Construction of partially degenerate Laguerre–Bernoulli polynomials of the first kind

ABSTRACT

In this paper, we introduce partially degenerate Laguerre–Bernoulli polynomials of the first kind and deduce some relevant properties by using a preliminary study of these polynomials. We derive some theorems on implicit summation formulae for partially degenerate Laguerre–Bernoulli polynomials of the first kind ${}_L{B}_j(\xi ,\eta |\lambda )$. Finally, we derive some symmetry identities for partially degenerate Laguerre–Bernoulli polynomials of the first kind.

KEYWORDS:

• Laguerre polynomials
• partially degenerate Bernoulli polynomials of the first kind
• partially degenerate Laguerre–Bernoulli polynomials of the first kind
• summation formulas
• symmetry identities

2010 MATHEMATICS SUBJECT CLASSIFICATIONS:

• 11B68
• 33C45
• 11S40
• 11S80

1. Introduction

Let p be a fixed odd prime number. Throughout this paper, ${\mathbb{Z}}_p$, ${\mathbb{Q}}_p$ and ${\mathbb{C}}_p$ will denote respectively, the ring of p-adic integers, the field of p-adic rational numbers and the completion of an algebraic closure of ${\mathbb{Q}}_p$. The p-adic norm $|.{|}_p$ is normally defined as $|p{|}_p=p^{-1}$. Let $f\left(\xi \right)$ be continuous function on ${\mathbb{Z}}_p$. Then the p-adic invariant integral on ${\mathbb{Z}}_p$ is defined by (see [1]) (1) ${\int }_{{\mathbb{Z}}_p}f\left(\xi \right)\mathrm{d}{\mu }_0\left(\xi \right)=\underset{N\to \mathrm{\infty }}{lim}\sum _{\xi =0}^{p^N-1}f\left(\xi \right){\mu }_0(\xi +p^N{\mathbb{Z}}_p)=\underset{N\to \mathrm{\infty }}{lim}\frac{1}{p^N}\sum _{\xi =0}^{p^N-1}f\left(\xi \right).$(1) It is apparent from (1(1) ${\int }_{{\mathbb{Z}}_p}f\left(\xi \right)\mathrm{d}{\mu }_0\left(\xi \right)=\underset{N\to \mathrm{\infty }}{lim}\sum _{\xi =0}^{p^N-1}f\left(\xi \right){\mu }_0(\xi +p^N{\mathbb{Z}}_p)=\underset{N\to \mathrm{\infty }}{lim}\frac{1}{p^N}\sum _{\xi =0}^{p^N-1}f\left(\xi \right).$(1) ) that (2) ${\int }_{{\mathbb{Z}}_p}{f}_j\left(\xi \right)\mathrm{d}{\mu }_0\left(\xi \right)-{\int }_{{\mathbb{Z}}_p}f\left(\xi \right)\mathrm{d}{\mu }_0\left(\xi \right)=\sum _{l=0}^{j-1}{f}^{\prime }\left(l\right),(j\in \mathbb{N}),$(2) where $f_j\left(\xi \right)=f(\xi +j)$, (see [1,2]).

The Bernoulli polynomials are defined by the generating function (3) ${\int }_{{\mathbb{Z}}_p}{\mathrm{e}}^{(\xi +\eta )z}\mathrm{d}{\mu }_0\left(\eta \right)=\frac{z}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}=\sum _{j=0}^{\mathrm{\infty }}{B}_j\left(\xi \right)\frac{z^j}{j!},(\mathrm{s}\mathrm{e}\mathrm{e}[3,4]).$(3) When $\xi =0$, $B_j=B_j\left(0\right)$ are called the Bernoulli numbers.

The two variable Laguerre polynomials (2-VLP) ${\text{Ł}}_n(x,y)$ [5] are defined by (4) $\frac{1}{(1-\eta z)}\exp \left(\frac{-\xi z}{1-\eta z}\right)=\sum _{j=0}^{\mathrm{\infty }}{\text{Ł}}_j(\xi ,\eta )z^j,(\mid \eta z\mid <1),$(4) which is equivalently [6,7] given by (5) $\exp \left(\xi z\right)C_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{\text{Ł}}_j(\xi ,\eta )\frac{z^n}{j!}.$(5) From (4(4) $\frac{1}{(1-\eta z)}\exp \left(\frac{-\xi z}{1-\eta z}\right)=\sum _{j=0}^{\mathrm{\infty }}{\text{Ł}}_j(\xi ,\eta )z^j,(\mid \eta z\mid <1),$(4) ) and (5(5) $\exp \left(\xi z\right)C_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{\text{Ł}}_j(\xi ,\eta )\frac{z^n}{j!}.$(5) ), we have (6) ${\text{Ł}}_j(\xi ,\eta )=j!\sum _{s=0}^j\frac{(-1{)}^s{\xi }^s{\eta }^{j-s}}{(s!{)}^2(j-s)!}={\eta }^j{\text{Ł}}_j\left(\xi /\eta \right).$(6) Thus we have (7) ${\text{Ł}}_j(\xi ,\eta )=\frac{(-1{)}^j{\xi }^j}{j!},{\text{Ł}}_j(0,\eta )={\eta }^j,{\text{Ł}}_j(\xi ,1)={\text{Ł}}_j\left(\xi \right),$(7) where $L_j\left(\xi \right)$ are the ordinary Laguerre polynomials [6].

The Daehee polynomials are defined by the generating function (8) $\frac{\log (1+z)}{z}(1+z{)}^{\xi }=\sum _{j=0}^{\mathrm{\infty }}{D}_j(\xi )\frac{z^j}{j!},(\mathrm{s}\mathrm{e}\mathrm{e}[8\left]\right)$(8) In the case $\xi =0$, $D_j=D_j\left(0\right)$ are the Daehee numbers.

The Bernoulli polynomials of the second kind are defined by the generating function (9) $\frac{z}{\log (1+z)}(1+z{)}^{\xi }=\sum _{j=0}^{\mathrm{\infty }}{b}_j(\xi )\frac{z^j}{j!},(\mathrm{s}\mathrm{e}\mathrm{e}[9,10\left]\right)$(9) At the point x = 0, $b_n=b_n\left(0\right)$ are called the Bernoulli numbers of the second kind.

For $\lambda ,z\in {\mathbb{C}}_p$ with $\mid \lambda z{\mid }_p<p^{-\frac{1}{p-1}}$, the partially degenerate Bernoulli polynomials of the first kind are defined by the generating function (see [11]) (10) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}=\sum _{j=0}^{\mathrm{\infty }}{B}_j\left(\xi |\lambda \right)\frac{z^j}{j!}.$(10) When $\xi =0$, $B_j\left(\lambda \right)=B_j\left(0|\lambda \right)$ are called the partially degenerate Bernoulli numbers of the first kind.

Kwon et al. [8] proved that $D_j^{\left(r\right)}\left(\xi |\lambda \right)=\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l})D_l^{\left(r\right)}\left(\lambda \right){\xi }^{j-l},$where (11) ${\left(\frac{\log (1+t)}{\log (1+\lambda t{)}^{\frac{1}{\lambda }}}\right)}^r{\mathrm{e}}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{D}_n^{\left(r\right)}\left(x|\lambda \right)\frac{t^n}{n!}.$(11) The falling factorial sequence is defined by $(\xi {)}_0=1,(\xi {)}_j=\xi (\xi -1)\cdots (\xi -j+1),(j\ge 1).$The first kind of Stirling numbers are defined by (12) $(\xi {)}_j=\sum _{k=0}^j{S}_1(j,k){\xi }^k,(j\ge 0),(\mathrm{s}\mathrm{e}\mathrm{e}[3-7,9,10,12-16\left]\right)$(12) and as an inversion formula of (12(12) $(\xi {)}_j=\sum _{k=0}^j{S}_1(j,k){\xi }^k,(j\ge 0),(\mathrm{s}\mathrm{e}\mathrm{e}[3-7,9,10,12-16\left]\right)$(12) ), the Stirling numbers of the second kind are given by (see [1,2,8,11,17–24]) (13) ${\xi }^j=\sum _{k=0}^j{S}_2(j,k)(\xi {)}_k.$(13) From (12(12) $(\xi {)}_j=\sum _{k=0}^j{S}_1(j,k){\xi }^k,(j\ge 0),(\mathrm{s}\mathrm{e}\mathrm{e}[3-7,9,10,12-16\left]\right)$(12) ) and (13(13) ${\xi }^j=\sum _{k=0}^j{S}_2(j,k)(\xi {)}_k.$(13) ), we note that the generating function of Stirling numbers of the first kind and that of the second kind are respectively given by (see [1,2,8,11,19–24]) (14) $\frac{1}{k!}(\log (1+z){)}^k=\sum _{j=k}^{\mathrm{\infty }}{S}_1(j,k)\frac{z^j}{j!}$(14) and (15) $\frac{1}{k!}({\mathrm{e}}^z-1{)}^k=\sum _{j=k}^{\mathrm{\infty }}{S}_2(j,k)\frac{z^j}{j!},(k\ge 0).$(15) For each $p\ge 0$, $S_p\left(j\right)$ [24] defined by (16) $S_p\left(j\right)=\sum _{l=0}^j{l}^p$(16) is called the sum of integer power sum or simply powers sum. The exponential generating function for $S_p\left(j\right)$ is (17) $\sum _{p=0}^{\mathrm{\infty }}{S}_p\left(j\right)\frac{z^p}{p!}=1+{\mathrm{e}}^z+{\mathrm{e}}^{2z}+\cdots +{\mathrm{e}}^{jz}=\frac{{\mathrm{e}}^{(j+1)z}-1}{{\mathrm{e}}^z-1}.$(17) For any nonnegative integer r, the r-Stirling numbers $S_r(j,k)$ of the second kind are defined by (see [13]) (18) $\frac{1}{k!}{\mathrm{e}}^{rz}({\mathrm{e}}^z-1{)}^k=\sum _{j=k}^{\mathrm{\infty }}{S}_r(j+r,k+r)\frac{z^j}{j!}.$(18)

Motivated by their importance and potential for applications in certain problems in number theory, combinatorics, classical and numerical analysis, and other fields of applied mathematics, a variety of polynomials and numbers with their variants and extensions have recently been introduced and investigated. In this paper, we aim to introduce partially degenerate Laguerre–Bernoulli polynomials of the first kind and investigate some properties such as explicit summation formulas, addition formulas, implicit formulas and symmetry identities. Relevant connections of the results presented here with those relatively simple numbers and polynomials are considered.

2. Partially degenerate Laguerre–Bernoulli polynomials of the first kind

In this section, we introduce a partially degenerate Laguerre–Bernoulli polynomials of the first kind and investigate some properties of these polynomials. First, we present the following definition.

Definition 2.1

Let us assume that $\lambda ,z\in {\mathbb{C}}_p$ such that $|\lambda z{|}_p<p^{-\frac{1}{p-1}}$. In view of (5(5) $\exp \left(\xi z\right)C_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{\text{Ł}}_j(\xi ,\eta )\frac{z^n}{j!}.$(5) ) and (10(10) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}=\sum _{j=0}^{\mathrm{\infty }}{B}_j\left(\xi |\lambda \right)\frac{z^j}{j!}.$(10) ), we introduce the partially degenerate Laguerre–Bernoulli polynomials of the first kind which are given by the generating function (19) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!},(\xi ,\eta \in \mathbb{R},\mid z\mid <\pi ).$(19) At the point $\xi =\eta =0$ in (19(19) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!},(\xi ,\eta \in \mathbb{R},\mid z\mid <\pi ).$(19) ), ${}_L{B}_j(0,0|\lambda )={}_L{B}_j\left(\lambda \right)$ are called the partially degenerate Laguerre–Bernoulli numbers of the first kind.

From (19(19) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!},(\xi ,\eta \in \mathbb{R},\mid z\mid <\pi ).$(19) ), we note that (20) $\begin{array}{rl}\sum _{j=0}^{\mathrm{\infty }}\underset{\lambda \to 0}{lim}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!}& =\underset{\lambda \to 0}{lim}\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)\\ & =\frac{z}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta )\frac{z^j}{j!}.\end{array}$(20) Thus we get (21) $\underset{\lambda \to 0}{lim}{}_L{B}_j(\xi ,\eta |\lambda )={}_L{B}_j(\xi ,\eta ),(j\ge 0),$(21) where ${}_L{B}_j(\xi ,\eta )$ are called the Laguerre–Bernoulli polynomials.

Theorem 2.1

Let $j\ge 0$. Then (22) ${}_L{B}_j(\xi ,\eta |\lambda )=\sum _{l=0}^j\left(\begin{array}{l}j\\ l\end{array}\right)B_{l,\lambda }{L}_{j-l}(\xi ,\eta ).$(22)

Proof.

Using (5(5) $\exp \left(\xi z\right)C_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{\text{Ł}}_j(\xi ,\eta )\frac{z^n}{j!}.$(5) ), (10(10) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}=\sum _{j=0}^{\mathrm{\infty }}{B}_j\left(\xi |\lambda \right)\frac{z^j}{j!}.$(10) ) and (19(19) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!},(\xi ,\eta \in \mathbb{R},\mid z\mid <\pi ).$(19) ), we have $\begin{array}{rl}\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!}& =\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)\\ & =\left(\sum _{l=0}^{\mathrm{\infty }}{B}_{l,\lambda }\frac{z^l}{l!}\right)\left(\sum _{j=0}^{\mathrm{\infty }}{L}_j\right(\xi ,\eta \left)\frac{z^j}{j!}\right)\\ & =\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j\left(\begin{array}{l}j\\ l\end{array}\right)B_{l,\lambda }{L}_{j-l}\right(\xi ,\eta \left)\right)\frac{z^j}{j!}.\end{array}$Comparing the coefficients of $\frac{z^j}{j!}$, we get (22(22) ${}_L{B}_j(\xi ,\eta |\lambda )=\sum _{l=0}^j\left(\begin{array}{l}j\\ l\end{array}\right)B_{l,\lambda }{L}_{j-l}(\xi ,\eta ).$(22) ).

Corollary 2.1

For $\eta =0$ in (19(19) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!},(\xi ,\eta \in \mathbb{R},\mid z\mid <\pi ).$(19) ), we have (23) $B_{j,\lambda }\left(\xi \right)=\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l})B_{j-l,\lambda }{\xi }^l.$(23)

Theorem 2.2

Let $j\ge 0$. Then (24) ${}_L{B}_j(\xi ,\eta |\lambda )=\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l})\frac{(-1{)}^{l}l!}{l+1}{\lambda }^l{}_L{B}_{j-l}(\xi ,\eta |\lambda ).$(24)

Proof.

Form (19(19) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!},(\xi ,\eta \in \mathbb{R},\mid z\mid <\pi ).$(19) ), we have (25) $\begin{array}{rl}\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)& =\left(\frac{\log (1+\lambda z}{\lambda z}\right)\left(\frac{z}{{\mathrm{e}}^z-1}\right){\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)\\ & =\left(\sum _{l=0}^{\mathrm{\infty }}\frac{(-1{)}^l}{l+1}{\lambda }^l{z}^l\right)\left(\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j\right(\xi ,\eta \left)\frac{z^j}{j!}\right)\\ & =\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l})\frac{(-1{)}^{l}l!}{l+1}{\lambda }^l{}_L{B}_{j-l}\right(\xi ,\eta \left)\right)\frac{z^j}{j!}.\end{array}$(25) In view of (19(19) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!},(\xi ,\eta \in \mathbb{R},\mid z\mid <\pi ).$(19) ) and (25(25) $\begin{array}{rl}\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)& =\left(\frac{\log (1+\lambda z}{\lambda z}\right)\left(\frac{z}{{\mathrm{e}}^z-1}\right){\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)\\ & =\left(\sum _{l=0}^{\mathrm{\infty }}\frac{(-1{)}^l}{l+1}{\lambda }^l{z}^l\right)\left(\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j\right(\xi ,\eta \left)\frac{z^j}{j!}\right)\\ & =\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l})\frac{(-1{)}^{l}l!}{l+1}{\lambda }^l{}_L{B}_{j-l}\right(\xi ,\eta \left)\right)\frac{z^j}{j!}.\end{array}$(25) ), we get the result (24(24) ${}_L{B}_j(\xi ,\eta |\lambda )=\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l})\frac{(-1{)}^{l}l!}{l+1}{\lambda }^l{}_L{B}_{j-l}(\xi ,\eta |\lambda ).$(24) ).

Corollary 2.2

For $\eta =0$ in Theorem 2.2, we get (26) $B_{j,\lambda }\left(\xi \right)=\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l})\frac{(-1{)}^{l}l!}{l+1}{\lambda }^l{B}_{j-l}\left(\xi \right).$(26)

Theorem 2.3

Let $j\ge 0$. Then (27) ${}_L{B}_j(\xi ,\eta |\lambda )=\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l}){}_L{B}_{j-l}(\xi ,\eta ){\lambda }^l{D}_l\left(0\right).$(27)

Proof.

Using (11(11) ${\left(\frac{\log (1+t)}{\log (1+\lambda t{)}^{\frac{1}{\lambda }}}\right)}^r{\mathrm{e}}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{D}_n^{\left(r\right)}\left(x|\lambda \right)\frac{t^n}{n!}.$(11) ) and (19(19) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!},(\xi ,\eta \in \mathbb{R},\mid z\mid <\pi ).$(19) ), we have (28) $\begin{array}{rl}\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!}& =\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)\\ & =\left(\frac{z}{{\mathrm{e}}^z-1}\right){\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)\left(\frac{\log (1+\lambda z}{\lambda z}\right)\\ & =\left(\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j\right(\xi ,\eta \left)\frac{z^j}{j!}\right)\left(\sum _{l=0}^{\mathrm{\infty }}{D}_l\right(0\left)\frac{(\lambda z{)}^l}{l!}\right)\\ & =\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l}){}_L{B}_{j-l}\right(\xi ,\eta \left){\lambda }^l{D}_l\right(0\left)\right)\frac{z^j}{j!}.\end{array}$(28) In view of (19(19) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!},(\xi ,\eta \in \mathbb{R},\mid z\mid <\pi ).$(19) ) and (28(28) $\begin{array}{rl}\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!}& =\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)\\ & =\left(\frac{z}{{\mathrm{e}}^z-1}\right){\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)\left(\frac{\log (1+\lambda z}{\lambda z}\right)\\ & =\left(\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j\right(\xi ,\eta \left)\frac{z^j}{j!}\right)\left(\sum _{l=0}^{\mathrm{\infty }}{D}_l\right(0\left)\frac{(\lambda z{)}^l}{l!}\right)\\ & =\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l}){}_L{B}_{j-l}\right(\xi ,\eta \left){\lambda }^l{D}_l\right(0\left)\right)\frac{z^j}{j!}.\end{array}$(28) ), we get the result (27(27) ${}_L{B}_j(\xi ,\eta |\lambda )=\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l}){}_L{B}_{j-l}(\xi ,\eta ){\lambda }^l{D}_l\left(0\right).$(27) ).

Corollary 2.3

On taking $\eta =0$ in Theorem 2.3, we acquire (29) $B_{j,\lambda }\left(\xi \right)=\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l})B_{j-l}\left(\xi \right){\lambda }^l{D}_l\left(0\right).$(29)

Theorem 2.4

Let $j\ge 0$. Then (30) $\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l})\left(\zeta {)}_l{}_L{B}_{j-l}\right(\xi ,\eta |\lambda )=\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l})D_l(\zeta \left){\lambda }^l{}_L{B}_{j-l}\right(\xi ,\eta ).$(30)

Proof.

Consider Equation (19(19) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!},(\xi ,\eta \in \mathbb{R},\mid z\mid <\pi ).$(19) ), we have (31) $\begin{array}{rl}& \sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!}=\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)\\ & (1+z{)}^{\zeta }\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!}=\left(\frac{z}{{\mathrm{e}}^z-1}\right)\left(\frac{\log (1+\lambda z}{\lambda z}\right)(1+z{)}^{\zeta }{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)\\ & =\sum _{l=0}^{\mathrm{\infty }}{D}_l\left(\zeta \right)\frac{(\lambda z{)}^l}{l!}\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta )\frac{z^j}{j!}\\ & =\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l}){}_L{B}_{j-l}\right(\xi ,\eta \left)D_l\right(\zeta \left){\lambda }^l\right)\frac{z^j}{j!}.\end{array}$(31) On the other hand, we have (32) $\sum _{l=0}^{\mathrm{\infty }}\left(\zeta {)}_l\frac{z^l}{l!}\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j\right(\xi ,\eta )\frac{z^j}{j!}=\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l}){}_L{B}_{j-l}\right(\xi ,\eta \left)\right(\zeta {)}_l)\frac{z^j}{j!}.$(32) Therefore, by (31(31) $\begin{array}{rl}& \sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!}=\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)\\ & (1+z{)}^{\zeta }\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!}=\left(\frac{z}{{\mathrm{e}}^z-1}\right)\left(\frac{\log (1+\lambda z}{\lambda z}\right)(1+z{)}^{\zeta }{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)\\ & =\sum _{l=0}^{\mathrm{\infty }}{D}_l\left(\zeta \right)\frac{(\lambda z{)}^l}{l!}\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta )\frac{z^j}{j!}\\ & =\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l}){}_L{B}_{j-l}\right(\xi ,\eta \left)D_l\right(\zeta \left){\lambda }^l\right)\frac{z^j}{j!}.\end{array}$(31) ) and (32(32) $\sum _{l=0}^{\mathrm{\infty }}\left(\zeta {)}_l\frac{z^l}{l!}\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j\right(\xi ,\eta )\frac{z^j}{j!}=\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l}){}_L{B}_{j-l}\right(\xi ,\eta \left)\right(\zeta {)}_l)\frac{z^j}{j!}.$(32) ), we get the result (30(30) $\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l})\left(\zeta {)}_l{}_L{B}_{j-l}\right(\xi ,\eta |\lambda )=\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l})D_l(\zeta \left){\lambda }^l{}_L{B}_{j-l}\right(\xi ,\eta ).$(30) ).

Theorem 2.5

Let $j\ge 0$. Then (33) ${}_L{B}_j(\xi ,\eta )=\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l}){\lambda }^l{b}_{l,\lambda }\left(0\right){}_L{B}_{j-l}(\xi ,\eta |\lambda ).$(33)

Proof.

By using (9(9) $\frac{z}{\log (1+z)}(1+z{)}^{\xi }=\sum _{j=0}^{\mathrm{\infty }}{b}_j(\xi )\frac{z^j}{j!},(\mathrm{s}\mathrm{e}\mathrm{e}[9,10\left]\right)$(9) ) and (19(19) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!},(\xi ,\eta \in \mathbb{R},\mid z\mid <\pi ).$(19) ), we note that (34) $\begin{array}{rl}& \frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!}\\ & \frac{z}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\left(\frac{\lambda z}{\log (1+\lambda z)}\right)\left(\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j\right(\xi ,\eta |\lambda \left)\frac{z^j}{j!}\right)\\ & \frac{z}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\left(\sum _{l=0}^{\mathrm{\infty }}{b}_{l,\lambda }\right(0\left)\frac{{\lambda }^l{z}^l}{l!}\right)\left(\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j\right(\xi ,\eta |\lambda \left)\frac{z^j}{j!}\right)\\ & =\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l}){\lambda }^l{b}_{l,\lambda }\right(0\left){}_L{B}_{j-l}\right(\xi ,\eta |\lambda \left)\right)\frac{z^j}{j!}.\end{array}$(34) Therefore, by (20(20) $\begin{array}{rl}\sum _{j=0}^{\mathrm{\infty }}\underset{\lambda \to 0}{lim}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!}& =\underset{\lambda \to 0}{lim}\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)\\ & =\frac{z}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta )\frac{z^j}{j!}.\end{array}$(20) ) and (34(34) $\begin{array}{rl}& \frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!}\\ & \frac{z}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\left(\frac{\lambda z}{\log (1+\lambda z)}\right)\left(\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j\right(\xi ,\eta |\lambda \left)\frac{z^j}{j!}\right)\\ & \frac{z}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\left(\sum _{l=0}^{\mathrm{\infty }}{b}_{l,\lambda }\right(0\left)\frac{{\lambda }^l{z}^l}{l!}\right)\left(\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j\right(\xi ,\eta |\lambda \left)\frac{z^j}{j!}\right)\\ & =\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l}){\lambda }^l{b}_{l,\lambda }\right(0\left){}_L{B}_{j-l}\right(\xi ,\eta |\lambda \left)\right)\frac{z^j}{j!}.\end{array}$(34) ), we get the result (33(33) ${}_L{B}_j(\xi ,\eta )=\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l}){\lambda }^l{b}_{l,\lambda }\left(0\right){}_L{B}_{j-l}(\xi ,\eta |\lambda ).$(33) ).

Theorem 2.6

Let $j\ge 0$ and $d\in \mathbb{N}$. Then (35) ${}_L{B}_j(\xi ,\eta |\lambda )=d^{j-1}\sum _{l=0}^{d-1}{}_L{B}_{j,\lambda /d}(\frac{l+\xi }{d},\eta ).$(35)

Proof.

From (19(19) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!},(\xi ,\eta \in \mathbb{R},\mid z\mid <\pi ).$(19) ) in the form (36) $\begin{array}{rl}& \sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!}=\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)\\ & =\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^{dz}-1}\sum _{l=0}^{d-1}{\mathrm{e}}^{(l+\xi )z}{C}_0\left(\eta z\right)\\ & =\sum _{j=0}^{\mathrm{\infty }}\left(d^{j-1}\sum _{l=0}^{d-1}{}_L{B}_{j,\lambda /d}(\frac{l+\xi }{d},\eta )\right)\frac{z^j}{j!}.\end{array}$(36) By (19(19) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!},(\xi ,\eta \in \mathbb{R},\mid z\mid <\pi ).$(19) ) and (36(36) $\begin{array}{rl}& \sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!}=\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)\\ & =\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^{dz}-1}\sum _{l=0}^{d-1}{\mathrm{e}}^{(l+\xi )z}{C}_0\left(\eta z\right)\\ & =\sum _{j=0}^{\mathrm{\infty }}\left(d^{j-1}\sum _{l=0}^{d-1}{}_L{B}_{j,\lambda /d}(\frac{l+\xi }{d},\eta )\right)\frac{z^j}{j!}.\end{array}$(36) ), we get (35(35) ${}_L{B}_j(\xi ,\eta |\lambda )=d^{j-1}\sum _{l=0}^{d-1}{}_L{B}_{j,\lambda /d}(\frac{l+\xi }{d},\eta ).$(35) ).

Corollary 2.4

Let $j\ge 0$ and $d\in \mathbb{N}$, we have (37) $B_{j,\lambda }\left(\xi \right)=d^{j-1}\sum _{l=0}^{d-1}{B}_{j,\lambda /d}\left(\frac{l+\xi }{d}\right).$(37)

Theorem 2.7

Let $j\ge 0$. Then (38) ${}_L{B}_j(\xi ,\eta |\lambda )=\sum _{k=0}^j\sum _{l=0}^k(\genfrac{}{}{0ex}{}{j}{k})\left(\xi {)}_l{S}_2\right(k,l\left){}_L{B}_{j-k}\right(0,\eta |\lambda ).$(38)

Proof.

From (19(19) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!},(\xi ,\eta \in \mathbb{R},\mid z\mid <\pi ).$(19) ), we have (39) $\begin{array}{rl}& \sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!}=\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)\\ & =\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{C}_0\left(\eta z\right)({\mathrm{e}}^z-1+1{)}^{\xi }\\ & =\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(0,\eta |\lambda )\frac{z^j}{j!}\sum _{l=0}^{\mathrm{\infty }}\left(\xi {)}_l\frac{1}{l!}\right({\mathrm{e}}^z-1{)}^l\\ & =\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(0,\eta |\lambda )\frac{z^j}{j!}\sum _{l=0}^{\mathrm{\infty }}\left(\xi {)}_l\sum _{k=l}^{\mathrm{\infty }}{S}_2\right(k,l)\frac{z^k}{k!}\\ & =\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(0,\eta |\lambda )\frac{z^j}{j!}\sum _{k=0}^{\mathrm{\infty }}\sum _{l=0}^k\left(\xi {)}_l{S}_2\right(k,l)\frac{z^k}{k!}\\ & =\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{k=0}^j\sum _{l=0}^k(\genfrac{}{}{0ex}{}{j}{k})\right(\xi {)}_l{S}_2(k,l){}_L{B}_{j-k}(0,\eta |\lambda ))\frac{z^j}{j!}.\end{array}$(39) Comparing the coefficients of z, we get (38(38) ${}_L{B}_j(\xi ,\eta |\lambda )=\sum _{k=0}^j\sum _{l=0}^k(\genfrac{}{}{0ex}{}{j}{k})\left(\xi {)}_l{S}_2\right(k,l\left){}_L{B}_{j-k}\right(0,\eta |\lambda ).$(38) ).

Theorem 2.8

Let $j\ge 0$. Then (40) ${}_L{B}_j(\xi +r,\eta |\lambda )=\sum _{k=0}^j\sum _{l=0}^k(\genfrac{}{}{0ex}{}{j}{k})\left(\xi {)}_l{S}_2^{\left(r\right)}\right(k+r,l+r\left){}_L{B}_{j-k}\right(0,\eta |\lambda ).$(40)

Proof.

By changing ξ by $\xi +r$ in (19(19) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!},(\xi ,\eta \in \mathbb{R},\mid z\mid <\pi ).$(19) ) and using (18(18) $\frac{1}{k!}{\mathrm{e}}^{rz}({\mathrm{e}}^z-1{)}^k=\sum _{j=k}^{\mathrm{\infty }}{S}_r(j+r,k+r)\frac{z^j}{j!}.$(18) ), we have (41) $\begin{array}{rl}& \sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi +r,\eta |\lambda )\frac{z^j}{j!}=\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right){\mathrm{e}}^{\alpha t}\\ & =\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{C}_0\left(\eta z\right){\mathrm{e}}^{rz}({\mathrm{e}}^z-1+1{)}^{\xi }\\ & =\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(0,\eta |\lambda )\frac{z^j}{j!}{\mathrm{e}}^{rz}\sum _{l=0}^{\mathrm{\infty }}\left(\xi {)}_l\frac{1}{l!}\right({\mathrm{e}}^z-1{)}^l\\ & =\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(0,\eta |\lambda )\frac{z^j}{j!}{\mathrm{e}}^{rz}\sum _{l=0}^{\mathrm{\infty }}\left(\xi {)}_l\sum _{k=l}^{\mathrm{\infty }}{S}_2\right(k,l)\frac{z^k}{k!}\\ & =\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(0,\eta |\lambda )\frac{z^j}{j!}\sum _{k=0}^{\mathrm{\infty }}\sum _{l=0}^k\left(\xi {)}_l{S}_2^{\left(r\right)}\right(k+r,l+r)\frac{z^k}{k!}\\ & =\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{k=0}^j\sum _{l=0}^k(\genfrac{}{}{0ex}{}{j}{k})\right(\xi {)}_l{S}_2^{\left(r\right)}(k+r,l+r){}_L{B}_{j-k}(0,\eta |\lambda ))\frac{z^j}{j!}.\end{array}$(41) In view of (19(19) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!},(\xi ,\eta \in \mathbb{R},\mid z\mid <\pi ).$(19) ) and (41(41) $\begin{array}{rl}& \sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi +r,\eta |\lambda )\frac{z^j}{j!}=\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right){\mathrm{e}}^{\alpha t}\\ & =\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{C}_0\left(\eta z\right){\mathrm{e}}^{rz}({\mathrm{e}}^z-1+1{)}^{\xi }\\ & =\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(0,\eta |\lambda )\frac{z^j}{j!}{\mathrm{e}}^{rz}\sum _{l=0}^{\mathrm{\infty }}\left(\xi {)}_l\frac{1}{l!}\right({\mathrm{e}}^z-1{)}^l\\ & =\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(0,\eta |\lambda )\frac{z^j}{j!}{\mathrm{e}}^{rz}\sum _{l=0}^{\mathrm{\infty }}\left(\xi {)}_l\sum _{k=l}^{\mathrm{\infty }}{S}_2\right(k,l)\frac{z^k}{k!}\\ & =\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(0,\eta |\lambda )\frac{z^j}{j!}\sum _{k=0}^{\mathrm{\infty }}\sum _{l=0}^k\left(\xi {)}_l{S}_2^{\left(r\right)}\right(k+r,l+r)\frac{z^k}{k!}\\ & =\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{k=0}^j\sum _{l=0}^k(\genfrac{}{}{0ex}{}{j}{k})\right(\xi {)}_l{S}_2^{\left(r\right)}(k+r,l+r){}_L{B}_{j-k}(0,\eta |\lambda ))\frac{z^j}{j!}.\end{array}$(41) ), we obtain the result (40(40) ${}_L{B}_j(\xi +r,\eta |\lambda )=\sum _{k=0}^j\sum _{l=0}^k(\genfrac{}{}{0ex}{}{j}{k})\left(\xi {)}_l{S}_2^{\left(r\right)}\right(k+r,l+r\left){}_L{B}_{j-k}\right(0,\eta |\lambda ).$(40) ).

Theorem 2.9

Let $j\ge 0$. Then (42) ${}_L{B}_{q+l}(\zeta ,\eta |\lambda )=\sum _{j,p=0}^{q,l}\left(\begin{array}{l}q\\ j\end{array}\right)\left(\begin{array}{l}l\\ p\end{array}\right)(\zeta -\xi {)}^{j+p}{}_L{B}_{q+l-p-j}(\xi ,\eta |\lambda ).$(42)

Proof.

Replacing z with z + u in (19(19) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!},(\xi ,\eta \in \mathbb{R},\mid z\mid <\pi ).$(19) ), we get (43) $\frac{\log (1+\lambda (z+u){)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^{z+u}-1}{C}_0\left(\eta \right(z+u\left)\right)={\mathrm{e}}^{-\xi (z+u)}\sum _{q,l=0}^{\mathrm{\infty }}{}_L{B}_{q+l}(\eta ,\xi |\lambda )\frac{z^q}{q!}\frac{u^l}{l!},\left(\mathrm{s}\mathrm{e}\mathrm{e}\right[24\left]\right).$(43) Changing ξ by ζ in the above equation, we get (44) $\begin{array}{rl}& {\mathrm{e}}^{(\zeta -\xi )(z+u)}\sum _{q,l=0}^{\mathrm{\infty }}{}_L{B}_{q+l}(\xi ,\eta |\lambda )\frac{z^q}{q!}\frac{u^l}{l!}=\sum _{q,l=0}^{\mathrm{\infty }}{}_L{B}_{q+l}(\zeta ,\eta |\lambda )\frac{z^q}{q!}\frac{u^l}{l!}.\end{array}$(44) (45) $\begin{array}{rl}& \sum _{N=0}^{\mathrm{\infty }}\frac{\left[\right(\zeta -\xi \left)\right(z+u){]}^N}{N!}\sum _{q,l=0}^{\mathrm{\infty }}{}_L{B}_{q+l}(\xi ,\eta |\lambda )\frac{z^q}{q!}\frac{u^l}{l!}=\sum _{q,l=0}^{\mathrm{\infty }}{}_L{B}_{q+l}(\zeta ,\eta |\lambda )\frac{z^q}{q!}\frac{u^l}{l!}.\end{array}$(45) By using Lemma [24], we have (46) $\begin{array}{rl}& \sum _{N=0}^{\mathrm{\infty }}f\left(N\right)\frac{(\xi +\eta {)}^N}{N!}=\sum _{j,m=0}^{\mathrm{\infty }}f(j+m)\frac{{\xi }^j}{j!}\frac{{\eta }^m}{m!},\end{array}$(46) (47) $\begin{array}{rl}& \sum _{j,p=0}^{\mathrm{\infty }}\frac{(\zeta -\xi {)}^{j+p}{z}^j{u}^p}{j!p!}\sum _{q,l=0}^{\mathrm{\infty }}{}_L{B}_{q+l}(\xi ,\eta |\lambda )\frac{z^q}{q!}\frac{u^l}{l!}=\sum _{q,l=0}^{\mathrm{\infty }}{}_L{B}_{q+l}(\zeta ,\eta |\lambda )\frac{z^q}{q!}\frac{u^l}{l!}.\end{array}$(47) (48) $\begin{array}{rl}& \sum _{q,l=0}^{\mathrm{\infty }}\sum _{j,p=0}^{q,l}\frac{(\zeta -\xi {)}^{j+p}}{j!p!}{}_L{B}_{q+l-j-p}(\xi ,\eta |\lambda )\frac{z^q}{(q-j)!}\frac{u^l}{(l-p)!}\\ & =\sum _{q,l=0}^{\mathrm{\infty }}{}_L{B}_{q+l}(\zeta ,\eta |\lambda )\frac{z^q}{q!}\frac{u^l}{l!}.\end{array}$(48) Equating the like powers of z and u in the above equation, we get the required result.

3. General identities

In our previous articles (Pathan and Khan [24] and Haroon and Khan [4] and Khan et al. [14–16]), it was shown that symmetric identities for Hermite-based generalized polynomials unify many properties and identities of Hermite–Bernoulli and Hermite–Euler polynomials. In this section, we give general symmetric identities for partially degenerate Laguerre–Bernoulli polynomials of the first kind ${}_L{B}_j(\xi ,\eta |\lambda )$ by applying the generating functions (10(10) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}=\sum _{j=0}^{\mathrm{\infty }}{B}_j\left(\xi |\lambda \right)\frac{z^j}{j!}.$(10) ) and (19(19) $\frac{\log (1+\lambda z{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^z-1}{\mathrm{e}}^{\xi z}{C}_0\left(\eta z\right)=\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(\xi ,\eta |\lambda )\frac{z^j}{j!},(\xi ,\eta \in \mathbb{R},\mid z\mid <\pi ).$(19) ).

Theorem 3.1

Let $a,b\in \mathbb{R}$ with $a\ne b$ and $j\ge 0$. Then (49) $\begin{array}{rl}& \sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l})a^{j-l}{b}^l{}_L{B}_{j-l}(b\xi ,b\eta |\lambda ){}_L{B}_l(a\xi ,a\eta |\lambda )\\ & =\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l})b^{j-l}{a}^l{}_L{B}_{j-l}(a\xi ,a\eta |\lambda ){}_L{B}_l(b\xi ,b\eta |\lambda ).\end{array}$(49)

Proof.

Suppose (50) $\begin{array}{rl}A\left(z\right)& =\frac{(\log (1+\lambda az{)}^{\frac{1}{\lambda }}\left)\right(\log (1+\lambda bz{)}^{\frac{1}{\lambda }})}{({\mathrm{e}}^{az}-1)({\mathrm{e}}^{bz}-1)}{\mathrm{e}}^{(a+b)\xi z}{C}_0\left(ab\eta z\right)C_0\left(ab\eta z\right)\end{array}$(50) (51) $\begin{array}{rl}A\left(z\right)& =\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l})a^{j-l}{b}^l{}_L{B}_{j-l}\right(b\xi ,b\eta |\lambda \left){}_L{B}_l\right(a\xi ,a\eta |\lambda \left)\right)\frac{z^j}{j!}.\end{array}$(51) On the similar lines we can show that (52) $A\left(z\right)=\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l})b^{j-l}{a}^l{}_L{B}_{j-l}\right(a\xi ,a\eta |\lambda \left){}_L{B}_l\right(b\xi ,b\eta |\lambda \left)\right)\frac{z^j}{j!}.$(52) In view of (51(51) $\begin{array}{rl}A\left(z\right)& =\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l})a^{j-l}{b}^l{}_L{B}_{j-l}\right(b\xi ,b\eta |\lambda \left){}_L{B}_l\right(a\xi ,a\eta |\lambda \left)\right)\frac{z^j}{j!}.\end{array}$(51) ) and (52(52) $A\left(z\right)=\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l})b^{j-l}{a}^l{}_L{B}_{j-l}\right(a\xi ,a\eta |\lambda \left){}_L{B}_l\right(b\xi ,b\eta |\lambda \left)\right)\frac{z^j}{j!}.$(52) ), we arrive at the desired result.

Corollary 3.1

By setting b = 1 in Theorem 3.1, we get $\begin{array}{rl}& \sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l})a^{j-l}{}_L{B}_{j-l}(\xi ,\eta |\lambda ){}_L{B}_l(a\xi ,a\eta |\lambda )\\ & =\sum _{l=0}^j(\genfrac{}{}{0ex}{}{j}{l})a^l{}_L{B}_{j-l}(a\xi ,a\eta ){}_L{B}_l^{\left(r\right)}(\xi ,\eta ).\end{array}$

Theorem 3.2

Let $a,b\in \mathbb{R}$ with $a\ne b$ and $j\ge 0$. Then (53) $\begin{array}{rl}& \sum _{l=0}^j\sum _{i=0}^{a-1}\sum _{p=0}^{b-1}(\genfrac{}{}{0ex}{}{j}{l})a^{j-l}{b}^l{}_L{B}_{j-l}(b\xi +\frac{b}{a}i+p,b\eta |\lambda ){}_L{B}_l(a\zeta ,a\eta ,|\lambda )\\ & =\sum _{l=0}^j\sum _{p=0}^{a-1}\sum _{i=0}^{b-1}(\genfrac{}{}{0ex}{}{j}{l})b^{j-l}{a}^l{}_L{B}_{j-l}(a\xi +\frac{a}{b}i+p,a\eta |\lambda ){}_L{B}_l(b\zeta ,b\eta |\lambda ).\end{array}$(53)

Proof.

Let (54) $\begin{array}{rl}& B\left(z\right)=\frac{(\log (1+\lambda az{)}^{\frac{1}{\lambda }}\left)\right(\log (1+\lambda bz{)}^{\frac{1}{\lambda }})({\mathrm{e}}^{abz}-1)}{({\mathrm{e}}^{az}-1{)}^2({\mathrm{e}}^{bz}-1{)}^2}{\mathrm{e}}^{ab(\xi +\zeta )z}{C}_0\left(ab\eta z\right)C_0\left(ab\eta z\right)\\ & =\frac{\log (1+\lambda az{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^{az}-1}{\mathrm{e}}^{ab\xi z}{C}_0\left(ab\eta z\right)\sum _{i=0}^{a-1}{\mathrm{e}}^{bzi}\frac{\log (1+\lambda bz{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^{bz}-1}{\mathrm{e}}^{ab\zeta z}{C}_0\left(ab\eta z\right)\sum _{p=0}^{b-1}{\mathrm{e}}^{azp}.\end{array}$(54) (55) $\begin{array}{rl}& B\left(z\right)=\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j\sum _{i=0}^{a-1}\sum _{p=0}^{b-1}(\genfrac{}{}{0ex}{}{j}{l})a^{j-l}{b}^l{}_L{B}_{j-l}(b\xi +\frac{b}{a}i+p,b\eta |\lambda ){}_L{B}_l\right(a\zeta ,a\eta ,|\lambda \left)\right)\frac{z^j}{j!}.\end{array}$(55) On the other hand, we have (56) $B\left(z\right)=\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j\sum _{p=0}^{a-1}\sum _{i=0}^{b-1}(\genfrac{}{}{0ex}{}{j}{l})b^{j-l}{a}^l{}_L{B}_{j-l}(a\xi +\frac{a}{b}i+p,a\eta |\lambda ){}_L{B}_l\right(b\zeta ,b\eta |\lambda \left)\right)\frac{z^j}{j!}.$(56) Therefore, by (55(55) $\begin{array}{rl}& B\left(z\right)=\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j\sum _{i=0}^{a-1}\sum _{p=0}^{b-1}(\genfrac{}{}{0ex}{}{j}{l})a^{j-l}{b}^l{}_L{B}_{j-l}(b\xi +\frac{b}{a}i+p,b\eta |\lambda ){}_L{B}_l\right(a\zeta ,a\eta ,|\lambda \left)\right)\frac{z^j}{j!}.\end{array}$(55) ) and (56(56) $B\left(z\right)=\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j\sum _{p=0}^{a-1}\sum _{i=0}^{b-1}(\genfrac{}{}{0ex}{}{j}{l})b^{j-l}{a}^l{}_L{B}_{j-l}(a\xi +\frac{a}{b}i+p,a\eta |\lambda ){}_L{B}_l\right(b\zeta ,b\eta |\lambda \left)\right)\frac{z^j}{j!}.$(56) ), we get (53(53) $\begin{array}{rl}& \sum _{l=0}^j\sum _{i=0}^{a-1}\sum _{p=0}^{b-1}(\genfrac{}{}{0ex}{}{j}{l})a^{j-l}{b}^l{}_L{B}_{j-l}(b\xi +\frac{b}{a}i+p,b\eta |\lambda ){}_L{B}_l(a\zeta ,a\eta ,|\lambda )\\ & =\sum _{l=0}^j\sum _{p=0}^{a-1}\sum _{i=0}^{b-1}(\genfrac{}{}{0ex}{}{j}{l})b^{j-l}{a}^l{}_L{B}_{j-l}(a\xi +\frac{a}{b}i+p,a\eta |\lambda ){}_L{B}_l(b\zeta ,b\eta |\lambda ).\end{array}$(53) ).

Theorem 3.3

Let $a,b\in \mathbb{R}$ with $a\ne b$ and $j\ge 0$. Then (57) $\begin{array}{rl}& \sum _{l=0}^j\sum _{i=0}^{a-1}\sum _{p=0}^{b-1}\left(\begin{array}{l}j\\ l\end{array}\right)a^{j-l}{b}^l{}_L{B}_{j-l}(b\xi +\frac{b}{a}i,b\eta |\lambda ){}_L{B}_l(a\zeta +\frac{a}{b}p,a\eta |\lambda )\\ & =\sum _{l=0}^j\sum _{p=0}^{a-1}\sum _{i=0}^{b-1}\left(\begin{array}{l}j\\ l\end{array}\right)b^{j-l}{a}^l{}_L{B}_{j-l}(a\xi +\frac{a}{b}i,a\eta |\lambda ){}_L{B}_l(b\zeta +\frac{b}{a}p,b\eta |\lambda ).\end{array}$(57)

Proof.

The proof is analogous to Theorem 3.2, but we need to write Equation (54(54) $\begin{array}{rl}& B\left(z\right)=\frac{(\log (1+\lambda az{)}^{\frac{1}{\lambda }}\left)\right(\log (1+\lambda bz{)}^{\frac{1}{\lambda }})({\mathrm{e}}^{abz}-1)}{({\mathrm{e}}^{az}-1{)}^2({\mathrm{e}}^{bz}-1{)}^2}{\mathrm{e}}^{ab(\xi +\zeta )z}{C}_0\left(ab\eta z\right)C_0\left(ab\eta z\right)\\ & =\frac{\log (1+\lambda az{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^{az}-1}{\mathrm{e}}^{ab\xi z}{C}_0\left(ab\eta z\right)\sum _{i=0}^{a-1}{\mathrm{e}}^{bzi}\frac{\log (1+\lambda bz{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^{bz}-1}{\mathrm{e}}^{ab\zeta z}{C}_0\left(ab\eta z\right)\sum _{p=0}^{b-1}{\mathrm{e}}^{azp}.\end{array}$(54) ) in the form (58) $B\left(z\right)=\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j\sum _{i=0}^{a-1}\sum _{p=0}^{b-1}\left(\begin{array}{l}j\\ l\end{array}\right)a^{j-l}{b}^l{}_L{B}_{j-l}(b\xi +\frac{b}{a}i,b\eta |\lambda ){}_L{B}_l(a\zeta +\frac{a}{b}p,a\eta |\lambda )\right)\frac{z^j}{j!}.$(58) On the other hand Equation (54(54) $\begin{array}{rl}& B\left(z\right)=\frac{(\log (1+\lambda az{)}^{\frac{1}{\lambda }}\left)\right(\log (1+\lambda bz{)}^{\frac{1}{\lambda }})({\mathrm{e}}^{abz}-1)}{({\mathrm{e}}^{az}-1{)}^2({\mathrm{e}}^{bz}-1{)}^2}{\mathrm{e}}^{ab(\xi +\zeta )z}{C}_0\left(ab\eta z\right)C_0\left(ab\eta z\right)\\ & =\frac{\log (1+\lambda az{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^{az}-1}{\mathrm{e}}^{ab\xi z}{C}_0\left(ab\eta z\right)\sum _{i=0}^{a-1}{\mathrm{e}}^{bzi}\frac{\log (1+\lambda bz{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^{bz}-1}{\mathrm{e}}^{ab\zeta z}{C}_0\left(ab\eta z\right)\sum _{p=0}^{b-1}{\mathrm{e}}^{azp}.\end{array}$(54) ) can be shown equal to (59) $B\left(z\right)=\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j\sum _{p=0}^{a-1}\sum _{i=0}^{b-1}\left(\begin{array}{l}j\\ l\end{array}\right)b^{j-l}{a}^l{}_L{B}_{j-l}(a\xi +\frac{a}{b}i,a\eta |\lambda ){}_L{B}_l(b\zeta +\frac{b}{a}p,b\eta |\lambda )\right)\frac{z^j}{j!}.$(59) By (58(58) $B\left(z\right)=\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j\sum _{i=0}^{a-1}\sum _{p=0}^{b-1}\left(\begin{array}{l}j\\ l\end{array}\right)a^{j-l}{b}^l{}_L{B}_{j-l}(b\xi +\frac{b}{a}i,b\eta |\lambda ){}_L{B}_l(a\zeta +\frac{a}{b}p,a\eta |\lambda )\right)\frac{z^j}{j!}.$(58) ) and (59(59) $B\left(z\right)=\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{l=0}^j\sum _{p=0}^{a-1}\sum _{i=0}^{b-1}\left(\begin{array}{l}j\\ l\end{array}\right)b^{j-l}{a}^l{}_L{B}_{j-l}(a\xi +\frac{a}{b}i,a\eta |\lambda ){}_L{B}_l(b\zeta +\frac{b}{a}p,b\eta |\lambda )\right)\frac{z^j}{j!}.$(59) ), we get (57(57) $\begin{array}{rl}& \sum _{l=0}^j\sum _{i=0}^{a-1}\sum _{p=0}^{b-1}\left(\begin{array}{l}j\\ l\end{array}\right)a^{j-l}{b}^l{}_L{B}_{j-l}(b\xi +\frac{b}{a}i,b\eta |\lambda ){}_L{B}_l(a\zeta +\frac{a}{b}p,a\eta |\lambda )\\ & =\sum _{l=0}^j\sum _{p=0}^{a-1}\sum _{i=0}^{b-1}\left(\begin{array}{l}j\\ l\end{array}\right)b^{j-l}{a}^l{}_L{B}_{j-l}(a\xi +\frac{a}{b}i,a\eta |\lambda ){}_L{B}_l(b\zeta +\frac{b}{a}p,b\eta |\lambda ).\end{array}$(57) ).

Now, we prove the following symmetric identity involving sum of integer powers $S_p\left(j\right)$ given by Equation (17(17) $\sum _{p=0}^{\mathrm{\infty }}{S}_p\left(j\right)\frac{z^p}{p!}=1+{\mathrm{e}}^z+{\mathrm{e}}^{2z}+\cdots +{\mathrm{e}}^{jz}=\frac{{\mathrm{e}}^{(j+1)z}-1}{{\mathrm{e}}^z-1}.$(17) ) and partially degenerate Laguerre–Bernoulli polynomials of the first kind ${}_L{B}_j(\xi ,\eta |\lambda )$.

Theorem 3.4

Let $a,b\in \mathbb{R}$ with $a\ne b$ and $j\ge 0$. Then (60) $\begin{array}{rl}& \sum _{k=0}^j\left(\begin{array}{l}j\\ k\end{array}\right)a^{j-k}{b}^k{}_L{B}_{j-k}(b\xi ,b\zeta |\lambda )\sum _{i=0}^k\left(\begin{array}{l}k\\ i\end{array}\right)S_i(b-1){}_L{B}_{k-i}(a\eta ,a\zeta |\lambda )\\ & \sum _{k=0}^j\left(\begin{array}{l}j\\ k\end{array}\right)b^{j-k}{a}^k{}_L{B}_{j-k}(a\xi ,a\zeta |\lambda )\sum _{i=0}^k\left(\begin{array}{l}k\\ i\end{array}\right)S_i(a-1){}_L{B}_{k-i}(b\eta ,b\zeta |\lambda ).\end{array}$(60)

Proof.

Consider $\begin{array}{rl}C\left(z\right)& =\frac{(\log (1+\lambda az{)}^{\frac{1}{\lambda }}\left)\right(\log (1+\lambda bz{)}^{\frac{1}{\lambda }})({\mathrm{e}}^{abz}-1)}{({\mathrm{e}}^{az}-1{)}^2({\mathrm{e}}^{bz}-1{)}^2}{\mathrm{e}}^{ab(\xi +\eta )z}{C}_0\left(ab\zeta z\right)C_0\left(ab\zeta z\right)\\ & =\frac{\log (1+\lambda az{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^{az}-1}{\mathrm{e}}^{ab\xi z}{C}_0\left(ab\zeta z\right)\frac{{\mathrm{e}}^{abz}-1}{{\mathrm{e}}^{az}-1}\frac{\log (1+\lambda bz{)}^{\frac{1}{\lambda }}}{{\mathrm{e}}^{bz}-1}{\mathrm{e}}^{ab\eta z}{C}_0\left(ab\zeta z\right)\\ & =\sum _{j=0}^{\mathrm{\infty }}{}_L{B}_j(b\xi ,b\zeta |\lambda )\frac{(az{)}^j}{j!}\sum _{i=0}^{\mathrm{\infty }}{S}_i(b-1)\frac{z^i}{i!}\sum _{k=0}^{\mathrm{\infty }}{}_L{B}_k(a\eta ,a\zeta |\lambda )\frac{(bz{)}^k}{k!}\\ C\left(z\right)& =\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{k=0}^j\left(\begin{array}{c}j\\ k\end{array}\right)a^{j-k}{b}^k{}_L{B}_{j-k}\right(b\xi ,b\zeta |\lambda )\\ & \sum _{i=0}^k\left(\begin{array}{c}k\\ i\end{array}\right)S_i(b-1){}_L{B}_{k-i}(a\eta ,a\zeta |\lambda ))\frac{z^j}{j!}.\end{array}$On the other hand, we have $\begin{array}{rl}C\left(z\right)& =\sum _{j=0}^{\mathrm{\infty }}\left(\sum _{k=0}^j\left(\begin{array}{c}j\\ k\end{array}\right)b^{j-k}{a}^k{}_L{B}_{j-k}\right(a\xi ,a\zeta |\lambda )\\ & \sum _{i=0}^k\left(\begin{array}{c}k\\ i\end{array}\right)S_i(a-1){}_L{B}_{k-i}(b\eta ,b\zeta |\lambda ))\frac{z^j}{j!}.\end{array}$By comparing the coefficients of $z^j$ on the right-hand sides of the last two equations, we obtain the result (60(60) $\begin{array}{rl}& \sum _{k=0}^j\left(\begin{array}{l}j\\ k\end{array}\right)a^{j-k}{b}^k{}_L{B}_{j-k}(b\xi ,b\zeta |\lambda )\sum _{i=0}^k\left(\begin{array}{l}k\\ i\end{array}\right)S_i(b-1){}_L{B}_{k-i}(a\eta ,a\zeta |\lambda )\\ & \sum _{k=0}^j\left(\begin{array}{l}j\\ k\end{array}\right)b^{j-k}{a}^k{}_L{B}_{j-k}(a\xi ,a\zeta |\lambda )\sum _{i=0}^k\left(\begin{array}{l}k\\ i\end{array}\right)S_i(a-1){}_L{B}_{k-i}(b\eta ,b\zeta |\lambda ).\end{array}$(60) ).

4. Concluding remarks

In this paper, we have presented the generalized partially degenerate Laguerre–Bernoulli polynomials of the first kind and discussed, in particular, some interesting series representations. We have deduced some relevant properties by using the structure and the relations satisfied by the recently generalized Laguerre polynomials incorporates the definition of partially degenerate Laguerre–Bernoulli polynomials of the first kind and a preliminary study of these polynomials. We derived some theorems on implicit summation formulae for partially degenerate Laguerre–Bernoulli polynomials of the first kind ${}_L{B}_j(\xi ,\eta |\lambda )$ and their special cases are given. Finally, we derived symmetry identities for type partially degenerate Laguerre–Bernoulli polynomials of the first kind.

Acknowledgments

The authors would like to thank the anonymous reviewers for their valuable comments and suggestions to improve the quality of the manuscript.

Disclosure statement

No potential conflict of interest was reported by the author(s).