1 Introduction, definitions and notations

Throughout this presentation, we use the following standard notions: N={1,2,}, N 0 ={0,1,2,}=N{0}, Z ={1,2,}. Also, as usual ℤ denotes the set of integers, ℝ denotes the set of real numbers and ℂ denotes the set of complex numbers. Furthermore, ( λ ) 0 =1 and

( λ ) k =λ(λ+1)(λ+2)(λ+k1),

where kN, λC.

The classical Frobenius-Euler polynomial H n ( α ) (x;u) of order α is defined by means of the following generating function:

( 1 u e t u ) α e x t = n = 0 H n ( α ) (x;u) t n n ! ,
(1)

where u is an algebraic number and αZ.

Observe that H n ( 1 ) (x;u)= H n (x;u), which denotes the Frobenius-Euler polynomials and H n ( α ) (0;u)= H n ( α ) (u), which denotes the Frobenius-Euler numbers of order α. H n (x;1)= E n (x), which denotes the Euler polynomials (cf. [124]).

Definition 1.1 (for details, see [16, 17])

Let a,b,c R + , ab, xR. The generalized Apostol-type Frobenius-Euler polynomials are defined by means of the following generating function:

( a t u λ b t u ) α c x t = n = 0 H n ( α ) (x;u;a,b,c;λ) t n n ! .
(2)

Remark 1.2 If we set x=0 and α=1 in (2), we get

a t u λ b t u = n = 0 H n (u;a,b,c;λ) t n n ! ,
(3)

where H n (u;λ;a,b,c) denotes the generalized Apostol-type Frobenius-Euler numbers (cf. [17]).

2 New identities

In this section, we derive many new identities related to the generalized Apostol-type Frobenius-Euler numbers and polynomials of order α.

Theorem 2.1 Let α,βZ. Each of the following relationships holds true:

(4)
(5)
(6)

and

H n ( α ) ( x ; u 2 ; a 2 , b 2 , c 2 ; λ 2 ) = k = 0 n ( n k ) H k ( α ) (x;u;a,b,c;λ) H n k ( α ) (x;u;a,b,c;λ).
(7)

Proof of (6) From (2),

n = 0 H n ( α ) (x;u;a,b,c;λ) t n n ! n = 0 H n ( α ) (y;u;a,b,c;λ) t n n ! = c ( x + y ) t .
(8)

Therefore,

n = 0 ( k = 0 n ( n k ) H n k ( α ) ( y ; u ; a , b , c ; λ ) H k ( α ) ( x ; u ; a , b , c ; λ ) ) t n n ! = n = 0 ( x ln c ) n t n n ! .

Thus, by using the Cauchy product in (8) and then equating the coefficients of t n n ! on both sides of the resulting equation, we obtain the desired result.

The proofs of (4), (5) and (7) are the same as that of (2), thus we omit them. □

Observe that in (6) we have

( ( x + y ) ln c ) n = ( H ( α ) ( y ; u ; a , b , c ; λ ) + H ( α ) ( x ; u ; a , b , c ; λ ) ) n ,

where ( H ( α ) ( y ; u ; a , b , c ; λ ) ) n is replaced by H n ( α ) (y;u;a,b,c;λ).

Theorem 2.2 Let αN. Then we have

k = 0 α ( α k ) ( u ) α k ( x ln c + k ln a ) n = p = 0 n k = 0 α ( n p ) ( α k ) ( u ) α k ( k ln b ) p H n p ( α ) (x;u;a,b,c;λ).

Proof By using (2), we get

n = 0 ( k = 0 α ( α k ) ( u ) α k ( x ln c + k ln a ) n ) t n n ! = n = 0 ( p = 0 n k = 0 α ( n p ) ( α k ) ( u ) α k ( k ln b ) p H n p ( α ) ( x ; u ; a , b , c ; λ ) ) t n n ! .

By equating the coefficients of t n n ! on both sides of the resulting equation, we obtain the desired result. □

Theorem 2.3 The following relationship holds true:

(9)

Proof We set

( 2 u 1 ) a t u λ b t u c x t a t ( 1 u ) λ b t ( 1 u ) c y t = ( a t u ) ( a t ( 1 u ) ) c ( x + y ) t ( 1 λ b t u 1 λ b t ( 1 u ) ) .

From the above equation, we see that

( 2 u 1 ) ( n = 0 H n ( x ; u ; a , b , c ; λ ) t n n ! ) ( n = 0 H n ( y ; 1 u ; a , b , c ; λ ) t n n ! ) = ( a t 1 + u ) n = 0 H n ( x + y ; u ; a , b , c ; λ ) t n n ! ( a t u ) n = 0 H n ( x + y ; 1 u ; a , b , c ; λ ) t n n ! .

Therefore,

( 2 u 1 ) n = 0 r = 0 n ( n r ) H r ( x ; u ; a , b , c ; λ ) H n r ( y ; 1 u ; a , b , c ; λ ) t n n ! = ( u 1 ) n = 0 H n ( x + y ; u ; a , b , c ; λ ) t n n ! + u n = 0 H n ( x + y ; 1 u ; a , b , c ; λ ) t n n ! + n = 0 r = 0 n ( n r ) ( ln a ) n r H r ( x + y ; u ; a , b , c ; λ ) t n n ! n = 0 r = 0 n ( n r ) ( ln a ) n r H r ( x + y ; 1 u ; a , b , c ; λ ) t n n ! .

Comparing the coefficients of t n n ! on both sides of the above equation, we arrive at the desired result. □

Remark 2.4 By substituting a=1, b=c=e, λ=1 into Theorem 2.3, we get Carlitz’s results (for details, see [[1], Eq. 2.19]) as follows:

( 2 u 1 ) r = 0 n ( n r ) H r ( x ; u ) H n r ( y ; 1 u ) = ( u 1 ) H n ( x + y ; u ) + u H n ( x + y ; 1 u ) + H n ( x + y ; u ) H n ( x + y ; 1 u ) .

We give the following generating function of the polynomials Y n (x;λ;a):

t λ a t 1 a x t = n = 0 Y n (x;λ;a) t n n ! (a1)
(10)

(cf. [16, 17]). We also note that

Y n (0;λ;a)= Y n (λ;a).

If we substitute x=0 and a=1 into (10), we see that

Y n (λ;1)= 1 λ 1 .

Theorem 2.5 The generalized Apostol-type Frobenius-Euler polynomial holds true as follows:

(11)

Proof Substituting c=b for α=1 into (2) and taking derivative with respect to t, we obtain

n = 0 H n + 1 ( x ; u ; a , b , b ; λ ) t n n ! = a t ln a a t u a t u λ b t u b x t + ln b λ b t a t u ( a t u λ b t u ) 2 b x t + ln ( b x ) a t u λ b t u b x t .

Using (10), we have

n = 0 H n + 1 ( x ; u ; a , b , b ; λ ) t n n ! = ln ( a 1 u ) t n = 0 k = 0 n ( n k ) Y n k ( 1 ; 1 u ; a ) H k ( x ; u ; a , b , b ; λ ) t n n ! + ln ( b λ u ) t n = 0 k = 0 n ( n k ) Y n k ( 1 u ; a ) H k ( 2 ) ( x ; u ; a , b , b ; λ ) t n n ! + ln ( b x ) n = 0 H n ( x ; u ; a , b , b ; λ ) t n n ! .

Thus, after some elementary calculations, we arrive at (11). □

Theorem 2.6 Let |u|<1 and mN. Then we have

H ( m ) (u;a,b,c;λ)= k = 0 n ( n k ) H k ( α ) (x;u;a,b,c;λ) H n k ( α m ) (x;u;a,b,c;λ).
(12)

Proof In (2), we replace α by −α, then we set

( a t u λ b t u ) α c ( x ) t n = 0 H n ( α m ) (x;u;a,b,c;λ) t n n ! = ( a t u λ b t u ) m .

By using (2), we get

n = 0 H n ( α ) (x;u;a,b,c;λ) t n n ! n = 0 H n ( α m ) (x;u;a,b,c;λ) t n n ! = n = 0 H n ( m ) (u;a,b,c;λ) t n n ! .

Therefore,

n = 0 k = 0 n ( n k ) H k ( α ) (x;u;a,b,c;λ) H n k ( α m ) (x;u;a,b,c;λ) t n n ! = n = 0 H n ( m ) (u;a,b,c;λ) t n n ! .

Comparing the coefficients of t n n ! on both sides of the above equation, we arrive at (12). □

3 Interpolation function

In this section, we give a recurrence relation between the generalized Frobenius-Euler polynomials and the Hurwitz-Lerch zeta function. Recently, many authors have studied not only the Hurwitz-Lerch zeta function, but also its generalizations, for example (among others), Srivastava [19], Srivastava and Choi [24] and also Garg et al. [6]. The generalization of the Hurwitz-Lerch zeta function Φ(z,s,a) is given as follows:

Φ μ , ν ( ρ , σ ) (z,s,a):= n = 0 ( μ ) ρ n ( ν ) σ n z n ( n + a ) s

(μC, a,υC Z 0 , ρ,σ R + , ρ<σ when s,zC (|z|<1); ρ=σ and (sμ+ν)>0 when |z|=1). It is obvious that

Φ μ , 1 ( 1 , 1 ) (z,s,a)= Φ μ (z,s,a)= n = 0 ( μ ) n n ! z n ( n + a ) s
(13)

and

Φ n (z,s,a)= n = 0 ( n ) n n ! z n ( n + a ) s =Φ(z,s,a),

where Φ(z,s,a) denotes the Lerch-Zeta function (cf. [6, 19, 21, 24]).

Relation between the generalized Apostol-type Frobenius-Euler polynomials and the Hurwitz-Lerch zeta function is given as follows.

Theorem 3.1 Let | λ u |<1. We have

H n ( α ) (x;u;a,b,c;λ)= k = 0 α ( α k ) ( u ) α k 1 G ( n ; x , λ u ; a , b , c ; α , k ) ,
(14)

where

G(s;x,β;a,b,c;α,j)= m = 0 ( m + α 1 m ) β m ( x ln c + j ln a + m ln b ) s ,|β|<1.

Proof From (2), we have

n = 0 H n ( α ) (x;u;a,b,c;λ) t n n ! = j = 0 α ( α j ) ( u ) α j 1 m = 0 ( m + α 1 m ) ( λ u ) m e α ( x ln c + k ln a + m ln b ) .

Therefore,

n = 0 H n ( α ) ( x ; u ; a , b , c ; λ ) t n n ! = n = 0 k = 0 α ( α k ) ( u ) α k 1 m = 0 ( m + α 1 m ) ( λ u ) m ( x ln c + k ln a + m ln b ) n t n n ! .

Comparing the coefficients of t n n ! on both sides of the above equation, we have arrive at (14). □

Remark 3.2 By substituting a=1, b=c=e into (14), we have

H n ( α ) (x;u;λ)= ( 1 u ) α u G ( n ; x , λ u ; 1 , e , e ; α , 1 ) = ( 1 u ) α u Φ ( λ u , n , x ) ,

where

G ( n ; x , λ u ; 1 , e , e ; α , 1 ) =Φ ( λ u , n , x ) .

Remark 3.3 The function G(s;x,β;a,b,c;α,j) is an interpolation function of the generalized Apostol-type Frobenius-Euler polynomials of order α at negative integers, which is given by the analytic continuation of the G(s;x,β;a,b,c;α,j) for s=n, nN.

4 Relations between Array-type polynomials, Apostol-Bernoulli polynomials and generalized Apostol-type Frobenius-Euler polynomial

In [17], Simsek constructed the generalized λ-Stirling type numbers of the second kind S(n,v;a,b;λ) by means of the following generating function:

f S , v (t;a,b;λ)= ( λ b t a t ) v v ! = n = 0 S(n,v;a,b;λ) t n n ! .
(15)

The generating function for these polynomials S v n (x;a,b;λ) is given by

g v (x,t;a,b;λ)= 1 v ! ( λ b t a t ) v b x t = n = 0 S v n (x;a,b;λ) t n n !
(16)

(cf. [17]).

The generalized Apostol-Bernoulli polynomials were defined by Srivastava et al. [[22], p.254, Eq. (20)] as follows.

Let a,b,c R + with ab, xR and n N 0 . Then the generalized Bernoulli polynomials B n ( α ) (x;λ;a,b,c) of order αZ are defined by means of the following generating functions:

f B (x,a,b,c;λ;α)= ( t λ b t a t ) α c x t = n = 0 B n ( α ) (x;λ;a,b,c) t n n ! ,
(17)

where

|tln ( a b ) +lnλ|<2π.

We note that B n ( 1 ) (x;λ;a,b,c)= B n (x;λ;a,b,c) and also B n (x;λ;1,e,e)= B n (x;λ), which denotes the Apostol-Bernoulli polynomials (cf. [124]).

Theorem 4.1 Let v be an integer. Then we have

H n v ( ν ) (x;u;a,b,c;λ)= ν ! u 2 ν ( n ) v k = 0 n ( n k ) S v n ( x , 1 , b ; λ u ) Y n k ( ν ) ( 1 u ; a ) .

Proof Replacing c by b in (2) and after some calculations, we have

n = 0 H n ( v ) (x;u;a,b,b;λ) t n + v n ! = ν ! u 2 ν n = 0 S ν n ( x , 1 , b ; λ u ) t n n ! n = 0 Y n ( ν ) ( 1 u ; a ) t n n ! .

Comparing the coefficients of t n n ! on both sides of the above equation, we arrive at the desired result. □

Corollary 4.2

H n v ( ν ) (x;u;a,b,c;λ)= ν ! u 2 ν ( n ) α k = 0 n ( n k ) S ( k , ν , 1 , b ; λ u ) B n k ( x , a , b ; λ u ) .

Proof Replacing c by b in (2) and after some calculations, we have

n = 0 H n v ( v ) (x;u;a,b,b;λ) t n + v n ! = ν ! u 2 ν n = 0 S ( n , ν , 1 , b ; λ u ) t n n ! n = 0 B n ( x , a , b ; λ u ) t n n ! .

Comparing the coefficients of t n n ! on both sides of the above equation, we arrive at the desired result. □