A NOTE on Q-ANALOGUE of POLY-BERNOULLI NUMBERS and POLYNOMIALS†

J. Appl. Math. & Informatics Vol. 35(2017), No. 5 - 6, pp. 611 - 621 http://dx.doi.org/10.14317/jami.2017.611

A NOTE ON q-ANALOGUE OF POLY-BERNOULLI NUMBERS AND POLYNOMIALS†

K.W. HWANG, B.R. NAM, N.S. JUNG∗

Abstract. In this paper, we deﬁne a q-analogue of the poly-Bernoulli numbers and polynomials which is generalization of the poly Bernoulli numbers and polynomials including q-polylogarithm function. We also give the relations between generalized poly-Bernoulli polynomials. We derive some relations that are connected with the Stirling numbers of second kind. By using special functions, we investigate some symmetric identities involving q-poly-Bernoulli polynomials.

AMS Mathematics Subject Classiﬁcation : 11B68, 11B73, 11B75. Key words and phrases : poly-Bernoulli polynomials, q-analogue of poly- Bernoulli polynomials, Stirling numbers of the second kind, q-polylogarithm function.

1. Introduction Many mathematicians are interested in the Bernoulli numbers and polynomials, Euler numbers and polynomials, Genocchi numbers and polynomials, tangent numbers and polynomials and their applications. They possess many in- teresting properties and are treated in many areas of mathematics and physics. Due to these reasons, many applications of Bernoulli numbers and polynomials, Euler numbers and polynomials, Genocchi numbers and polynomials, tangent numbers and polynomials have been studied, and recently various analogues for the above numbers and polynomials was introduced(see [1-14]). In this paper, we use the following notations. N = {1, 2, 3,... } denotes the set of natural numbers, N0 = {0, 1, 2,... } denotes the set of nonnegative inte- ger, Z denotes the set of integers, and C denotes the set of complex numbers, respectively.

Received March 22, 2017, Revised August 20, 2017. Accepted August 28, 2017 ∗Corresponding author. †This work was supported by the Dong-A university research fund. ⃝c 2017 Korean SIGCAM and KSCAM. 611 612 K. W. Hwang, B. R. Nam, N. S. Jung

The ordinary Bernoulli polynomials Bn(x) are given by the generating functions: ∞ t ∑ tn ext = B (x) (see[1, 2, 6]). (1.1) et − 1 n n! n=0 (k) (k) When x = 0,Bn,q = Bn,q(0) are called poly-Bernoulli numbers. The polylogarithm function Lik is defined by ∞ ∑ xn Li (x) = , (k ∈ Z)(see[1, 2, 3, 6, 7, 8, 14]) (1.2) k nk n=1 For k ≤ 1, the polylogarithm functions are as follows x x Li1(x) = −log(1 − x), Li0(x) = , Li−1(x) = , ··· . 1 − x (1 − x)2 By using polylogarithm function, Kaneko defined a sequence of rational numbers, which is refered to as poly-Bernoulli numbers, In [3] and [13], the k-th q-analogue of polylogarithm function Lik,q is introduced by ∞ ∑ xn Li (x) = , (k ∈ Z). (1.3) k,q [n]k n=1 q The q-analogue of polylogarithm function for k ≤ 1 is represented by a rational function, ( ) 1 ∑k k qlx Li (x) = (−1)l . k,q (1 − q)k l 1 − qlx l=0 In [4] and [12], the Stirling number of the first kind is given by

∑n m (x)n = S1(n, m)x , (n ≥ 0) m=0 and ∞ ∑ tn (log(1 + t))m S (n, m) = 1 n! m! n=m where − − ··· − n − − (x)n = x(x 1)(x 2) (x n + 1) = Πk=1(x (k 1)) (1.4) is falling factorial. The Stirling numbers of the second kind is deﬁned by

∞ ∑ tn (et − 1)m S (n, m) = . (1.5) 2 n! m! n=m In this paper, we consider a q-analogue of the poly Bernoulli polynomials containing Equation(1.3). We also ﬁnd some relations between q-poly-Bernoulli polynomials and ordinary Bernoulli polynomials. And we derive several properties that are connected with the Stirling numbers of the second kind. Finally, we A note on q-analogue of poly-Bernoulli numbers and polynomials 613

ﬁnd some symmetric identities of the q-degenerate poly Bernoulli polynomials by using special functions.

2. A q-analogue of the poly-Bernoulli polynomials (k) In this section, we define a q-analogue of poly-Bernoulli numbers Bn,q and (k) polynomials Bn,q(x) by the generating functions. From the definition, we get some identities that is similar to the ordinary Bernoulli polynomials. Definition 2.1. For n ≥ 0, n, k ∈ Z, 0 ≤ p < 1, we introduce a q-analogue of poly-Bernoulli polynomials by: ∞ Li (1 − e−t) ∑ tn k,q ext = B(k)(x) (2.1) et − 1 n,q n! n=0 where ∞ ∑ tn Li (t) = k,q [n]k n=0 q is the k-th q-polylogarithm function. (k) (k) When x = 0, Bn,q = Bn,q(0) are called a q-analogue of poly-Bernoulli numbers. (k) (k) Note that limq→1[n]q = n, and limq→1Bn,q(x) = Bn (x). From Equation(2.1), we have the following relation between q-poly-Bernoulli numbers and q-poly-polynomials. Theorem 2.2. Let n ≥ 0, n, k ∈ Z, 0 < p < 1. We have ( ) ∑n n B(k)(x) = B(k)xn−l. (2.2) n,q l l,q l=0 Proof. For n ≥ 0, n, k ∈ Z, 0 < p < 1, we easily get: ∞ ∑ tn Li (1 − e−t) B(k)(x) = k,q ext n,q n! et − 1 n=0 ( ) ∞ ( ) ∑ ∑n n tn = B(k)xn−l . l l,q n! n=0 l=0 Therefore, we have ( ) ∑n n B(k)(x) = B(k)xn−l. n,q l l,q l=0 By Equation(2.2), we obtain an addition theorem. Theorem 2.3. For n ≥ 0, n, k ∈ Z, 0 < p < 1, we have ( ) ∑n n B(k)(x + y) = B(k)(x)yn−l. n,q l l,q l=0 614 K. W. Hwang, B. R. Nam, N. S. Jung

Proof. Let n ≥ 0, n, k ∈ Z. Then we obtain ∞ ∑ tn Li (1 − e−t) B(k)(x + y) = k,q e(x+y)t n,q n! et − 1 n=0 ( ) ∞ ( ) ∑ ∑n n tn = B(k)(x)yn−l . l l,q n! n=0 l=0 Thus, we get ( ) ∑n n B(k)(x + y) = B(k)(x)yn−l. n,q l l,q l=0

By using the deﬁnition of the q-analogue of polylogarithm function Lik,q in Equation(1.3), we have next relation which is connected with the ordinary Bernoulli polynomials.

Theorem 2.4. Let n ≥ 0, n, k ∈ Z, 0 ≤ p < 1. We obtain ∞ ( ) ∑ 1 ∑l+1 l + 1 B(k)(x) = (−1)aB (x − a). n,q [l + 1]k a n l=0 q a=0 Proof. For n ≥ 0, n, k ∈ Z, 0 ≤ p < 1, ∞ Li (1 − e−t) ∑ (1 − e−t)l ext k,q ext = et − 1 [l]k et − 1 l=1 q ∞ ( ) ∑ 1 ∑l+1 l + 1 e(x−a)t = (−1)a k t − [l + 1]q a e 1 n=0 ( a=0 ) ∞ ∞ ( ) ∑ ∑ 1 ∑l+1 l + 1 B (x − a) tn = (−1)a n+1 [l + 1]k a n + 1 n! n=0 l=0 q a=0 Comparing the coeﬃcient of the result, we easily get next equation: ∞ ( ) ∑ 1 ∑l+1 l + 1 B (x − a) B(k)(x) = (−1)a n+1 . n,q [l + 1]k a n + 1 l=0 q a=0

From the binomials series and the Equation(1.3), we derive the following Theorem. Theorem 2.5. If n, k ∈ Z, n ≥ 0and 0 ≤ p < 1, then we have ∞ ( ) ∑ ∑l m∑+1 m + 1 (−1)a(l − m − a + x)n B(k)(x) = . n,q a [m + 1]n l=0 m=0 a=0 q A note on q-analogue of poly-Bernoulli numbers and polynomials 615

Proof. Let n, k ∈ Z, n ≥ 0 and 0 ≤ p < 1. Using the Equation(1.3), we have ( )( ) ∞ ∞ Li (1 − e−t) ∑ ∑ (1 − e−t)l+1 k,q ext = emt ext t − k e 1 [l + 1]q (m=0 l=0 )( ) ∞ ( ) ∑ ∑l e(l−m)t m∑+1 m + 1 = (−1)ae(x−a)t [m + 1]k a l=0 m=0 q a=0 ∞ ( ) ∑ ∑l m∑+1 m + 1 (−1)ae(l−m−a+x)t = a [m + 1]k l=0 m=0 a=0 q ∞ ∞ ( ) ∑ ∑ ∑l m∑+1 m + 1 (−1)a(l − m − a + x)n tn = a [m + 1]k n! n=0 l=0 m=0 a=0 q Hence, we have ∞ ( ) ∑ ∑l m∑+1 m + 1 (−1)a(l − m − a + x)n B(k)(x) = . n,q a [m + 1]n l=0 m=0 a=0 q

3. Some relation involving the Stirling numbers of the second kind In this section, using well-known Stirling numbers, we ﬁnd several identities of the q-poly-Bernoulli polynomials. Note that the Stirling numbers of the second kind is deﬁned ∞ (et − 1)m ∑ tn = S (n, m) (see[4, 12, 13, 14]). (3.1) m! 2 n! n=m

The q-analogue of polylogarithm function Lik,q, in the Equation(1.3), is represented as below ∞ ∑ (1 − e−t)l Li (1 − e−t) = k,q [l]k l=1 q ∞ ∑ (e−t − 1)l = (−1)l [l]k n=1 q ∞ ∑ ∑n (−1)l+n tn = l!S (n, l) [l]k 2 n! n=1 l=1 q Therefore, we obtain

∞ 1 ∑ n∑+1 (−1)l+n+1 S (n + 1, l) tn Li (1 − e−t) = l! 2 . (3.2) t k,q [l]k n + 1 n! n=0 l=1 q Using the Equation(3.2), we get next theorem. 616 K. W. Hwang, B. R. Nam, N. S. Jung

Theorem 3.1. For n, k ∈ Z, n ≥ 0, we have ( ) ∑n a∑+1 − l+a+1 (k) n ( 1) l!S2(a + 1, l) B (x) = B − (x). n,q a [l]k(a + 1) n a a=0 l=1 q Proof. Let n, k ∈ Z, n ≥ 0. By the recomposition of q-polylogarithm function in Equation(3.2), the relation between the q-poly-Bernoulli polynomials and the ordinary Bernoulli polynomials is derived: ∞ ∑ tn Li (1 − e−t) B(k)(x) = k,q ext n,q n! et − 1 n=0 ( ) ∞ ∞ ∑ n∑+1 (−1)l+n+1 S (n + 1, l) tn ∑ tn = l! 2 B (x) [l]k n + 1 n! n n! n=0 l=1 q n=0 ( ) ∑∞ ∑n a∑+1 l+a+1 n n (−1) l!S2(a + 1, l) t = B − (x) . a [l]k(a + 1) n a n! n=0 a=0 l=1 q Thus, we have ( ) ∑n a∑+1 − l+a+1 (k) n ( 1) l!S2(a + 1, l) B (x) = B − (x). n,q a [l]k(a + 1) n a a=0 l=1 q

By the Equation(3.2), we have the following result which is connected with q-poly-Bernoulli numbers. Theorem 3.2. For n ≥ 0, n, k ∈ Z, we have ∞ ( ) ∑ ∑n n B(k)(x) = (x) S (a, l)B(k) . n,q a l 2 n−a,q l=0 a=l Proof. Let n ≥ 0, n, k ∈ Z. We obtain a relation between q-poly-Bernoulli numbers and q-poly-Bernoulli polynomials by the Equation(3.1): ∞ ∑ tn Li (1 − e−t) B(k)(x) = k,q ext n,q n! et − 1 n=0 ∞ Li (1 − e−t) ∑ (et − 1)l = k,q (x) et − 1 l l! l=0 ∞ ∞ ∞ ∑ tn ∑ ∑ tn = B(k) (x) S (a, l) n,q n! l 2 n! n=0 ( l=0 a=0 ) ∞ ∞ ( ) ∑ ∑ ∑n n tn = (x) S (a, l)B(k) . a l 2 n−a,q n! n=0 l=0 a=l A note on q-analogue of poly-Bernoulli numbers and polynomials 617

Thus, we have ∞ ( ) ∑ ∑n n B(k)(x) = (x) S (a, l)B(k) . n,q a l 2 n−a,q l=0 a=l where (x)l = x(x − 1)(x − 2) ··· (x − l + 1) is falling factorial. From Deﬁnition 2.1, we have the recurrence formula with the Stirling numbers of the second kind. Theorem 3.3. For n ≥ 1, n, k ∈ Z. and 0 < p < 1, we get B(k)(x + 1) − B(k)(x) n,q n,q( ) ( ) − ∑n n ∑r 1 (−1)l+1+r = (l + 1)!S (r, l + 1) xn−r. r [l + 1]k 2 r=1 l=0 q Proof. Let n ≥ 0, n, k ∈ Z, 0 < p < 1. Using the deﬁnition of the q-poly-Bernoulli polynomials, we have a equation as follows: ∞ ∞ ∑ tn ∑ tn B(k)(x + 1) − B(k)(x) n,q n! n,q n! n=0 n=0 Li (1 − e−t) Li (1 − e−t) = k,q e(x+1)t − k,q ext et − 1 et − 1 ∑∞ n∑−1 − n+l+1 n ∑∞ m ( 1) t m t = k (l + 1)!S2(n, l + 1) x [l + 1]q n! m! n=0 l=0 ( m=0 ) ∞ ( ) − ∑ ∑n n ∑r 1 (−1)r+l+1 tn = (l + 1)!S (r, l + 1) xn−r . r [l + 1]k 2 n! n=0 r=1 l=0 q Therefore, the equation is appeared by B(k)(x + 1) − B(k)(x) n,q n,q( ) ( ) − ∑n n ∑r 1 (−1)r+l+1 = (l + 1)!S (r, l + 1) xn−r. r [l + 1]k 2 r=1 l=0 q

In [4], Carlitz deﬁned the weighted Stirling numbers of the second kind as follows ∞ (et − 1)m ∑ tn ext = S (n, m, x) . (3.3) m! 2 n! n=m By using the Equation(3.3), we get the next result. Theorem 3.4. For n ≥ 1, k ∈ Z, 0 < p < 1, we have ∑n (−1)m+nm! B(k)(x) = S (n, m, x). n,q [m + 1]k 2 m=0 q 618 K. W. Hwang, B. R. Nam, N. S. Jung

Proof. Let n ≥ 1, k ∈ Z, 0 < p < 1. ∞ Li (1 − e−t) ∑ (1 − e−t)m k,q ext = ext et − 1 [m + 1]k m=0 q ∞ ∞ ∑ (−1)m+nm! ∑ tn = S (n, m, x) [m + 1]k 2 n! n=0 ( q n=m ) ∞ ∑ ∑n (−1)m+nm! tn = S (n, m, x) . [m + 1]k 2 n! n=0 m=0 q Therefore, the formula is expressed as below: ∑n (−1)m+nm! B(k)(x) = S (n, m, x) n,q [m + 1]k 2 m=0 q

4. Symmetric identities for the q-analogue of the poly-Bernoulli polynomials In this section, we investigate several symmetric identities for the q-poly- Bernoulli polynomials by given special functions. Theorem 4.1. For n ≥ 0, n, k ∈ Z, a, b > 0(a ≠ b), the following identity is obtained: ( ) ∑n n an−mbmB(k) (ax)B(k) (bx) m m,q n−m,q m=0 ( ) ∑n n = ambn−mB(k) (ax)B(k) (bx). m n−m,q m,q m=0 Proof. Let n, k ∈ Z, n ≥ 0 and a, b > 0(a ≠ b). If we start with the function, Li (1 − e−at)Li (1 − e−bt) F (t) = k,q k,q e2abxt, (4.1) (eat − 1)(ebt − 1) then the Equation(4.1) is written by ∞ ∞ ∑ (at)n ∑ (bt)m F (t) = B(k)(bx) B(k) (ax) n,q n! m,q m! n=0 m=0 ∞ ( ) (4.2) ∑ ∑n n tn = an−mbmB(k) (ax)B(k) (bx) . m m,q n−m,q n! n=0 m=0 In similarly, we can see that ( ) ∑∞ ∑n n n m n−m (k) (k) t F (t) = a b B − (ax)B (bx) . (4.3) m n m,q m,q n! n=0 m=0 From Equation(4.2) and (4.3), it is clear to Theorem 4.1. A note on q-analogue of poly-Bernoulli numbers and polynomials 619

By substituting b = 1, we easily get next corollary. Corollary 4.2. For a > 0 anf n ≥ 0, n ∈ Z, we have ( ) ( ) ∑n n ∑n n an−mB(k) (ax)B(k) (x) = amB(k) (ax)B(k) (x). m m,q n−m,q m n−m,q m,q m=0 m=0 ∑ a m Note that Sm(a) = k=0 k is a power sum polynomials(cf [15]). We consider the following exponential generating function with indeterminate t, ∞ e(a+1)t − 1 ∑ tm = S (a) . (4.4) et − 1 m m! m=0 Using the generating function, we get a symmetric relation of the q-poly-Bernoulli polynomials.

Theorem 4.3. For n ∈ Z, n ≥ 0, a, b > 0 and a ≠ b, we have ( ) ∑n n n−m m−1 a b B (ax)S − (b − 1) m m n m m=0 ( ) ∑n n m−1 n−m = a b B (bx)S − (a − 1) m m n m m=0 Proof. Let n ∈ Z, n ≥ 0, a, b > 0 and a ≠ b. We consider the generating function: Li (1 − e−at)Li (1 − e−bt)(eabt − 1)(eabxt)t F (t) = k,q k,q . (4.5) (eat − 1)2(ebt − 1)2 The Equation(4.5) follows as below Li (1 − e−at)Li (1 − e−bt)(eabt − 1)(eabxt)t F (t) = k,q k,q (eat − 1)2(ebt − 1)2 ∞ ∞ ∞ ∞ ∑ (at)n ∑ (bt)n ∑ (at)m ∑ (bt)n = B(k) B(k) S (b − 1) b−1 B (ax) n,q n! n,q n! m m! n n! n=0 n=0 m=0 ( ) n=0 ∑∞ n ∑∞ n ∑∞ ∑n n (k) (at) (k) (bt) n n−m m−1 t = B B a b B (ax)S − (b − 1) . n,q n! n,q n! m m n m n! n=0 n=0 n=0 m=0 In similar method, the Equation(4.5) is written by ∞ ∞ ∞ ∞ ∑ (at)n ∑ (bt)n ∑ (bt)m ∑ (at)n F (t) = B(k) B(k) S (a − 1) a−1 B (bx) n,q n! n,q n! m m! n n! n=0 n=0 m=0 ( ) n=0 ∑∞ n ∑∞ n ∑∞ ∑n n (k) (at) (k) (bt) n m−1 n−m t = B B a b B (bx)S − (a − 1) . n,q n! n,q n! m m n m n! n=0 n=0 n=0 m=0 620 K. W. Hwang, B. R. Nam, N. S. Jung

Now, comparing the coeﬃcient of tn, then it gives the symmetric identity, ( ) ( ) ∑n ∑n n n−m m−1 n m−1 n−m a b B (ax)S − (b − 1) = a b B (bx)S − (a − 1). m m n m m m n m m=0 m=0

Theorem 4.4. For n ∈ Z, n ≥ 0, a, b > 0 (a ≠ b), we have ( ) ∑n −bt n m n−m (k) Li (1 − e ) a b B (bx)S − (a − 1) k,q m m,q n m m=0 ( ) ∑n −at n n−m m (k) = Li (1 − e ) a b B (ax)S − (b − 1) k,q m m,q n m m=0 Proof. Let n ∈ Z, n ≥ 0, a, b > 0 and a ≠ b. We consider a function that is given below: Li (1 − e−at)Li (1 − e−bt)(eabt − 1)(eabxt) F (t) = k,q k,q . (4.6) (eat − 1)(ebt − 1) The Equation(4.6) follows: Li (1 − e−at)Li (1 − e−bt)(eabt − 1)(eabxt) F (t) = k,q k,q (eat − 1)(ebt − 1) ∞ ∞ ∑ (at)n ∑ (bt)n = Li (1 − e−bt) B(k)(bx) S (a − 1) k,q n,q n! n n! n=0 ( ) m=0 ∑∞ ∑n n −bt n m n−m (k) t = Li (1 − e ) a b B (bx)S − (a − 1) . k,q m m,q n m n! n=0 m=0 In similar method, we have ∞ ∞ ∑ (bt)n ∑ (at)n F (t) = Li (1 − e−at) B(k)(ax) S (b − 1) k,q n,q n! n n! n=0 ( ) n=0 ∑∞ ∑n n −at n n−m m (k) t = Li (1 − e ) a b B (ax)S − (b − 1) . k,q m m,q n m n! n=0 m=0 Now, comparing the coeﬃcient of tn, then it gives the symmetric identity, ( ) ∑n −bt n m n−m (k) Li (1 − e ) a b B (bx)S − (a − 1) k,q m m,q n m m=0 ( ) ∑n −at n n−m m (k) = Li (1 − e ) a b B (ax)S − (b − 1). k,q m m,q n m m=0 A note on q-analogue of poly-Bernoulli numbers and polynomials 621

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Kyung Won Hwang received Ph.D. degree from University of Illinois at Urbana- Champaign. His research interests focus on the combinatorics, graph theory, discrete geom- etry, scientiﬁc computing and p-adic functional analysis. Department of Mathematics, Dong-A University, Pusan 49315, Korea e-mail:[email protected]

Bo Ryeong Nam received M.Sc. from Hannam University. Since 2012, she has been a senior reseacher at Korea Institute of Oriental Medicine. Her current research interests include complex analysis, Korean medicine information and healthcare. Department of Mathematics, Hannam University, Daejeon 34430, Korea. e-mail: [email protected]

Nam-Soon Jung received Ph.D. from Hannam University. Her research interests concen- trate on algebraic analysis and number theory. College of Talmage Liberal Arts, Hannam University, Daejeon 34430, Korea. e-mail: [email protected]