A NOTE on the GENERALIZED BERNOULLI POLYNOMIALS with (P, Q)-POLYLOGARITHM FUNCTION†

J. Appl. Math. & Informatics Vol. 38(2020), No. 1 - 2, pp. 145 - 157 https://doi.org/10.14317/jami.2020.145

A NOTE ON THE GENERALIZED BERNOULLI POLYNOMIALS WITH (p, q)-POLYLOGARITHM FUNCTION†

N.S. JUNG

Abstract. In this article, we define a generating function of the generalized (p, q)-poly-Bernoulli polynomials with variable a by using the polylogarithm function. From the definition, we derive some properties that is concerned with other numbers and polynomials. Furthermore, we construct a special functions and give some symmetric identities involving the generalized (p, q)-poly-Bernoulli polynomials and power sums of the first integers.

AMS Mathematics Subject Classiﬁcation : 11B68, 11B73, 11B75. Key words and phrases : polylogarithm function, generalized (p, q)–poly- Bernoulli polynomials with variable a, Stirling numbers of the second kind, power sums of ﬁrst integers.

1. Introduction

Throughout this paper, we use the following standard notations. N = {1, 2, 3,... } denotes the set of natural numbers, Z+ denotes the set of nonnegative integers, Z denotes the set of integers, and C denotes the set of complex numbers, respectively. The ordinary Bernoulli polynomials Bn(x) are deﬁned by the generating functions ∞ t X tn ext = B (x) (see [1,2,4,5]). et − 1 n n! n=0 (k) (k) When x = 0,Bn,q = Bn,q(0) are called poly-Bernoulli numbers. In [4], we introduced a generalization of the ordinary Bernoulli polynomials Bn(x; a) with variable a that are deﬁned by −t ∞ n Lik(1 − e ) X t ext = B(k)(x; a) eat − 1 n n! n=0

Received November 6, 2019. Revised January 9, 2020. Accepted January 14, 2020. †This work was supported by 2019 Hannam University Research Fund. c 2020 KSCAM. 145 146 N.S. Jung where ∞ X xn Li (x) = , (k ∈ ) (see [1,2,4,5,6,7]) k nk Z n=1 is polylogarithm function. When a = 1, it is equal to the poly-Bernoulli poly- −t nomials. If k = 1, then we have Li1(x) = −log(1 − x) and Li1(1 − e ) = t. Using the result of polylogarithm function, we observe that the poly-Bernoulli polynomials is identical to the ordinary Bernoulli polynomials Bn(x). For n ∈ C, the (p, q)-integer [n]p,q is deﬁned as follows pn − qn [n] = . p,q p − q

Note that limp→1[n]p,q = [n]q and limq→1[n]q = n. From the deﬁnition of the (p, q)-integer, the (p, q)-analogue of the polylogarithm function Lik,p,q is given by

∞ X xn Li (x) = , (k ∈ ) (see [6]). (1.1) k,p,q [n]k Z n=1 p,q

The Stirling numbers of the second kind S2(n, m) are deﬁned as below

n n X x = S2(n, m)(x)m, m=0 where (x)n = x(x − 1)(x − 2) ··· (x − n + 1) is falling factorial. The generating function of the Stirling numbers of the second kind is given by

∞ X tn (et − 1)m S (n, m) = (see [3,4,5,8]). (1.2) 2 n! m! n=m In this paper, by using the (p, q)-polylogarithm function, we define a generalized (p, q)-poly-Bernoulli polynomials with variable a. We show that is related with the other numbers and polynomials. We also find some identities that are concerned with the Stirling numbers and the weighted Stirling numbers of the second kind. Moreover, by using special functions and power sums of first integers, we derive some symmetric properties of the generalized (p, q)-poly-Bernoulli numbers and polynomials.

2. Generalized of (p, q)-poly-Bernoulli polynomials with variable a In this section, by using the Equation (1.2), we construct a generalized (p, q)- (k) poly-Bernoulli polynomials Bn,p,q(x; a) with variable a by the following generating functions. We investigate some properties of the polynomials and ﬁnd several relations that are connected with other polynomials. A note on the generalized Bernoulli polynomials with (p, q)-polylogarithm function 147

Deﬁnition 2.1. For n ∈ Z+, k ∈ Z and 0 < q < p ≤ 1, we deﬁne a generalized (p, q)-poly-Bernoulli polynomials with variable a by

−t ∞ n Lik,p,q(1 − e ) X t ext = B(k) (x; a) (2.1) eat − 1 n,p,q n! n=0 where ∞ X tn Li (t) = k,p,q [n]k n=1 p,q

(k) (k) is the k-th (p, q)-polylogarithm function. When x = 0, Bn,p,q(0; a) = Bn,p,q(a) are called the generalized (p, q)-poly-Bernoulli numbers with variable a.

From the Equation (2.1), we have a relation between the generalized poly- Bernoulli numbers and polynomials.

Theorem 2.2. Let n ≥ 0, m ≥ 1 and k ∈ Z. We have n X n (k) B(k) (mx; a) = mn−1B (a)xn−l. n,p,q l l,p,q l=0

Proof. For n ≥ 0, m ≥ 1 and k ∈ Z, we get

∞ n −t X t Lik,p,q(1 − e ) B(k) (mx; a) = emxt n,p,q n! eat − 1 n=0 ∞ n ! n X X n (k) t = mn−1B (a)xn−l . l l,p,q n! n=0 l=0

Thus, we obtain the above result.

When m = 1, it satisﬁes

n X n (k) B(k) (x; a) = B (a)xn−l. n,p,q l l,p,q l=0

Theorem 2.3. Let n ∈ Z+ and k ∈ Z. Then we get n X n (k) B(k) (x + y; a) = B (x; a)yn−l. (2.2) n,p,q l l,p,q l=0 148 N.S. Jung

Proof. For n ∈ Z+, k ∈ Z, we have

∞ n −t X t Lik,p,q(1 − e ) B(k) (x + y; a) = e(x+y)t n,p,q n! eat − 1 n=0 ∞ ∞ X tn X tn = B(k) (x; a) yn n,p,q n! n! n=0 n=0 ∞ n ! n X X n (k) t = B (x; a)yn−l . l l,p,q n! n=0 l=0

Therefore, we have the obvious result.

If mx is replaced x + y in Theorem 2.3, the next theorem is obtained.

Corollary 2.4. For n > 0, m ≥ 1 and k ∈ Z, we get

n X n (k) B(k) (mx; a) = (m − 1)n−1B (x; a)xn−l. n,p,q l l,p,q l=0

Proof. Let n > 0, m ≥ 1 and k ∈ Z. We derive

∞ n −t X t Lik,p,q(1 − e ) B(k) (mx; a) = emxt n,p,q n! eat − 1 n=0 ∞ n ∞ n X t X n t = B(k) (x; a) ((m − 1)x) n,p,q n! n! n=0 n=0 ∞ n ! n X X n (k) t = (m − 1)n−1B (x; a)xn−l . l l,p,q n! n=0 l=0

From the Equation (2.2), we ﬁnd a recurrence relation as below.

Theorem 2.5. Let n ∈ Z+ and k ∈ Z. We obtain

n−1 X n (k) B(k) (x + 1; a) − B(k) (x; a) = B (x; a). n,p,q n,p,q l l,p,q l=0 A note on the generalized Bernoulli polynomials with (p, q)-polylogarithm function 149

Proof. For n ∈ Z+, k ∈ Z, we get ∞ ∞ X tn X tn B(k) (x + 1; a) − B(k) (x; a) n,p,q n! n,p,q n! n=0 n=0 Li (1 − e−t) = k,p,q ext(et − 1) eat − 1 ∞ ∞ X tn X tn = B(k) (x; a) n,p,q n! n! n=0 n=1 ∞ n−1 n X X n (k) t = B (x; a) . l l,p,q n! n=1 l=0 Comparing the coeﬃcient on both sides, we have the above theorem. By using the binomials series and the deﬁnition of polylogarithm function, we derive next result.

Theorem 2.6. For n ∈ Z+ and k ∈ Z, we have ∞ l m+1 X X X m + 1(−1)r+1(x − r + al − am))n B(k) (x; a) = . n,p,q r [m + 1]n l=0 m=0 r=0 p,q

Proof. Let n ∈ Z+, k ∈ Z. From the Equation (1.4), we obtain

−t ∞ ! ∞ −t n+1 ! Lik,p,q(1 − e ) X X (1 − e ) ext = − elat ext eat − 1 [n + 1]k l=0 n=0 p,q ∞ l X X (1 − e−t)m+1 = − e(l−m)at ext [m + 1]k l=0 m=0 p,q ∞ l ! m+1 ! X X e(l−m)at X m + 1 = − (−1)re(x−r)t [m + 1]k r l=0 m=0 p,q r=0 ∞ ∞ l m+1 X X X X m + 1(−1)r+1(x − r + al − am)n tn = r [m + 1]k n! n=0 l=0 m=0 r=0 p,q So, we get the desired result.

∞ l m+1 X X X m + 1(−1)r+1(x − r + al − am)n B(k) (x; a) = . n,p,q r [m + 1]k l=0 m=0 r=0 p,q In similar method, we get result that is related with the generalized classical Bernoulli polynomials with variable a. 150 N.S. Jung

Theorem 2.7. Let n ∈ Z+ and k ∈ Z. Then we have ∞ l+1 X 1 X l + 1 Bn+1(x − r; a) B(k) (x; a) = (−1)r . n,p,q [l + 1]k r n + 1 l=0 p,q r=0

Proof. For n ∈ Z+ and k ∈ Z, −t ∞ −t l xt Lik,p,q(1 − e ) X (1 − e ) e ext = eat − 1 [l]k eat − 1 l=1 p,q ∞ l+1 X 1 X l + 1 e(x−r)t = (−1)r [l + 1]k r eat − 1 n=0 p,q r=0 ∞ ∞ l+1 ! n X X 1 X l + 1 Bn+1(x − r; a) t = (−1)r . [l + 1]k r n + 1 n! n=0 l=0 p,q r=0 Hence, we obtain the result. ∞ l+1 X 1 X l + 1 Bn+1(x − r; a) B(k) (x; a) = (−1)r . n,p,q [l + 1]k r n + 1 l=0 p,q r=0

3. Relations with the Stirling numbers of the second kind In this section, we investigate some identities that is concerned with the Stir- ling numbers of the second kind by using the generating function. Furthermore, the deﬁnition of the weighted Stirling numbers of the second kind gives some interesting results that is associated with the generalized poly Bernoulli polynomials with variable a. From the Equation (1.2), the polylogarithm function Lik,p,q(x) is represented by the following formula. ∞ n 1 1 X X (−1)n+m tn Li (1 − e−t) = m!S (n, m) t k,p,q t [m]k 2 n! n=1 m=1 p,q (3.1) ∞ n+1 n+1+m n X X (−1) S2(n + 1, m) t = m! [m]k n + 1 n! n=0 m=1 p,q The above equation gives the following identity that is related with the Stirling numbers of the second kind.

Theorem 3.1. Let n ∈ Z+, k ∈ Z. Then we have n r+1 r+1+l X X n (−1) l!S2(r + 1, l) B(k) (x; a) = B (x; a). n,p,q r [l]k (r + 1) n−r r=0 l=1 p,q A note on the generalized Bernoulli polynomials with (p, q)-polylogarithm function 151

Proof. Let n ∈ Z+, k ∈ Z. From the Equation (3.1), the generalized (p, q)-poly- (k) Bernoulli polynomials Bn,p,q(x; a) is represented with the Stirling numbers and the generalized ordinary Bernoulli polynomials with variable a

∞ n −t xt X t Lik,p,q(1 − e ) te B(k) (x; a) = n,p,q n! t eat − 1 n=0 ∞ n+1 n+1+m ! n ∞ n X X (−1) S2(n + 1, m) t X t = m! B (x; a) [m]k n + 1 n! n n! n=0 m=1 p,q n=0 ∞ n l+1 l+1+m n X X X n (−1) m!S2(l + 1, m) t = B (x; a) . l [m]k (l + 1) n−l n! n=0 l=0 m=1 p,q

tn By comparing the coeﬃcients of n! on both sides, the proof is complete.

Theorem 3.2. If n ∈ Z+ and k ∈ Z, then we get n m X X n (k) B(k) (x; a) = (x) S (m, l)B (a). (3.2) n,p,q m l 2 n−m,p,q m=0 l=0 where (x)l = x(x − 1)(x − 2) ··· (x − l + 1) is falling factorial.

Proof. Let n ∈ Z+ and k ∈ Z. The generalized (p, q)-poly-Bernoulli numbers and polynomials can be indicated by the following formula that is related with the Stirling numbers.

∞ n −t ∞ t l X t Lik,p,q(1 − e ) X (e − 1) B(k) (x; a) = (x) n,p,q n! eat − 1 l l! n=0 l=0 ∞ ∞ ∞ X tn X X tr = B(k) (a) (x) S (r, l) n,p,q n! l 2 r! n=0 l=0 r=l ∞ ∞ n X tn X X tn = B(k) (a) (x) S (n, l) n,p,q n! l 2 n! n=0 n=0 l=0 ∞ n m ! n X X X n (k) t = (x) S (m, l)B (a) . m l 2 n−m,p,q n! n=0 m=0 l=0 Comparing the coefficient on both sides, we have the explicit result. n m X X n (k) B(k) (x; a) = (x) S (m, l)B (a). n,p,q m l 2 n−m,p,q m=0 l=0 From Definition 2.1 and the Equation (3.2), we have the another recurrence formula that is different from the result of Theorem 2.5. 152 N.S. Jung

Theorem 3.3. For n ≥ 1, k ∈ Z, we get (k) (k) Bn,p,q(x + a; a) − Bn,p,q(x; a) n−r r r+l X X n (−1) l!S2(r, l) = xn−r. r [l]k r=1 l=0 p,q Proof. Let n ≥ 1, k ∈ Z. From the deﬁnition of the generalized (p, q)-ploy- Bernoulli polynomials, we obtain ∞ ∞ X tn X tn B(k) (x + a; a) − B(k) (x; a) n,p,q n! n,p,q n! n=0 n=0 Li (1 − e−t) Li (1 − e−t) = k,p,q e(x+a)t − k,p,q ext eat − 1 eat − 1 ∞ n−1 ∞ X X (−1)n+l+1 tn X tm = (l + 1)!S (n, l + 1) xm [l + 1]k 2 n! m! n=1 l=0 p,q m=0 ∞ n ∞ X X (−1)n+l tn X tm = l!S (n, l) xm [l]k 2 n! m! n=1 l=1 p,q m=0 ∞ n−r r X X X n(−1)r+l tn = l!S (r, l)xn−r . r [l]k 2 n! n=1 r=1 l=1 p,q

Therefore, we get the following recurrence formula.

n−r r r+l X X n (−1) l!S2(r, l) B(k) (x + a; a) − B(k) (x; a) = xn−r. n,p,q n,p,q r [l]k r=1 l=1 p,q

Theorem 3.4. For n ≥ 1, k ∈ Z, we get (k) (k) (k) (k) En,p,q(x + 1) + En,p,q(x) = 2Bn,p,q(x + a; a) − 2Bn,p,q(x; a). Proof. Let n ≥ 1, k ∈ Z. From the following equation 2Li (1 − e−t) 2Li (1 − e−t) k,p,q (1 + et)ext = k,p,q (eat − 1)ext, et + 1 eat − 1 we have ∞ ∞ X tn X tn E(k) (x + 1) + E(k) (x) n,p,q n! n,p,q n! n=0 n=0 ∞ ∞ X tn X tn = 2 B(k) (x + a; a) − 2 B(k) (x; a) . n,p,q n! n,p,q n! n=0 n=0 A note on the generalized Bernoulli polynomials with (p, q)-polylogarithm function 153

In [3], Carlitz introduced the weighted Stirling numbers of the second kind S2(n, m, x) as follows

∞ (et − 1)m X tn ext = S (n, m, x) (3.3) m! 2 n! n=m

Theorem 3.5. If n ∈ N, k ∈ Z. then we have

n X (m + 1)! S2(n + 1, m + 1, x − (m + 1)) B(k) (x; a) = . n,p,q [m + 1]k n + 1 m=0 p,q

Proof. Let n ∈ N, k ∈ Z. From the deﬁnition 2.1, we observe that

∞ n −t X t Lik,p,q(1 − e ) B(k) (x; a) = ext n,p,q n! eat − 1 n=0 ∞ X (1 − e−t)m+1 ext = [m + 1]k eat − 1 m=0 p,q ∞ ∞ X X (m + 1)! 1 tn = S (n, m + 1, x − (m + 1)) [m + 1]k eat − 1 2 n! m=0 n=m+1 p,q ∞ n n X X (m + 1)! S2(n + 1, m + 1, x − (m + 1)) t = B (a) [m + 1]k n,p,q n + 1 n! n=0 m=0 p,q

Comparing the coeﬃcient on both sides, we get the above result.

Corollary 3.6. Let n ∈ N, k ∈ Z. We obtain n l X X n (m + 1)! S2(l + 1, m + 1) B(k) (x; a) = B (a) (x − (m + 1))n−l. n,p,q l [m + 1]k l,p,q l + 1 l=1 m=0 p,q

Proof. For n ∈ N, k ∈ Z, the generalized (p, q)-poly-Bernoulli polynomials are satisﬁed as follows

∞ n ∞ −t m+1 xt X (k) t X (1 − e ) e B (x; a) = n,p,q n! [m + 1]k eat − 1 n=0 m=0 p,q ∞ ∞ n ∞ n X (m + 1)! 1 X t X n t = B (a) S (n, m + 1) (x − (m + 1)) [m + 1]k n,p,q t 2 n! n! m=0 p,q n=m+1 n=0

∞ n l ! n X X X n (m + 1)! S2(l + 1, m + 1) n−l t = B (a) (x − (m + 1)) . l [m + 1]k l,p,q l + 1 n! n=0 m=0 m=0 p,q

Thus, we obtain the desired result. 154 N.S. Jung

4. Symmetric identities of the generalized (p, q)-poly-Bernoulli polynomials with variable a In this section, we construct special functions and ﬁnd symmetric properties of the generalized (p, q)-poly-Bernoulli polynomials with variable a from the special functions.

Theorem 4.1. For m1, m2 > 0 (m1 6= m2), n ∈ Z+, k ∈ Z, we have n X n (k) mn−rmrB(k) (m x; a)B (m x; a) r 1 2 r,p,q 1 n−r,p,q 2 r=0 n X n (k) = mrmn−rB (m x; a)B(k) (m x; a). r 1 2 n−r,p,q 1 r,p,q 2 r=0

Proof. For n ∈ Z+, k ∈ Z, m1, m2 > 0 and m1 6= m2, we construct a special function as below

−m1t −m2t Lik,p,q(1 − e )Likp,q(1 − e ) F (t) = e2m1m2xt. (eam1t − 1)(eam2t − 1) From the special function F (t), we derive the following equation.

−m1t −m2t Lik,p,q(1 − e ) Lik,p,q(1 − e ) F (t) = em1m2xt em1m2xt (eam1t − 1) (eam2t − 1) ∞ n ∞ r X (m1t) X (m2t) = B(k) (m x; a) B(k) (m x; a) n,p,q 2 n! r,p,q 1 r! (4.1) n=0 r=0 ∞ n n X X n (k) t = mn−rmrB(k) (m x; a)B (m x; a) . r 1 2 r,p,q 1 n−r,p,q 2 n! n=0 r=0 In similar method, we obtain

∞ n n X X n (k) t F (t) = mrmn−rB (m x; a)B(k) (m x; a) . (4.2) r 1 2 n−r,p,q 1 r,p,q 2 n! n=0 r=0 By the Equation (4.1) and (4.2), it is easy to get the above theorem. n n m Pm−1 n Note that Sen(x, m) = x +(x+1) +···+(x−(m−1)) = k=0 (x + k) , x ∈ C is the power sums on an arithmetic progression. The power sums of the Pm−1 n ﬁrst integers are expressed by Sen(m) = Sen(0, m) = k=0 k (see [8]). The exponential generating function of the power sums are expressed by ∞ n mt X t e − 1 xt Sen(x, m) = e n! et − 1 n=0 From the generating function of the power sums of ﬁrst integers, we have the symmetric identity of the generalized (p, q)-poly-Bernoulli polynomials with variable a. A note on the generalized Bernoulli polynomials with (p, q)-polylogarithm function 155

Theorem 4.2. For n ∈ Z+, k ∈ Z, m1, m2 ∈ N and m1 6= m2, we get n n −m2t X n−r r n−r (k) Lik,p,q(1 − e ) a m m B (m2x; a)Sen−r(m1) r 1 2 r,p,q r=0 n n −m1t X n−r n−r r (k) = Lik,p,q(1 − e ) a m m B (m1x; a)Sen−r(m2) r 1 2 r,p,q r=0

Proof. Let n ∈ Z+ , m1, m2 > 0 and m1 6= m2. If we consider a special function that is given below, then we get Li (1 − e−m1t)Li (1 − e−m2t)(em1m2t − 1)(eam1m2xt) F (t) = k,p,q k,p,q (eam1t − 1)(eam2t − 1) ∞ n n tn X −m2t X n−r r n−r (k) = Lik,p,q(1 − e ) a m m B (m2x; a)Sen−r(m1) . r 1 2 r,p,q n! n=0 r=0 In analogous method, we have ∞ (m t)n ∞ (am t)r −m1t X (k) 2 X 1 F (t) = Lik,p,q(1 − e ) B (m1x) Ser(m2) n,p,q n! r! n=0 r=0 ∞ n n tn X −m1t X n−r n−r r (k) = Lik,p,q(1 − e ) a m m B (m1x; a)Sen−r(m2) . r 1 2 r,p,q n! n=0 r=0 tn Comparing the coeﬃcient of n! , then we get the above symmetric identity.

Theorem 4.3. If n ∈ Z+, k ∈ Z, m1, m2 (m1 6= m2) ∈ N, then we get n n m −m2t X n−r r n−r (k) 1 Lik,p,q(1 − e ) a m m B (a)Sen−r( x, m1) r 1 2 r,p,q a r=0 n n m −m1t X n−r n−r r (k) 2 = Lik,p,q(1 − e ) a m m B (a)Sen−r( x, m2) r 1 2 r,p,q a r=0

Proof. For n ∈ Z+ , m1, m2 ∈ N and m1 6= m2. From the special function that is given in theorem 4.2, then we obtain Li (1 − e−m1t)Li (1 − e−m2t)(em1m2t − 1)(eam1m2xt) F (t) = k,p,q k,p,q (eam1t − 1)(eam2t − 1) ∞ n n m tn X −m2t X n−r r n−r (k) 1 = Lik,p,q(1 − e ) a m m B (a)Sen−r( x, m1) . r 1 2 r,p,q a n! n=0 r=0 In similar method, we get ∞ (m t)n ∞ (am t)r −m1t X (k) 2 X 1 F (t) = Lik,p,q(1 − e ) B (m1x) Ser(m2) n,p,q n! r! n=0 r=0 ∞ n n m tn X −m1t X n−r n−r r (k) 2 = Lik,p,q(1 − e ) a m m B (a)Sen−r( x, m2) . r 1 2 r,p,q a n! n=0 r=0 156 N.S. Jung

Therefore, we have the above result.

Theorem 4.4. Let n ∈ Z+, k ∈ Z and m1, m2 ∈ N (m1 6= m2), we have n X n n−r n−r r−1 a m m Br,p,q(m1x)Sen−r(m2) r 1 2 r=0 n X n n−r r−1 n−r = a m m Br,p,q(m2x)Sen−r(m1). r 1 2 r=0

Proof. Let n ∈ Z+, k ∈ Z and m1, m2 ∈ N (m1 6= m2). Consider a special function F (t) as follows, then we get

−m1t −m2t am1m2t am1m2xt Lik,p,q(1 − e )Lik,p,q(1 − e )(e − 1)(e )t F (t) = 2 2 (eam1t − 1) (eam2t − 1) ∞ n ∞ n X (m1t) X (m2t) = B(k) (a) B(k) (a) n,p,q n! n,p,q n! n=0 n=0 ∞ r ∞ n X (am1t) −1 −1 X (am2t) × Ser(m2) a m Bn,p,q(m1x) r! 2 n! r=0 n=0 ∞ n ∞ n X (m1t) X (m2t) = B(k) (a) B(k) (a) n,p,q n! n,p,q n! n=0 n=0 ∞ n n X X n n−1 n−r r−1 t × a m m Br,p,q(m1x)Sen−r(m2) . r 1 2 n! n=0 r=0 In similar method, F (t) is expressed by

∞ n ∞ n X (m1t) X (m2t) F (t) = B(k) (a) B(k) (a) n,p,q n! n,p,q n! n=0 n=0 ∞ n n X X n n−1 n−r r−1 t × a m m Br,p,q(m2x)Sen−r(m1) . r 2 1 n! n=0 r=0 Comparing the coeﬃcient of both sides, we ﬁnd the symmetric identity: n X n n−l n−r r−1 a m m Br,p,q(m1x)Sen−r(m2) r 1 2 r=0 n X n n−l r−1 n−l = a m m Br(m2x)Sen−r(m1). r 1 2 r=0 A note on the generalized Bernoulli polynomials with (p, q)-polylogarithm function 157

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N.S. Jung received Ph.D. from Hannam University. Her research interests include combi- natorial analysis, number theory and applied mathematics. College of Talmage Liberal Arts, Hannam University, Daejeon 34430, Korea. e-mail: [email protected]