides of (82) by a b n ab^n a b n
In particular, taking x = y = 0 x=y=0 x = y = 0 Corollary 4.4 , we see that for a , b , n , q 1 , q 2 ∈ N a,b,n,q_1,q_2\in\mathbb{N} a , b , n , q 1 , q 2 ∈ N q 1 ≥ 2 q_1\ge2 q 1 ≥ 2 q 2 ≥ 2 q_2\ge2 q 2 ≥ 2 q 1 ≠ q 2 q_1\ne q_2 q 1 = q 2 χ 1 \chi_1 χ 1 χ 2 \chi_2 χ 2 q 1 q_1 q 1 q 2 q_2 q 2 ( − 1 ) n + 1 χ 1 ( − 1 ) χ 2 ( − 1 ) = 1 (-1)^{n+1}\chi_1(-1)\chi_2(-1)=1 ( − 1 ) n + 1 χ 1 ( − 1 ) χ 2 ( − 1 ) = 1
a b n s ^ n ( a , b ; χ 1 , χ 2 ‾ ; 0 , 0 ) + b a n s ^ n ( b , a ; χ 2 ‾ , χ 1 ; 0 , 0 ) = 1 n + 1 ∑ j = 1 n ( n + 1 j ) a j b n + 1 − j B ‾ j ( χ 1 ) B n + 1 − j ( χ 2 ‾ ) + n ( a q 1 , b q 2 ) n + 1 q 1 q 2 ( n + 1 ) ∑ k = 1 q 1 ∑ j = 1 q 2 χ 1 ( k ) χ 2 ‾ ( j ) B ‾ n + 1 ( a k + b j ( a q 1 , b q 2 ) ) . (83) ab^n\widehat{s}_n(a,b;\chi_1,\overline{\chi_2};0,0)+ba^n\widehat{s}_n(b,a;\overline{\chi_2},\chi_1;0,0)
= \frac{1}{n+1}\sum_{j=1}^{n}\binom{n+1}{j}a^jb^{n+1-j}\overline{B}_j(\chi_1)B_{n+1-j}(\overline{\chi_2})
+ \frac{n(aq_1,bq_2)^{n+1}}{q_1q_2(n+1)}\sum_{k=1}^{q_1}\sum_{j=1}^{q_2}\chi_1(k)\overline{\chi_2}(j)\overline{B}_{n+1}\left(\frac{ak+bj}{(aq_1,bq_2)}\right). \tag*{(83)} a b n s n ( a , b ; χ 1 , χ 2 ; 0 , 0 ) + b a n s n ( b , a ; χ 2 , χ 1 ; 0 , 0 ) = n + 1 1 j = 1 ∑ n ( j n + 1 ) a j b n + 1 − j B j ( χ 1 ) B n + 1 − j ( χ 2 ) + q 1 q 2 ( n + 1 ) n ( a q 1 , b q 2 ) n + 1 k = 1 ∑ q 1 j = 1 ∑ q 2 χ 1 ( k ) χ 2 ( j ) B n + 1 ( ( a q 1 , b q 2 ) ak + bj ) . (83) This indicates an error in the reciprocity formula of Daǧlı and Can [11 Corollary 3] . Specifically, their formula omits the last term appearing on the right-hand side of (83) .
School of Mathematics and Big Data, Neijiang Normal University, Neijiang 641100, Sichuan, People’s Republic of China
Email address: hyyhe@aliyun.com
