EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 12, No. 2, 2019, 605-621 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Combinatorial Identities with Generalized Higher-order Genocchi Sequences Tian Hao1,∗, Wuyungaowa1 1 Department of Mathematical Sciences, Inner Mongolia University, Hohhot, Inner Mongolia, P.R.China Abstract. In this paper, we make use of the probabilistic method to calculate the moment rep- resentation of generalized higher-order Genocchi polynomials. We obtain the moment expression of the generalized higher-order Genocchi numbers with a and b parameters. Some characteriza- tions and identities of generalized higher-order Genocchi polynomials are given by the proof of the moment expression. As far as properties given by predecessors are concerned, we prove them by the probabilistic method. Finally, new identities of relationships involving generalized higher-order Genocchi numbers and harmonic numbers, derangement numbers, Fibonacci numbers, Bell num- bers, Bernoulli numbers, Euler numbers, Cauchy numbers and Stirling numbers of the second kind are established. 2010 Mathematics Subject Classifications: 11B68, 60E07, 11B83, 62E15. Key Words and Phrases: Moment, Generating function, Generalized higher-order Genocchi numbers and polynomials, Laplace distribution. 1. Introduction and Preliminaries The classical Genocchi numbers and polynomials play important roles in combinatorics. It is widely used in combinatorial mathematics, function theory, graph theory, approximate calculation and theoretical physics, such as the diffusion of matter. In the present paper, a further investigation for the generalized higher-order Genocchi polynomials with a, b and c parameters is performed. We depend on the Laplace distribution to calculate the moment representation of the generalized higher-order Genocchi polynomials. Hence the moment representation can also be used to give some relative identities and characterizations of the generalized higher-order Genocchi polynomials. Recently, symmetry identities of the generalized Bernoulli, Euler and Genocchi polynomials are investigated by Waseem A. Khan and other predecessors[6][7][8][9]. Here the probabilistic method provides great convenience to investigate symmetry identities of the generalized Genocchi polynomials ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v12i2.3403 Email addresses: 1179827028@qq.com (T. Hao), wuyungw@163.com (Wuyungaowa) http://www.ejpam.com 605 c© 2019 EJPAM All rights reserved. T. Hao, Wuyungaowa / Eur. J. Pure Appl. Math, 12 (2) (2019), 605-621 606 and relations between the generalized higher-order Genocchi numbers and combinatorial numbers. Throughout this paper, let t ∈ C and | t |<π/|lnb− lna|, α ∈ N+, x ∈ R, a, b and c are positive integers, a 6= b, we now turn to the generalized higher-order Genocchi polynomials G (α) n (x; a, b, c) with a, b and c parameters for nonnegative integer n, which are usually defined by means of the following generating function[2][4, 5]: ( 2t at + bt )αcxt = ∞∑ n=0 G(α) n (x; a, b, c) tn n! . (1) Taking x = 0 in Eq.(1), we get the generating function of the generalized higher-order Genocchi numbers G (α) n (a, b) with a and b parameters[4]: ( 2t at + bt )α = ∞∑ n=0 G(α) n (a, b) tn n! . (2) Setting a = 1, b = e, c = e in Eq.(1), the generating function of the classical higher- order Genocchi polynomials is as follows[3][5][9][12]: ( 2t et + 1 )αext = ∞∑ n=0 G(α) n (x) tn n! , (3) then setting x = 0 in Eq.(3), we get the generating function of the classical higher-order Genocchi numbers[1][12]: ( 2t et + 1 )α = ∞∑ n=0 G(α) n tn n! . (4) In this paper, we make use of the special combinatorial sequences of the generalized Eu- ler polynomials En(x; a, b, c) with a, b and c, which are defined by the following generating function[10][11]: 2cxt at + bt = ∞∑ n=0 En(x; a, b, c) tn n! . (5) For α ∈ N+, we get the generalized higher-order Euler polynomials E (α) n (x; a, b, c) with a, b and c[11]: ( 2 at + bt )αcxt = ∞∑ n=0 E(α) n (x; a, b, c) tn n! . (6) We also need the following notations: r.v denotes a random variable, i.i.d shows that a sequence of random variables are independent and identically distributed. The notation E denotes an expectation operator and definition is as follows: When f(x) is a measurable function with continuous random variables X and p(x) is a density function of X, we have Ef(X) = ∫ +∞ −∞ f(x)p(x)dx. T. Hao, Wuyungaowa / Eur. J. Pure Appl. Math, 12 (2) (2019), 605-621 607 Especially, setting f(x) = xn ,we obtain the moment of n-th order EXn of random vari- ables. Remark 1. [see 15] If f and g are exponential generating functions, and fg = ( ∞∑ r=0 art r r! )( ∞∑ s=0 bst s s! ), then the coefficients of tn n! in fg are given by [ tn n! ](fg) = n∑ r=0 ( n r ) arbn−r. Next we shall introduce several moment representations of some combinatorial se- quences. Lemma 1. [see 14] Suppose that r.v u1, u2, i.i.d ∼ U [0, 1], then harmonic numbers Hn =∑n k=1 1 k have the following moment representation, Hn = nE(1− u1u2)n−1, n > 1. (7) Lemma 2. [see 14] Suppose that r.v X ∼ Γ(1, 1), then derangement numbers dn = n! ∑n k=0 (−1)k k! satisfy the following moment representation, dn = E(X − 1)n, n > 0. (8) Lemma 3. [see 14] Suppose that r.v u ∼ U [0, 1], then Fibonacci numbers Fn have the following moment representation, Fn = (n+ 1)E( √ 5u+ (1− √ 5) 2 )n, n > 0. (9) Lemma 4. [see 14] Suppose that r.v X ∼ P (1), then Bell numbers bn have the following moment representation, bn = E(X)n, n > 0, bn = E(X + 1)n−1, n > 1. (10) Lemma 5. [see 13] Suppose that r.v L1, L2, · · · , i.i.d ∼ L[0, 1] and r.v Le = ∑ k≥1 Lk 2kπ , then Bernoulli numbers Bn satisfy the following moment representation, Bn = E(iLe − 1 2 )n, n > 0. (11) Lemma 6. [see 13] Suppose that r.v L1, L2, · · · , i.i.d ∼ L[0, 1] and r.v L = ∑ k≥1 Lk (2k−1)π , then Euler numbers En have the following moment representation, En = 2nE(iL)n, n > 0. (12) T. Hao, Wuyungaowa / Eur. J. Pure Appl. Math, 12 (2) (2019), 605-621 608 Lemma 7. [see 14] Suppose that r.v X ∼ Γ(u, 1), u ∼ U [0, 1], and X, u are independent, then Cauchy numbers of the second kind cn satisfy the following moment representation, cn = EXn, n > 0. (13) Lemma 8. [see 14] Suppose that r.v u1, u2, · · · , i.i.d ∼ U [0, 1] for all i, when n, k > 1, then Stirling numbers of the second kind have the following moment representation, S(n, k) = ( n k ) E(u1 + u2 + · · ·+ uk) n−k. (14) It is demanded that S(n, 0) = S(0, k) = 0, S(0, 0) = 1. 2. Moment Representations of the Generalized Higher-order Genocchi polynomials In this section, we derive the moment representation of the generalized higher-order Genocchi polynomials by means of the probabilistic method. Some properties and iden- tities of generalized higher-order Genocchi polynomials and numbers are given on the foundation of the moment expression. Theorem 1. Suppose that r.v L1, L2, · · · , i.i.d ∼ L[0, 1], i2 = −1, let L = ∑ k≥1 Lk (2k−1)π and for all j, L(j) = ∑ k≥1 L (j) k (2k−1)π be random variables and r.v {L(j)}1≤ j ≤α are indepen- dent and obey the same distribution as L. Then let a, b and c be positive integers with conditions a 6= b, ab 6= 1 and c 6= 1, for α ∈ N+, x ∈ R, n ≥ α, the generalized higher-order Genocchi polynomials G (α) n (x; a, b, c) satisfy the following moment representation: G(α) n (x; a, b, c) = n! (n− α)! E[i(L(1) + · · ·+L(α))ln b a + xlnc− α 2 lnab]n−α, (n ≥ α). (15) Proof. From the generating function of the generalized higher-order Genocchi polyno- mials we discover ∞∑ n=0 G(α) n (x; a, b, c) tn n! = ( 2t bt + at )αcxt = tα( 2 ( ba)t + 1 )αa−tαcxt = tα( 2e t 2 ln b a etln b a + 1 )αet(xlnc− α 2 ln b a −αlna) = tα( 2e t 2 ln b a etln b a + 1 )αet(xlnc− α 2 lnab). (16) Suppose that r.v L1, L2, · · · , i.i.d ∼ L[0, 1], i2 = −1, L = ∑ k≥1 Lk (2k−1)π [13], 2e t 2 et + 1 = 1 e t 2+e− t 2 2 = 1 cosh( t2) = ∏ k≥1 [1 + ( t (2k − 1)π )2]−1. (17) T. Hao, Wuyungaowa / Eur. J. Pure Appl. Math, 12 (2) (2019), 605-621 609 Because [1+( t (2k−1)π )2]−1 is the characteristic function of r.v Lk (2k−1)π , we discover that∏ k≥1[1 + ( t (2k−1)π )2]−1 is the characteristic function of L = ∑ k≥1 Lk (2k−1)π . All moments of the random variables exist, hence the discovery depends on the following identity: ∞∑ n=0 ELn (it)n n! = ∏ k≥1 [1 + ( t (2k − 1)π )2]−1 = 2e t 2 et + 1 , (i2 = −1). (18) Let L(j) = ∑ k≥1 L (j) k (2k−1)π be a random variable and for all j r.v {L(j)}1≤ j ≤α are independent and obey the same distribution as L. According to Eq.(16) and Eq.(18), we have tα( ∑ m≥0 ELm (itln ba)m m! )αet(xlnc− α 2 lnab) = tα{ ∑ n≥0 ∑ m1+···+mα=n ( n m1, · · · ,mα ) E(iL(1)ln b a )m1 · · ·E(iL(α)ln b a )mα tn n! } × { ∑ n≥0 (xlnc− α 2 lnab)n tn n! } = tα ∑ n≥0 E(iL(1)ln b a + · · ·+ iL(α)ln b a )n tn n! ∑ n≥0 (xlnc− α 2 lnab)n tn n! = ∑ n≥0 n∑ k=0 ( n k ) E[i(L(1) + · · ·+ L(α))ln b a ]k(xlnc− α 2 lnab)n−k tn+α n! = ∑ n≥0 E[i(L(1) + · · ·+ L(α))ln b a + xlnc− α 2 lnab]n tn+α n! = ∑ n≥α n! (n− α)! E[i(L(1) + · · ·+ L(α))ln b a + xlnc− α 2 lnab]n−α tn n! By comparing the coefficients tn n! , we arrive at the moment expression of G (α) n (x; a, b, c). Corollary 1. Taking x = 0 in Eq.(15), we get the moment representation of the higher- order Genocchi numbers with a and b parameters G (α) n (a, b), G(α) n (0; a, b) = G(α) n (a, b) = n! (n− α)! E[i(L(1) + · · ·+ L(α))ln b a − α 2 lnab]n−α, (n ≥ α). (19) For a = 1, b = e in Eq.(19), we get the moment representation of the higher-order Genocchi numbers G (α) n , G(α) n (1, e) = G(α) n = n! (n− α)! E(iL(1) + · · ·+ iL(α) − α 2 )n−α, (n ≥ α). (20) T. Hao, Wuyungaowa / Eur. J. Pure Appl. Math, 12 (2) (2019), 605-621 610 Corollary 2. For α = 1 in Eq.(15), we get the moment representation of the generalized Genocchi polynomials Gn(x; a, b, c), G(1) n (x; a, b, c) = Gn(x; a, b, c) = nE(iLln b a + xlnc− 1 2 lnab)n−1, (n ≥ 1). (21) Corollary 3. Setting a = 1, b = e, c = e in Eq.(15), we get the moment representation of the higher-order Genocchi polynomials G (α) n (x), G(α) n (x; 1, e, e) = G(α) n (x) = n! (n− α)! E[iL(1) + · · ·+ iL(α) + x− α 2 ]n−α, (n ≥ α). (22) Corollary 4. We can easily obtain the moment representation of the generalized higher- order Euler polynomials and get the relation between generalized higher-order Euler poly- nomials and generalized higher-order Gennocchi polynomials in the proof of theorem 1, E(α) n (x; a, b, c) = E[i(L(1) + · · ·+ L(α))ln b a + xlnc− α 2 lnab]n, (23) G(α) n (x; a, b, c) = n! (n− α)! E (α) n−α(x; a, b, c), (n ≥ α). (24) The generalized higher-order Euler polynomials and the generalized higher-order Genoc- chi polynomials have similar properties and forms, we only aim at the properties of gen- eralized higher-order Genocchi polynomials here. Theorem 2. Let a, b and c be positive integers with conditions a 6= b, ab 6= 1 and c 6= 1, for α ∈ N+, x ∈ R, n ≥ α, then we get G(α) n (x+ α; a, b, c) = G(α) n (x; a c , b c , c). (25) Proof. In light of the theorem 1 we have G(α) n (x+ α; a, b, c) = n! (n− α)! E[i(L(1) + · · ·+ L(α))ln b a + (x+ α)lnc− α 2 lnab]n−α = n! (n− α)! E[i(L(1) + · · ·+ L(α))ln b a + xlnc+ α 2 lnc2 − α 2 lnab]n−α = n! (n− α)! E[i(L(1) + · · ·+ L(α))ln b c a c + xlnc− α 2 ln a c b c ]n−α = G(α) n (x; a c , b c , c). This concludes the proof. Theorem 3. Let a, b and c be positive integers with conditions a 6= b, ab 6= 1 and c 6= 1, for α ∈ N+, x ∈ R, n ≥ α, then we get G(α) n (α− x; a, b, c) = (−1)n−αG(α) n (x; c a , c b , c) = G(α) n (−x; a c , b c , c). (26) T. Hao, Wuyungaowa / Eur. J. Pure Appl. Math, 12 (2) (2019), 605-621 611 Proof. According to theorem 1, we have G(α) n (α− x; a, b, c) = n! (n− α)! E[i(L(1) + · · ·+ L(α))ln b a + (α− x)lnc− α 2 lnab]n−α = n! (n− α)! (−1)n−αE[i(L(1) + · · ·+ L(α))ln 1 b 1 a + xlnc− α 2 lnc2 − α 2 ln 1 ab ]n−α = n! (n− α)! (−1)n−αE[i(L(1) + · · ·+ L(α))ln c b c a + xlnc− α 2 ln c a c b ]n−α = (−1)n−αG(α) n (x; c a , c b , c). On substituting x with -x in Eq.(25), then Eq.(26) arises. Theorem 4. Let a, b and c be positive integers with conditions a 6= b, ab 6= 1 and c 6= 1, for α ∈ N+, x ∈ R, n ≥ α, then we get G(α) n (−x; a, b, c) = G(α) n (x; a, b, 1 c ). (27) Proof. With the help of theorem 1 we have G(α) n (−x; a, b, c) = n! (n− α)! E[i(L(1) + · · ·+ L(α))ln b a − xlnc− α 2 lnab]n−α = n! (n− α)! E[i(L(1) + · · ·+ L(α))ln b a + xln 1 c − α 2 lnab]n−α = G(α) n (x; a, b, 1 c ). This concludes the proof. Corollary 5. Taking α = 1 in Eq.(25), Eq.(26), Eq.(27), we can easily get the following identities, Gn(x+ 1; a, b, c) = Gn(x; a c , b c , c), (28) Gn(1− x; a, b, c) = (−1)n−1Gn(x; c a , c b , c) = Gn(−x; a c , b c , c), (29) Gn(−x; a, b, c) = Gn(x; a, b, 1 c ). (30) Theorem 5. Let a, b and c be positive integers with conditions a 6= b, ab 6= 1 and c 6= 1, for α ∈ N+, x ∈ R, n ≥ α, then we get G(α) n (x; a, b, c) = (ln b a )n−αG(α) n ( xlnc− αlna lnb− lna ). (31) T. Hao, Wuyungaowa / Eur. J. Pure Appl. Math, 12 (2) (2019), 605-621 612 Proof. From theorem 1 we have G(α) n (x; a, b, c) = n! (n− α)! E[i(L(1) + · · ·+ L(α))ln b a + xlnc− α 2 lnab]n−α = (ln b a )n−α n! (n− α)! E[i(L(1) + · · ·+ L(α)) + xlnc lnb− lna − α 2 lnb+ α 2 lna lnb− lna ]n−α = (ln b a )n−α n! (n− α)! E[i(L(1) + · · ·+ L(α)) + xlnc lnb− lna − α 2 lnb− α 2 lna+ αlna lnb− lna ]n−α = (ln b a )n−α n! (n− α)! E[i(L(1) + · · ·+ L(α)) + xlnc− αlna lnb− lna − α 2 ]n−α = (ln b a )n−αG(α) n ( xlnc− αlna lnb− lna ). Thus we arrive at the desired result. Theorem 6. Let a, b and c be positive integers with conditions a 6= b, ab 6= 1 and c 6= 1, for α ∈ N+, x ∈ R, n ≥ α, then we get (−1)n−αG(α) n (x; a, b, c) = n∑ k=α ( n k ) G (α) k (−x; b, a, c)(αlnab)n−k. (32) Proof. From Eq.(15) in theorem 1, we have (−1)n−αG(α) n (x; a, b, c) = n! (n− α)! E[i(L(1) + · · ·+ L(α))ln a b − xlnc− α 2 ln 1 ab ]n−α = n! (n− α)! E[i(L(1) + · · ·+ L(α))ln a b − xlnc− α 2 ln 1 ab − α 2 ln(ab)2 + α 2 ln(ab)2]n−α = n! (n− α)! E[i(L(1) + · · ·+ L(α))ln a b − xlnc− α 2 lnab+ αlnab]n−α = n! (n− α)! n∑ k=α ( n− α k − α ) E[i(L(1) + · · ·+ L(α))ln a b − xlnc− α 2 lnab]k−α(αlnab)n−k = n∑ k=α ( n k ) G (α) k (−x; b, a, c)(αlnab)n−k. Therefor we derive the Eq.(32). Theorem 7. Let a, b and c be positive integers with conditions a 6= b, ab 6= 1 and c 6= 1, for α ∈ N+, x ∈ R, n ≥ α, then we get n∑ k=α ( n k ) G (α) k (a, b)xn−k(lnc)n−k = n∑ k=α ( n k ) (−αlnc)n−kG(α) k (x; a c , b c , c). (33) T. Hao, Wuyungaowa / Eur. J. Pure Appl. Math, 12 (2) (2019), 605-621 613 Proof. By theorem 1 we have G(α) n (x; a, b, c) = n! (n− α)! E[i(L(1) + · · ·+ L(α))ln b a + xlnc− α 2 lnab]n−α = n! (n− α)! n∑ k=α ( n− α k − α ) E[i(L(1) + · · ·+ L(α))ln b a − α 2 lnab]k−α(xlnc)n−k = n∑ k=α ( n k ) G (α) k (a, b)xn−k(lnc)n−k, (34) and G(α) n (x; a, b, c) = n! (n− α)! E[i(L(1) + · · ·+ L(α))ln b a + xlnc− α 2 lnab]n−α = n! (n− α)! n∑ k=α ( n− α k − α ) E[i(L(1) + · · ·+ L(α))ln b c a c + xlnc− α 2 ln ab c2 ]k−α(−αlnc)n−k = n∑ k=α ( n k ) (−αlnc)n−kG(α) k (x; a c , b c , c). (35) Thus, combining Eq.(34) and Eq.(35) gives the theorem 7. Theorem 8. Let a, b and c be positive integers with conditions a 6= b, ab 6= 1 and c 6= 1, for α ∈ N+, x, y ∈ R, n ≥ α, then we get G(α) n (x+ y; a, b, c) = n∑ k=α ( n k ) G (α) k (x; a, b, c)yn−k(lnc)n−k. (36) Proof. It follows from Eq.(15) that G(α) n (x+ y; a, b, c) = n! (n− α)! E[i(L(1) + · · ·+ L(α))ln b a + (x+ y)lnc− α 2 lnab]n−α = n! (n− α)! n∑ k=α ( n− α k − α ) E[i(L(1) + · · ·+ L(α))ln b a + xlnc− α 2 lnab]k−α(ylnc)n−k = n! (n− α)! n∑ k=α ( n− α k − α ) (k − α)! k! G (α) k (x; a, b, c)(ylnc)n−k = n∑ k=α ( n k ) G (α) k (x; a, b, c)yn−k(lnc)n−k. This concludes the proof. T. Hao, Wuyungaowa / Eur. J. Pure Appl. Math, 12 (2) (2019), 605-621 614 Theorem 9. Let a, b and c be positive integers with conditions a 6= b, ab 6= 1 and c 6= 1, for α ∈ N+, x, y, z ∈ R, n ≥ α, then we get∫ y x G(α) n (z; a, b, c)dz = 1 (n+ 1)lnc [G (α) n+1(y; a, b, c)−G(α) n+1(x; a, b, c)], (37) ∂lG (α) n (x; a, b, c) ∂xl = (n)l(lnc) lG (α) n−l(x; a, b, c). (38) Proof. By Eq.(15), we have∫ y x G(α) n (z; a, b, c)dz = ∫ y x n! (n− α)! E[i(L(1) + · · ·+ L(α))ln b a + zlnc− α 2 lnab]n−αdz = n! (n− α)! E ∫ y x [i(L(1) + · · ·+ L(α))ln b a + zlnc− α 2 lnab]n−αdz = n! (n− α)! 1 (n− α+ 1)lnc {E[i(L(1) + · · ·+ L(α))ln b a + ylnc− α 2 lnab]n−α+1 − E[i(L(1) + · · ·+ L(α))ln b a + xlnc− α 2 lnab]n−α+1} = n! (n− α)! 1 (n− α+ 1)lnc (n− α+ 1)! (n+ 1)! [G (α) n+1(y; a, b, c)−G(α) n+1(x; a, b, c)] = 1 (n+ 1)lnc [G (α) n+1(y; a, b, c)−G(α) n+1(x; a, b, c)]. As for Eq.(38), we have ∂lG (α) n (x; a, b, c) ∂xl = ∂l{ n! (n−α)!E[i(L(1) + · · ·+ L(α))ln ba + xlnc− α 2 lnab] n−α} ∂xl = n! (n− α)! (lnc)l(n− α)(n− α− 1) · · · (n− α− l + 1) × E[i(L(1) + · · ·+ L(α))ln b a + xlnc− α 2 lnab]n−α−l = n!(lnc)l (n− α)! (n− α)! (n− α− l)! (n− α− l)! (n− l)! G (α) n−l(x; a, b, c) = (n)l(lnc) lG (α) n−l(x; a, b, c). This concludes the proof. Theorem 10. Let a, b and c be positive integers with conditions a 6= b, ab 6= 1 and c 6= 1, for α ∈ N+, x ∈ R, n ≥ α, for all d, l ∈ N , then we get n−α∑ k=α k∑ j=α ( n k )( k j ) dn−klk(−αlnc)k−jG(α) n−k(lx; a, b, c)G (α) j (dx; a c , b c , c) = n−α∑ k=α k∑ j=α ( n k )( k j ) ln−kdk(−αlnc)k−jG(α) n−k(dx; a, b, c)G (α) j (lx; a c , b c , c). (39) T. Hao, Wuyungaowa / Eur. J. Pure Appl. Math, 12 (2) (2019), 605-621 615 Proof. Firstly, we have n! (n− 2α)! E[i(d+ l)(L(1) + · · ·+ L(α))ln b a + 2dlxlnc− α 2 (d+ l)lnab]n−2α = n! (n− 2α)! n−2α∑ k=α ( n− 2α k − α ) E[id(L(1) + · · ·+ L(α))ln b a + dlxlnc− αd 2 lnab]n−k−α × E[il(L(1) + · · ·+ L(α))ln b a + dlxlnc− αl 2 lnab]k−α = (dl)−αn! (n− 2α)! n−α∑ k=α ( n− 2α k − α ) (n− k − α)!dn−kG (α) n−k(lx; a, b, c) (n− k)! (k − α)!lkG (α) k (dx; a, b, c) k! = (dl)−α n−α∑ k=α ( n k ) dn−klkG (α) n−k(lx; a, b, c)G (α) k (dx; a, b, c), according to Eq.(35), (dl)−α n−α∑ k=α ( n k ) dn−klkG (α) n−k(lx; a, b, c)G (α) k (dx; a, b, c) = (dl)−α n−α∑ k=α ( n k ) dn−klkG (α) n−k(lx; a, b, c) k∑ j=α ( k j ) (−αlnc)k−jG(α) j (dx; a c , b c , c). (40) Secondly, we have n! (n− 2α)! E[i(d+ l)(L(1) + · · ·+ L(α))ln b a + 2dlxlnc− α 2 (d+ l)lnab]n−2α = n! (n− 2α)! n−2α∑ k=α ( n− 2α k − α ) E[il(L(1) + · · ·+ L(α))ln b a + dlxlnc− αl 2 lnab]n−k−α × E[id(L(1) + · · ·+ L(α))ln b a + dlxlnc− αd 2 lnab]k−α = (ld)−αn! (n− 2α)! n−α∑ k=α ( n− 2α k − α ) (n− k − α)!ln−kG (α) n−k(dx; a, b, c) (n− k)! (k − α)!dkG (α) k (lx; a, b, c) k! = (ld)−α n−α∑ k=α ( n k ) ln−kdkG (α) n−k(dx; a, b, c)G (α) k (lx; a, b, c) = (ld)−α n−α∑ k=α ( n k ) ln−kdkG (α) n−k(dx; a, b, c) k∑ j=α ( k j ) (−αlnc)k−jG(α) j (lx; a c , b c , c). (41) By equating Eq.(40) and Eq.(41), we arrive at the desired result. T. Hao, Wuyungaowa / Eur. J. Pure Appl. Math, 12 (2) (2019), 605-621 616 From the proof of theorem 10, we get the following identity: n−α∑ k=α ( n k ) dn−klkG (α) n−k(lx; a, b, c)G (α) k (dx; a, b, c) = n−α∑ k=α ( n k ) ln−kdkG (α) n−k(dx; a, b, c)G (α) k (lx; a, b, c). (42) Corollary 6. Setting α = 1 in Eq.(39), we obtain n−1∑ k=1 k∑ j=1 ( n k )( k j ) dn−klk(−lnc)k−jGn−k(lx; a, b, c)Gj(dx; a c , b c , c) = n−1∑ k=1 k∑ j=1 ( n k )( k j ) ln−kdk(−lnc)k−jGn−k(dx; a, b, c)Gj(lx; a c , b c , c). (43) Then taking a = 1, b = e, c = e in Eq.(43), we get n−1∑ k=1 k∑ j=1 ( n k )( k j ) dn−klkGn−k(lx)Gj(dx) = n−1∑ k=1 k∑ j=1 ( n k )( k j ) ln−kdkGn−k(dx)Gj(lx). (44) 3. Identities of Generalized Higher-order Genocchi Numbers and Special Combinatorial Sequences In this section, we establish some new identities involving the generalized higher-order Genocchi numbers and harmonic numbers, derangement numbers, Fibonacci numbers, Bell numbers, Bernoulli numbers, Euler numbers, Cauchy numbers and Stirling numbers of second kind by means of the moment representation of the generalized higher-order Genocchi polynomials. Theorem 11. Let a, b and c be positive integers with conditions a 6= b, ab 6= 1 and c 6= 1, for α ∈ N+, x ∈ R, n ≥ α, then we get G(α) n (x; a, b, c) = n∑ k=α ( n k ) (lnc)n−kxn−kG (α) k (a, b), (45) G(α) n (x; a, b, c) = n∑ k=α ( n k ) (lnc)n−k(ln b a )k−αxn−kG (α) k ( αlna lna− lnb ), (46) G(α) n (x; a, b, c) = n∑ k=α k∑ j=α ( n k )( k j ) (lnc)n−k(ln b a )j−α(−αlna)k−jxn−kG (α) j . (47) T. Hao, Wuyungaowa / Eur. J. Pure Appl. Math, 12 (2) (2019), 605-621 617 Proof. By the Eq.(34) we can arrive at the Eq.(45) directly. It follows from Eq.(15) that G(α) n (x; a, b, c) = n! (n− α)! E[i(L(1) + · · ·+ L(α))ln b a + xlnc− α 2 lnab]n−α = n! (n− α)! n−α∑ k=α ( n− α k − α ) E[i(L(1) + · · ·+ L(α))ln b a − α 2 lnab]k−α(xlnc)n−k = n! (n− α)! n∑ k=α ( n− α k − α ) (ln b a )k−αE[i(L(1) + · · ·+ L(α))− αlna lnb− lna − α 2 ]k−α(xlnc)n−k = n! (n− α)! n∑ k=α ( n− α k − α ) (k − α)! k! G (α) k ( αlna lna− lnb )xn−k(lnc)n−k(ln b a )k−α = n∑ k=α ( n k ) (lnc)n−k(ln b a )k−αxn−kG (α) k ( αlna lna− lnb ), from which we see that E[i(L(1) + · · ·+ L(α))ln b a − α 2 lnab]k−α = (ln b a )k−αE[i(L(1) + · · ·+ L(α))− αlna lnb− lna − α 2 ]k−α = (ln b a )k−α k−α∑ j=α ( k − α j − α ) E[i(L(1) + · · ·+ L(α))− α 2 ]j−α(−αlna)k−j(ln b a )j−k = k∑ j=α ( k − α j − α ) (j − α)! j! G (α) j (ln b a )j−α(−αlna)k−j . Hence we complete the proof of the theorem 11. According to Eq.(45), Eq.(46) and Eq.(47) in theorem 11, we get the following theorem. Theorem 12. Let a, b and c be positive integers with conditions a 6= b, ab 6= 1 and c 6= 1, for α ∈ N+, x ∈ R, n ≥ α, then we get n∑ k=α ( n k ) (lnc)n−kxn−kG (α) k (a, b) = n∑ k=α ( n k ) (lnc)n−k(ln b a )k−αxn−kG (α) k ( αlna lna− lnb ), (48) n∑ k=α ( n k ) (lnc)n−kxn−kG (α) k (a, b) = n∑ k=α k∑ j=α ( n k )( k j ) (lnc)n−k(ln ba)j−α (−αlna)j−k xn−kG (α) j . (49) We now turn to the above identities of the generalized higher-order Genocchi numbers and x. For the case x is a constant variable, it seems normal. It is worth noticing T. Hao, Wuyungaowa / Eur. J. Pure Appl. Math, 12 (2) (2019), 605-621 618 that we can see x as a random variable which obeys to a certain distribution and then take mathematical expectation on the two sides of the equation of random variables. It turns out that some interesting identities including the generalized higher-order Genocchi numbers and some combinatorial sequences are derived. The probabilistic method is much simpler and more effective here. (1)Suppose that r.v x = 1− u1u2, r.v u1, u2, i.i.d ∼ U [0, 1], for the moment represen- tation of the harmonic numbers Hn in Eq.(7), the following identities hold true: n∑ k=α ( n k ) (lnc)n−k Hn−k+1G (α) k (a, b) n− k + 1 = n∑ k=α ( n k ) (lnc)n−k(ln b a )k−α Hn−k+1G (α) k ( αlna lna−lnb) n− k + 1 , n∑ k=α ( n k ) (lnc)n−k Hn−k+1G (α) k (a, b) n− k + 1 = n∑ k=α k∑ j=α ( n k )( k j ) (lnc)n−k(ln ba)j−α (−αlna)j−k Hn−k+1G (α) j n− k + 1 . (2)Suppose that r.v x = Γ − 1, Γ ∼ Γ(1, 1), for the moment representation of the derangement numbers dn in Eq.(8), the following identities hold true: n∑ k=α ( n k ) (lnc)n−kdn−kG (α) k (a, b) = n∑ k=α ( n k ) (lnc)n−k(ln b a )k−αdn−kG (α) k ( αlna lna− lnb ), n∑ k=α ( n k ) (lnc)n−kdn−kG (α) k (a, b) = n∑ k=α k∑ j=α ( n k )( k j ) (lnc)n−k(ln ba)j−α (−αlna)j−k dn−kG (α) j . (3)Suppose that r.v x = √ 5u+ (1− √ 5) 2 , r.v u ∼ U [0, 1], for the moment representation of the Fibonacci numbers Fn in Eq.(9), the following identities hold true: n∑ k=α ( n k ) (lnc)n−k Fn−kG (α) k (a, b) n− k + 1 = n∑ k=α ( n k ) (lnc)n−k(ln b a )k−α Fn−kG (α) k ( αlna lna−lnb) n− k + 1 , n∑ k=α ( n k ) (lnc)n−k Fn−kG (α) k (a, b) n− k + 1 = n∑ k=α k∑ j=α ( n k )( k j ) (lnc)n−k(ln ba)j−α (−αlna)j−k Fn−kG (α) j n− k + 1 . (4)Suppose that r.v x = X, r.v X ∼ P (1), for the moment representation of the Bell numbers bn in Eq.(10), the following identities hold true: n∑ k=α ( n k ) (lnc)n−kbn−kG (α) k (a, b) = n∑ k=α ( n k ) (lnc)n−k(ln b a )k−αbn−kG (α) k ( αlna lna− lnb ), T. Hao, Wuyungaowa / Eur. J. Pure Appl. Math, 12 (2) (2019), 605-621 619 n∑ k=α ( n k ) (lnc)n−kbn−kG (α) k (a, b) = n∑ k=α k∑ j=α ( n k )( k j ) (lnc)n−k(ln ba)j−α (−αlna)j−k bn−kG (α) j . (5)Suppose that r.v x = iLe− 1 2 , r.v L1, L2, · · · , i.i.d ∼ L[0, 1], let Le = ∑ k≥1 Lk 2kπ be a random variable, for the moment representation of the Bernoulli numbers Bn in Eq.(11), the following identities hold true: n∑ k=α ( n k ) (lnc)n−kBn−kG (α) k (a, b) = n∑ k=α ( n k ) (lnc)n−k(ln b a )k−αBn−kG (α) k ( αlna lna− lnb ), n∑ k=α ( n k ) (lnc)n−kBn−kG (α) k (a, b) = n∑ k=α k∑ j=α ( n k )( k j ) (lnc)n−k(ln ba)j−α (−αlna)j−k Bn−kG (α) j . (6)Suppose that r.v x = iL, L1, L2, · · · , i.i.d ∼ L[0, 1], let r.v L = ∑ k≥1 Lk (2k−1)π be a random variable, for the moment representation of the Euler numbers En in Eq.(12), the following identities hold true: n∑ k=α ( n k ) (lnc)n−k En−kG (α) k (a, b) 2n−k = n∑ k=α ( n k ) (lnc)n−k(ln b a )k−α En−kG (α) k ( αlna lna−lnb) 2n−k , n∑ k=α ( n k ) (lnc)n−k En−kG (α) k (a, b) 2n−k = n∑ k=α k∑ j=α ( n k )( k j ) (lnc)n−k(ln ba)j−α (−αlna)j−k En−kG (α) j 2n−k . (7)Suppose that r.v x = X, r.v X ∼ Γ(u, 1), u ∼ U [0, 1], X and u are independent, for the moment representation of the Cauchy numbers cn in Eq.(13), the following identities hold true: n∑ k=α ( n k ) (lnc)n−kcn−kG (α) k (a, b) = n∑ k=α ( n k ) (lnc)n−k(ln b a )k−αcn−kG (α) k ( αlna lna− lnb ), n∑ k=α ( n k ) (lnc)n−kcn−kG (α) k (a, b) = n∑ k=α k∑ j=α ( n k )( k j ) (lnc)n−k(ln ba)j−α (−αlna)j−k cn−kG (α) j . (8)Suppose that r.v x = u1 + u2 + · · ·+ uk, r.v u1, u2, · · · , i.i.d ∼ U [0, 1], for all i and n, k > 1, for the moment representation of the Stirling numbers of second kind S(n, k) in Eq.(14), the following identities hold true: n∑ k=α (lnc)n−kS(n, k)G (α) k (a, b) = n∑ k=α (lnc)n−k(ln b a )k−αS(n, k)G (α) k ( αlna lna− lnb ), n∑ k=α (lnc)n−kS(n, k)G (α) k (a, b) = n∑ k=α k∑ j=α ( k j ) (lnc)n−k(ln ba)j−α (−αlna)j−k S(n, k)G (α) j . 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