Frequency of hereditary hemochromatosis gene mutations and their effects on iron overload among beta thalassemia patients of Chennai residents

Bhuvana Selvaraj; Sangeetha Soundararajan; Shettu Narayanasamy; Ganesan Subramanian; Senthil Kumar Ramanathan; Bhuvana Selvaraj; Sangeetha Soundararajan; Shettu Narayanasamy; Ganesan Subramanian; Senthil Kumar Ramanathan

doi:10.3934/molsci.2021018

AIMS Molecular Science

2021, Volume 8, Issue 4: 233-247. doi: 10.3934/molsci.2021018

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Research article

Frequency of hereditary hemochromatosis gene mutations and their effects on iron overload among beta thalassemia patients of Chennai residents

1.
Department of Molecular Biology, Dr. Ganesan's Hitech Diagnostic centre, Chennai, Tamil Nadu, India
2.
PG and Research Department of Zoology, Pachaiyappa's College, University of Madras, Chennai, Tamil Nadu, India
3.
Department of Pathology, Dr. Ganesan's Hitech Diagnostic centre, Chennai, Tamil Nadu, India
4.
Department of Molecular Biology, Dr. Ganesan's Hitech Diagnostic centre, Chennai, Tamil Nadu, India

Received: 09 September 2021 Accepted: 09 November 2021 Published: 12 November 2021

Hereditary Hemochromatosis (HH) is an autosomal recessive disorder of iron metabolism associated with HFE gene mutations, characterized by increased iron absorption and accumulation leading to multi-organ damage caused by iron overload toxicity. Beta thalassemia is caused by a mutation in the human beta globin gene. Imbalanced production of globin chain results in beta thalassemia, where the unpaired alpha chains precipitates in red cell precursors leading to ineffective erythropoiesis and reduced RBC survival. Both HH and beta thalassemia condition results in rapid accumulation of iron lead to iron overload in tissues and organs. The study aims to analyze the frequency of HFE variants among beta thalassemia cases and their effect on iron overload. The frequency of three HFE variants C282Y, H63D, S65C was analyzed by PCR RFLP method among Beta Thalassemia Trait (BTT) (n = 203), Beta Thalassemia Major (BTM) (n = 19) and age and sex-matched control samples (n = 200). The present study furnished allele frequency of H63D variant in BTT, BTM and controls 8.13, 15.8 and 6% respectively. Ten out of 33 heterozygous H63D variants exhibited iron overload with higher ferritin levels indicating HFE variant might aggravate the absorption of iron. The C282Y variant was present in heterozygous state in 1 case among beta thalassemia carriers. The C282Y variant was absent among BTM and control cases. S65C HFE variant was absent in the present study. Iron overload was completely absent in the control cases among all three HFE genotypes. Hence it is inferred from the present investigation, analysis of HFE genes and iron status will remarkably help to reason out the probable reason behind the iron status and support in proper management of beta thalassemia cases.

Keywords:

Citation: Bhuvana Selvaraj, Sangeetha Soundararajan, Shettu Narayanasamy, Ganesan Subramanian, Senthil Kumar Ramanathan. Frequency of hereditary hemochromatosis gene mutations and their effects on iron overload among beta thalassemia patients of Chennai residents[J]. AIMS Molecular Science, 2021, 8(4): 233-247. doi: 10.3934/molsci.2021018

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Abstract

Abbreviations

BTH:

Beta Thalassemia Homozygous;

BTM:

Beta Thalassemia Major;

BTT:

Beta Thalassemia Trait;

HBB:

Hemoglobin Subunit Beta gene;

HH:

Hereditary Hemochromatosis;

SNPs:

Single Nucleotide Polymorphism;

WT:

Wild type allele

1. Introduction

In ^[1,2], Carlitz initiated study of the degenerate Bernoulli and Euler polynomials and obtained some arithmetic and combinatorial results on them. In recent years, many mathematicians have drawn their attention to various degenerate versions of some old and new polynomials and numbers, namely some degenerate versions of Bernoulli numbers and polynomials of the second kind, Changhee numbers of the second kind, Daehee numbers of the second kind, Bernstein polynomials, central Bell numbers and polynomials, central factorial numbers of the second kind, Cauchy numbers, Eulerian numbers and polynomials, Fubini polynomials, Stirling numbers of the first kind, Stirling polynomials of the second kind, central complete Bell polynomials, Bell numbers and polynomials, type 2 Bernoulli numbers and polynomials, type 2 Bernoulli polynomials of the second kind, poly-Bernoulli numbers and polynomials, poly-Cauchy polynomials, and of Frobenius-Euler polynomials, to name a few ^{[3,14,16,17,18]} and the references therein. They have studied those polynomials and numbers with their interest not only in combinatorial and arithmetic properties but also in differential equations and certain symmetric identities ^[4,5] and references therein, and found many interesting results related to them ^{[12,19,20,21,22,23,24,25,26,27,28]}. It is remarkable that studying degenerate versions is not only limited to polynomials but also extended to transcendental functions.

The Bernoulli polynomials of the second are defined by as follows (see ^[9,13])

$\begin{align} \frac{z}{\log(1+z)}(1+z)^x = \sum\limits_{q = 0}^{\infty}b_q(x)\frac{z^q}{q!}. \end{align}$

(1.1)

When $x = 0$ , $b_q(0) = b_q$ are called the Bernoulli numbers of the second kind.

The degenerate exponential function $e_{\lambda}^{x}(z)$ is defined by (see ^{[6,7,8,9,10,11,12,13,14,15,16,17,18,19]})

$\begin{align} e_{\lambda}^{x}(z) = (1+\lambda z)^{\frac{x}{\lambda}}, \; \; e_{\lambda}(z) = (1+\lambda z)^{\frac{1}{\lambda}}, \; \lambda\in \mathbb{C}\backslash \{0\}. \end{align}$

(1.2)

We note that

$\begin{align} e_{\lambda}^{x}(z) = \sum\limits_{q = 0}^{\infty}(x)_{q, \lambda}\frac{z^q}{q!}, {\rm{(see \; [4, 21])}}, \end{align}$

(1.3)

where $(x)_{q, \lambda} = x(x-\lambda)\cdots(x-(q-1)\lambda), (q\geq1)$ , $(x)_{0, \lambda} = 1$ .

Note that

$\lim\limits_{\lambda\rightarrow 0}e_{\lambda}^{x}(z) = \sum\limits_{q = 0}^{\infty}x^q\frac{z^q}{q!} = e^{xz}.$

The degenerate Bernoulli polynomials which are defined by Carlitz's as follows (see ^[1,2])

$\begin{align} \frac{z}{e_{\lambda}(z)-1}e_{\lambda}^{x}(z) = \frac{z}{(1+\lambda z)^{\frac{1}{\lambda}}-1}(1+\lambda z)^{\frac{x}{\lambda}} = \sum\limits_{q = 0}^{\infty}\beta_{q}(x;\lambda)\frac{z^q}{q!}. \end{align}$

(1.4)

At the point $x = 0$ , $\beta_{q}(\lambda) = \beta_{q}(0;\lambda)$ are called the degenerate Bernoulli numbers.

Note that

$\lim\limits_{\lambda \longrightarrow 0}\beta_{q}(x;\lambda) = B_{q}(x).$

The polylogarithm function is defined by

$\begin{align} {\rm Li}_{k}(x) = \sum\limits_{q = 1}^{\infty}\frac{x^q}{q^k}\; \; \; (k\in\mathbb Z, \; \mid x\mid < 1), {\rm{(see \; [7])}}. \end{align}$

(1.5)

Note that

$\begin{align} {\rm Li}_{1}(x) = \sum\limits_{q = 1}^{\infty}\frac{x^q}{q} = -\log(1-x). \end{align}$

(1.6)

The poly-Bernoulli polynomials of the second are defined by (see ^[13])

$\begin{align} \frac{{\rm Li}_{k}(1-e^{-z})}{\log(1+z)}(1+z)^{x} = \sum\limits_{q = 0}^{\infty}b_{q}^{(k)}(x)\frac{z^q}{q!}. \end{align}$

(1.7)

In the case when $x = 0$ , $b_{q}^{(k)} = b_{q}^{(k)}(0)$ are called the poly-Bernoulli numbers of the second kind.

The modified degenerate polyexponential function is defined by (see ^[14])

$\begin{align} {\rm Ei}_{k, \lambda}(x) = \sum\limits_{q = 1}^{\infty}\frac{(1)_{q, \lambda}}{(q-1)!q^k}x^q. \end{align}$

(1.8)

It is noteworthy to mention that

${\rm Ei}_{1, \lambda}(x) = \sum\limits_{q = 1}^{\infty}\frac{(1)_{q, \lambda}}{q!}x^q = e_{\lambda}(x)-1.$

The degenerate poly-Genocchi polynomials which are defined by Kim et al. as follows (see ^[14])

$\begin{align} \frac{2{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right)}{e_{\lambda}(z)+1}e_{\lambda}^{x}(z) = \sum\limits_{q = 0}^{\infty}G_{q, \lambda}^{(k)}(x)\frac{z^q}{q!}\; \; \; (k\in\mathbb Z). \end{align}$

(1.9)

When $x = 0$ , $G_{q, \lambda}^{(k)} = G_{q, \lambda}^{(k)}(0)$ are called the degenerate poly-Genocchi numbers.

For $\lambda\in\mathbb R$ , Kim-Kim defined the degenerate version of the logarithm function, denoted by $\log_{\lambda}(1+t)$ as follows (see ^[11])

$\begin{align} \log_{\lambda}(1+z) = \sum\limits_{q = 1}^{\infty}\lambda^{\lambda-1}(1)_{q, 1/\lambda}\frac{z^q}{q!}, \end{align}$

(1.10)

being the inverse of the degenerate version of the exponential function $e_{\lambda}(z)$ as has been shown below

$e_{\lambda}(\log_{\lambda}(z)) = \log_{\lambda}(e_{\lambda}(z)) = z.$

It is noteworthy to mention that

$\lim\limits_{\lambda\rightarrow 0}\log_{\lambda}(1+z) = \sum\limits_{q = 1}^{\infty}(-1)^{q-1}\frac{z^q}{q!} = \log(1+z).$

The degenerate Daehee polynomials are defined by (see ^[15])

$\begin{align} \frac{\log_{\lambda}(1+z)}{z}(1+z)^{x} = \sum\limits_{q = 0}^{\infty}D_{q, \lambda}(x)\frac{z^q}{q!}. \end{align}$

(1.11)

In the case when $x = 0$ , $D_{q, \lambda} = D_{q, \lambda}(0)$ denotes the degenerate Daehee numbers.

The degenerate Bernoulli polynomials of the second kind which are defined by Kim et al. as follows (see ^[9])

$\begin{align} \frac{z}{\log_{\lambda}(1+z)}(1+z)^{x} = \sum\limits_{q = 0}^{\infty}b_{q, \lambda}(x)\frac{z^q}{q!}. \end{align}$

(1.12)

When $x = 0$ , $b_{q, \lambda} = b_{q, \lambda}(0)$ are called the degenerate Bernoulli numbers of the second kind.

Note here that $\lim_{\lambda \rightarrow 0}b_{q, \lambda}(x) = b_q(x), (q\geq 0)$ .

The degenerate Stirling numbers of the first kind are defined by

$\begin{align} \frac{1}{k!}(\log_{\lambda}(1+z))^k = \sum\limits_{q = k}^{\infty}S_{1, \lambda}(q, k)\frac{z^q}{q!}\; \; (k\geq 0), \rm{(see \; [11, 12])}. \end{align}$

(1.13)

It is noticed that

$\lim\limits_{\lambda\rightarrow 0}S_{1, \lambda}(q, k) = S_1(q, k),$

are the Stirling numbers of the first kind presented by

$\frac{1}{k!}(\log(1+z))^k = \sum\limits_{q = k}^{\infty}S_{1}(q, k)\frac{z^q}{q!}\; \; (k\geq 0), {\rm{(see \; [7, 17])}}.$

The degenerate Stirling numbers of the second kind are defined by (see ^[8])

$\begin{align} \frac{1}{k!}(e_{\lambda}(z)-1)^k = \sum\limits_{q = k}^{\infty}S_{2, \lambda}(q, k)\frac{z^q}{q!}\; \; (k\geq 0). \end{align}$

(1.14)

It is clear that

$\lim\limits_{\lambda\rightarrow 0}S_{2, \lambda}(q, k) = S_2(q, k),$

are the Stirling numbers of the second kind specified by

$\frac{1}{k!}(e^z-1)^k = \sum\limits_{q = k}^{\infty}S_{2}(q, k)\frac{z^q}{q!}\; \; (k\geq 0), {\rm{(see \; [1-28])}}.$

Motivated by the works of Kim et al. ^[11,14], in this paper, we study the type 2 degenerate poly-Bernoulli polynomials of the second kind arising from modified degenerate polyexponential function and obtain some related identities and explicit expressions. Also, we establish the type 2 degenerate unipoly-Bernoulli polynomials of the second kind attached to an arithmetic function by using modified degenerate polyexponential function and discuss some properties of them.

2. Type 2 degenerate poly-Bernoulli polynomials of the second kind

Here, the type 2 degenerate poly-Bernoulli polynomials of the second kind are defined by using the modified degenerate polyexponential function which is called the degenerate poly-Bernoulli polynomials of the second kind as

$\begin{align} \frac{{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right)}{\log_{\lambda}(1+z)}(1+z)^{x} = \sum\limits_{j = 0}^{\infty}Pb_{j, \lambda}^{(k)}(x)\frac{z^j}{j!}, (k\in\mathbb Z). \end{align}$

(2.1)

When $x = 0$ , $Pb_{j, \lambda}^{(k)} = Pb_{j, \lambda}^{(k)}(0)$ are called the type 2 degenerate poly-Bernoulli numbers of the second kind.

Note that

$\lim\limits_{\lambda\rightarrow 0}\frac{{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right)}{\log_{\lambda}(1+z)}(1+z)^{x} = \sum\limits_{j = 0}^{\infty}\lim\limits_{\lambda\rightarrow 0}Pb_{j, \lambda}^{(k)}(x)\frac{z^j}{j!}$

$\begin{align} = \frac{{\rm Ei}_{k}\left(\log(1+z)\right)}{\log(1+z)}(1+z)^{x} = \sum\limits_{j = 0}^{\infty}Pb_{j}^{(k)}(x)\frac{z^j}{j!}, (k\in\mathbb Z), \end{align}$

(2.2)

where $Pb_{j}^{(k)}(x)$ are called the type 2 poly-Bernoulli polynomials of the second kind (see ^[9]).

First, we note that

${\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right) = \sum\limits_{q = 1}^{\infty}\frac{(1)_{q, \lambda}(\log_{\lambda}(1+z))^q}{(q-1)!q^k}$

$= \sum\limits_{q = 0}^{\infty}\frac{(1)_{q+1, \lambda}(\log_{\lambda}(1+z))^{q+1}}{(q+1)^kq!}$

$= \sum\limits_{q = 0}^{\infty}\frac{(1)_{q+1, \lambda}}{(q+1)^{k-1}}\frac{1}{(q+1)!}(\log_{\lambda}(1+z))^{q+1}$

$\begin{align} = \sum\limits_{q = 0}^{\infty}\frac{(1)_{q+1, \lambda}}{(q+1)^{k-1}}\sum\limits_{r = q+1}^{\infty}S_{1, \lambda}(r, q+1)\frac{z^r}{r!}. \end{align}$

(2.3)

By making use of (2.1) and (2.3), we see that

$\frac{z}{\log_{\lambda}(1+z)}(1+z)^{x}{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right) \quad$

$= \frac{z}{\log_{\lambda}(1+z)}(1+z)^{x}\sum\limits_{q = 0}^{\infty}\frac{(1)_{q+1, \lambda}}{(q+1)^{k-1}} \sum\limits_{r = q}^{\infty}\frac{S_{1, \lambda}(r+1, q+1)}{r+1}\frac{z^r}{r!}$

$= \sum\limits_{j = 0}^{\infty}b_{j, \lambda}(x)\frac{z^j}{j!}\sum\limits_{q = 0}^{\infty}\frac{(1)_{q+1, \lambda}}{(q+1)^{k-1}}\sum\limits_{r = q}^{\infty} \frac{S_{1, \lambda}(r+1, q+1)}{r+1}\frac{z^r}{r!}$

$\begin{align} = \sum\limits_{j = 0}^{\infty}\left(\sum\limits_{r = 0}^{j}\binom{j}{r}\sum\limits_{q = 0}^{r}\frac{(1)_{q+1, \lambda}}{(q+1)^{k-1}}\frac{S_{1, \lambda}(r+1, q+1)}{r+1}b_{j-r, \lambda}(x)\right)\frac{z^j}{j!}. \end{align}$

(2.4)

Therefore, by (2.3) and (2.4), we obtain the following theorem.

Theorem 2.1. For $k\in\mathbb Z$ and $j\geq 0$ , we have

$Pb_{j, \lambda}^{(k)}(x) = \sum\limits_{r = 0}^{j}\binom{j}{r}\sum\limits_{q = 0}^{r}\frac{(1)_{q+1, \lambda}}{(q+1)^{k-1}}\frac{S_{1, \lambda}(r+1, q+1)}{r+1} b_{j-r, \lambda}(x).$

Corollary 2.1. Putting $k = 1$ in Theorem 2.1 yields

$Pb_{j, \lambda}(x) = \sum\limits_{r = 0}^{j}\binom{j}{r}\sum\limits_{q = 0}^{r}\frac{(1)_{q+1, \lambda}S_{1, \lambda}(r+1, q+1)}{r+1}b_{j-r, \lambda}(x).$

Let $1\leq k\in\mathbb{Z}$ . For $s\in\mathbb C$ , the function $\chi_{k, \lambda}(s)$ is given as

$\begin{align} \chi_{k, \lambda}(s) = \frac{1}{\Gamma (s)}\int_{0}^{\infty}\frac{z^{s-1}}{\log_{\lambda}(1+z)}{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right)dz. \end{align}$

(2.5)

From Eq (2.5), we have

$\chi_{k, \lambda}(s) = \frac{1}{\Gamma (s)}\int_{0}^{\infty}\frac{z^{s-1}}{\log_{\lambda}(1+z)}{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right)dz$

$= \frac{1}{\Gamma (s)}\int_{0}^{1}\frac{z^{s-1}}{\log_{\lambda}(1+z)}{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right)dz$

$\begin{align} +\frac{1}{\Gamma (s)}\int_{1}^{\infty}\frac{z^{s-1}}{\log_{\lambda}(1+z)}{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right)dz. \end{align}$

(2.6)

For any $s\in\mathbb C$ , the second integral is absolutely convergent and thus, the second term on the r.h.s. vanishes at non-positive integers. That is,

$\begin{align} \lim\limits_{s\rightarrow -m}\left|\frac{1}{\Gamma (s)}\int_{1}^{\infty}\frac{z^{s-1}}{\log_{\lambda}(1+z)}{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right)dz\right|\leq\frac{1}{\Gamma(-m)}M = 0. \end{align}$

(2.7)

On the other hand, the first integral in Eq (2.7), for $\Re(s) > 0$ can be written as

$\frac{1}{\Gamma (s)}\sum\limits_{r = 0}^{\infty}\frac{Pb_{r, \lambda}^{(k)}}{r!}\frac{1}{s+r},$

which defines an entire function of $s$ . Thus, we may include that $\chi_{k, \lambda}(s)$ can be continued to an entire function of $s$ .

Further, from (2.6) and (2.7), we obtain

$\chi_{k, \lambda}(-m) = \lim\limits_{s\rightarrow -m}\frac{1}{\Gamma (s)}\int_{0}^{1}\frac{z^{s-1}}{\log_{\lambda}(1+z)}{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right)dz$

$= \lim\limits_{s\rightarrow -m}\frac{1}{\Gamma (s)}\int_{0}^{1}z^{s-1}\sum\limits_{r = 0}^{\infty}\frac{Pb_{r, \lambda}^{(k)}z^r}{r!}dz = \lim\limits_{s\rightarrow -m}\frac{1}{\Gamma (s)}\sum\limits_{r = 0}^{\infty}\frac{Pb_{r, \lambda}^{(k)}}{s+r}\frac{1}{r!}$

$\begin{align} = \cdots+0+\cdots+0+\lim\limits_{s\rightarrow -m}\frac{1}{\Gamma (s)}\frac{1}{s+m}\frac{Pb_{m, \lambda}^{(k)}}{m!}+0+0+\cdots \end{align}$

(2.8)

$= \lim\limits_{s\rightarrow -m}\frac{\left(\frac{\Gamma(1-s)\sin\pi s}{\pi}\right)}{s+m}\frac{Pb_{m, \lambda}^{(k)}}{m!} = \Gamma(1+m)\cos(\pi m)\frac{Pb_{m, \lambda}^{(k)}}{m!}$

$= (-1)^mPb_{m, \lambda}^{(k)}.$

In view of (2.8), we obtain the following theorem.

Theorem 2.2. Let $k\geq 1$ and $m\in\mathbb N\bigcup\{0\}$ , $s\in\mathbb C$ , we have

$\chi_{k, \lambda}(-m) = (-1)^mPb_{m, \lambda}^{(k)}.$

Using (1.8), we observe that

$\frac{\rm{d}}{\rm{dx}}{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+x)\right) = \frac{\rm{d}}{\rm{dx}}\sum\limits_{j = 1}^{\infty}\frac{(1)_{j, \lambda}(\log_{\lambda}(1+x))^{j}}{j^{k}(j-1)!}$

$\begin{align} = \frac{(1+x)^{\lambda-1}}{\log_{\lambda}(1+x)}\sum\limits_{j = 1}^{\infty}\frac{(1)_{j, \lambda}(\log_{\lambda}(1+x))^{j}}{j^{k-1}(j-1)!} = \frac{(1+x)^{\lambda-1}}{\log_{\lambda}(1+x)}{\rm Ei}_{k-1, \lambda}(\log_{\lambda}(1+x)). \end{align}$

(2.9)

Thus, by (2.9), for $k\geq 2$ , we get

${\rm Ei}_{k, \lambda}(\log_{\lambda}(1+x)) = \int_{0}^{x}\frac{(1+z)^{\lambda-1}}{\log(1+z)}{\rm Ei}_{k-1, \lambda}(\log_{\lambda}(1+z))dz$

$= \int_{0}^{x}{\underbrace {\frac{(1+z)^{\lambda-1}}{\log_{\lambda}(1+z)}\int_{0}^{z}\cdots\frac{(1+z)^{\lambda-1}}{\log_{\lambda}(1+z)} \int_{0}^{z}\frac{(1+z)^{\lambda-1}}{\log_{\lambda}(1+z)}}_{(k-2)- \mbox{times}}}dz\cdots dz$

$\times {\rm Ei}_{1, \lambda}(\log_{\lambda}(1+z))dz\cdots dz$

$\begin{align} = \int_{0}^{x}{\underbrace {\frac{(1+z)^{\lambda-1}}{\log_{\lambda}(1+z)}\int_{0}^{z}\cdots\frac{(1+z)^{\lambda-1}}{\log_{\lambda}(1+z)}\int_{0}^{z}\frac{(1+z)^{\lambda-1}}{\log_{\lambda}(1+z)}}_{(k-2)- \mbox{times}}}zdz\cdots dz. \end{align}$

(2.10)

From (2.1) and (2.10), we get

$\sum\limits_{j = 0}^{\infty}Pb_{j, \lambda}^{(k)}\frac{x^j}{j!} = \frac{{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+x)\right)}{\log_{\lambda}(1+x)} = \frac{1}{\log_{\lambda}(1+x)}$

$\begin{align} \times\int_{0}^{x}{\underbrace {\frac{(1+z)^{\lambda-1}}{\log_{\lambda}(1+z)}\int_{0}^{z}\frac{(1+z)^{\lambda-1}}{\log_{\lambda}(1+z)}\cdots\int_{0}^{z}}_{(k-2)- \mbox{times}}}\frac{(1+z)^{\lambda-1}}{\log_{\lambda}(1+z)}zdz\cdots dz. \end{align}$

(2.11)

$= \frac{x}{\log_{\lambda}(1+x)}\sum\limits_{q = 0}^{\infty}\sum\limits_{q_1+\cdots+q_{k-1} = q}\binom{q}{q_1, \cdots, q_{k-1}}$

$\times \frac{b_{q_1, \lambda}(\lambda-1)}{q_1+1}\frac{b_{q_2, \lambda}(\lambda-1)}{q_1+q_2+1}\cdots\frac{b_{q_{k-1}, \lambda}(\lambda-1)}{q_1+\cdots+q_{k-1}+1}\frac{x^q}{q!}$

$= \sum\limits_{j = 0}^{\infty}\sum\limits_{q = 0}^{j}\binom{j}{q}\sum\limits_{q_1+\cdots+q_{k-1} = q}\binom{q}{q_1, \cdots, q_{k-1}}b_{j-q, \lambda}$

$\begin{align} \times \frac{b_{q_1, \lambda}(\lambda-1)}{q_1+1}\frac{b_{q_2, \lambda}(\lambda-1)}{q_1+q_2+1}\cdots\frac{b_{q_{k-1}, \lambda}(\lambda-1)}{q_1+\cdots+q_{k-1}+1}\frac{x^j}{j!}. \end{align}$

(2.12)

Therefore, by (2.12), we obtain the following theorem.

Theorem 2.3. For $j\in\mathbb N$ and $k\in\mathbb Z$ , we have

$Pb_{j, \lambda}^{(k)} = \sum\limits_{q = 0}^{j}\binom{j}{q}\sum\limits_{q_1+\cdots+q_{k-1} = q}\binom{q}{q_1, \cdots, q_{k-1}}b_{j-q, \lambda}$

$\times \frac{b_{q_1, \lambda}(\lambda-1)}{q_1+1}\frac{b_{q_2, \lambda}(\lambda-1)}{q_1+q_2+1}\cdots\frac{b_{q_{k-1}, \lambda}(\lambda-1)}{q_1+\cdots+q_{k-1}+1}.$

Corollary 2.2. Taking $k = 2$ in Theorem 2.3 yields

$Pb_{j, \lambda}^{(2)} = \sum\limits_{q = 0}^{j}\binom{j}{q}\frac{b_{q, \lambda}(\lambda-1)}{q+1}b_{j-q, \lambda}.$

Replacing $z$ by $e_{\lambda}(z)-1$ in (2.1), we get

$\sum\limits_{q = 0}^{\infty}Pb_{q, \lambda}^{(k)}(x)\frac{(e_{\lambda}(z)-1)^q}{q!} = \frac{{\rm Ei}_{k, \lambda}\left(z\right)}{z}e_{\lambda}^{x}(z)$

$\begin{align} = \sum\limits_{j = 0}^{\infty}\frac{(x)_{j, \lambda}z^j}{j!}\sum\limits_{r = 0}^{\infty}\frac{(1)_{r+1, \lambda}z^r}{(r+1)^kr!} = \sum\limits_{j = 0}^{\infty} \left(\sum\limits_{r = 0}^{j}\binom{j}{r}\frac{(1)_{r+1, \lambda}(x)_{j-r, \lambda}}{(r+1)^k}\right)\frac{z^j}{j!}. \end{align}$

(2.13)

On the other hand,

$\sum\limits_{q = 0}^{\infty}Pb_{q, \lambda}^{(k)}(x)\frac{(e_{\lambda}(z)-1)^q}{q!} = \sum\limits_{q = 0}^{\infty}Pb_{q, \lambda}^{(k)}(x) \sum\limits_{j = q}^{\infty}S_{2, \lambda}(j, q)\frac{z^j}{j!}$

$\begin{align} = \sum\limits_{j = 0}^{\infty}\left(\sum\limits_{q = 0}^{j}Pb_{q, \lambda}^{(k)}(x)S_{2, \lambda}(j, q)\right)\frac{z^j}{j!}. \end{align}$

(2.14)

In view of (2.13) and (2.14), we get the following theorem.

Theorem 2.4. For $k\in\mathbb Z$ and $j\geq 0$ , we have

$\sum\limits_{q = 0}^{j}Pb_{q, \lambda}^{(k)}(x)S_{2, \lambda}(j, q) = \sum\limits_{r = 0}^{j}\binom{j}{r}\frac{(1)_{r+1, \lambda}(x)_{j-r, \lambda}}{(r+1)^k}.$

By using (2.1), we get

$\sum\limits_{j = 1}^{\infty}\left[Pb_{j, \lambda}^{(k)}(x+1)-Pb_{j, \lambda}^{(k)}(x)\right]\frac{z^j}{j!} = \frac{{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right)}{\log_{\lambda}(1+z)}(1+z)^{x+1}-\frac{{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right)}{\log_{\lambda}(1+z)}(1+z)^{x}$

$= \frac{z{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right)}{\log_{\lambda}(1+z)}(1+z)^{x} = \left(\frac{z}{\log_{\lambda}(1+z)}(1+z)^{x}\right) \left({\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right)\right)$

$= \left(\sum\limits_{j = 0}^{\infty}b_{j, \lambda}(x)\frac{z^j}{j!}\right)\left(\sum\limits_{q = 1}^{\infty}\frac{(1)_{q, \lambda}(\log_{\lambda}(1+z))^q}{(q-1)!q^k}\right)$

$= \left(\sum\limits_{j = 0}^{\infty}b_{j, \lambda}(x)\frac{z^j}{j!}\right)\left(\sum\limits_{q = 1}^{\infty}\frac{(1)_{q, \lambda}(\log_{\lambda}(1+z))^{q}}{(q-1)!q^k}\right)$

$= \left(\sum\limits_{j = 1}^{\infty}b_{j, \lambda}(x)\frac{z^j}{j!}\right)\left(\sum\limits_{r = 1}^{\infty}\sum\limits_{q = 1}^{r}\frac{(1)_{q, \lambda}}{q^{k-1}}S_{1, \lambda}(r, q)\frac{z^r}{r!}\right)$

$\begin{align} = \sum\limits_{j = 1}^{\infty}\left(\sum\limits_{r = 1}^{j}\binom{j}{r}\sum\limits_{q = 1}^{r}\frac{(1)_{q, \lambda}}{q^{k-1}}S_{1, \lambda}(r, q)b_{j-r, \lambda}(x)\right)\frac{z^j}{j!}. \end{align}$

(2.15)

Therefore, by comparing the coefficients on both sides of (2.15), we obtain the following theorem.

Theorem 2.5. For $j\geq 0$ , we have

$Pb_{j, \lambda}^{(k)}(x+1)-Pb_{j, \lambda}^{(k)}(x) = \sum\limits_{r = 1}^{j}\binom{j}{r}\sum\limits_{q = 1}^{r}\frac{(1)_{q, \lambda}}{q^{k-1}}S_{1, \lambda}(r, q)b_{j-r, \lambda}(x).$

By making use of (1.3) and (2.1), we have

$\sum\limits_{j = 0}^{\infty}Pb_{j, \lambda}^{(k)}(x+\eta)\frac{z^j}{j!} = \frac{{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right)}{\log_{\lambda}(1+z)}(1+z)^{x+\eta}$

$= \frac{{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right)}{\log_{\lambda}(1+z)}(1+z)^{x}(1+z)^{\eta} = \left(\sum\limits_{j = 0}^{\infty}Pb_{j, \lambda}^{(k)}(x)\frac{z^j}{j!}\right) \left(\sum\limits_{q = 0}^{\infty}(\eta)_q\frac{z^q}{q!}\right)$

$\begin{align} = \sum\limits_{j = 0}^{\infty}\left(\sum\limits_{q = 0}^{j}\binom{j}{q}Pb_{j-q, \lambda}^{(k)}(x)(\eta)_q\right)\frac{z^j}{j!}. \end{align}$

(2.16)

Therefore, by Eq (2.16), we obtain the following theorem.

Theorem 2.6. For $j\geq 0$ , we have

$Pb_{j, \lambda}^{(k)}(x+\eta) = \sum\limits_{q = 0}^{j}\binom{j}{q}Pb_{j-q, \lambda}^{(k)}(x)(\eta)_q.$

By using (2.1), we have

$\frac{{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right)}{\log_{\lambda}(1+z)} = \sum\limits_{j = 0}^{\infty}Pb_{j, \lambda}^{(k)}\frac{z^j}{j!}$

${\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right) = \log_{\lambda}(1+z)\sum\limits_{j = 0}^{\infty}Pb_{j, \lambda}^{(k)}\frac{z^j}{j!}$

$\frac{{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right)}{z} = \frac{\log_{\lambda}(1+z)}{z}\sum\limits_{j = 0}^{\infty}Pb_{j, \lambda}^{(k)}\frac{z^j}{j!}$

$= \left(\sum\limits_{q = 0}^{\infty}D_{q, \lambda}\frac{t^q}{q!}\right)\left(\sum\limits_{j = 0}^{\infty}Pb_{j, \lambda}^{(k)}\frac{z^j}{j!}\right)$

$\begin{align} = \sum\limits_{j = 0}^{\infty}\left(\sum\limits_{q = 0}^{j}\binom{j}{q}Pb_{j-q, \lambda}^{(k)}D_{q, \lambda}\right)\frac{z^j}{j!}. \end{align}$

(2.17)

On the other hand,

$\frac{{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right)}{z} = \frac{1}{z}\sum\limits_{q = 1}^{\infty}\frac{(1)_{q, \lambda}(\log_{\lambda}(1+z))^q}{(q-1)!q^k}$

$= \frac{1}{z}\sum\limits_{q = 0}^{\infty}\frac{(1)_{q+1, \lambda}(\log_{\lambda}(1+z))^{q+1}}{q^k!(q+1)}$

$= \frac{1}{z}\sum\limits_{q = 0}^{\infty}\frac{(1)_{q+1, \lambda}}{(q+1)^{k-1}}\frac{1}{(q+1)!}(\log_{\lambda}(1+z))^{q+1}$

$\begin{align} = \sum\limits_{j = 0}^{\infty}\left(\sum\limits_{q = 0}^{j}\frac{(1)_{q+1, \lambda}}{(q+1)^{k-1}}\frac{S_{1, \lambda}(j+1, q+1)}{j+1}\right)\frac{z^j}{j!}. \end{align}$

(2.18)

Thus, by equations (2.17) and (2.18), we get the following theorem.

Theorem 2.7. For $j\geq 0$ , we have

$\sum\limits_{q = 0}^{j}\binom{j}{q}Pb_{j-q, \lambda}^{(k)}D_{q, \lambda} = \sum\limits_{q = 0}^{j}\frac{(1)_{q+1, \lambda}}{(q+1)^{k-1}}\frac{S_{1, \lambda}(j+1, q+1)}{j+1}.$

From (2.1), we have

$\sum\limits_{n = 0}^{\infty}Pb_{j, \lambda}^{(k)}(x)\frac{z^j}{j!} = \frac{{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right)} {\log_{\lambda}(1+z)}(1+z)^{x}$

$= \frac{{\rm Ei}_{k, \lambda}\left(\log_{\lambda}(1+z)\right)}{\log_{\lambda}(1+z)}e_{\lambda}^{x}(\log_{\lambda}(1+z))$

$= \sum\limits_{j = 0}^{\infty}Pb_{j, \lambda}^{(k)}(x)\frac{z^j}{j!}\sum\limits_{q = 0}^{\infty}(x)_{q, \lambda}\sum\limits_{r = q}^{\infty}S_{1, \lambda}(r, q)\frac{z^r}{r!}$

$= \sum\limits_{j = 0}^{\infty}Pb_{j, \lambda}^{(k)}(x)\frac{z^j}{j!}\sum\limits_{r = 0}^{\infty}\sum\limits_{q = 0}^{r}(x)_{q, \lambda}S_{1, \lambda}(r, q)\frac{z^r}{r!}$

$\begin{align} = \sum\limits_{j = 0}^{\infty}\left(\sum\limits_{r = 0}^{j}\binom{j}{r}Pb_{j-r, \lambda}^{(k)}(x)_{q, \lambda}S_{1, \lambda}(r, q)\right)\frac{z^j}{j!}. \end{align}$

(2.19)

Therefore, by comparing the coefficients on both sides of (2.19), we obtain the following theorem.

Theorem 2.8. For $j\geq 0$ , we have

$Pb_{j, \lambda}^{(k)}(x) = \sum\limits_{r = 0}^{j}\binom{j}{r}Pb_{j-r, \lambda}^{(k)}(x)_{q, \lambda}S_{1, \lambda}(r, q).$

3. The degenerate unipoly-Bernoulli polynomials of the second kind

Let $p$ be any arithmetic real or complex valued function defined on $\mathbb N$ . Kim-Kim ^[7] presented the unipoly function attached to polynomials $p(x)$ as

$\begin{align} u_k(x|p) = \sum\limits_{j = 1}^{\infty}\frac{p(j)}{j^k}x^n, (k\in\mathbb Z). \end{align}$

(3.1)

Moreover,

$\begin{align} u_k(x|1) = \sum\limits_{j = 1}^{\infty}\frac{x^j}{j^k} = {\rm Li}_{k}(x), \rm{(see \; [10, 14])}, \end{align}$

(3.2)

represent the known ordinary polylogarithm function.

Dolgy and Khan ^[3] introduced the degenerate unipoly function attached to polynomials $p(x)$ are considered as follows

$\begin{align} u_{k, \lambda}(x|p) = \sum\limits_{j = 1}^{\infty}p(j)\frac{(1)_{j, \lambda}x^j}{j^k}. \end{align}$

(3.3)

We see that

$\begin{align} u_{k, \lambda}\left(x|\frac{1}{\Gamma}\right) = {\rm Ei}_{k, \lambda}(x), \rm{{(see[14])}} \end{align}$

(3.4)

is the modified degenerate polyexponential function.

Now, we introduce the degenerate unipoly-Bernoulli polynomials of the second kind attached to polynomials $p(x)$ as

$\begin{align} \frac{u_{k, \lambda}\left(\log_{\lambda}(1+z)|p\right)}{\log_{\lambda}(1+z)}(1+z)^{x} = \sum\limits_{j = 0}^{\infty}Pb_{j, \lambda, p}^{(k)}(x)\frac{z^j}{j!}. \end{align}$

(3.5)

When $x = 0$ , $Pb_{j, \lambda, p}^{(k)} = Pb_{j, \lambda, p}^{(k)}(0)$ are called the degenerate unipoly-Bernoulli numbers of the second kind attached to $p$ .

If we take $p(j) = \frac{1}{\Gamma (j)}$ . Then, we have

$\sum\limits_{j = 0}^{\infty}Pb_{j, \lambda, \frac{1}{\Gamma}}^{(k)}(x)\frac{z^j}{j!} = \frac{1}{\log_{\lambda}(1+z)}(1+z)^{x}u_{k, \lambda}\left(\log_{\lambda}(1+z)|\frac{1}{\Gamma}\right)$

$\begin{align} = \frac{1}{\log_{\lambda}(1+z)}(1+z)^{x}\sum\limits_{q = 1}^{\infty}\frac{(1)_{q, \lambda}(\log_{\lambda}(1+z))^q}{q^k(q-1)!}. \end{align}$

(3.6)

For $k = 1$ , we have

$\begin{align} \sum\limits_{j = 0}^{\infty}Pb_{j, \lambda, \frac{1}{\Gamma}}^{(1)}(x)\frac{z^j}{j!} = \frac{1}{\log_{\lambda}(1+z)}(1+z)^{x}\sum\limits_{q = 1}^{\infty} \frac{(1)_{q, \lambda}(\log_{\lambda}(1+z))^q}{q!} = \frac{z}{\log_{\lambda}(1+z)}(1+z)^{x}. \end{align}$

(3.7)

Thus, we have

$\begin{align} Pb_{j, \lambda, \frac{1}{\Gamma}}^{(1)}(x) = b_{j, \lambda}(x), (j\geq 0). \end{align}$

(3.8)

By making use of (1.12) and (3.3), we note that

$u_{k, \lambda}\left(\log_{\lambda}(1+z)|p\right) = \sum\limits_{q = 1}^{\infty}p(q)\frac{(1)_{q, \lambda}(\log_{\lambda}(1+z))^q}{q^k}$

$= \sum\limits_{q = 1}^{\infty}\frac{p(q)(1)_{q, \lambda}(\log_{\lambda}(1+z))^q}{q^k}\frac{q!}{q!}$

$= \sum\limits_{q = 1}^{\infty}\frac{p(q)(1)_{q, \lambda}q!}{q^k}\frac{(\log_{\lambda}(1+z))^q}{q!}$

$= \sum\limits_{q = 1}^{\infty}\frac{p(q)(1)_{q, \lambda}q!}{q^k}\sum\limits_{r = q}^{\infty}S_{1, \lambda}(r, q)\frac{z^r}{r!}$

$= \sum\limits_{r = 1}^{\infty}\left(\sum\limits_{q = 1}^{r}\frac{p(q)(1)_{q, \lambda}q!}{q^k}S_{1, \lambda}(r, q)\right)\frac{z^r}{r!}.$

Thus, we have the required result.

Lemma 3.1. For $k\in\mathbb Z$ , we have

$u_{k, \lambda}\left(\log_{\lambda}(1+z)|p\right) = \sum\limits_{r = 1}^{\infty}\left(\sum\limits_{q = 1}^{r}\frac{p(q)(1)_{q, \lambda}q!}{q^k}S_{1, \lambda}(r, q)\right)\frac{z^r}{r!}.$

Recalling from (3.5), we have

$\sum\limits_{j = 0}^{\infty}Pb_{j, \lambda, p}^{(k)}(x)\frac{z^j}{j!} = \frac{1}{\log_{\lambda}(1+z)}(1+z)^{x} u_{k, \lambda}\left(\log_{\lambda}(1+z)|p\right)$

$= \frac{1}{\log_{\lambda}(1+z)}(1+z)^{x}\sum\limits_{q = 1}^{\infty}\frac{(1)_{q, \lambda}p(q)}{q^k}(\log_{\lambda}(1+z))^q$

$= \frac{1}{\log_{\lambda}(1+z)}(1+z)^{x}\sum\limits_{q = 0}^{\infty}\frac{(1)_{q+1, \lambda}p(q+1)}{(q+1)^k}(\log_{\lambda}(1+z))^{q+1}$

$= \frac{1}{\log_{\lambda}(1+z)}(1+z)^{x}\sum\limits_{q = 0}^{\infty}\frac{(1)_{q+1, \lambda}p(q+1)(q+1)!}{(q+1)^k}\sum\limits_{r = q+1}^{\infty}S_{r, \lambda}(r, q+1)\frac{z^r}{r!}$

$= \frac{z}{\log_{\lambda}(1+z)}(1+z)^{x}\sum\limits_{q = 0}^{\infty}\frac{(1)_{q+1, \lambda}p(q+1)(q+1)!}{(q+1)^k}\sum\limits_{r = q}^{\infty}\frac{S_{1, \lambda}(r+1, q+1)}{r+1}\frac{z^r}{r!}$

$= \sum\limits_{j = 0}^{\infty}b_{j, \lambda}(x)\frac{z^j}{j!}\sum\limits_{r = 0}^{\infty} \left(\sum\limits_{q = 0}^{r}\frac{(1)_{q+1, \lambda}p(q+1)(q+1)!}{(q+1)^k}\frac{S_{1, \lambda}(r+1, q+1)}{r+1}\right)\frac{z^r}{r!}$

$\begin{align} = \sum\limits_{j = 0}^{\infty}\left(\sum\limits_{r = 0}^{j}\sum\limits_{q = 0}^{r}\binom{j}{r}\frac{(1)_{q+1, \lambda}p(q+1)(q+1)!}{(q+1)^k} \frac{S_{1, \lambda}(r+1, q+1)}{r+1}b_{j-r, \lambda}(x)\right)\frac{z^j}{j!}. \end{align}$

(3.9)

Therefore, by comparing the coefficients on both sides of (3.9), we obtain the following theorem.

Theorem 3.1. For $j\geq 0$ and $k\in\mathbb Z$ . Then

$Pb_{j, \lambda, p}^{(k)}(x) = \sum\limits_{r = 0}^{j}\sum\limits_{q = 0}^{r}\binom{j}{r}\frac{(1)_{q+1, \lambda}p(q+1)(q+1)!}{(q+1)^k} \frac{S_{1, \lambda}(r+1, q+1)}{r+1}b_{j-r, \lambda}(x).$

Moreover,

$Pb_{j, \lambda, \frac{1}{\Gamma}}^{(k)}(x) = \sum\limits_{r = 0}^{j}\sum\limits_{q = 0}^{r}\binom{j}{r}\frac{b_{j-r, \lambda}(x)}{(q+1)^{k-1}} \frac{S_{1, \lambda}(r+1, q+1)}{r+1}.$

Using (3.5), we have

$\sum\limits_{j = 0}^{\infty}Pb_{j, \lambda, p}^{(k)}(x)\frac{z^j}{j!} = \frac{1}{\log_{\lambda}(1+z)}u_{k, \lambda}(\log_{\lambda}(1+z)|p)(1+z)^{x}$

$= \frac{u_{k, \lambda}\left(\log_{\lambda}(1+z)|p\right)}{\log_{\lambda}(1+z)}\sum\limits_{j = 0}^{\infty}(x)_{j}\frac{z^j}{j!}$

$= \sum\limits_{i = 0}^{\infty}Pb_{i, \lambda, p}^{(k)}\frac{z^i}{i!}\sum\limits_{j = 0}^{\infty}(x)_{j}\frac{z^j}{j!}$

$\begin{align} = \sum\limits_{j = 0}^{\infty}\left(\sum\limits_{i = 0}^{j}\binom{j}{i}Pb_{i, \lambda, p}^{(k)}(x)_{j-i}\right)\frac{z^j}{j!}. \end{align}$

(3.10)

Upon comparing the coefficients on both sides of Eq (3.10), we get the following theorem.

Theorem 3.2. For $j\geq 0$ and $k\in\mathbb Z$ . Then

$Pb_{j, \lambda, p}^{(k)}(x) = \sum\limits_{i = 0}^{j}\binom{j}{i}Pb_{i, \lambda, p}^{(k)}(x)_{j-i}.$

By making use of (1.11), (1.12) and (3.5), we have

$\sum\limits_{j = 0}^{\infty}Pb_{j, \lambda, p}^{(k)}\frac{z^j}{j!} = \frac{1}{\log_{\lambda}(1+z)}u_{k}\left(\log_{\lambda}(1+z)|p\right)$

$= \frac{1}{\log_{\lambda}(1+z)}\sum\limits_{q = 1}^{\infty}\frac{(1)_{q, \lambda}p(q)}{q^k}(\log_{\lambda}(1+z))^q$

$= \sum\limits_{q = 0}^{\infty}\frac{(1)_{q, \lambda}p(q+1)}{(q+1)^k}(\log_{\lambda}(1+z))^{q+1}$

$= \frac{z}{\log_{\lambda}(1+z)}\frac{\log_{\lambda}(1+z)}{z}\sum\limits_{q = 0}^{\infty}\frac{(1)_{q+1, \lambda}p(q+1)q!}{(q+1)^k} \frac{(\log_{\lambda}(1+z))^{q}}{q!}$

$= \sum\limits_{j = 0}^{\infty}D_{j, \lambda}\frac{z^j}{j!}\sum\limits_{i = 0}^{\infty}b_{i, \lambda}\frac{z^i}{i!} \sum\limits_{q = 0}^{\infty}\frac{(1)_{q+1, \lambda}p(q+1)q!}{(q+1)^k}\sum\limits_{r = q}^{\infty}S_{1, \lambda}(r, q)\frac{z^r}{r!}$

$= \sum\limits_{j = 0}^{\infty}D_{j, \lambda}\frac{z^j}{j!}\sum\limits_{i = 0}^{\infty}b_{i, \lambda}\frac{z^i}{i!}\sum\limits_{r = 0}^{\infty} \sum\limits_{q = 0}^{r}\frac{(1)_{q+1, \lambda}p(q+1)q!}{(q+1)^k}S_{1, \lambda}(r, q)\frac{z^r}{r!}$

$= \sum\limits_{j = 0}^{\infty}\sum\limits_{i = 0}^{j}\binom{j}{i}D_{j-i, \lambda}b_{i, \lambda}\frac{z^j}{j!}\sum\limits_{r = 0}^{\infty} \sum\limits_{q = 0}^{r}\frac{(1)_{q+1, \lambda}p(q+1)q!}{(q+1)^k}S_{1, \lambda}(r, q)\frac{z^r}{r!}$

$\begin{align} = \sum\limits_{j = 0}^{\infty}\left(\sum\limits_{r = 0}^{j} \sum\limits_{q = 0}^{r}\sum\limits_{i = 0}^{j-r}\binom{j-r}{i}\binom{j}{r}D_{j-i-r, \lambda}b_{i, \lambda}\frac{(1)_{q+1, \lambda}p(q+1)q!}{(q+1)^k} S_{1, \lambda}(r, q)\right)\frac{z^j}{j!}. \end{align}$

(3.11)

Thus, by comparing the coefficients on both sides of (3.11), we obtain the following theorem.

Theorem 3.3. For $j\geq 0$ and $k\in\mathbb Z$ . Then

$Pb_{j, \lambda, p}^{(k)} = \sum\limits_{r = 0}^{j} \sum\limits_{q = 0}^{r}\sum\limits_{i = 0}^{j-r}\binom{j-r}{i}\binom{j}{r}D_{j-i-r, \lambda}b_{i, \lambda}\frac{(1)_{q+1, \lambda}p(q+1)q!}{(q+1)^k}S_{1, \lambda}(r, q).$

4. Numerical computations

In this section, certain numerical computations are done to calculate certain zeros of the degenerate poly-Bernoulli polynomials of the second kind and show some graphical representations. The first five members of $Pb_{j, \lambda}^{(k)}(x)$ are calculated and given as:

$\begin{align*} &Pb_{0, \lambda}^{(k)}(x) = 1, \\&Pb_{1, \lambda}^{(k)}(x) = \frac{1}{2}+x-\frac{1}{8 \log 3}-\frac{\log 81}{8 \log 3}, \\&Pb_{2, \lambda}^{(k)}(x) = \frac{1}{2}+x^2+\frac{10}{81 (\log 3)^2}+\frac{1}{8 \log 3}-\frac{x}{4 \log 3}-\frac{\log 81}{8 \log 3}-\frac{x \log 81}{4 \log 3}, \\&Pb_{3, \lambda}^{(k)}(x) = -\frac{1}{4}+2 x-\frac{3 x^2}{2}+x^3-\frac{5}{16 (\log 3)^3}-\frac{10}{27 (\log 3)^2}+\frac{10 x}{27 (\log 3)^2}\\& \quad -\frac{1}{4 \log 3}+\frac{3 x}{4 \log 3}-\frac{3 x^2}{8 \log 3}+\frac{\log 81}{16 \log 3}-\frac{3 x^2 \log 81}{8 \log 3}, \\&Pb_{4, \lambda}^{(k)}(x) = \frac{1}{2}-6 x+8 x^2-4 x^3+x^4+\frac{176}{125 (\log 3)^4}+\frac{15}{8 (\log 3)^3}-\frac{5 x}{4 (\log 3)^3}\\& \quad +\frac{110}{81 (\log 3)^2}-\frac{20 x}{9 (\log 3)^2}+\frac{20 x^2}{27 (\log 3)^2}+\frac{3}{4 \log 3}-\frac{11 x}{4 \log 3}\\& \quad +\frac{9 x^2}{4 \log 3}-\frac{x^3}{2 \log 3}-\frac{\log 81}{8 \log 3}+\frac{3 x^2 \log 81}{4 \log 3}-\frac{x^3 \log 81}{2 \log 3}. \end{align*}$

To show the behavior of $Pb_{j, \lambda}^{(k)}(x)$ , we display the graph $Pb_{j, \lambda}^{(k)}(x)$ for $k = 4$ and $\lambda = 3$ , this graph is presented in Figure 1.

Figure 1. Graph of

$Pb_{j, \lambda}^{(k)}(x)$ .

DownLoad: Full-Size Img PowerPoint

Next, the approximate solutions of $Pb_{j, \lambda}^{(k)}(x) = 0$ when $k = 4$ and $\lambda = 3$ , are calculated and listed in Table 1.

Table 1. Approximate solutions of

$Pb_{j, \lambda}^{(k)}(x) = 0$ .

$j$	Real zeros	Complex zeros
1	$0.11378$	-
2	$0.212959, \; 1.0146$	-
3	$0.468628, \; 0.788431, \; 2.08428$	-
4	$2.27482, \; 3.00114$	$0.589582 - 0.515659\; i, 0.589582 + 0.515659\; i$
5	$4.09322$	$0.470967 - 0.872952\; i, \; 0.470967 + 0.872952\; i,$
		$2.76687 - 0.464588\; i, \; 2.76687 + 0.464588\; i$
6	$4.47754, \; 4.94352$	$0.270509 - 1.2071\; i, \; 0.270509 + 1.2071\; i$
		$2.8603 - 1.06554\; i, \; 2.8603 + 1.06554\; i$
7	$6.12953$	$-0.00407237 - 1.52417\; i, \; -0.00407237 + 1.52417\; i,$
		$2.8544 - 1.67974\; i, \; 2.8544 +1.67974\; i$
		$4.98314 - 0.749479\; i, \; 4.98314 + 0.749479\; i$
8	-	$-0.344872 - 1.82511 \; i, \; -0.344872 + 1.82511 \; i,$
		$2.7537 - 2.30093\; i, \; 2.7537 + 2.30093\; i,$
		$5.21262 - 1.46596 \; i, \; 5.21262 + 1.46596 \; i,$
		$6.83367 - 0.248836 \; i, \; 6.83367 + 0.248836 \; i$

| Show Table

DownLoad: CSV

The zeros of $Pb_{j, \lambda}^{(k)}(x)$ for $\lambda \in \mathbb{C}, j = 12$ are plotted in Figure 2.

Figure 2. Zeros of

$Pb_{12, \lambda}^{(k)}(x)$ .

DownLoad: Full-Size Img PowerPoint

The stacking structure of approximate zeros of $Pb_{j, \lambda}^{(k)}(x) = 0$ for $\lambda = 4, j = 1, 2, ..., 12$ is given in Figure 3.

Figure 3. Stacking structure of zeros

$Pb_{j, \lambda}^{(k)}(x)$ .

DownLoad: Full-Size Img PowerPoint

5. Conclusions

In this article, we introduced the type 2 degenerate poly-Bernoulli polynomials of the second kind and derived many related interesting properties. Furthermore, we defined the degenerate unipoly Bernoulli polynomials of the second kind and established some considerable results. Finally, certain related beautiful zeros and graphs are shown.

Acknowledgments

The authors would like to express the gratitude to Deanship of Scientific Research at King Khalid University, Saudi Arabia for providing funding research group under the research grant number R G P.1/162/42.

Conflict of interest

The authors declare no conflict of interest.

Acknowledgments

The authors thank HOD of Zoology, Principal, Pachaiyappa's college, Chennai, for their support. The authors are very grateful to the authorities of Hitech Diagnostic Centre, Kilpauk, Chennai for their support and valuable suggestions.

Conflict of interest

Authors declare no conflict of interest in this manuscript.

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AIMS Molecular Science

Frequency of hereditary hemochromatosis gene mutations and their effects on iron overload among beta thalassemia patients of Chennai residents

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1. Introduction

2. Type 2 degenerate poly-Bernoulli polynomials of the second kind

3. The degenerate unipoly-Bernoulli polynomials of the second kind

4. Numerical computations

5. Conclusions

Acknowledgments

Conflict of interest

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AIMS Molecular Science

Frequency of hereditary hemochromatosis gene mutations and their effects on iron overload among beta thalassemia patients of Chennai residents

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Abstract

1. Introduction

2. Type 2 degenerate poly-Bernoulli polynomials of the second kind

3. The degenerate unipoly-Bernoulli polynomials of the second kind

4. Numerical computations

5. Conclusions

Acknowledgments

Conflict of interest

References

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