# Central factorial numbers associated with sequences of polynomials

Dae San Kim  Department of Mathematics, Sogang University, Seoul 121-742,
Republic of Korea<EMAIL_ADDRESS>and  Taekyun Kim  Department of
Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea
<EMAIL_ADDRESS>
###### Abstract.

Many important special numbers appear in the expansions of some polynomials in
terms of central factorials and vice versa, for example central factorial
numbers, degenerate central factorial numbers, and central Lah numbers which
are recently introduced. Here we generalize this to any sequence of
polynomials. Let $\mathbb{P}=\left\\{p_{n}(x)\right\\}_{n=0}^{\infty}$ be a
sequence of polynomials such that deg $p_{n}(x)=n,p_{0}(x)=1$. The aim of this
paper is to study the central factorial numbers of the second associated with
$\mathbb{P}$ and of the first kind associated with $\mathbb{P}$, in a unified
and systematic way with the help of umbral calculus technique. The central
factorial numbers associated with $\mathbb{P}$ enjoy orthogonality and inverse
relations. We illustrate our results with many examples and obtain interesting
orthogonality and inverse relations by applying such relations for the central
factorial numbers associated with $\mathbb{P}$ to each of our examples.

###### Key words and phrases:

central factorial numbers of the first kind associated with sequence of
polynomials; central factorial numbers of the second kind associated with
sequence of polynomials; umbral calculus

###### 2000 Mathematics Subject Classification:

05A19, 05A40; 11B73; 11B83

##  1\. Introduction

The central factorial numbers of the second $T_{2}(n,k)$ appear as the
coefficients in the expansion of the powers of $x$ in terms of the central
factorials (see (2.23), (3.2)), whereas the central factorial numbers of the
first kind $T_{1}(n,k)$ appear as the coefficients in the expansion of the
central factorials in terms of powers of $x$ (see (3.1)). Combinatorial
interpretations of central factorial numbers of the second kind are as
follows. $T_{2}(n,k)$ is equal to the number of partitions of the set
$\left\\{1,1^{\prime},2,2^{\prime},...,n,n^{\prime}\right\\}$ into $k$
disjoint nonempty subsets $A_{1},...,A_{k}$ such that, for each $1\leq j\leq
k$, if $i$ is the least integer such that either $i$ or $i^{\prime}$ belongs
to $V_{j}$ then $\left\\{i,i^{\prime}\right\\}$ is a subset of $A_{j}$. For
example, $T_{2}(3,2)=5$. Indeed, they are given by
$\left\\{1,1^{\prime},2,2^{\prime}\right\\}\left\\{3,3^{\prime}\right\\},\left\\{1,1^{\prime},3,3^{\prime}\right\\}\left\\{2,2^{\prime}\right\\},\left\\{1,1^{\prime}\right\\}\left\\{2,2^{\prime},3,3^{\prime}\right\\}$,
$\left\\{1,1^{\prime},3\right\\}\left\\{2,2^{\prime},3^{\prime}\right\\}$, and
$\left\\{1,1^{\prime},3^{\prime}\right\\}\left\\{2,2^{\prime},3\right\\}$. It
is also known that $T_{2}(2n,2n–-2k)$ enumerates the number of ways to pace
$k$ rooks on a 3$D$-triangle board of size $(n-1)$ (see [28]). Further, the
number of ways to place $k$ rooks on a size $n$ triangle board is
$T_{2}(n+1,n+1-k)$ (see [28]). Even though the central factorial numbers are
less known than Stirling numbers, they are as important as Stirling numbers.
Let $\mathbb{P}=\left\\{p_{n}(x)\right\\}_{n=0}^{\infty}$ be a sequence of
polynomials such that deg $p_{n}(x)=n,p_{0}(x)=1$. The aim of this paper is to
study the central factorial numbers of the second $T_{2}(n,k;\mathbb{P})$
associated with $\mathbb{P}$, which are defined as the coefficients in the
expansion of $p_{n}(x)$ in terms of the central factorials $x^{[k]}$. This is
done with the help of umbral calculus technique. We illustrate our results
with twenty examples. Moreover, we also introduce the central factorial
numbers of the first kind $T_{1}(n,k;\mathbb{P})$ associated with
$\mathbb{P}$, which are defined as the coefficients in the expansion of
$x^{[n]}$ in terms of the polynomials $p_{k}(x)$. Again, we study them by
means of umbral calculus and illustrate our results with the same twenty
examples for the central factorial numbers of the second kind. We note here
that the orthogonality and inverse relations for the ordinary central
factorial numbers also hold for the central factorial numbers associated with
sequences of polynomials $\mathbb{P}$. This observation applied to each
individual case yields interesting orthogonality and inverse relations for the
twenty examples, which are explicitly stated. For example, in the case of the
sequence $\mathbb{P}=\left\\{\langle{x\rangle}_{n}\right\\}$ of rising
factorials, we obtain:

$\displaystyle\sum_{k=l}^{n}\sum_{m=k}^{n}\sum_{j=l}^{k}T_{1}(n,m)(-1)^{m-k}S_{2}(m,k)(-1)^{k-j}S_{1}(k,j)T_{2}(j,l)=\delta_{n,l},$
$\displaystyle\sum_{k=l}^{n}\sum_{j=k}^{n}\sum_{m=l}^{k}(-1)^{n-j}S_{1}(n,j)T_{2}(j,k)T_{1}(k,m)(-1)^{m-l}S_{2}(m,l)=\delta_{n,l},$
$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}(-1)^{n-l}S_{1}(n,l)T_{2}(l,k)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}(-1)^{l-k}T_{1}(n,l)S_{2}(l,k)a_{k},$
$\displaystyle
a_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}(-1)^{k-l}S_{1}(k,l)T_{2}(l,n)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}(-1)^{l-n}T_{1}(k,l)S_{2}(l,n)a_{k},$

where $S_{1}(n,k),S_{2}(n,k)$ are the Stirling numbers of the first kind and
the Stirling numbers of the second kind, respectively. As another example, in
the case of the sequence $\mathbb{P}=\left\\{L_{n}(x)\right\\}$ of Laguerre
polynomials of order -1, we have:

$\displaystyle\sum_{k=l}^{n}\sum_{m=k}^{n}\sum_{j=l}^{k}(-1)^{k-j}T_{1}(n,m)L(m,k)L(k,j)T_{2}(j,l)=\delta_{n,l},$
$\displaystyle\sum_{k=l}^{n}\sum_{j=k}^{n}\sum_{m=l}^{k}(-1)^{j-l}L(n,j)T_{2}(j,k)T_{1}(k,m)L(m,l)=\delta_{n,l},$
$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}(-1)^{l}L(n,l)T_{2}(l,k)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}(-1)^{k}\sum_{l=k}^{n}T_{1}(n,l)L(l,k)a_{k},$
$\displaystyle
a_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}(-1)^{l}L(k,l)T_{2}(l,n)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}(-1)^{n}\sum_{l=n}^{k}T_{1}(k,l)L(l,n)a_{k},$

where $L(n,k)$ are the (unsigned) Lah numbers. We would like to remark that in
this investigation of central factorial numbers we are led to discover natural
definitions for central logarithm and central exponential associated to any
delta series. The novelty of this paper is that it is the first paper which
studies the central factorial numbers of both kinds associated with any
sequence of polynomials in a unified and systematic way with the help of
umbral calculus.
In [26], Koutras generalized the classical Eulerian numbers and polynomials to
$p_{n}$-associated Eulerian numbers and polynomials with motivation of
providing a unified approach to the study of Eulerian-related numbers and with
combinatorial, probabilistic and statistical applications in mind (see
[2,9,10,29,33]). In addition, he mentioned in the same paper that a lot of
special numbers, like the Stirling numbers of the second, the Lah numbers and
Gould-Hopper numbers, appear in the expansions of some polynomials in terms of
the falling factorials. This observation is the motivation and impetus for the
present research on the central factorial numbers of both kinds associated
with sequences of polynomials (see [26,27]).
The outline of this paper is as follows. In Section 2, we will briefly go over
umbral calculus. In Section 3, we introduce the central factorial numbers of
the second associated with sequences of polynomials. We illustrate our results
with twenty examples in Section 4. In Section 5, we introduce the central
factorial numbers of the first kind associated with sequences of polynomials.
We illustrate our results in Section 6 with the same examples from Section 4
and thereby get interesting orthogonality and inverse relations for each case.
Finally, we conclude our paper in Section 7.

##  2\. Review of umbral calculus

Here we will briefly go over very basic facts about umbral calculus. For more
details on this, we recommend the reader to refer to [1,8,31-32]. We remark
that recently umbral calculus has been extended to the case of degenerate
umbral calculus in order to treat degenerate special polynomials and numbers,
which involves degenerate exponentials (see [12]). Let $\mathbb{C}$ be the
field of complex numbers. Then $\mathcal{F}$ denotes the algebra of formal
power series in $t$ over $\mathbb{C}$, given by

$\mathcal{F}=\bigg{\\{}f(t)=\sum_{k=0}^{\infty}a_{k}\frac{t^{k}}{k!}~{}\bigg{|}~{}a_{k}\in\mathbb{C}\bigg{\\}},$

and $\mathbb{P}=\mathbb{C}[x]$ indicates the algebra of polynomials in $x$
with coefficients in $\mathbb{C}$.

Let $\mathbb{P}^{*}$ be the vector space of all linear functionals on
$\mathbb{P}$. If $\langle L|p(x)\rangle$ denotes the action of the linear
functional $L$ on the polynomial $p(x)$, then the vector space operations on
$\mathbb{P}^{*}$ are defined by

$\langle L+M|p(x)\rangle=\langle L|p(x)\rangle+\langle
M|p(x)\rangle,\quad\langle cL|p(x)\rangle=c\langle L|p(x)\rangle,$

where $c$ is a complex number.

For $f(t)\in\mathcal{F}$ with $\displaystyle
f(t)=\sum_{k=0}^{\infty}a_{k}\frac{t^{k}}{k!}\displaystyle$, we define the
linear functional on $\mathbb{P}$ by

(2.1) $\langle f(t)|x^{k}\rangle=a_{k}.$

From (2.1), we note that

$\langle t^{k}|x^{n}\rangle=n!\delta_{n,k},\quad(n,k\geq 0),$

where $\delta_{n,k}$ is the Kronecker’s symbol.

Some remarkable linear functionals are as follows:

$\displaystyle\langle e^{yt}|p(x)\rangle=p(y),$ (2.2) $\displaystyle\langle
e^{yt}-1|p(x)\rangle=p(y)-p(0),$
$\displaystyle\bigg{\langle}\frac{e^{yt}-1}{t}\bigg{|}p(x)\bigg{\rangle}=\int_{0}^{y}p(u)du.$

Let

(2.3) $f_{L}(t)=\sum_{k=0}^{\infty}\langle L|x^{k}\rangle\frac{t^{k}}{k!}.$

Then, by (2.1) and (2.3), we get

$\langle f_{L}(t)|x^{n}\rangle=\langle L|x^{n}\rangle.$

That is, $f_{L}(t)=L$, as linear functionals on $\mathbb{P}$. In fact, the map
$L\longmapsto f_{L}(t)$ is a vector space isomorphism from $\mathbb{P}^{*}$
onto $\mathcal{F}$.

Henceforth, $\mathcal{F}$ denotes both the algebra of formal power series in
$t$ and the vector space of all linear functionals on $\mathbb{P}$.
$\mathcal{F}$ is called the umbral algebra and the umbral calculus is the
study of umbral algebra. For each nonnegative integer $k$, the differential
operator $t^{k}$ on $\mathbb{P}$ is defined by

(2.4) $t^{k}x^{n}=\left\\{\begin{array}[]{cc}(n)_{k}x^{n-k},&\textrm{if $k\leq
n$,}\\\ 0,&\textrm{if $k>n$.}\end{array}\right.$

Extending (2.4) linearly, any power series

$f(t)=\sum_{k=0}^{\infty}\frac{a_{k}}{k!}t^{k}\in\mathcal{F}$

gives the differential operator on $\mathbb{P}$ defined by

(2.5) $f(t)x^{n}=\sum_{k=0}^{n}\binom{n}{k}a_{k}x^{n-k},\quad(n\geq 0).$

It should be observed that, for any formal power series $f(t)$ and any
polynomial $p(x)$, we have

(2.6) $\langle f(t)|p(x)\rangle=\langle 1|f(t)p(x)\rangle=f(t)p(x)|_{x=0}.$

Here we note that an element $f(t)$ of $\mathcal{F}$ is a formal power series,
a linear functional and a differential operator. Some notable differential
operators are as follows:

$\displaystyle e^{yt}p(x)=p(x+y),$ (2.7)
$\displaystyle(e^{yt}-1)p(x)=p(x+y)-p(x),$
$\displaystyle\frac{e^{yt}-1}{t}p(x)=\int_{x}^{x+y}p(u)du.$

The order $o(f(t))$ of the power series $f(t)(\neq 0)$ is the smallest integer
for which $a_{k}$ does not vanish. If $o(f(t))=0$, then $f(t)$ is called an
invertible series. If $o(f(t))=1$, then $f(t)$ is called a delta series.

For $f(t),g(t)\in\mathcal{F}$ with $o(f(t))=1$ and $o(g(t))=0$, there exists a
unique sequence $s_{n}(x)$ (deg $s_{n}(x)=n$) of polynomials such that

(2.8)
$\big{\langle}g(t)f(t)^{k}|s_{n}(x)\big{\rangle}=n!\delta_{n,k},\quad(n,k\geq
0).$

The sequence $s_{n}(x)$ is said to be the Sheffer sequence for $(g(t),f(t))$,
which is denoted by $s_{n}(x)\sim(g(t),f(t))$. We observe from (2.8) that

(2.9) $s_{n}(x)=\frac{1}{g(t)}q_{n}(x),$

where $q_{n}(x)=g(t)s_{n}(x)\sim(1,f(t))$.

In particular, if $s_{n}(x)\sim(g(t),t)$, then $q_{n}(x)=x^{n}$, and hence

(2.10) $s_{n}(x)=\frac{1}{g(t)}x^{n}.$

It is well known that $s_{n}(x)\sim(g(t),f(t))$ if and only if

(2.11)
$\frac{1}{g\big{(}\overline{f}(t)\big{)}}e^{x\overline{f}(t)}=\sum_{k=0}^{\infty}\frac{s_{k}(x)}{k!}t^{k},$

for all $x\in\mathbb{C}$, where $\overline{f}(t)$ is the compositional inverse
of $f(t)$ such that $\overline{f}(f(t))=f(\overline{f}(t))=t$.

The following equations (2.12), (2.13), and (2.14) are equivalent to the fact
that $s_{n}\left(x\right)$ is Sheffer for
$\left(g\left(t\right),f\left(t\right)\right)$, for some invertible $g(t)$:

(2.12) $\displaystyle f\left(t\right)s_{n}\left(x\right)$
$\displaystyle=ns_{n-1}\left(x\right),\quad\left(n\geq 0\right),$ (2.13)
$\displaystyle s_{n}\left(x+y\right)$
$\displaystyle=\sum_{j=0}^{n}\binom{n}{j}s_{j}\left(x\right)q_{n-j}\left(y\right),$

with $q_{n}\left(x\right)=g\left(t\right)s_{n}\left(x\right),$

(2.14)
$s_{n}\left(x\right)=\sum_{j=0}^{n}\frac{1}{j!}\big{\langle}{g\left(\overline{f}\left(t\right)\right)^{-1}\overline{f}\left(t\right)^{j}}\big{|}{x^{n}\big{\rangle}}x^{j}.$

If $s_{n}(x)\sim(g(t),f(t))$, then the following recurrence relation holds:

(2.15)
$s_{n+1}(x)=\bigg{(}x-\frac{g^{\prime}(t)}{g(t)}\bigg{)}\frac{1}{f^{\prime}(t)}s_{n}(x).$

Let $s_{n}(x)\sim(g(t),f(t))$. Then, for any polynomial $p(x)$, we have the
polynomial exapnsion given by

(2.16) $\displaystyle p(x)=\sum_{k\geq
0}\frac{1}{k!}\big{\langle}{g(t)f(t)^{k}|p(x)\big{\rangle}}s_{k}(x).$

and, for any formal power series $h(t)$, we have

(2.17) $\displaystyle
h(t)=\sum_{k=0}^{\infty}\frac{1}{k!}\langle{h(t)|s_{k}\rangle}g(t)f(t)^{k}.$

For any formal power series $h(t)$ and any polynomial $p(x)$, we have

(2.18) $\displaystyle\langle{h(at)|p(x)\rangle}=\langle{h(t)|p(ax)\rangle}.$

The falling factorials $(x)_{n}$ are defined by

(2.19) $\displaystyle(x)_{n}=x(x-1)\cdots(x-n+1),\,\,(n\geq
1),\quad(x)_{0}=1.$

More generally, for any real number $\lambda$, the generalized falling
factorials are defined by

(2.20)
$\displaystyle(x)_{n,\lambda}=x(x-\lambda)\cdots(x-(n-1)\lambda),\,\,(n\geq
1),\quad(x)_{0,\lambda}=1$

For any nonzero real number $\lambda$, the degenerate exponentials
$e_{\lambda}^{x}(t)$ and the degenerate logarithms $\log_{\lambda}t$ are
defined by

(2.21) $\displaystyle e_{\lambda}^{x}(t)=(1+\lambda
t)^{\frac{x}{\lambda}}=\sum_{n=0}^{\infty}(x)_{n,\lambda}\frac{t^{n}}{n!},\quad
e_{\lambda}(t)=e_{\lambda}^{1}(t)=\sum_{n=0}^{\infty}(1)_{n,\lambda}\frac{t^{n}}{n!},$
(2.22) $\displaystyle\log_{\lambda}t=\frac{1}{\lambda}(t^{\lambda}-1).$

We recall that the Stirling numbers of the first kind $S_{1}(n,k)$ and of the
second kind $S_{2}(n,k)$ are respectively given by

(2.23)
$\displaystyle\frac{1}{k!}\big{(}\log(1+t)\big{)}^{k}=\sum_{n=k}^{\infty}S_{1}(n,k)\frac{t^{n}}{n!},\quad(x)_{n}=\sum_{k=0}^{n}S_{1}(n,k)x^{k},$
(2.24)
$\displaystyle\frac{1}{k!}\big{(}e^{t}-1\big{)}^{k}=\sum_{n=k}^{\infty}S_{2}(n,k)\frac{t^{n}}{n!},\quad
x^{n}=\sum_{k=0}^{n}S_{2}(n,k)(x)_{k}.$

As degenerate versions of the Stirling numbers, the degenerate Stirling
numbers of the first kind $S_{1,\lambda}(n,k)$ and of the second kind
$S_{2,\lambda}(n,k)$ are respectively given by

(2.25)
$\displaystyle\frac{1}{k!}\big{(}\log_{\lambda}(1+t)\big{)}^{k}=\sum_{n=k}^{\infty}S_{1,\lambda}(n,k)\frac{t^{n}}{n!},\quad(x)_{n}=\sum_{k=0}^{n}S_{1,\lambda}(n,k)(x)_{k,\lambda},$
(2.26)
$\displaystyle\frac{1}{k!}\big{(}e_{\lambda}(t)-1\big{)}^{k}=\sum_{n=k}^{\infty}S_{2,\lambda}(n,k)\frac{t^{n}}{n!},\quad(x)_{n,\lambda}=\sum_{k=0}^{n}S_{2,\lambda}(n,k)(x)_{k}.$

The central factorials $x^{[n]}$ are defined by

(2.27) $\displaystyle x^{[n]}=x(x+\frac{1}{2}n-1)_{n-1},\,\,(n\geq 1),\quad
x^{[0]}=1.$

Then, we note that (see [30,31])

(2.28) $\displaystyle x^{[n]}\sim(1,f(t)=e^{\frac{t}{2}}-e^{-\frac{t}{2}}).$

The following facts are well known (see [30,31]).

######  Proposition 2.1.

Let $f_{1}(t)=e^{\frac{t}{2}}-e^{-\frac{t}{2}}$, and let
$f_{2}(t)=e_{\lambda}^{\frac{1}{2}}(t)-e_{\lambda}^{-\frac{1}{2}}(t)$. Then
the following hold true.

$\displaystyle(a)\,\,\bar{f_{1}}(t)=2\log\bigg{(}\frac{t+\sqrt{t^{2}+4}}{2}\bigg{)}=\log\big{(}1+\frac{t}{2}(t+\sqrt{t^{2}+4}\big{)}\big{)}.$
$\displaystyle(b)\,\,\bar{f_{2}}(t)=\log_{\lambda}\bigg{(}\frac{t+\sqrt{t^{2}+4}}{2}\bigg{)}^{2}=\frac{1}{\lambda}\bigg{(}\bigg{(}\frac{t+\sqrt{t^{2}+4}}{2}\bigg{)}^{2\lambda}-1\bigg{)}.$
$\displaystyle(c)\,\,\sum_{n=0}^{\infty}x^{[n]}\frac{t^{n}}{n!}=\bigg{(}\frac{t+\sqrt{t^{2}+4}}{2}\bigg{)}^{2x}=\big{(}1+\frac{t}{2}(t+\sqrt{t^{2}+4}\big{)}\big{)}^{x}.$
$\displaystyle(d)\,\,e^{yt}=\sum_{k=0}^{\infty}\frac{1}{k!}(e^{\frac{t}{2}}-e^{-\frac{t}{2}})^{k}y^{[k]}.$

###### Proof.

(a), (b) These can be easily checked.
(c) By (2.28), $\sum_{n=0}^{\infty}x^{[n]}\frac{t^{n}}{n!}=e^{x\bar{f}(t)}$,
with $\bar{f}(t)$ as in (a).
(d) This follows from (2.17) and (2.28). ∎

##  3\. Central factorial numbers of the second kind associated with sequences
of polynomials

The central factorial numbers of the first kind $T_{1}(n,k)$ and of second
kind $T_{2}(n,k)$ are respectively defined by

(3.1)
$\displaystyle\frac{1}{k!}\bigg{(}2\log\bigg{(}\frac{t+\sqrt{t^{2}+4}}{2}\bigg{)}\bigg{)}^{k}=\frac{1}{k!}\Big{(}\log\big{(}1+\frac{t}{2}(t+\sqrt{t^{2}+4}\big{)}\big{)}\Big{)}^{k}$
$\displaystyle=\sum_{n=k}^{\infty}T_{1}(n,k)\frac{t^{n}}{n!},\quad
x^{[n]}=\sum_{k=0}^{n}T_{1}(n,k)x^{k},$ (3.2)
$\displaystyle\frac{1}{k!}\big{(}e^{\frac{t}{2}}-e^{-\frac{t}{2}}\big{)}^{k}=\sum_{n=k}^{\infty}T_{2}(n,k)\frac{t^{n}}{n!},\quad
x^{n}=\sum_{k=0}^{n}T_{2}(n,k)x^{[k]}.$

We recommend [3-6,30,31] for general references on central factorial numbers.
For recent works on central factorial numbers, one refers to [11-13,19-23]. As
degenerate versions of the central factorial numbers, the degenerate central
factorial numbers of the first kind $T_{1,\lambda}(n,k)$ and of the second
kind $T_{2,\lambda}(n,k)$ are respectively defined by

(3.3)
$\displaystyle\frac{1}{k!}\left\\{\frac{1}{\lambda}\bigg{(}\bigg{(}\frac{t+\sqrt{t^{2}+4}}{2}\bigg{)}^{2\lambda}-1\bigg{)}\right\\}^{k}=\frac{1}{k!}\bigg{(}\log_{\lambda}\bigg{(}\frac{t+\sqrt{t^{2}+4}}{2}\bigg{)}^{2}\bigg{)}^{k}$
$\displaystyle\quad=\sum_{n=k}^{\infty}T_{1,\lambda}(n,k)\frac{t^{n}}{n!},\quad
x^{[n]}=\sum_{k=0}^{n}T_{1,\lambda}(n,k)(x)_{k,\lambda},$ (3.4)
$\displaystyle\frac{1}{k!}\big{(}e_{\lambda}^{\frac{1}{2}}(t)-e_{\lambda}^{-\frac{1}{2}}(t)\big{)}^{k}=\sum_{n=k}^{\infty}T_{2,\lambda}(n,k)\frac{t^{n}}{n!},\quad(x)_{n,\lambda}=\sum_{k=0}^{n}T_{2,\lambda}(n,k)x^{[k]}.$

Let $\mathbb{P}=\left\\{p_{n}(x)\right\\}_{n=0}^{\infty}$ be a sequence of
polynomials such that deg $p_{n}(x)=n,p_{0}(x)=1$. In (3.2) and (3.4), we
observe that $T_{2}(n,k)$ and $T_{2,\lambda}(n,k)$ arise as the coefficients
respectively when we expand $x^{n}$ and $(x)_{n,\lambda}$ in terms of
$x^{[k]}$. In view of this observation, it seems natural to define the central
factorial numbers of the second associated with
$\mathbb{P}=\left\\{p_{n}(x)\right\\}_{n=0}^{\infty}$ as the coefficients when
we expand $p_{n}(x)$ in terms of $x^{[k]}$:

(3.5) $\displaystyle p_{n}(x)=\sum_{k=0}^{n}T_{2}(n,k;\mathbb{P})x^{[k]}.$

######  Theorem 3.1.

Let $\mathbb{P}=\left\\{p_{n}(x)\right\\}_{n=0}^{\infty}$ be a sequence of
polynomials such that deg $p_{n}(x)=n,p_{0}(x)=1$, with
$p_{n}(x)=\sum_{l=0}^{n}p_{n,l}x^{l}$. We let

$\displaystyle p_{n}(x)=\sum_{k=0}^{n}T_{2}(n,k;\mathbb{P})x^{[k]}.$

(a) Then the central factorial numbers of the second kind associated with
$\mathbb{P}$ are given by

$\displaystyle
T_{2}(n,k;\mathbb{P})=\frac{1}{k!}\big{\langle}{(e^{\frac{t}{2}}-e^{-\frac{t}{2}})^{k}|p_{n}(x)\big{\rangle}}.$

More explicitly, it is given by

$\displaystyle T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}T_{2}(l,k)p_{n,l}.$

(b) $p_{n,l}=\sum_{k=l}^{n}T_{1}(k,l)T_{2}(n,k;\mathbb{P})$.
(c) $T_{2}(n,k;\mathbb{P})$ are alternatively given by

$\displaystyle
T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}S_{2}(l,k)\frac{1}{l!}\Big{(}\frac{d}{dx}\Big{)}^{l}p_{n}(x)\big{|}_{x=-\frac{k}{2}}.$

(d) Let $p_{n}(x)$ be Sheffer for the pair $(g(t),f(t))$. Then the generating
function of $T_{2}(n,k;\mathbb{P})$ is given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}=\frac{1}{g(\bar{f}(t))}\frac{1}{k!}\big{(}e^{\frac{1}{2}\bar{f}(t)}-e^{-\frac{1}{2}\bar{f}(t)}\big{)}^{k}=\frac{1}{g(\bar{f}(t))}\sum_{n=k}^{\infty}T_{2}(n,k)\frac{\bar{f}(t)^{n}}{n!}.$

(e), Let $\bar{\mathbb{P}}=\left\\{\bar{p}_{n}(x)\right\\}$, where
$\bar{p}_{n}(x)$ is the sequence of polynomials defined by
$\bar{p}_{0}(x)=1,\,\,\bar{p}_{n}(x)=xp_{n-1}(x),\,\,(n\geq 1)$. Then we have

$\displaystyle
T_{2}(n+1,k;\bar{\mathbb{P}})=T_{2}(n,k-1;\mathbb{P})+\frac{k}{2}T_{2}(n,k;\mathbb{P}),\,\,(0\leq
k\leq n+1).$

###### Proof.

(a) By noting $x^{[n]}\sim(1,e^{\frac{t}{2}}-e^{-\frac{t}{2}})$, this follows
from (2.16).
From (2.6), (2.16), and (3.2), we have

$\displaystyle T_{2}(n,k;\mathbb{P})$
$\displaystyle=\frac{1}{k!}\big{\langle}{(e^{\frac{t}{2}}-e^{-\frac{t}{2}})^{k}|p_{n}(x)\big{\rangle}}$
(3.6)
$\displaystyle=\sum_{l=k}^{n}T_{2}(l,k)\frac{1}{l!}\big{(}\frac{d}{dx}\big{)}^{l}p_{n}(x)|_{x=0}$
$\displaystyle=\sum_{l=k}^{n}T_{2}(l,k)p_{n,l}.$

(b) The identity follows from the next observation:

$\displaystyle\sum_{l=0}^{n}p_{n,l}x^{l}$
$\displaystyle=\sum_{k=0}^{n}T_{2}(n,k;\mathbb{P})x^{[k]}$
$\displaystyle=\sum_{k=0}^{n}T_{2}(n,k;\mathbb{P})\sum_{l=0}^{k}T_{1}(k,l)x^{l}$
$\displaystyle=\sum_{l=0}^{n}\sum_{k=l}^{n}T_{1}(k,l)T_{2}(n,k;\mathbb{P})x^{l}.$

Note here that the explicit expression of $T_{2}(n,k;\mathbb{P})$ also follows
from this by inversion.
(c) From (a) and (2.2), we have

$\displaystyle T_{2}(n,k;\mathbb{P})$
$\displaystyle=\frac{1}{k!}\big{\langle}{(e^{\frac{t}{2}}-e^{-\frac{t}{2}})^{k}|p_{n}(x)\big{\rangle}}=\frac{1}{k!}\big{\langle}{(e^{t}-1)^{k}e^{-\frac{k}{2}t}\big{|}p_{n}(x)\rangle}$
$\displaystyle=\sum_{l=k}^{n}S_{2}(l,k)\frac{1}{l!}\langle{e^{-\frac{k}{2}t}|t^{l}p_{n}(x)\rangle}$
$\displaystyle=\sum_{l=k}^{n}S_{2}(l,k)\frac{1}{l!}\Big{(}\frac{d}{dx}\Big{)}^{l}p_{n}(x)\big{|}_{x=-\frac{k}{2}}.$

(d) From (2.11) and (3.5), and recalling that
$e^{yt}=\sum_{k=0}^{\infty}\frac{1}{k!}(e^{\frac{1}{2}t}-e^{-\frac{1}{2}t})^{k}y^{[k]}$
from Proposition 2.1 (d), we have

$\displaystyle\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}u^{[k]}$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{k=0}^{n}T_{2}(n,k;\mathbb{P})u^{[k]}\frac{t^{n}}{n!}$
(3.7)
$\displaystyle=\sum_{n=0}^{\infty}p_{n}(u)\frac{t^{n}}{n!}=\frac{1}{g(\bar{f}(t))}e^{u\bar{f}(t)}$
$\displaystyle=\sum_{k=0}^{\infty}\frac{1}{g(\bar{f}(t))}\frac{1}{k!}(e^{\frac{1}{2}\bar{f}(t)}-e^{-\frac{1}{2}\bar{f}(t)})^{k}u^{[k]}.$

(e) This follows from the next observation:

$\displaystyle\sum_{k=0}^{n+1}T_{2}(n+1,k;\bar{\mathbb{P}})x^{[k]}$
$\displaystyle=xp_{n}(x)=\sum_{k=0}^{n}(x-\frac{k}{2}+\frac{k}{2})T_{2}(n,k;\mathbb{P})x^{[k]}$
$\displaystyle=\sum_{k=0}^{n}T_{2}(n,k;\mathbb{P})x^{[k+1]}+\sum_{k=0}^{n}\frac{k}{2}T_{2}(n,k;\mathbb{P})x^{[k]}$
$\displaystyle=\sum_{k=0}^{n+1}T_{2}(n,k-1;\mathbb{P})x^{[k]}+\sum_{k=0}^{n+1}\frac{k}{2}T_{2}(n,k;\mathbb{P})x^{[k]}.$

∎

##  4\. Examples on central factorial numbers of the second kind associated
with sequences of polynomials

(a) Let $\mathbb{P}=\left\\{x^{n}\right\\}$. Then $x^{n}\sim(1,t)$. By the
definition in (3.5) and Theorem 3.1, we have (see [7,30,31])

$\displaystyle
T_{2}(n,k;\mathbb{P})=T_{2}(n,k),\quad\sum_{n=k}^{\infty}T_{2}(n,k)\frac{t^{n}}{n!}=\frac{1}{k!}(e^{\frac{t}{2}}-e^{-\frac{t}{2}})^{k},$

(see (3.2)).
(b) Let $\mathbb{P}=\left\\{(x)_{n,\lambda}\right\\}$ be the sequence of
generalized falling factorials. Then
$(x)_{n,\lambda}\sim\big{(}1,f(t)=\frac{1}{\lambda}(e^{\lambda t}-1)\big{)}$,
with $\bar{f}(t)=\frac{1}{\lambda}\log(1+\lambda t)$ (see
[13,14,16,18,23-25]). By Theorem 3.1, the generating function is given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}=\frac{1}{k!}\big{(}e_{\lambda}^{\frac{1}{2}}(t)-e_{\lambda}^{-\frac{1}{2}}(t)\big{)}=\sum_{n=k}^{\infty}T_{2,\lambda}(n,k)\frac{t^{n}}{n!},$

(see (3.4)). Thus $T_{2}(n,k;\mathbb{P})=T_{2,\lambda}(n,k)$. Alternatively,
by Theorem 3.1 it is also given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}$
$\displaystyle=\sum_{l=k}^{\infty}T_{2}(l,k)\frac{1}{l!}\big{(}\frac{1}{\lambda}\log(1+\lambda
t)\big{)}^{l}$
$\displaystyle=\sum_{n=k}^{\infty}\sum_{l=k}^{n}\lambda^{n-l}S_{1}(n,l)T_{2}(l,k)\frac{t^{n}}{n!}.$

Thus $T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}\lambda^{n-l}S_{1}(n,l)T_{2}(l,k)$.
(c) The rising factorials $\langle x\rangle_{n}$ are defined by (see
[7,15,30,31])

$\displaystyle\langle x\rangle_{n}=x(x+1)\cdots(x+n-1),\,\,(n\geq
1),\quad\langle x\rangle_{0}=1.$

Let $\mathbb{P}=\left\\{\langle{x\rangle}_{n}\right\\}$. Then
$\langle{x\rangle}_{n}\sim(1,1-e^{-t}),$ with $\bar{f}(t)=-\log(1-t)$. From
Theorem 3.1, the generating function is given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}=\frac{1}{k!}\big{(}(1-t)^{-\frac{1}{2}}-(1-t)^{\frac{1}{2}}\big{)}^{k}.$

More explicitly, from Theorem 3.1, it is given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}$
$\displaystyle=\sum_{l=k}^{\infty}T_{2}(l,k)\frac{1}{l!}\big{(}-\log(1-t)\big{)}^{l}$
$\displaystyle=\sum_{n=k}^{\infty}\sum_{l=k}^{n}(-1)^{n-l}S_{1}(n,l)T_{2}(l,k)\frac{t^{n}}{n!}.$

Hence we have

$\displaystyle
T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}(-1)^{n-l}S_{1}(n,l)T_{2}(l,k).$

(d) The generalized rising factorials are defined by (see [18])

$\displaystyle\langle
x\rangle_{n,\lambda}=x(x+\lambda)\cdots(x+(n-1)\lambda),\,\,(n\geq
1),\quad\langle x\rangle_{0,\lambda}=1.$

Let $\mathbb{P}=\left\\{\langle{x\rangle}_{n,\lambda}\right\\}$. Then
$\langle{x\rangle}_{n,\lambda}\sim\big{(}1,f(t)=\frac{1}{\lambda}(1-e^{-\lambda
t})\big{)}$, with $\bar{f}(t)=-\frac{1}{\lambda}\log(1-\lambda t)$. From
Theorem 3.1, the generating function is given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}=\frac{1}{k!}\Big{(}(1-\lambda
t)^{-\frac{1}{2\lambda}}-(1-\lambda t)^{\frac{1}{2\lambda}}\Big{)}^{k}.$

In terms of central factorial numbers of the second, from Theorem 3.1, it is
also given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}$
$\displaystyle=\sum_{l=k}^{\infty}T_{2}(l,k)\frac{1}{l!}\big{(}-\frac{1}{\lambda}\log(1-\lambda
t)\big{)}^{l}$
$\displaystyle=\sum_{n=k}^{\infty}\sum_{l=k}^{n}(-\lambda)^{n-l}S_{1}(n,l)T_{2}(l,k)\frac{t^{n}}{n!}.$

Hence we have

$\displaystyle
T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}(-\lambda)^{n-l}S_{1}(n,l)T_{2}(l,k).$

(e) Let $\alpha(t)$ be the delta series given by

(4.1)
$\displaystyle\alpha(t)=2\Big{(}\frac{1}{2-t}-\frac{1}{2+t}\Big{)}=\sum_{n=1}^{\infty}2^{2-2n}t^{2n-1}.$

Then its compositional inverse is given by

(4.2)
$\displaystyle\bar{\alpha}(t)=\frac{2}{t}\big{(}\sqrt{t^{2}+1}-1\big{)}=2\sum_{n=1}^{\infty}\big{(}\frac{1}{2}\big{)}_{n}\frac{1}{n!}t^{2n-1}.$

H. K. Kim (see [19]) defined the central Lah numbers, which we call the
central Lah numbers of the second kind, by

(4.3)
$\displaystyle\frac{1}{k!}(\alpha(t))^{k}=\sum_{n=k}^{\infty}L_{2}^{(c)}(n,k)\frac{t^{n}}{n!}.$

We also define the central Lah numbers of the first kind by

(4.4)
$\displaystyle\frac{1}{k!}(\bar{\alpha}(t))^{k}=\sum_{n=k}^{\infty}L_{1}^{(c)}(n,k)\frac{t^{n}}{n!}.$

The central Lah-Bell polynomials of the second, which is called the central
Lah-Bell polynomials (see [19]), and the central Lah-Bell polynomials of the
first kind are respectively defined by

$\displaystyle LB_{n,2}^{(c)}(x)=\sum_{k=0}^{n}L_{2}^{(c)}(n,k)x^{k},\quad
LB_{n,1}^{(c)}(x)=\sum_{k=0}^{n}L_{1}^{(c)}(n,k)x^{k}.$

Then it is easy to see that their generating functions are given by

$\displaystyle
e^{x\alpha(t)}=\sum_{n=0}^{\infty}LB_{n,2}^{(c)}(x)\frac{t^{n}}{n!},\quad
e^{x\bar{\alpha}(t)}=\sum_{n=0}^{\infty}LB_{n,1}^{(c)}(x)\frac{t^{n}}{n!}.$

Let $f(t)=e^{\frac{t}{2}}-e^{-\frac{t}{2}}$. Then its compositional inverse is
given by $\bar{f}(t)=2\log\Big{(}\frac{t+\sqrt{t^{2}+4}}{2}\Big{)}$. Define
the central factorial Lah numbers of the first kind by

(4.5) $\displaystyle\frac{1}{k!}\big{(}\bar{f}(\alpha(t))\big{)}^{k}$
$\displaystyle=\frac{1}{k!}\Big{(}2\log\Big{(}\frac{1}{2-t}-\frac{1}{2+t}+\sqrt{\big{(}\frac{1}{2-t}-\frac{1}{2+t}\big{)}^{2}+1}\Big{)}\Big{)}^{k}$
$\displaystyle=\sum_{n=k}^{\infty}TL_{1}(n,k)\frac{t^{n}}{n!}.$

Now, we observe from (3.1) and (4.3) that

$\displaystyle\sum_{n=k}^{\infty}TL_{1}(n,k)\frac{t^{n}}{n!}$
$\displaystyle=\sum_{l=k}^{\infty}T_{1}(l,k)\frac{1}{l!}(\alpha(t))^{l}=\sum_{l=k}^{\infty}T_{1}(l,k)\sum_{n=l}^{\infty}L_{2}^{(c)}(n,l)\frac{t^{n}}{n!}$
$\displaystyle=\sum_{n=k}^{\infty}\sum_{l=k}^{n}T_{1}(l,k)L_{2}^{(c)}(n,l)\frac{t^{n}}{n!}.$

Thus we have shown that

$\displaystyle TL_{1}(n,k)=\sum_{l=k}^{n}L_{2}^{(c)}(n,l)T_{1}(l,k).$

Then we define the central factorial Lah Bell polynomials of the first kind
$TLB_{n,1}(x)$ as

$\displaystyle TLB_{n,1}(x)=\sum_{k=0}^{n}TL_{1}(n,k)x^{k}.$

Then it is immediate to see from (4.5) that

(4.6)
$\displaystyle\sum_{n=0}^{\infty}TLB_{n,1}(x)\frac{t^{n}}{n!}=e^{x\bar{f}(\alpha(t))}.$

Now, we observe that

$\displaystyle TLB_{n,1}(x)$
$\displaystyle=\sum_{k=0}^{n}\sum_{l=k}^{n}L_{2}^{(c)}(n,l)T_{1}(l,k)x^{k}$
$\displaystyle=\sum_{l=0}^{n}L_{2}^{(c)}(n,l)\sum_{k=0}^{l}T_{1}(l,k)x^{k}=\sum_{l=0}^{n}L_{2}^{(c)}(n,l)x^{[l]}.$

Hence we have shown that $TLB_{n,1}(x)=\sum_{l=0}^{n}L_{2}^{(c)}(n,l)x^{[l]}$,
which says that $T_{2}(n,k;\mathbb{P_{1}})=L_{2}^{(c)}(n,k)$, with
$\mathbb{P_{1}}=\left\\{TLB_{n,1}(x)\right\\}_{n=0}^{\infty}$. We note that an
explicit expression of this polynomial is obtained in Theorem 8 of [19].
To proceed further, we define the central factorial Lah numbers of the second
kind as

(4.7) $\displaystyle\frac{1}{k!}(\bar{\alpha}(f(t)))^{k}$
$\displaystyle=\frac{1}{k!}\bigg{(}\frac{2}{e^{\frac{t}{2}}-e^{-\frac{t}{2}}}\Big{(}\sqrt{\big{(}e^{\frac{t}{2}}-e^{-\frac{t}{2}}\big{)}^{2}+1}-1\Big{)}\bigg{)}^{k}$
$\displaystyle=\frac{1}{k!}\bigg{(}\frac{2}{e^{\frac{t}{2}}-e^{-\frac{t}{2}}}\Big{(}\sqrt{e^{t}+e^{-t}-1}-1\Big{)}\bigg{)}^{k}$
$\displaystyle=\sum_{n=k}^{\infty}TL_{2}(n,k)\frac{t^{n}}{n!}.$

Now, we observe from (3.2) and (4.4) that

$\displaystyle\sum_{n=k}^{\infty}TL_{2}(n,k)\frac{t^{n}}{n!}$
$\displaystyle=\sum_{l=k}^{\infty}L_{1}^{(c)}(l,k)\frac{1}{l!}(f(t))^{l}=\sum_{l=k}^{\infty}L_{1}^{(c)}(l,k)\sum_{n=l}^{\infty}T_{2}(n,l)\frac{t^{n}}{n!}$
$\displaystyle=\sum_{n=k}^{\infty}\sum_{l=k}^{n}T_{2}(n,l)L_{1}^{(c)}(l,k)\frac{t^{n}}{n!}.$

Thus we have shown that

$\displaystyle TL_{2}(n,k)=\sum_{l=k}^{n}T_{2}(n,l)L_{1}^{(c)}(l,k).$

We define the central factorial Lah Bell polynomials of the second kind
$TLB_{n,2}(x)$ as

$\displaystyle TLB_{n,2}(x)=\sum_{k=0}^{n}TL_{2}(n,k)x^{k}.$

Then it is easy to see from (4.7) that

(4.8)
$\displaystyle\sum_{n=0}^{\infty}TLB_{n,2}(x)\frac{t^{n}}{n!}=e^{x\bar{\alpha}(f(t))}.$

Now, we let $\mathbb{P_{2}}=\left\\{TLB_{n,2}(x)\right\\}_{n=0}^{\infty}$.
Then, from Theorem 3.1 we have
$T_{2}(n,k;\mathbb{P_{2}})=\sum_{l=k}^{n}T_{2}(l,k)TL_{2}(n,l)$.

(f) The central Bell polynomials $\mathrm{Bel}_{n}^{(c)}(x)$ and the central
Bell numbers $\mathrm{Bel}_{n}^{(c)}$ are respectively defined by (see [22])

$\displaystyle
e^{x(e^{\frac{t}{2}}-e^{-\frac{t}{2}})}=\sum_{n=0}^{\infty}\mathrm{Bel}_{n}^{(c)}(x)\frac{t^{n}}{n!},\quad\mathrm{Bel}_{n}^{(c)}=\mathrm{Bel}_{n}^{(c)}(1).$

Then we see that

$\displaystyle\mathrm{Bel}_{n}^{(c)}(x)\sim\Big{(}1,f(t)=2\log\bigg{(}\frac{t+\sqrt{t^{2}+4}}{2}\bigg{)}\Big{)},\quad\mathrm{Bel}_{n}^{(c)}(x)=\sum_{k=0}^{n}T_{2}(n,k)x^{k}.$

Let $\mathbb{P}=\left\\{\mathrm{Bel}_{n}^{(c)}(x)\right\\}$. Then, by Theorem
3.1, we see that

$\displaystyle T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}T_{2}(n,l)T_{2}(l,k).$

Note here that the generating function is given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}=\frac{1}{k!}\Big{(}e^{\frac{1}{2}(e^{\frac{t}{2}}-e^{-\frac{t}{2}})}-e^{-\frac{1}{2}(e^{\frac{t}{2}}-e^{-\frac{t}{2}})}\Big{)}^{k}.$

(g) The numbers $T_{1,\lambda}(n,k)$ and $T_{2,\lambda}(n,k)$ are respectively
called the degenerate central factorial numbers of the first kind and of the
second kind, and defined by (3.3) and (3.4) (see [23]). The degenerate central
Bell polynomials $\mathrm{Bel}_{n,\lambda}^{(c)}(x)$ and the degenerate
central Bell numbers $\mathrm{Bel}_{n,\lambda}^{(c)}$ are respectively defined
by (see [23])

$\displaystyle
e^{x(e_{\lambda}^{\frac{1}{2}}(t)-e_{\lambda}^{-\frac{1}{2}}(t))}=\sum_{n=0}^{\infty}\mathrm{Bel}_{n,\lambda}^{(c)}(x)\frac{t^{n}}{n!},\quad\mathrm{Bel}_{n,\lambda}^{(c)}=\mathrm{Bel}_{n,\lambda}^{(c)}(1).$

Then we see that

$\displaystyle\mathrm{Bel}_{n,\lambda}^{(c)}(x)\sim\Big{(}1,f(t)=\frac{1}{\lambda}\bigg{(}\bigg{(}\frac{t+\sqrt{t^{2}+4}}{2}\bigg{)}^{2\lambda}-1\bigg{)}\Big{)},\quad\mathrm{Bel}_{n,\lambda}^{(c)}(x)=\sum_{k=0}^{n}T_{2,\lambda}(n,k)x^{k}.$

Let $\mathbb{P}=\left\\{\mathrm{Bel}_{n}^{(c)}(x)\right\\}$. Then, by Theorem
3.1, we see that

$\displaystyle
T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}T_{2}(l,k)T_{2,\lambda}(n,l).$

We note here that the generating function is given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}=\frac{1}{k!}\Big{(}e^{\frac{1}{2}\big{(}e_{\lambda}^{\frac{1}{2}}(t)-e_{\lambda}^{-\frac{1}{2}}(t)\big{)}}-e^{-\frac{1}{2}\big{(}e_{\lambda}^{\frac{1}{2}}(t)-e_{\lambda}^{-\frac{1}{2}}(t)\big{)}}\Big{)}^{k}.$

(h) Let $x^{[n,\lambda]}$ be defined by

$\displaystyle
x^{[n,\lambda]}=x\big{(}x+(\frac{1}{2}n-1)\lambda\big{)}_{n-1,\lambda},\,\,(n\geq
1),\quad x^{[0,\lambda]}=1.$

where $x^{[n,\lambda]}\sim\big{(}1,f(t)=\frac{1}{\lambda}(e^{\frac{\lambda
t}{2}}-e^{-\frac{\lambda t}{2}})\big{)}$. Here we note that

$\displaystyle\bar{f}(t)=\frac{2}{\lambda}\log\bigg{(}\frac{\lambda
t+\sqrt{\lambda^{2}t^{2}+4}}{2}\bigg{)}=\frac{1}{\lambda}\log\big{(}1+\frac{\lambda
t}{2}(\lambda t+\sqrt{\lambda^{2}t^{2}+4})\big{)}.$

As analogies to the numbers $T_{1,\lambda}(n,k)$ and $T_{2,\lambda}(n,k)$, we
may introduce the numbers $R_{1,\lambda}(n,k)$ and $R_{2,\lambda}(n,k)$ which
are defined by

$\displaystyle
x^{[n,\lambda]}=\sum_{k=0}^{n}R_{1,\lambda}(n,k)x^{k},\quad\sum_{n=k}^{\infty}R_{1,\lambda}(n,k)\frac{t^{n}}{n!}=\frac{1}{k!}\bigg{(}\frac{2}{\lambda}\log\bigg{(}\frac{\lambda
t+\sqrt{\lambda^{2}t^{2}+4}}{2}\bigg{)}\bigg{)}^{k},$ $\displaystyle
x^{n}=\sum_{k=0}^{n}R_{2,\lambda}(n,k)x^{[k,\lambda]},\quad\sum_{n=k}^{\infty}R_{2,\lambda}(n,k)\frac{t^{n}}{n!}=\frac{1}{k!}\Big{(}\frac{1}{\lambda}\big{(}e^{\frac{\lambda
t}{2}}-e^{-\frac{\lambda t}{2}}\big{)}\Big{)}^{k}.$

Let $\mathbb{P}=\left\\{x^{[n,\lambda]}\right\\}$. From Theorem 3.1, we get

$\displaystyle
T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}T_{2}(l,k)R_{1,\lambda}(n,l).$

From Theorem 3.1, the generating function is given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}=\frac{1}{k!}\bigg{(}\bigg{(}\frac{\lambda
t+\sqrt{\lambda^{2}t^{2}+4}}{2}\bigg{)}^{\frac{1}{\lambda}}-\bigg{(}\frac{\lambda
t+\sqrt{\lambda^{2}t^{2}+4}}{2}\bigg{)}^{-\frac{1}{\lambda}}\bigg{)}^{k}.$

(i) Let $\mathbb{P}=\left\\{B_{n}^{L}(x)\right\\}$ be the sequence of Lah-Bell
polynomials. Then it is given by
$e^{x(\frac{t}{1-t})}=\sum_{n=0}^{\infty}B_{n}^{L}(x)\frac{t^{n}}{n!}$, so
that $B_{n}^{L}(x)\sim(1,\frac{t}{1+t})$, and
$B_{n}^{L}(x)=\sum_{k=0}^{n}L(n,k)x^{k}$. Here
$L(n,k)=\binom{n-1}{k-1}\frac{n!}{k!}$ are the (unsigned) Lah numbers (see
[7,15,30,31]) whose generating function is given by
$\sum_{n=k}^{\infty}L(n,k)\frac{t^{n}}{n!}=\frac{1}{k!}\big{(}\frac{t}{1-t}\big{)}^{k}$.
By Theorem 3.1, we have

$\displaystyle T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}T_{2}(l,k)L(n,l).$

In addition, by Theorem 3.1, the generating function is given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}=\frac{1}{k!}(e^{\frac{t}{2(1-t)}}-e^{-\frac{t}{2(1-t)}})^{k}.$

(j) Let $\mathbb{P}=\left\\{B_{n,\lambda}^{L}(x)\right\\}$ be the sequence of
degenerate Lah-Bell polynomials. Then it is given by
$e_{\lambda}^{x}(\frac{t}{1-t})=\sum_{n=0}^{\infty}B_{n,\lambda}^{L}(x)\frac{t^{n}}{n!}$.
Thus we see that $B_{n,\lambda}^{L}(x)\sim(1,f(t)=\frac{e^{\lambda
t}-1}{\lambda+e^{\lambda t}-1})$, with
$\bar{f}(t)=\frac{1}{\lambda}\log(1+\frac{\lambda t}{1-t})$ (see [18]). By
Theorem 3.1, we have

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}$
$\displaystyle=\sum_{l=k}^{\infty}T_{2}(l,k)\frac{1}{l!}\Big{(}\frac{1}{\lambda}\log\big{(}1+\frac{\lambda
t}{1-t}\big{)}\Big{)}^{l}$
$\displaystyle=\sum_{l=k}^{\infty}T_{2}(l,k)\frac{1}{\lambda^{l}}\sum_{m=l}^{\infty}S_{1}(m,l)\frac{1}{m!}\big{(}\frac{\lambda
t}{1-t}\big{)}^{m}$
$\displaystyle=\sum_{l=k}^{\infty}\sum_{m=l}^{\infty}\lambda^{m-l}S_{1}(m,l)T_{2}(l,k)\sum_{n=m}^{\infty}L(n,m)\frac{t^{n}}{n!}$
$\displaystyle=\sum_{n=k}^{\infty}\sum_{l=k}^{n}\sum_{m=l}^{n}\lambda^{m-l}S_{1}(m,l)T_{2}(l,k)L(n,m)\frac{t^{n}}{n!}.$

Thus we obtain
$T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}\sum_{m=l}^{n}\lambda^{m-l}L(n,m)S_{1}(m,l)T_{2}(l,k).$
Alternatively, the generating function is also given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}=\frac{1}{k!}\Big{(}\Big{(}1+\frac{\lambda
t}{1-t}\Big{)}^{\frac{1}{2\lambda}}-\Big{(}1+\frac{\lambda
t}{1-t}\Big{)}^{-\frac{1}{2\lambda}}\Big{)}^{k}.$

(k) Let $\mathbb{P}=\left\\{\mathrm{Bel}_{n}(x)\right\\}$ be the sequence of
Bell polynomials given by
$e^{x(e^{t}-1)}=\sum_{n=0}^{\infty}\mathrm{Bel}_{n}(x)\frac{t^{n}}{n!}$. Then
$\mathrm{Bel}_{n}(x)\sim(1,\log(1+t))$, and
$\mathrm{Bel}_{n}(x)=\sum_{k=0}^{\infty}S_{2}(n,k)x^{k}$ (see [7,30,31]). By
Theorem 3.1, the generating function is given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}$
$\displaystyle=\frac{1}{k!}\big{(}e^{\frac{1}{2}(e^{t}-1)}-e^{-\frac{1}{2}(e^{t}-1)}\big{)}^{k}.$

More explicitly, from Theorem 3.1, we get

$\displaystyle T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}S_{2}(n,l)T_{2}(l,k).$

(l) Let $\mathbb{P_{1}}=\left\\{\mathrm{Bel}_{n,\lambda}(x)\right\\}$ be the
sequence of partially degenerate Bell polynomials given by
$e^{x(e_{\lambda}(t)-1)}=\sum_{n=0}^{\infty}\mathrm{Bel}_{n,\lambda}(x)\frac{t^{n}}{n!}$.
Then $\mathrm{Bel}_{n,\lambda}(x)\sim(1,\log_{\lambda}(1+t))$, and
$\mathrm{Bel}_{n,\lambda}(x)=\sum_{k=0}^{n}S_{2,\lambda}(n,k)x^{k}$ (see
[24]). From Theorem 3.1, it is immediate to see that

$\displaystyle
T_{2}(n,k;\mathbb{P_{1}})=\sum_{l=k}^{n}S_{2,\lambda}(n,l)T_{2}(l,k).$

The generating function is given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P_{1}})\frac{t^{n}}{n!}=\frac{1}{k!}\big{(}e^{\frac{1}{2}(e_{\lambda}(t)-1)}-e^{-\frac{1}{2}(e_{\lambda}(t)-1)}\big{)}^{k}.$

There are several other degenerate versions of Bell polynomials, one of which
is the so called fully degenerate Bell polynomials $\phi_{n,\lambda}(x)$ given
by
$e_{\lambda}^{x}(e_{\lambda}(t)-1)=\sum_{n=0}^{\infty}\phi_{n,\lambda}(x)\frac{t^{n}}{n!}.$
Then
$\phi_{n,\lambda}(x)\sim\big{(}1,f(t)=\log_{\lambda}(1+\frac{1}{\lambda}(e^{\lambda
t}-1))\big{)}$, with
$\bar{f}(t)=\frac{1}{\lambda}\log(1+\lambda(e_{\lambda}(t)-1)\big{)}$ (see
[25]). Let $\mathbb{P_{2}}=\left\\{\phi_{n,\lambda}(x)\right\\}$. From Theorem
3.1, we have

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P_{2}})\frac{t^{n}}{n!}$
$\displaystyle=\sum_{l=k}^{\infty}T_{2}(l,k)\frac{1}{l!}\Big{(}\frac{1}{\lambda}\log\big{(}1+\lambda(e_{\lambda}(t)-1)\big{)}\Big{)}^{l}$
$\displaystyle=\sum_{l=k}^{\infty}T_{2}(l,k)\frac{1}{\lambda^{l}}\sum_{m=l}^{\infty}S_{1}(m,l)\frac{1}{m!}\big{(}\lambda(e_{\lambda}(t)-1)\big{)}^{m}$
$\displaystyle=\sum_{l=k}^{\infty}\sum_{m=l}^{\infty}T_{2}(l,k)S_{1}(m,l)\lambda^{m-l}\sum_{n=m}^{\infty}S_{2,\lambda}(n,m)\frac{t^{n}}{n!}$
$\displaystyle=\sum_{n=k}^{\infty}\sum_{l=k}^{n}\sum_{m=l}^{n}\lambda^{m-l}S_{2,\lambda}(n,m)S_{1}(m,l)T_{2}(l,k)\frac{t^{n}}{n!},$

which verifies that we have
$T_{2}(n,k;\mathbb{P_{2}})=\sum_{l=k}^{n}\sum_{m=l}^{n}\lambda^{m-l}S_{2,\lambda}(n,m)S_{1}(m,l)T_{2}(l,k).$
Alternatively, the generating function is also given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P_{2}})\frac{t^{n}}{n!}=\frac{1}{k!}\Big{(}\big{(}1+\lambda(e_{\lambda}(t)-1)\big{)}^{\frac{1}{2\lambda}}-\big{(}1+\lambda(e_{\lambda}(t)-1)\big{)}^{-\frac{1}{2\lambda}}\Big{)}^{k}.$

(m) Let $\mathbb{P}=\left\\{M_{n}(x)\right\\}$ be the sequence of Mittag-
Leffler polynomials. That is, $M_{n}(x)\sim(1,f(t)=\frac{e^{t}-1}{e^{t}+1})$,
with $\bar{f}(t)=\log\big{(}\frac{1+t}{1-t}\big{)}$ (see [31]). By Theorem
3.1, we have

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}=\frac{1}{k!}\Big{(}\Big{(}\frac{1+t}{1-t}\Big{)}^{\frac{1}{2}}-\Big{(}\frac{1+t}{1-t}\Big{)}^{-\frac{1}{2}}\Big{)}^{k}.$

Alternatively, we also have

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}$
$\displaystyle=\sum_{l=k}^{\infty}T_{2}(l,k)\frac{1}{l!}\Big{(}\log\Big{(}1+\frac{2t}{1-t}\Big{)}\Big{)}^{l}$
$\displaystyle=\sum_{l=k}^{\infty}T_{2}(l,k)\sum_{m=l}^{\infty}S_{1}(m,l)2^{m}\frac{1}{m!}\Big{(}\frac{t}{1-t}\Big{)}^{m}$
$\displaystyle=\sum_{l=k}^{\infty}T_{2}(l,k)\sum_{m=l}^{\infty}S_{1}(m,l)2^{m}\sum_{n=m}^{\infty}L(n,m)\frac{t^{n}}{n!}$
$\displaystyle=\sum_{n=k}^{\infty}\sum_{l=k}^{n}\sum_{m=l}^{n}2^{m}L(n,m)S_{1}(m,l)T_{2}(l,k)\frac{t^{n}}{n!}.$

This shows that
$T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}\sum_{m=l}^{n}2^{m}L(n,m)S_{1}(m,l)T_{2}(l,k)$.

(n) Let $\mathbb{P}=\left\\{L_{n}(x)\right\\}$ be the sequence of Laguerre
polynomials of order -1 (see [31]). That is,
$L_{n}(x)\sim(1,f(t)=\frac{t}{t-1})$, with $\bar{f}(t)=\frac{t}{t-1}$. By
Theorem 3.1, we have

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}$
$\displaystyle=\sum_{l=k}^{\infty}T_{2}(l,k)(-1)^{l}\frac{1}{l!}\Big{(}\frac{t}{1-t}\Big{)}^{l}$
$\displaystyle=\sum_{l=k}^{\infty}T_{2}(l,k)(-1)^{l}\sum_{n=l}^{\infty}L(n,l)\frac{t^{n}}{n!}$
$\displaystyle=\sum_{n=k}^{\infty}\sum_{l=k}^{n}(-1)^{l}L(n,l)T_{2}(l,k)\frac{t^{n}}{n!}.$

Thus we have $T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}(-1)^{l}L(n,l)T_{2}(l,k)$.
Moreover, the generating function is also given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}=\frac{1}{k!}\big{(}e^{\frac{t}{2(t-1)}}-e^{-\frac{t}{2(t-1)}}\big{)}^{k}.$

(o) Let $\mathbb{P}=\left\\{B_{n}(x)\right\\}$ be the sequence of Bernoulli
polynomials. Then $B_{n}(x)\sim\big{(}\frac{e^{t}-1}{t},t\big{)}$, and
$B_{n}(x)=\sum_{k=0}^{n}\binom{n}{k}B_{n-k}x^{k}$ (see [31]). By Theorem 3.1,
we have

$\displaystyle T_{2}(n,k;\mathbb{P})$
$\displaystyle=\sum_{l=k}^{n}T_{2}(l,k)\binom{n}{l}B_{n-l}=\sum_{l=0}^{n-k}\binom{n}{l}T_{2}(n-l,k)B_{l},$
$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}=\frac{t}{e^{t}-1}\frac{1}{k!}(e^{\frac{t}{2}}-e^{-\frac{t}{2}})^{k}.$

(p) Let $\mathbb{P}=\left\\{E_{n}(x)\right\\}$ be the sequence of Euler
polynomials. Then $E_{n}(x)\sim\big{(}\frac{e^{t}+1}{2},t\big{)}$ (see [31]).
Analogously to (c), by Theorem 3.1 we see that

$\displaystyle
T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}T_{2}(l,k)\binom{n}{l}E_{n-l}=\sum_{l=0}^{n-k}\binom{n}{l}T_{2}(n-l,k)E_{l},$
$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}=\frac{2}{e^{t}+1}\frac{1}{k!}(e^{\frac{t}{2}}-e^{-\frac{t}{2}})^{k}.$

(q) Let $\mathbb{P}=\left\\{(rx+s)_{n}\right\\}$, with $r\neq 0$. Then

$\displaystyle(rx+s)_{n}=\sum_{k=0}^{n}G(n,k;r,s)(x)_{k},$

where $G(n,k;r,s)$ are called the Gould-Hopper numbers (see [26]). We note
that $(rx+s)_{n}\sim\big{(}e^{-\frac{s}{r}t},f(t)=e^{\frac{t}{r}}-1\big{)}$,
where $\bar{f}(t)=r\log(1+t)$. By Theorem 3.1, the generating function is
given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}=(1+t)^{s}\frac{1}{k!}\big{(}(1+t)^{\frac{r}{2}}-(1+t)^{-\frac{r}{2}}\big{)}^{k}.$

Alternatively, from Theorem 3.1 the generating function is also given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}$
$\displaystyle=(1+t)^{s}\sum_{l=k}^{\infty}T_{2}(l,k)r^{l}\frac{1}{l!}(\log(1+t))^{l}$
$\displaystyle=(1+t)^{s}\sum_{l=k}^{\infty}T_{2}(l,k)r^{l}\sum_{m=l}^{\infty}S_{1}(m,l)\frac{t^{m}}{m!}$
$\displaystyle=\sum_{j=0}^{\infty}(s)_{j}\frac{t^{j}}{j!}\sum_{m=k}^{\infty}\sum_{l=k}^{m}r^{l}S_{1}(m,l)T_{2}(l,k)\frac{t^{m}}{m!}$
$\displaystyle=\sum_{n=k}^{\infty}\sum_{l=k}^{n}\sum_{m=l}^{n}\binom{n}{m}r^{l}(s)_{n-m}S_{1}(m,l)T_{2}(l,k)\frac{t^{n}}{n!}.$

Thus we have

$\displaystyle
T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}\sum_{m=l}^{n}\binom{n}{m}r^{l}(s)_{n-m}S_{1}(m,l)T_{2}(l,k).$

(r) Let $\mathbb{P}=\left\\{b_{n}(x)\right\\}$ be the sequence of Bernoulli
polynomials of the second kind. Then
$b_{n}(x)\sim\big{(}\frac{t}{e^{t}-1},f(t)=e^{t}-1\big{)}$, with
$\bar{f}(t)=\log(1+t)$, so that
$\sum_{n=0}^{\infty}b_{n}(x)\frac{t^{n}}{n!}=\frac{t}{\log(1+t)}(1+t)^{x}$
(see [31]). From Theorem 3.1, the generating function is

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}=\frac{t}{\log(1+t)}\frac{1}{k!}\big{(}(1+t)^{\frac{1}{2}}-(1+t)^{-\frac{1}{2}}\big{)}^{k}.$

Alternatively, we also have

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}$
$\displaystyle=\frac{t}{\log(1+t)}\sum_{l=k}^{\infty}T_{2}(l,k)\frac{1}{l!}(\log(1+t))^{l}$
$\displaystyle=\frac{t}{\log(1+t)}\sum_{l=k}^{\infty}T_{2}(l,k)\sum_{m=l}^{\infty}S_{1}(m,l)\frac{t^{m}}{m!}$
$\displaystyle=\sum_{j=0}^{\infty}b_{j}\frac{t^{j}}{j!}\sum_{m=k}^{\infty}\sum_{l=k}^{m}S_{1}(m,l)T_{2}(l,k)\frac{t^{m}}{m!}$
$\displaystyle=\sum_{n=k}^{\infty}\sum_{l=k}^{n}\sum_{m=l}^{n}\binom{n}{m}b_{n-m}S_{1}(m,l)T_{2}(l,k)\frac{t^{n}}{n!}.$

Thus we get

$\displaystyle
T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}\sum_{m=l}^{n}\binom{n}{m}b_{n-m}S_{1}(m,l)T_{2}(l,k),$

where $b_{n}=b_{n}(0)$ are the Bernoulli numbers of the second.
(s) Let $\mathbb{P}=\left\\{C_{n}(x;a)\right\\}$ be the sequence of Poisson-
Charlier polynomials. Here
$C_{n}(x;a)\sim\big{(}e^{a(e^{t}-1)},f(t)=a(e^{t}-1)\big{)}$, with $a\neq 0$.
Then $\bar{f}(t)=\log(1+\frac{t}{a})$, and
$\sum_{n=0}^{\infty}C_{n}(x;a)\frac{t^{n}}{n!}=e^{-t}\big{(}1+\frac{t}{a}\big{)}^{x}$
(see [31]). From Theorem 3.1, the generating function is given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}=e^{-t}\frac{1}{k!}\Big{(}\Big{(}1+\frac{t}{a}\Big{)}^{\frac{1}{2}}-\Big{(}1+\frac{t}{a}\Big{)}^{-\frac{1}{2}}\Big{)}^{k}.$

Another expression for generating function follows from Theorem 3.1, which is
given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}$
$\displaystyle=e^{-t}\sum_{l=k}^{\infty}T_{2}(l,k)\frac{1}{l!}\big{(}\log(1+\frac{t}{a})\big{)}^{l}$
$\displaystyle=e^{-t}\sum_{l=k}^{\infty}T_{2}(l,k)\sum_{m=l}^{\infty}S_{1}(m,l)a^{-m}\frac{t^{m}}{m!}$
$\displaystyle=\sum_{j=0}^{\infty}(-1)^{j}\frac{t^{j}}{j!}\sum_{m=k}^{\infty}\sum_{l=k}^{m}a^{-m}S_{1}(m,l)T_{2}(l,k)\frac{t^{m}}{m!}$
$\displaystyle=\sum_{n=k}^{\infty}\sum_{l=k}^{n}\sum_{m=l}^{n}\binom{n}{m}(-1)^{n-m}a^{-m}S_{1}(m,l)T_{2}(l,k)\frac{t^{n}}{n!}.$

Thus we see that

$\displaystyle
T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}\sum_{m=l}^{n}\binom{n}{m}(-1)^{n-m}a^{-m}S_{1}(m,l)T_{2}(l,k).$

(t) Let $\mathbb{P}=\left\\{p_{n}(x)\right\\}$, with
$p_{n}(x)=\sum_{k=0}^{n}B_{k}(x)B_{n-k}(x)$. This is not a Sheffer sequence.
For this, we recall from [17] that

$\displaystyle
p_{n}(x)=\frac{2}{n+2}\sum_{m=0}^{n-2}\binom{n+2}{m}B_{n-m}B_{m}(x)+(n+1)B_{n}(x).$

By (2.12) and (3.2), we note that

$\displaystyle\frac{1}{k!}\langle{(e^{\frac{t}{2}}-e^{-\frac{t}{2}})^{k}|B_{n}(x)\rangle}$
$\displaystyle=\sum_{l=0}^{n-k}\binom{n}{l}T_{2}(n-l,k)B_{l}.$

By Theorem 3.1, we see that

$\displaystyle
T_{2}(n,k;\mathbb{P})=\frac{1}{k!}\langle{(e^{\frac{t}{2}}-e^{-\frac{t}{2}})^{k}|p_{n}(x)\rangle}$
$\displaystyle=\frac{2}{n+2}\sum_{m=0}^{n-2}\binom{n+2}{m}B_{n-m}\frac{1}{k!}\langle{(e^{\frac{t}{2}}-e^{-\frac{t}{2}})^{k}|B_{m}(x)\rangle}+(n+1)\frac{1}{k!}\langle{(e^{\frac{t}{2}}-e^{-\frac{t}{2}})^{k}|B_{n}(x)\rangle}$
$\displaystyle=\frac{2}{n+2}\sum_{m=0}^{n-2}\binom{n+2}{m}B_{n-m}\sum_{l=0}^{m-k}\binom{m}{l}T_{2}(m-l,k)B_{l}+(n+1)\sum_{l=0}^{n-k}\binom{n}{l}T_{2}(n-l,k)B_{l}.$

This says that the following identity holds:

$\displaystyle\sum_{k=0}^{n}B_{k}(x)B_{n-k}(x)$
$\displaystyle=\sum_{k=0}^{n}\left\\{\frac{2}{n+2}\sum_{m=0}^{n-2}\binom{n+2}{m}B_{n-m}\sum_{l=0}^{m-k}\binom{m}{l}T_{2}(m-l,k)B_{l}\right.$
$\displaystyle\left.\quad\quad\quad+(n+1)\sum_{l=0}^{n-k}\binom{n}{l}T_{2}(n-l,k)B_{l}\right\\}x^{[k]}.$

##  5\. Central factorial numbers of the first kind associated with sequences
of polynomials

As in the ‘classical’ case, we would like to introduce the central factorial
numbers of the first kind associated with sequences of polynomials. Let
$\mathbb{P}=\left\\{p_{n}(x)\right\\}_{n=0}^{\infty}$ be a sequence of
polynomials such that deg $p_{n}(x)=n,p_{0}(x)=1$. In view of our definition
of the central factorial numbers of the second kind associated with
$\mathbb{P}=\left\\{p_{n}(x)\right\\}_{n=0}^{\infty}$, it is natural to define
the central factorial numbers of the first kind associated with
$\mathbb{P}=\left\\{p_{n}(x)\right\\}_{n=0}^{\infty}$ as the coefficients when
we expand $x^{[n]}$ in terms of $p_{k}(x)$:

(5.1) $\displaystyle x^{[n]}=\sum_{k=0}^{n}T_{1}(n,k;\mathbb{P})p_{k}(x).$

######  Theorem 5.1.

Let $\mathbb{P}=\left\\{p_{n}(x)\right\\}_{n=0}^{\infty}$ be a sequence of
polynomials such that deg $p_{n}(x)=n,p_{0}(x)=1$. We let

$\displaystyle x^{[n]}=\sum_{k=0}^{n}T_{1}(n,k;\mathbb{P})p_{k}(x).$

(a) Let $\bar{\mathbb{P}}=\left\\{\bar{p}_{n}(x)\right\\}$, where
$\bar{p}_{n}(x)$ is the sequence of polynomials defined by
$\bar{p}_{0}(x)=1,\,\,\bar{p}_{n}(x)=xp_{n-1}(x),\,\,(n\geq 1)$. Then we have

$\displaystyle
T_{1}(n+1,k;\bar{\mathbb{P}})=T_{1}(n,k-1;\mathbb{P})-\frac{n}{2}T_{1}(n,k;\bar{\mathbb{P}}).$

(b) Assume that $p_{n}(x)\sim(g(t),f(t))$. Then the central factorial numbers
of the first kind associated with $\mathbb{P}$ are given by

$\displaystyle
T_{1}(n,k;\mathbb{P})=\frac{1}{k!}\langle{g(t)(f(t))^{k}|x^{[n]}\rangle}=\frac{1}{k!}\sum_{l=k}^{n}T_{1}(n,l)\langle{g(t)f(t)^{k}|x^{l}\rangle}.$

(c)
$\sum_{k=0}^{n}T_{1}(n,k;\mathbb{P})p_{k}(1)=(\frac{n}{2})_{n-1},\,\,(n\geq
1).$

###### Proof.

(a) This follows from the next observation:

$\displaystyle\sum_{k=0}^{n+1}T_{1}(n+1,k;\bar{\mathbb{P}})\bar{p}_{k}(x)$
$\displaystyle=x^{[n]}\big{(}x-\frac{n}{2}\big{)}$
$\displaystyle=x\sum_{k=0}^{n}T_{1}(n,k;\mathbb{P})p_{k}(x)-\frac{n}{2}\sum_{k=0}^{n}T_{1}(n,k;\bar{\mathbb{P}})\bar{p}_{k}(x)$
$\displaystyle=\sum_{k=0}^{n}T_{1}(n,k;\mathbb{P})\bar{p}_{k+1}(x)-\frac{n}{2}\sum_{k=0}^{n}T_{1}(n,k;\bar{\mathbb{P}})\bar{p}_{k}(x)$
$\displaystyle=\sum_{k=0}^{n+1}T_{1}(n,k-1;\mathbb{P})\bar{p}_{k}(x)-\frac{n}{2}\sum_{k=0}^{n+1}T_{1}(n,k;\bar{\mathbb{P}})\bar{p}_{k}(x).$

(b) This follows from (2.16) and (3.1).
(c) This is clear by definition in (5.1) and (2.27). ∎

For any delta series $f(t)$, we define the central logarithm associated to
$f(t)$ by

$\displaystyle
LC_{f(t)}t=f\big{(}2\log\Big{(}\frac{t+\sqrt{t^{2}+4}}{2}\Big{)}\big{)},$

In case $p_{n}(x)\sim(1,f(t))$, $LC_{f(t)}t$ may be also denoted by
$LC_{\mathbb{P}}t$ and called the central logarithm associated to
$\mathbb{P}=\left\\{p_{n}(x)\right\\}$. Recalling that
$x^{[n]}\sim(1,l(t)=e^{\frac{t}{2}}-e^{-\frac{t}{2}})$, we remark that
$LC_{f(t)}t=f(\bar{l}(t))$.

######  Theorem 5.2.

Let $p_{n}(x)\sim(1,f(t)),$ and let $\mathbb{P}=\left\\{p_{n}(x)\right\\}$.
Then the generating function of the central factorial numbers of the first
kind associated with $\mathbb{P}$ is given by

$\displaystyle\sum_{n=k}^{\infty}T_{1}(n,k;\mathbb{P})\frac{t^{n}}{n!}=\frac{1}{k!}(LC_{f(t)}t)^{k},$

where
$LC_{f(t)}t=LC_{\mathbb{P}}t=f\big{(}2\log\Big{(}\frac{t+\sqrt{t^{2}+4}}{2}\Big{)}\big{)}$
is the central logarithm associated to $f(t)$.

###### Proof.

Assume now that $p_{n}(x)\sim(1,f(t))$. Then we observe that

$\displaystyle\sum_{k=0}^{\infty}p_{k}(x)\sum_{n=k}^{\infty}T_{1}(n,k;\mathbb{P})\frac{t^{n}}{n!}$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{k=0}^{n}T_{1}(n,k;\mathbb{P})p_{k}(x)\frac{t^{n}}{n!}$
$\displaystyle=\sum_{n=0}^{\infty}x^{[n]}\frac{t^{n}}{n!}=\Big{(}\frac{t+\sqrt{t^{2}+4}}{2}\Big{)}^{2x}.$

On the other hand, we also have

$\displaystyle\sum_{k=0}^{\infty}p_{k}(x)\frac{1}{k!}\big{(}LC_{f(t)}t\big{)}^{k}$
$\displaystyle=e^{x\bar{f}\big{(}f\big{(}2\log\big{(}\frac{t+\sqrt{t^{2}+4}}{2}\big{)}\big{)}\big{)}}=\Big{(}\frac{t+\sqrt{t^{2}+4}}{2}\Big{)}^{2x},$

where we used the fact that
$\sum_{k=0}^{\infty}p_{k}(x)\frac{t^{k}}{k!}=e^{x\bar{f}(t)}$. This shows the
assertion.
We would like to give another proof for this by using Theorem 5.1 (b).
Here we have to show that

$\displaystyle\sum_{n=k}^{\infty}\frac{1}{k!}\langle{f(t)^{k}|x^{[n]}\rangle}\frac{t^{n}}{n!}=\frac{1}{k!}\big{(}f\big{(}2\log\Big{(}\frac{t+\sqrt{t^{2}+4}}{2}\Big{)}\big{)}\big{)}^{k}.$

Equivalently, we need to verify that

(5.2)
$\displaystyle\sum_{n=k}^{\infty}\langle{f(t)^{k}|x^{[n]}\rangle}\frac{(e^{\frac{t}{2}}-e^{-\frac{t}{2}})^{n}}{n!}=(f(t))^{k}.$

The left hand side of (5.2) is equal to

$\displaystyle\sum_{n=k}^{\infty}\sum_{l=k}^{n}T_{1}(n,l)\langle{f(t)^{k}|x^{l}\rangle}\frac{(e^{\frac{t}{2}}-e^{-\frac{t}{2}})^{n}}{n!}$
$\displaystyle=\sum_{l=k}^{\infty}\langle{f(t)^{k}|x^{l}\rangle}\sum_{n=l}^{\infty}T_{1}(n,l)\frac{(e^{\frac{t}{2}}-e^{-\frac{t}{2}})^{n}}{n!}$
$\displaystyle=\sum_{l=0}^{\infty}\langle{f(t)^{k}|x^{l}\rangle}\frac{t^{l}}{l!}=(f(t))^{k}.$

∎

The following orthogonality and inverse relations can be easily shown from the
definitions of central factorial numbers associated with $\mathbb{P}$.

######  Theorem 5.3.

Let $\mathbb{P}=\left\\{p_{n}(x)\right\\}_{n=0}^{\infty}$ be a sequence of
polynomials such that deg $p_{n}(x)=n,p_{0}(x)=1$. Then we have

$\displaystyle(a)\,\,\sum_{k=l}^{n}T_{1}(n,k;\mathbb{P})T_{2}(k,l;\mathbb{P})=\delta_{n,l},\,\,\sum_{k=l}^{n}T_{2}(n,k;\mathbb{P})T_{1}(k,l;\mathbb{P})=\delta_{n,l},\,\,(0\leq
l\leq n).$
$\displaystyle(b)\,\,a_{n}=\sum_{k=0}^{n}T_{2}(n,k;\mathbb{P})c_{k}\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}T_{1}(n,k;\mathbb{P})a_{k}.$
$\displaystyle(c)\,\,a_{n}=\sum_{k=n}^{m}T_{2}(k,n;\mathbb{P})c_{k}\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}T_{1}(k,n;\mathbb{P})a_{k}.$

######  Remark 5.4.

Let $p_{n}(x)$ be Sheffer for the pair $(g(t),f(t))$. Then we saw in Theorem
3.2 that the generating function of $T_{2}(n,k;\mathbb{P})$ is given by

$\displaystyle\sum_{n=k}^{\infty}T_{2}(n,k;\mathbb{P})\frac{t^{n}}{n!}=\frac{1}{g(\bar{f}(t))}\frac{1}{k!}(e^{\frac{1}{2}\bar{f}(t)}-e^{-\frac{1}{2}\bar{f}(t)})^{k}.$

Let $f(t)$ be any delta series. After introducing the central logarithm
associated to $f(t)$, it seems natural to introduce the central exponential
associated to $f(t)$ as

$\displaystyle
EC_{f(t)}t=e^{\frac{1}{2}\bar{f}(t)}-e^{-\frac{1}{2}\bar{f}(t)}.$

When $p_{n}(x)\sim(g(t),f(t))$, $EC_{f(t)}t$ may be also denoted by
$EC_{\mathbb{P}}t$ and called the central exponential associated to
$\mathbb{P}$. Recalling that
$x^{[n]}\sim(1,l(t)=e^{\frac{t}{2}}-e^{-\frac{t}{2}})$, we observe that
$EC_{f(t)}t=l(\bar{f}(t))$.
As an example, consider the delta series $f(t)=\frac{1}{\lambda}(e^{\lambda
t}-1)$. Here $\bar{f}(t)=\frac{1}{\lambda}\log(1+\lambda t)$. Thus we see that

(5.3) $\displaystyle
LC_{f(t)}t=\frac{1}{\lambda}\bigg{(}\bigg{(}\frac{t+\sqrt{t^{2}+4}}{2}\bigg{)}^{2\lambda}-1\bigg{)}=\log_{\lambda}\bigg{(}\frac{t+\sqrt{t^{2}+4}}{2}\bigg{)}^{2},$
(5.4) $\displaystyle EC_{f(t)}t=(1+\lambda t)^{\frac{1}{2\lambda}}-(1+\lambda
t)^{-\frac{1}{2\lambda}}=e_{\lambda}^{\frac{1}{2}}(t)-e_{\lambda}^{-\frac{1}{2}}(t).$

Here we note that, in many of the papers on degenerate special polynomials and
numbers, $\frac{1}{\lambda}\big{(}t^{\lambda}-1\big{)}$ is called the
degenerate logarithm and denoted by $\log_{\lambda}t$, and $(1+\lambda
t)^{\frac{1}{\lambda}}$ is called the degenerate exponential and denoted by
$e_{\lambda}(t)$.

##  6\. Examples on central factorial numbers of the first kind associated
with sequences of polynomials

(a) Let $\mathbb{P}=\left\\{x^{n}\right\\}$. Then $x^{n}\sim(1,t)$, and
$T_{2}(n,k;\mathbb{P})=T_{2}(n,k)$. As
$x^{[n]}=\sum_{k=0}^{n}T_{1}(n,k;\mathbb{P})x^{k}$, we have (see [7,30,31])

$\displaystyle
T_{1}(n,k;\mathbb{P})=T_{1}(n,k),\quad\sum_{n=k}^{\infty}T_{1}(n,k)\frac{t^{n}}{n!}=\frac{1}{k!}\bigg{(}2\log\bigg{(}\frac{t+\sqrt{t^{2}+4}}{2}\bigg{)}\bigg{)}^{k},$

(see (3.1)). Now, by Theorem 5.3, we get

$\displaystyle\sum_{k=l}^{n}T_{1}(n,k)T_{2}(k,l)=\delta_{n,l},\,\,\sum_{k=l}^{n}T_{2}(n,k)T_{1}(k,l)=\delta_{n,l},\,\,(0\leq
l\leq n),$ $\displaystyle
a_{n}=\sum_{k=0}^{n}T_{2}(n,k)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}T_{1}(n,k)a_{k},$ $\displaystyle
a_{n}=\sum_{k=n}^{m}T_{2}(k,n)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}T_{1}(k,n)a_{k}.$

(b) Let $\mathbb{P}=\left\\{(x)_{n,\lambda}\right\\}$ be the sequence of
generalized falling factorials. Then
$(x)_{n,\lambda}\sim\big{(}1,f(t)=\frac{1}{\lambda}(e^{\lambda t}-1)\big{)}$,
and $T_{2}(n,k;\mathbb{P})=T_{2,\lambda}(n,k)$. As
$x^{[n]}=\sum_{k=0}^{n}T_{1}(n,k;\mathbb{P})(x)_{k,\lambda}$, we have (see
[13,14,16,18,23-25])

$\displaystyle
T_{1}(n,k;\mathbb{P})=T_{1,\lambda}(n,k),\quad\sum_{n=k}^{\infty}T_{1,\lambda}(n,k)\frac{t^{n}}{n!}=\frac{1}{k!}\bigg{(}\log_{\lambda}\bigg{(}\frac{t+\sqrt{t^{2}+4}}{2}\bigg{)}^{2}\bigg{)}^{k},$

(see (3.3)). Now, by Theorem 5.3, we get

$\displaystyle\sum_{k=l}^{n}T_{1,\lambda}(n,k)T_{2,\lambda}(k,l)=\delta_{n,l},\,\,\sum_{k=l}^{n}T_{2,\lambda}(n,k)T_{1,\lambda}(k,l)=\delta_{n,l},\,\,(0\leq
l\leq n),$ $\displaystyle
a_{n}=\sum_{k=0}^{n}T_{2,\lambda}(n,k)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}T_{1,\lambda}(n,k)a_{k},$ $\displaystyle
a_{n}=\sum_{k=n}^{m}T_{2,\lambda}(k,n)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}T_{1,\lambda}(k,n)a_{k}.$

(c) Let $\mathbb{P}=\left\\{\langle{x\rangle}_{n}\right\\}$ be the sequence of
rising factorials. Then $\langle{x\rangle}_{n}\sim(1,1-e^{-t})$, and
$T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}(-1)^{n-l}S_{1}(n,l)T_{2}(l,k)$ (see
[7,15,30,31]). In addition, by Theorem 5.1, we have

$\displaystyle T_{1}(n,k;\mathbb{P})$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)(-1)^{k}\big{\langle}{\frac{1}{k!}(e^{-t}-1)^{k}|x^{l}\big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)(-1)^{k}\sum_{j=k}^{l}S_{2}(j,k)(-1)^{j}\frac{1}{j!}\langle{t^{j}|x^{l}\rangle}$
$\displaystyle=\sum_{l=k}^{n}(-1)^{l-k}T_{1}(n,l)S_{2}(l,k).$

Thus $T_{1}(n,k;\mathbb{P})=\sum_{l=k}^{n}(-1)^{l-k}T_{1}(n,l)S_{2}(l,k)$.
Hence, by Theorem 5.3, we have the following identities given by:

$\displaystyle\sum_{k=l}^{n}\sum_{m=k}^{n}\sum_{j=l}^{k}T_{1}(n,m)(-1)^{m-k}S_{2}(m,k)(-1)^{k-j}S_{1}(k,j)T_{2}(j,l)=\delta_{n,l},$
$\displaystyle\sum_{k=l}^{n}\sum_{j=k}^{n}\sum_{m=l}^{k}(-1)^{n-j}S_{1}(n,j)T_{2}(j,k)T_{1}(k,m)(-1)^{m-l}S_{2}(m,l)=\delta_{n,l},$
$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}(-1)^{n-l}S_{1}(n,l)T_{2}(l,k)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}(-1)^{l-k}T_{1}(n,l)S_{2}(l,k)a_{k},$
$\displaystyle
a_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}(-1)^{k-l}S_{1}(k,l)T_{2}(l,n)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}(-1)^{l-n}T_{1}(k,l)S_{2}(l,n)a_{k}.$

(d) Let $\mathbb{P}=\left\\{\langle{x\rangle}_{n,\lambda}\right\\}$ be the
sequence of generalized rising factorials. Then
$\langle{x\rangle}_{n,\lambda}\sim\big{(}1,\frac{1}{\lambda}(1-e^{-\lambda
t})\big{)}$, and
$T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}(-\lambda)^{n-l}S_{1}(n,l)T_{2}(l,k)$
(see [18]). By Theorem 5.1, we have

$\displaystyle T_{1}(n,k;\mathbb{P})$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\frac{1}{(-\lambda)^{k}}\big{\langle}{\frac{1}{k!}(e^{-\lambda
t}-1)^{k}|x^{l}\big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\frac{1}{(-\lambda)^{k}}\sum_{j=k}^{l}S_{2}(j,k)\frac{1}{j!}(-\lambda)^{j}\langle{t^{j}|x^{l}\rangle}$
$\displaystyle=\sum_{l=k}^{n}(-\lambda)^{l-k}T_{1}(n,l)S_{2}(l,k).$

Thus
$T_{1}(n,k;\mathbb{P})=\sum_{l=k}^{n}(-\lambda)^{l-k}T_{1}(n,l)S_{2}(l,k)$,
(see [14]). Hence, by Theorem 5.3, we get

$\displaystyle\sum_{k=l}^{n}\sum_{m=k}^{n}\sum_{j=l}^{k}T_{1}(n,m)(-\lambda)^{m-k}S_{2}(m,k)(-\lambda)^{k-j}S_{1}(k,j)T_{2}(j,l)=\delta_{n,l},$
$\displaystyle\sum_{k=l}^{n}\sum_{j=k}^{n}\sum_{m=l}^{k}(-\lambda)^{n-j}S_{1}(n,j)T_{2}(j,k)T_{1}(k,m)(-\lambda)^{m-l}S_{2}(m,l)=\delta_{n,l},$
$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}(-\lambda)^{n-l}S_{1}(n,l)T_{2}(l,k)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}(-\lambda)^{l-k}T_{1}(n,l)S_{2}(l,k)a_{k},$
$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=n}^{k}(-\lambda)^{k-l}S_{1}(k,l)T_{2}(l,n)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}\sum_{l=n}^{k}(-\lambda)^{l-n}T_{1}(k,l)S_{2}(l,n)a_{k}.$

(e) Let $\mathbb{P_{1}}=\left\\{TLB_{n,1}(x)\right\\}_{n=0}^{\infty}$ be the
sequence of central factorial Lah Bell polynomials of the first kind. Then we
saw that $T_{2}(n,k;\mathbb{P_{1}})=L_{2}^{(c)}(n,k)$, where
$L_{2}^{(c)}(n,k)$ are the central Lah numbers of the second kind (see [19]).
We note from (4.6) that

$\displaystyle\sum_{n=0}^{\infty}x^{[n]}\frac{t^{n}}{n!}$
$\displaystyle=e^{x\bar{f}(t)}=\sum_{k=0}^{\infty}TLB_{k,1}(x)\frac{1}{k!}(\bar{\alpha}(t))^{k}$
$\displaystyle=\sum_{k=0}^{\infty}TLB_{k,1}(x)\sum_{n=k}^{\infty}L_{1}^{(c)}(n,k)\frac{t^{n}}{n!}$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{k=0}^{n}L_{1}^{(c)}(n,k)TLB_{k,1}(x)\frac{t^{n}}{n!}.$

Thus we have shown that

$\displaystyle x^{[n]}=\sum_{k=0}^{n}L_{1}^{(c)}(n,k)TLB_{k,1}(x),$

which verifies that $T_{1}(n,k;\mathbb{P_{1}})=L_{1}^{(c)}(n,k)$. Now, by
Theorem 5.3, we get

$\displaystyle\sum_{k=l}^{n}L_{1}^{(c)}(n,k)L_{2}^{(c)}(k,l)=\delta_{n,l},\,\,\sum_{k=l}^{n}L_{2}^{(c)}(n,k)L_{1}^{(c)}(k,l)=\delta_{n,l},\,\,(0\leq
l\leq n),$ $\displaystyle
a_{n}=\sum_{k=0}^{n}L_{2}^{(c)}(n,k)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}L_{1}^{(c)}(n,k)a_{k},$ $\displaystyle
a_{n}=\sum_{k=n}^{m}L_{2}^{(c)}(k,n)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}L_{1}^{(c)}(k,n)a_{k}.$

Now, we let $\mathbb{P_{2}}=\left\\{TLB_{n,2}(x)\right\\}_{n=0}^{\infty}$ (see
(4.8)). Then we saw that
$T_{2}(n,k;\mathbb{P_{2}})=\sum_{l=k}^{n}TL_{2}(n,l)T_{2}(l,k)$, and
$TLB_{n,2}(x)\sim(1,\bar{f}(\alpha(t))$. Recalling from (4.5) that
$\frac{1}{k!}\big{(}\bar{f}(\alpha(t))\big{)}^{k}=\sum_{n=k}^{\infty}TL_{1}(n,k)\frac{t^{n}}{n!}$,
from Theorem 5.1 we have

$\displaystyle T_{1}(n,k;\mathbb{P_{2}})$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\big{\langle}{\frac{1}{k!}\big{(}\bar{f}(\alpha(t))\big{)}^{k}\big{|}x^{l}\big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\big{\langle}{\sum_{m=k}^{\infty}TL_{1}(m,k)\frac{t^{m}}{m!}\big{|}x^{l}\big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\sum_{m=k}^{l}TL_{1}(m,k)\frac{1}{m!}\langle{t^{m}|x^{l}\rangle}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)TL_{1}(l,k).$

Thus we have shown that
$T_{1}(n,k;\mathbb{P_{2}})=\sum_{l=k}^{n}T_{1}(n,l)TL_{1}(l,k)$. Now, by
Theorem 5.3 we obtain

$\displaystyle\sum_{k=l}^{n}\sum_{m=k}^{n}\sum_{j=l}^{k}T_{1}(n,m)TL_{1}(m,k)TL_{2}(k,j)T_{2}(j,l)=\delta_{n,l},$
$\displaystyle\sum_{k=l}^{n}\sum_{j=k}^{n}\sum_{m=l}^{k}TL_{2}(n,j)T_{2}(j,k)T_{1}(k,m)TL_{1}(m,l)=\delta_{n,l},\,\,(0\leq
l\leq n),$ $\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}TL_{2}(n,l)T_{2}(l,k)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}T_{1}(n,l)TL_{1}(l,k)a_{k},$ $\displaystyle
a_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}TL_{2}(k,l)T_{2}(l,n)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}T_{1}(k,l)TL_{1}(l,n)a_{k}.$

(f) Let $\mathbb{P}=\left\\{\mathrm{Bel}_{n}^{(c)}(x)\right\\}$ be the
sequence of central Bell polynomials. Then
$\mathrm{Bel}_{n}^{(c)}(x)\sim\Big{(}1,f(t)=2\log\big{(}\frac{t+\sqrt{t^{2}+4}}{2}\big{)}\Big{)}$,
and $T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}T_{2}(n,l)T_{2}(l,k)$ (see [22]).
Now, by Theorem 5.1 and (3.1), we get

$\displaystyle T_{1}(n,k;\mathbb{P})$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\big{\langle}{\frac{1}{k!}\Big{(}2\log\Big{(}\frac{t+\sqrt{t^{2}+4}}{2}\Big{)}\Big{)}^{k}|x^{l}\big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\sum_{j=k}^{l}T_{1}(j,k)\frac{1}{j!}\langle{t^{j}|x^{l}\rangle}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)T_{1}(l,k).$

Therefore $T_{1}(n,k;\mathbb{P})=\sum_{l=k}^{n}T_{1}(n,l)T_{1}(l,k)$ (see
[15]). From Theorem 5.3, we have:

$\displaystyle\sum_{k=l}^{n}\sum_{m=k}^{n}\sum_{j=l}^{k}T_{1}(n,m)T_{1}(m,k)T_{2}(k,j)T_{2}(j,l)=\delta_{n,l},$
$\displaystyle\sum_{k=l}^{n}\sum_{j=k}^{n}\sum_{m=l}^{k}T_{2}(n,j)T_{2}(j,k)T_{1}(k,m)T_{1}(m,l)=\delta_{n,l},$
$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}T_{2}(n,l)T_{2}(l,k)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}T_{1}(n,l)T_{1}(l,k)a_{k},$ $\displaystyle
a_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}T_{2}(k,l)T_{2}(l,n)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}T_{1}(k,l)T_{1}(l,n)a_{k}.$

(g) Let $\mathbb{P}=\left\\{\mathrm{Bel}_{n,\lambda}^{(c)}(x)\right\\}$ be the
sequence of degenerate central Bell polynomials. Then
$\mathrm{Bel}_{n,\lambda}^{(c)}(x)\sim\Big{(}1,f(t)=\log_{\lambda}\Big{(}\frac{t+\sqrt{t^{2}+4}}{2}\Big{)}^{2}\Big{)},$
and $T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}T_{2,\lambda}(n,l)T_{2}(l,k)$ (see
[23]). Now, by Theorem 5.1 and (3.3), we have

$\displaystyle T_{1}(n,k;\mathbb{P})$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\big{\langle}{\frac{1}{k!}\Big{(}\log_{\lambda}\Big{(}\frac{t+\sqrt{t^{2}+4}}{2}\Big{)}^{2}\Big{)}^{k}|x^{l}\big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\sum_{j=k}^{l}T_{1,\lambda}(j,k)\frac{1}{j!}\langle{t^{j}|x^{l}\rangle}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)T_{1,\lambda}(l,k).$

Thus we have
$T_{1}(n,k;\mathbb{P})=\sum_{l=k}^{n}T_{1}(n,l)T_{1,\lambda}(l,k)$ (see [16]).
From Theorem 5.3, we have:

$\displaystyle\sum_{k=l}^{n}\sum_{m=k}^{n}\sum_{j=l}^{k}T_{1}(n,m)T_{1,\lambda}(m,k)T_{2,\lambda}(k,j)T_{2}(j,l)=\delta_{n,l},$
$\displaystyle\sum_{k=l}^{n}\sum_{j=k}^{n}\sum_{m=l}^{k}T_{2,\lambda}(n,j)T_{2}(j,k)T_{1}(k,m)T_{1,\lambda}(m,l)=\delta_{n,l},$
$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}T_{2,\lambda}(n,l)T_{2}(l,k)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}T_{1}(n,l)T_{1,\lambda}(l,k)a_{k},$
$\displaystyle
a_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}T_{2,\lambda}(k,l)T_{2}(l,n)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}T_{1}(k,l)T_{1,\lambda}(l,n)a_{k}.$

(h) Let $x^{[n,\lambda]}$ be the sequence of polynomials defined by
$x^{[0,\lambda]}=1,\quad
x^{[n,\lambda]}=x\big{(}x+(\frac{1}{2}n-1)\lambda\big{)}_{n-1,\lambda},\,\,(n\geq
1)$. Then $x^{[n,\lambda]}\sim\big{(}1,f(t)=\frac{1}{\lambda}(e^{\frac{\lambda
t}{2}}-e^{-\frac{\lambda t}{2}})\big{)}$, and
$T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}R_{1,\lambda}(n,l)T_{2}(l,k)$, where
$R_{1,\lambda}(n,k)$ are defined by
$x^{[n,\lambda]}=\sum_{k=0}^{n}R_{1,\lambda}(n,k)x^{k}$. Here we recall that

$\displaystyle\sum_{n=k}^{\infty}R_{2,\lambda}(n,k)\frac{t^{n}}{n!}=\frac{1}{k!}\Big{(}\frac{1}{\lambda}\big{(}e^{\frac{\lambda
t}{2}}-e^{-\frac{\lambda t}{2}}\big{)}\Big{)}^{k}.$

Let $\mathbb{P}=\left\\{x^{[n,\lambda]}\right\\}$. Now, by Theorem 5.1, we
have

$\displaystyle T_{1}(n,k;\mathbb{P})$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\big{\langle}{\frac{1}{k!}\big{(}\frac{1}{\lambda}\big{(}e^{\frac{\lambda
t}{2}}-e^{-\frac{\lambda t}{2}}\big{)}\big{)}^{k}|x^{l}\big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\sum_{j=k}^{l}R_{2,\lambda}(j,k)\frac{1}{j!}\langle{t^{j}|x^{l}\rangle}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)R_{2,\lambda}(l,k).$

Hence $T_{1}(n,k;\mathbb{P})=\sum_{l=k}^{n}T_{1}(n,l)R_{2,\lambda}(l,k)$. From
Theorem 5.3, we get

$\displaystyle\sum_{k=l}^{n}\sum_{m=k}^{n}\sum_{j=l}^{k}T_{1}(n,m)R_{2,\lambda}(m,k)R_{1,\lambda}(k,j)T_{2}(j,l)=\delta_{n,l},$
$\displaystyle\sum_{k=l}^{n}\sum_{j=k}^{n}\sum_{m=l}^{k}R_{1,\lambda}(n,j)T_{2}(j,k)T_{1}(k,m)R_{2,\lambda}(m,l)=\delta_{n,l},$
$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}R_{1,\lambda}(n,l)T_{2}(l,k)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}T_{1}(n,l)R_{2,\lambda}(l,k)a_{k},$
$\displaystyle
a_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}R_{1,\lambda}(k,l)T_{2}(l,n)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}T_{1}(k,l)R_{2,\lambda}(l,n)a_{k}.$

(i) Let $\mathbb{P}=\left\\{B_{n}^{L}(x)\right\\}$ be the sequence of Lah-Bell
polynomials. Then $B_{n}^{L}(x)\sim(1,\frac{t}{1+t})$, and
$T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}L(n,l)T_{2}(l,k)$ (see [7,15,30,31]). By
Theorem 5.1, we see that

$\displaystyle T_{1}(n,k;\mathbb{P})$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)(-1)^{k}\big{\langle}{\frac{1}{k!}\Big{(}\frac{-t}{1-(-t)}\Big{)}^{k}|x^{l}\big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)(-1)^{k}\sum_{j=k}^{l}L(j,k)(-1)^{j}\frac{1}{j!}\langle{t^{j}|x^{l}\rangle}$
$\displaystyle=\sum_{l=k}^{n}(-1)^{l-k}T_{1}(n,l)L(l,k).$

Therefore we have
$T_{1}(n,k;\mathbb{P})=\sum_{l=k}^{n}(-1)^{l-k}T_{1}(n,l)L(l,k)$ (see [11]).
Now, from Theorem 5.3, we obtain

$\displaystyle\sum_{k=l}^{n}\sum_{m=k}^{n}\sum_{j=l}^{k}(-1)^{m-k}T_{1}(n,m)L(m,k)L(k,j)T_{2}(j,l)=\delta_{n,l},$
$\displaystyle\sum_{k=l}^{n}\sum_{j=k}^{n}\sum_{m=l}^{k}(-1)^{m-l}L(n,j)T_{2}(j,k)T_{1}(k,m)L(m,l)=\delta_{n,l},$
$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}L(n,l)T_{2}(l,k)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}(-1)^{l-k}T_{1}(n,l)L(l,k)a_{k},$
$\displaystyle
a_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}L(k,l)T_{2}(l,n)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=n}^{l}\sum_{l=n}^{k}(-1)^{l-n}T_{1}(k,l)L(l,n)a_{k}.$

(j) Let $\mathbb{P}=\left\\{B_{n,\lambda}^{L}(x)\right\\}$ be the sequence of
degenerate Lah-Bell polynomials. Then
$B_{n,\lambda}^{L}(x)\sim(1,\frac{e^{\lambda t}-1}{\lambda+e^{\lambda t}-1})$,
and
$T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}\sum_{m=l}^{n}\lambda^{m-l}L(n,m)S_{1}(m,l)T_{2}(l,k).$
Now, by Theorem 5.1, we get

$\displaystyle T_{1}(n,k;\mathbb{P})$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)(-1)^{k}\Big{\langle}{\frac{1}{k!}\Big{(}\frac{-\frac{1}{\lambda}(e^{\lambda
t}-1)}{1-(-\frac{1}{\lambda}(e^{\lambda
t}-1))}\Big{)}^{k}|x^{l}\Big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)(-1)^{k}\sum_{m=k}^{l}L(m,k)\big{(}-\frac{1}{\lambda}\big{)}^{m}\big{\langle}{\frac{1}{m!}(e^{\lambda
t}-1)^{m}|x^{l}\big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)(-1)^{k}\sum_{m=k}^{l}L(m,k)\big{(}-\frac{1}{\lambda}\big{)}^{m}\sum_{j=m}^{l}S_{2}(j,m)\lambda^{j}\frac{1}{j!}\langle{t^{j}|x^{l}\rangle}$
$\displaystyle=\sum_{l=k}^{n}\sum_{m=k}^{l}(-1)^{m-k}\lambda^{l-m}T_{1}(n,l)S_{2}(l,m)L(m,k).$

Thus we have
$T_{1}(n,k;\mathbb{P})=\sum_{l=k}^{n}\sum_{m=k}^{l}(-1)^{m-k}\lambda^{l-m}T_{1}(n,l)S_{2}(l,m)L(m,k)$,
(see [14]). Now, by Theorem 5.3, we obtain

$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}\sum_{m=l}^{n}\lambda^{m-l}L(n,m)S_{1}(m,l)T_{2}(l,k)c_{k}$
$\displaystyle\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}\sum_{m=k}^{l}(-1)^{m-k}\lambda^{l-m}T_{1}(n,l)S_{2}(l,m)L(m,k)a_{k},$
$\displaystyle
a_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}\sum_{m=l}^{k}\lambda^{m-l}L(k,m)S_{1}(m,l)T_{2}(l,n)c_{k}$
$\displaystyle\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}\sum_{m=n}^{l}(-1)^{m-n}\lambda^{l-m}T_{1}(k,l)S_{2}(l,m)L(m,n)a_{k}.$

(k) Let $\mathbb{P}=\left\\{\mathrm{Bel}_{n}(x)\right\\}$ be the sequence of
Bell polynomials. Then $\mathrm{Bel}_{n}(x)\sim(1,\log(1+t))$, and
$T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}S_{2}(n,l)T_{2}(l,k)$ (see [7,30,31]). By
Theorem 5.1, we get

$\displaystyle T_{1}(n,k;\mathbb{P})$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\big{\langle}{\frac{1}{k!}(\log(1+t))^{k}|x^{l}\big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\sum_{j=k}^{l}S_{1}(j,k)\frac{1}{j!}\langle{t^{j}|x^{l}\rangle}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)S_{1}(l,k).$

Thus we see that $T_{1}(n,k;\mathbb{P})=\sum_{l=k}^{n}T_{1}(n,l)S_{1}(l,k)$
(see [4,22,23]). Now, by Theorem 5.3, we have

$\displaystyle\sum_{k=l}^{n}\sum_{m=k}^{n}\sum_{j=l}^{k}T_{1}(n,m)S_{1}(m,k)S_{2}(k,j)T_{2}(j,l)=\delta_{n,l},$
$\displaystyle\sum_{k=l}^{n}\sum_{j=k}^{n}\sum_{m=l}^{k}S_{2}(n,j)T_{2}(j,k)T_{1}(k,m)S_{1}(m,l)=\delta_{n,l},$
$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}S_{2}(n,l)T_{2}(l,k)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}T_{1}(n,l)S_{1}(l,k)a_{k},$ $\displaystyle
a_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}S_{2}(k,l)T_{2}(l,n)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}T_{1}(k,l)S_{1}(l,n)a_{k}.$

(l) Let $\mathbb{P_{1}}=\left\\{\mathrm{Bel}_{n,\lambda}(x)\right\\}$ be the
sequence of partially degenerate Bell polynomials. Then
$\mathrm{Bel}_{n,\lambda}(x)\sim(1,\log_{\lambda}(1+t))$, and
$T_{2}(n,k;\mathbb{P_{1}})=\sum_{l=k}^{n}S_{2,\lambda}(n,l)T_{2}(l,k)$ (see
[24]). By Theorem 5.1, we have

$\displaystyle T_{1}(n,k;\mathbb{P_{1}})$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\big{\langle}{\frac{1}{k!}(\log_{\lambda}(1+t))^{k}|x^{l}\big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\sum_{j=k}^{l}S_{1,\lambda}(j,k)\frac{1}{j!}\langle{t^{j}|x^{l}\rangle}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)S_{1,\lambda}(l,k).$

Hence we have
$T_{1}(n,k;\mathbb{P_{1}})=\sum_{l=k}^{n}T_{1}(n,l)S_{1,\lambda}(l,k)$. Now,
by Theorem 5.3, we obtain

$\displaystyle\sum_{k=l}^{n}\sum_{m=k}^{n}\sum_{j=l}^{k}T_{1}(n,m)S_{1,\lambda}(m,k)S_{2,\lambda}(k,j)T_{2}(j,l)=\delta_{n,l},$
$\displaystyle\sum_{k=l}^{n}\sum_{j=k}^{n}\sum_{m=l}^{k}S_{2,\lambda}(n,j)T_{2}(j,k)T_{1}(k,m)S_{1,\lambda}(m,l)=\delta_{n,l},$
$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}S_{2,\lambda}(n,l)T_{2}(l,k)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}T_{1}(n,l)S_{1,\lambda}(l,k)a_{k},$
$\displaystyle
a_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}S_{2,\lambda}(k,l)T_{2}(l,n)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}T_{1}(k,l)S_{1,\lambda}(l,n)a_{k}.$

Let $\mathbb{P_{2}}=\left\\{\phi_{n,\lambda}(x)\right\\}$ be the sequence of
fully degenerate Bell polynomials. Then $\phi_{n,\lambda}(x)$
$\sim\big{(}1,\log_{\lambda}(1+\frac{1}{\lambda}(e^{\lambda t}-1))\big{)}$,
and
$T_{2}(n,k;\mathbb{P_{2}})=\sum_{l=k}^{n}\sum_{m=l}^{n}\lambda^{m-l}S_{2,\lambda}(n,m)S_{1}(m,l)T_{2}(l,k)$
(see [25]). By Theorem 5.1, we obtain

$\displaystyle T_{1}(n,k;\mathbb{P_{2}})$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\big{\langle}{\frac{1}{k!}(\log_{\lambda}(1+\frac{1}{\lambda}(e^{\lambda
t}-1)))^{k}|x^{l}\rangle}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\sum_{m=k}^{l}S_{1,\lambda}(m,k)\frac{1}{\lambda^{m}}\langle{\frac{1}{m!}(e^{\lambda
t}-1)^{m}|x^{l}\rangle}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\sum_{m=k}^{l}S_{1,\lambda}(m,k)\frac{1}{\lambda^{m}}\sum_{j=m}^{l}S_{1}(j,m)\lambda^{j}\frac{1}{j!}\langle{t^{j}|x^{l}\rangle}$
$\displaystyle=\sum_{l=k}^{n}\sum_{m=k}^{l}\lambda^{l-m}T_{1}(n,l)S_{1}(l,m)S_{1,\lambda}(m,k).$

Therefore we get
$T_{1}(n,k;\mathbb{P_{2}})=\sum_{l=k}^{n}\sum_{m=k}^{l}\lambda^{l-m}T_{1}(n,l)S_{1}(l,m)S_{1,\lambda}(m,k)$
(see [18]). Now, by Theorem 5.3, we obtain

$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}\sum_{m=l}^{n}\lambda^{m-l}S_{2,\lambda}(n,m)S_{1}(m,l)T_{2}(l,k)c_{k}$
$\displaystyle\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}\sum_{m=k}^{l}\lambda^{l-m}T_{1}(n,l)S_{1}(l,m)S_{1,\lambda}(m,k)a_{k},$
$\displaystyle
a_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}\sum_{m=l}^{k}\lambda^{m-l}S_{2,\lambda}(k,m)S_{1}(m,l)T_{2}(l,n)c_{k}$
$\displaystyle\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}\sum_{m=n}^{l}\lambda^{l-m}T_{1}(k,l)S_{1}(l,m)S_{1,\lambda}(m,n)a_{k}.$

(m) Let $\mathbb{P}=\left\\{M_{n}(x)\right\\}$ be the sequence of Mittag-
Leffler polynomials. Then $M_{n}(x)\sim(1,f(t)=\frac{e^{t}-1}{e^{t}+1})$, and
$T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}\sum_{m=l}^{n}2^{m}L(n,m)S_{1}(m,l)T_{2}(l,k)$
(see [31]). Then, from Theorem 5.1, we have

$\displaystyle T_{1}(n,k;\mathbb{P})$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\big{\langle}{\frac{1}{k!}\Big{(}\frac{e^{t}-1}{e^{t}+1}\Big{)}^{k}|x^{l}\big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)(-1)^{k}\Big{\langle}{\frac{1}{k!}\Big{(}\frac{-\frac{1}{2}(e^{t}-1)}{1-(-\frac{1}{2}(e^{t}-1))}\Big{)}^{k}\Big{|}x^{l}\Big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)(-1)^{k}\sum_{m=k}^{l}L(m,k)(-\frac{1}{2})^{m}\big{\langle}{\frac{1}{m!}(e^{t}-1)^{m}|x^{l}\big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)(-1)^{k}\sum_{m=k}^{l}L(m,k)(-\frac{1}{2})^{m}\sum_{j=m}^{l}S_{2}(j,m)\frac{1}{j!}\langle{t^{j}|x^{l}\rangle}$
$\displaystyle=\sum_{l=k}^{n}\sum_{m=k}^{l}(-1)^{m-k}2^{-m}T_{1}(n,l)S_{2}(l,m)L(m,k).$

Hence we have
$T_{1}(n,k;\mathbb{P})=\sum_{l=k}^{n}\sum_{m=k}^{l}(-1)^{m-k}2^{-m}T_{1}(n,l)S_{2}(l,m)L(m,k)$.
By Theorem 5.3, we get

$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}\sum_{m=l}^{n}2^{m}L(n,m)S_{1}(m,l)T_{2}(l,k)c_{k}$
$\displaystyle\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}\sum_{m=k}^{l}(-1)^{m-k}2^{-m}T_{1}(n,l)S_{2}(l,m)L(m,k)a_{k},$
$\displaystyle
a_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}\sum_{m=l}^{k}2^{m}L(k,m)S_{1}(m,l)T_{2}(l,n)c_{k}$
$\displaystyle\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}\sum_{m=n}^{l}(-1)^{m-n}2^{-m}T_{1}(k,l)S_{2}(l,m)L(m,n)a_{k}.$

(n) Let $\mathbb{P}=\left\\{L_{n}(x)\right\\}$ be the sequence of Laguerre
polynomials of order -1. Then $L_{n}(x)\sim(1,f(t)=\frac{t}{t-1})$, and
$T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}(-1)^{l}L(n,l)T_{2}(l,k)$ (see [31]). By
Theorem 5.1, we see that

$\displaystyle T_{1}(n,k;\mathbb{P})$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\big{\langle}{\frac{1}{k!}\Big{(}\frac{t}{t-1}\Big{)}^{k}|x^{l}\big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)(-1)^{k}\big{\langle}{\frac{1}{k!}\Big{(}\frac{t}{1-t}\Big{)}^{k}|x^{l}\big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)(-1)^{k}\sum_{j=k}^{l}L(j,k)\frac{1}{j!}\langle{t^{j}|x^{l}\rangle}$
$\displaystyle=(-1)^{k}\sum_{l=k}^{n}T_{1}(n,l)L(l,k).$

Thus we have $T_{1}(n,k;\mathbb{P})=(-1)^{k}\sum_{l=k}^{n}T_{1}(n,l)L(l,k)$
(see [23]). Now, from Theorem 5.3, we get

$\displaystyle\sum_{k=l}^{n}\sum_{m=k}^{n}\sum_{j=l}^{k}(-1)^{k-j}T_{1}(n,m)L(m,k)L(k,j)T_{2}(j,l)=\delta_{n,l},$
$\displaystyle\sum_{k=l}^{n}\sum_{j=k}^{n}\sum_{m=l}^{k}(-1)^{j-l}L(n,j)T_{2}(j,k)T_{1}(k,m)L(m,l)=\delta_{n,l},$
$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}(-1)^{l}L(n,l)T_{2}(l,k)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}(-1)^{k}\sum_{l=k}^{n}T_{1}(n,l)L(l,k)a_{k},$
$\displaystyle
a_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}(-1)^{l}L(k,l)T_{2}(l,n)c_{k}\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}(-1)^{n}\sum_{l=n}^{k}T_{1}(k,l)L(l,n)a_{k}.$

(o) Let $\mathbb{P}=\left\\{B_{n}(x)\right\\}$ be the sequence of Bernoulli
polynomials. Then $B_{n}(x)\sim\big{(}\frac{e^{t}-1}{t},t\big{)}$, and
$T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}T_{2}(l,k)\binom{n}{l}B_{n-l}$ (see
[31]). By using (2.7) and Theorem 5.1, we see that

$\displaystyle T_{1}(n,k;\mathbb{P})$
$\displaystyle=\frac{1}{k!}\sum_{l=k}^{n}T_{1}(n,l)\Big{\langle}{t^{k}\Big{|}\frac{e^{t}-1}{t}x^{l}\Big{\rangle}}$
$\displaystyle=\frac{1}{k!}\sum_{l=k}^{n}T_{1}(n,l)\Big{\langle}{t^{k}\Big{|}\int_{x}^{x+1}u^{l}du\Big{\rangle}}$
$\displaystyle=\frac{1}{k!}\sum_{l=k}^{n}T_{1}(n,l)\frac{1}{l+1}(l+1)_{k}\Big{\langle}{1\Big{|}(x+1)^{l+1-k}-x^{l+1-k}\Big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}\frac{1}{l+1}\binom{l+1}{k}T_{1}(n,l).$

Thus we have
$T_{1}(n,k;\mathbb{P})=\sum_{l=k}^{n}\frac{1}{l+1}\binom{l+1}{k}T_{1}(n,l)$.
Now, by Theorem 5.3, we obtain

$\displaystyle\sum_{k=l}^{n}\sum_{m=k}^{n}\sum_{j=l}^{k}\frac{1}{m+1}\binom{k}{j}\binom{m+1}{k}T_{1}(n,m)T_{2}(j,l)B_{k-j}=\delta_{n,l},$
$\displaystyle\sum_{k=l}^{n}\sum_{j=k}^{n}\sum_{m=l}^{k}\frac{1}{m+1}\binom{n}{j}\binom{m+1}{l}T_{2}(j,k)T_{1}(k,m)B_{n-j}=\delta_{n,l},$
$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}T_{2}(l,k)\binom{n}{l}B_{n-l}c_{k}\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}\frac{1}{l+1}\binom{l+1}{k}T_{1}(n,l)a_{k},$
$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=n}^{k}T_{2}(l,n)\binom{k}{l}B_{k-l}c_{k}\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}\sum_{l=n}^{k}\frac{1}{l+1}\binom{l+1}{n}T_{1}(k,l)a_{k}.$

(p) Let $\mathbb{P}=\left\\{E_{n}(x)\right\\}$ be the sequence of Euler
polynomials. Then $E_{n}(x)\sim\big{(}\frac{e^{t}+1}{2},t\big{)}$, and
$T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}\binom{n}{l}T_{2}(l,k)E_{n-l}$ (see
[31]). By Theorem 5.1, we note that

$\displaystyle T_{1}(n,k;\mathbb{P})$
$\displaystyle=\frac{1}{2k!}\sum_{l=k}^{n}T_{1}(n,l)\big{\langle}{e^{t}+1\big{|}t^{k}x^{l}\big{\rangle}}=\frac{1}{2k!}\sum_{l=k}^{n}T_{1}(n,l)(l)_{k}\big{\langle}{e^{t}+1\big{|}x^{l-k}\rangle}$
$\displaystyle=\frac{1}{2k!}\sum_{l=k}^{n}T_{1}(n,l)(l)_{k}(1+\delta_{l,k})=\frac{1}{2}\sum_{l=k}^{n}\binom{l}{k}T_{1}(n,l)+\frac{1}{2}T_{1}(n,k).$

Thus we have
$T_{1}(n,k;\mathbb{P})=\frac{1}{2}\sum_{l=k}^{n}\binom{l}{k}T_{1}(n,l)+\frac{1}{2}T_{1}(n,k)$.
By Theorem 5.3, we obtain

$\displaystyle\sum_{k=l}^{n}\sum_{m=k}^{n}\sum_{j=l}^{k}\frac{1}{2}\binom{m}{k}\binom{k}{j}T_{1}(n,m)T_{2}(j,l)E_{k-j}$
$\displaystyle+\sum_{k=l}^{n}\sum_{j=l}^{k}\frac{1}{2}\binom{k}{j}T_{1}(n,k)T_{2}(j,l)E_{k-j}=\delta_{n,l},$
$\displaystyle\sum_{k=l}^{n}\sum_{j=k}^{n}\sum_{m=l}^{k}\frac{1}{2}\binom{m}{l}\binom{n}{j}T_{2}(j,k)T_{1}(k,m)E_{n-j}$
$\displaystyle+\sum_{k=l}^{n}\sum_{j=k}^{n}\frac{1}{2}\binom{n}{j}T_{2}(j,k)T_{1}(k,l)E_{n-j}=\delta_{n,l},$
$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}\binom{n}{l}T_{2}(l,k)E_{n-l}c_{k}\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}\left\\{\frac{1}{2}\sum_{l=k}^{n}\binom{l}{k}T_{1}(n,l)+\frac{1}{2}T_{1}(n,k)\right\\}a_{k},$
$\displaystyle
a_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}\binom{k}{l}T_{2}(l,n)E_{k-l}c_{k}\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}\left\\{\frac{1}{2}\sum_{l=n}^{k}\binom{l}{n}T_{1}(k,l)+\frac{1}{2}T_{1}(k,n)\right\\}a_{k}.$

(q) Let $\mathbb{P}=\left\\{(rx+s)_{n}\right\\}$, with $r\neq 0$. Then
$(rx+s)_{n}\sim\big{(}e^{-\frac{s}{r}t},e^{\frac{t}{r}}-1\big{)}$, and
$T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}\sum_{m=l}^{n}\binom{n}{m}r^{l}(s)_{n-m}S_{1}(m,l)T_{2}(l,k)$
(see [26]). By (2.7), (2.18) and Theorem 5.1, we see that

$\displaystyle T_{1}(n,k;\mathbb{P})$
$\displaystyle=\frac{1}{k!}\sum_{l=k}^{n}T_{1}(n,l)\big{\langle}{e^{-\frac{s}{r}t}\big{(}e^{\frac{t}{r}}-1\big{)}^{k}\big{|}x^{l}\big{\rangle}}$
$\displaystyle=\frac{1}{k!}\sum_{l=k}^{n}T_{1}(n,l)\Big{\langle}{(e^{t}-1)^{k}\Big{|}\Big{(}\frac{x-s}{r}\Big{)}^{l}\Big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\frac{1}{r^{l}}\frac{1}{k!}\big{\langle}{(e^{t}-1)^{k}|(x-s)^{l}\big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\frac{1}{r^{l}}\sum_{m=k}^{l}S_{2}(m,k)\frac{1}{m!}\big{\langle}{t^{m}|(x-s)^{l}\big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}\sum_{m=k}^{l}\binom{l}{m}r^{-l}(-s)^{l-m}T_{1}(n,l)S_{2}(m,k).$

Thus we have
$T_{1}(n,k;\mathbb{P})=\sum_{l=k}^{n}\sum_{m=k}^{l}\binom{l}{m}r^{-l}(-s)^{l-m}T_{1}(n,l)S_{2}(m,k)$.
From Theorem 5.3, we obtain

$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}\sum_{m=l}^{n}\binom{n}{m}r^{l}(s)_{n-m}S_{1}(m,l)T_{2}(l,k)c_{k}$
$\displaystyle\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}\sum_{m=k}^{l}\binom{l}{m}r^{-l}(-s)^{l-m}T_{1}(n,l)S_{2}(m,k)a_{k},$
$\displaystyle
a_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}\sum_{m=l}^{k}\binom{k}{m}r^{l}(s)_{k-m}S_{1}(m,l)T_{2}(l,n)c_{k}$
$\displaystyle\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}\sum_{m=n}^{l}\binom{l}{m}r^{-l}(-s)^{l-m}T_{1}(k,l)S_{2}(m,n)a_{k}.$

(r) Let $\mathbb{P}=\left\\{b_{n}(x)\right\\}$ be the sequence of Bernoulli
polynomials of the second kind. Then
$b_{n}(x)\sim\big{(}\frac{t}{e^{t}-1},e^{t}-1\big{)}$, and
$T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}\sum_{m=l}^{n}\binom{n}{m}b_{n-m}S_{1}(m,l)T_{2}(l,k)$,
where $b_{n}=b_{n}(0)$ are the Bernoulli numbers of the second (see [31]). By
using (2.10), (2.12) and Theorem 5.1 and noting
$B_{n}(x)\sim(\frac{e^{t}-1}{t},t)$, we have

$\displaystyle T_{1}(n,k;\mathbb{P})$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\Big{\langle}{\frac{1}{k!}(e^{t}-1)^{k}\Big{|}\frac{t}{e^{t}-1}x^{l}\Big{\rangle}}=\sum_{l=k}^{n}T_{1}(n,l)\Big{\langle}{\frac{1}{k!}(e^{t}-1)^{k}\Big{|}B_{l}(x)\Big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)\sum_{m=k}^{l}S_{2}(m,k)\frac{1}{m!}\big{\langle}{t^{m}|B_{l}(x)\rangle}=\sum_{l=k}^{n}\sum_{m=k}^{l}\binom{l}{m}B_{l-m}T_{1}(n,l)S_{2}(m,k),$

where $B_{l}$ are the Bernoulli numbers. Hence
$T_{1}(n,k;\mathbb{P})=\sum_{l=k}^{n}\sum_{m=k}^{l}\binom{l}{m}B_{l-m}T_{1}(n,l)S_{2}(m,k)$.
Now, by Theorem 5.3, we get

$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}\sum_{m=l}^{n}\binom{n}{m}b_{n-m}S_{1}(m,l)T_{2}(l,k)c_{k}$
$\displaystyle\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}\sum_{m=k}^{l}\binom{l}{m}B_{l-m}T_{1}(n,l)S_{2}(m,k)a_{k},$
$\displaystyle
a_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}\sum_{m=l}^{k}\binom{k}{m}b_{k-m}S_{1}(m,l)T_{2}(l,n)c_{k}$
$\displaystyle\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}\sum_{m=n}^{l}\binom{l}{m}B_{l-m}T_{1}(k,l)S_{2}(m,n)a_{k}.$

(s) Let $\mathbb{P}=\left\\{C_{n}(x;a)\right\\}$ be the sequence of Poisson-
Charlier polynomials. Then
$C_{n}(x;a)\sim\big{(}e^{a(e^{t}-1)},a(e^{t}-1)\big{)}$, with $a\neq 0$, and
$T_{2}(n,k;\mathbb{P})=\sum_{l=k}^{n}\sum_{m=l}^{n}\binom{n}{m}(-1)^{n-m}a^{-m}S_{1}(m,l)T_{2}(l,k)$
(see [31]). By Theorem 5.1, we get

$\displaystyle T_{1}(n,k;\mathbb{P})$
$\displaystyle=\frac{1}{k!}\sum_{l=k}^{n}T_{1}(n,l)a^{k}\langle{(e^{t}-1)^{k}|e^{a(e^{t}-1)}x^{l}\rangle}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)a^{k}\sum_{m=k}^{l}\binom{l}{m}\mathrm{Bel}_{l-m}(a)\big{\langle}{\frac{1}{k!}(e^{t}-1)^{k}|x^{m}\big{\rangle}}$
$\displaystyle=\sum_{l=k}^{n}T_{1}(n,l)a^{k}\sum_{m=k}^{l}\binom{l}{m}\mathrm{Bel}_{l-m}(a)\sum_{j=k}^{m}S_{2}(j,k)\frac{1}{j!}\langle{t^{j}|x^{m}\rangle}$
$\displaystyle=\sum_{l=k}^{n}\sum_{m=k}^{l}a^{k}\binom{l}{m}\mathrm{Bel}_{l-m}(a)T_{1}(n,l)S_{2}(m,k).$

where $\mathrm{Bel}_{n}(x)$ are the Bell polynomials given by
$e^{x(e^{t}-1)}=\sum_{n=0}^{\infty}\mathrm{Bel}_{n}(x)\frac{t^{n}}{n!}$.
Thus we have
$T_{1}(n,k;\mathbb{P})=\sum_{l=k}^{n}\sum_{m=0}^{l}a^{k}\binom{l}{m}\mathrm{Bel}_{l-m}(a)T_{1}(n,l)S_{2}(m,k)$.

Now, by Theorem 5.3, we obtain

$\displaystyle
a_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}\sum_{m=l}^{n}\binom{n}{m}(-1)^{n-m}a^{-m}S_{1}(m,l)T_{2}(l,k)c_{k}$
$\displaystyle\Longleftrightarrow
c_{n}=\sum_{k=0}^{n}\sum_{l=k}^{n}\sum_{m=0}^{l}a^{k}\binom{l}{m}\mathrm{Bel}_{l-m}(a)T_{1}(n,l)S_{2}(m,k)a_{k},$
$\displaystyle
a_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}\sum_{m=l}^{k}\binom{k}{m}(-1)^{k-m}a^{-m}S_{1}(m,l)T_{2}(l,n)c_{k}$
$\displaystyle\Longleftrightarrow
c_{n}=\sum_{k=n}^{m}\sum_{l=n}^{k}\sum_{m=0}^{l}a^{n}\binom{l}{m}\mathrm{Bel}_{l-m}(a)T_{1}(k,l)S_{2}(m,n)a_{k}.$

(t) Let $\mathbb{P}=\left\\{p_{n}(x)\right\\}$, with
$p_{n}(x)=\sum_{k=0}^{n}B_{k}(x)B_{n-k}(x)$. This is not a Sheffer sequence.
Here we would like to describe how to determine $T_{1}(n,k;\mathbb{P})$. For
this, we recall from [17] that

(6.1) $\displaystyle
p_{n}(x)=\frac{2}{n+2}\sum_{m=0}^{n-2}\binom{n+2}{m}B_{n-m}B_{m}(x)+(n+1)B_{n}(x).$

On the one hand, by (o) we have

(6.2) $\displaystyle x^{[n]}=\sum_{m=0}^{n}\gamma_{m}B_{m}(x),$

where $\gamma_{m}=\sum_{l=m}^{n}\frac{1}{l+1}\binom{l+1}{m}T_{1}(n,l)$.
On the other hand, by definition and from (6.1) we also have

$\displaystyle x^{[n]}$
$\displaystyle=\sum_{k=0}^{n}T_{1}(n,k;\mathbb{P})p_{k}(x)$
$\displaystyle=\sum_{k=2}^{n}T_{1}(n,k;\mathbb{P})\frac{2}{k+2}\sum_{m=0}^{k-2}\binom{k+2}{m}B_{k-m}B_{m}(x)$
(6.3)
$\displaystyle\quad\quad+\sum_{k=0}^{n}T_{1}(n,k;\mathbb{P})(k+1)B_{k}(x)$
$\displaystyle=\sum_{m=0}^{n-2}\sum_{k=m+2}^{n}\epsilon_{m,k}T_{1}(n,k;\mathbb{P})B_{m}(x)$
$\displaystyle\quad\quad+\sum_{m=0}^{n}(m+1)T_{1}(n,m;\mathbb{P})B_{m}(x),$

where $\epsilon_{m,k}=\frac{2}{k+2}\binom{k+2}{m}B_{k-m},\quad(0\leq m\leq
n-2,m+2\leq k\leq n)$.
By comparing (6.2) and (6.3), we have

(6.4)
$\displaystyle\gamma_{m}=(m+1)T_{1}(n,m;\mathbb{P})+\sum_{k=m+2}^{n}\epsilon_{m,k}T_{1}(n,k;\mathbb{P}),\quad(0\leq
m\leq n),$

where we understand that the sum is zero for $m=n-1$ and $m=n$. Let $\Gamma$
be the column vector consisting of $\gamma_{0},\gamma_{1},\dots,\gamma_{n}$,
and let $S$ be the column vector consisting of
$T_{1}(n,0;\mathbb{P}),T_{1}(n,1;\mathbb{P}),\dots,T_{1}(n,n;\mathbb{P})$.
Then, in matrix form, (6.4) can be written as $\Gamma=AS$, where $A$ is an
$(n+1)\times(n+1)$ upper triangular matrix with diagonal entries
$1,2,\dots,n+1$. Thus $S=A^{-1}\Gamma$.

##  7\. Conclusion

In this paper, we studied central factorial numbers of both kinds associated
with sequences of polynomilas by using umbral calculus techniques, which was
motivated by the observation that many important special numbers appear in the
expansions of some polynomials in terms of central factorials and vice versa.
Our results were illustrated by twenty examples both for the second kind and
for the first kind. In all cases of these examples, some interesting
orthogonality and inverse relations were obtained as immediate consequences of
our results. The novelty of this paper is that it is the first paper which
studies the central factorial numbers of both kinds associated with any
sequence of polynomials in a unified and systematic way with the help of
umbral calculus.
It is one of our future projects to study special numbers and polynomials by
making use of various tools, including combinatorial methods, generating
functions, umbral calculus, $p$-adic analysis, probability theory,
differential equations, analytic number theory, operator theory, special
functions and so on.

## References

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