In this section, we define poly-central factorial sequences and poly-central-Bell polynomials respectively by using the degenerate polylogarithm functions and give explicit expressions and recurrence formula of poly-central Bell polynomials.
First, we consider the poly-central factorial sequences \(x^{[n](k)}\), which are derived from the polyexponential function to be
$$ \begin{aligned} 1+ J_{k} \biggl(2x\log \biggl(\frac{t}{2}+\sqrt{1+ \frac{t^{2}}{4}} \biggr) \biggr) = \sum _{n=0}^{\infty }x^{[n](k)} \frac{t^{n}}{n!} \quad \text{and} \quad x^{[0](k)}=1. \end{aligned} $$
(21)
When \(k=1\), since \(J_{1}(x)=e^{x}-1\), we note that
$$ \begin{aligned} 1+ J_{1} \biggl(2x\log \biggl(\frac{t}{2}+\sqrt{1+ \frac{t^{2}}{4}} \biggr) \biggr) = \biggl( \frac{t}{2}+\sqrt{1+ \frac{t^{2}}{4}} \biggr)^{2x}= \sum _{n=0}^{\infty }x^{[n]} \frac{t^{n}}{n!}. \end{aligned} $$
(22)
Therefore, by (22), we have \(x^{[n](1)}=x^{[n]}\).
Second, we define the poly-central-Bell polynomials \(B_{n}^{(c,k)}(x)\), which arise from the polyexponential function to be
$$ \begin{aligned} 1+ J_{k}\bigl(x \bigl(e^{\frac{t}{2}}-e^{-\frac{t}{2}}\bigr)\bigr) = \sum _{n=0}^{\infty }B_{n}^{(c,k)}(x) \frac{t^{n}}{n!} \quad \text{and} \quad B_{0}^{(c,k)}(x)=1. \end{aligned} $$
(23)
When \(k=1\), since \(J_{1}(x)=e^{x}-1\), we note that
$$ \begin{aligned} 1+ J_{1}\bigl(x \bigl(e^{\frac{t}{2}}-e^{-\frac{t}{2}}\bigr)\bigr) &= 1+ e^{x (e^{ \frac{t}{2}}-e^{-\frac{t}{2}})}-1= \sum _{n=0}^{\infty }B_{n}^{(c)}(x) \frac{t^{n}}{n!}. \end{aligned} $$
(24)
By (24), we have \(B_{n}^{(c,1)}(x) = B_{n}^{(c)}(x)\).
First, we observe relations of poly-falling factorial sequences and powers of x.
Theorem 1
For \(k \in \mathbb{Z}\) and \(n\geq 1\), we have
$$ \begin{aligned} x^{[n](k)} = \sum_{l=1}^{n} \frac{1}{l^{k-1}} t(n,l)x^{l}, \end{aligned} $$
where \(t(n,l)\) is the central factorial numbers of the first kind.
Proof
By (4) and (21), we observe that
$$ \begin{aligned} 1+J_{k} \biggl(2x\log \biggl( \frac{t}{2}+\sqrt{1+ \frac{t^{2}}{4}} \biggr) \biggr) &=1+ \sum _{l=1}^{\infty }\frac{(2x)^{l} (\log (\frac{t}{2}+\sqrt{1+\frac{t^{2}}{4}}))^{l}}{(l-1)! l^{k}} \\ &= 1+\sum_{l=1}^{\infty }\frac{x^{l}}{l^{k-1}} \frac{ 1}{l!} \biggl(2 \log \biggl(\frac{t}{2}+\sqrt{1+ \frac{t^{2}}{4}} \biggr) \biggr)^{l} \\ &= 1+\sum_{l=1}^{\infty }\frac{x^{l}}{l^{k-1}}\sum _{n=l}^{\infty }t(n,l) \frac{t^{n}}{n!} \\ &= 1+\sum_{n=1}^{\infty } \Biggl(\sum _{l=1}^{n} \frac{1}{l^{k-1}} t(n,l)x^{l} \Biggr)\frac{t^{n}}{n!}. \end{aligned} $$
(25)
Combining (21) with (25), we get the desired result. □
In Theorem 1, when \(k=1\), we note that
$$ \begin{aligned} x^{[n]}=\sum_{l=0}^{n} t(n,l)x^{l} . \end{aligned} $$
Theorem 2
For \(k \in \mathbb{Z}\) and \(n\geq 1\), we have
$$ \begin{aligned} x^{n}= n^{k-1}\sum _{l=0}^{n} T(n,l)x^{[l](k)}. \end{aligned} $$
Proof
By replacing t with \(e^{\frac{t}{2}}-e^{-\frac{t}{2}}\) in (21), from (18), the left-hand side of (21) is
$$ \begin{aligned} 1+ J_{k}(xt) = 1+\sum _{n=1}^{\infty }\frac{x^{n} t^{n}}{(n-1)! n^{k}} = 1+\sum _{n=1}^{\infty }\frac{x^{n}}{n^{k-1}} \frac{t^{n}}{n!}. \end{aligned} $$
(26)
On the other hand, from (6), the right-hand side of (21) is
$$ \begin{aligned} \sum_{m=0}^{\infty }x^{[m](k)} \frac{(e^{\frac{t}{2}}-e^{-\frac{t}{2}})^{m}}{m!} &= \sum_{m=0}^{\infty }x^{[m](k)} \sum_{n=m}^{\infty }T(n,m)\frac{t^{n}}{n!} \\ &= \sum_{n=0}^{\infty }\sum _{m=0}^{n} x^{[m](k)} T(n,m) \frac{t^{n}}{n!}. \end{aligned} $$
(27)
Comparing with the coefficients of (26) and (27), we have the desired result. □
In Theorem 2, when \(k=1\), we note that
$$ \begin{aligned} x^{n} =\sum_{l=0}^{n} t(n,l)x^{[l](k)}. \end{aligned} $$
Theorem 3
For \(k \in \mathbb{Z}\) and \(n\geq 1\), we have
$$ \begin{aligned} B_{n}^{(c,k)}(x) = \sum _{l=1}^{n} \frac{1}{l^{k-1}}T(n,l)x^{l}. \end{aligned} $$
Proof
From (6) and (23), we observe that
$$ \begin{aligned} 1+J_{k}\bigl(x \bigl(e^{\frac{t}{2}}-e^{-\frac{t}{2}}\bigr)\bigr) &=1+ \sum _{l=1}^{\infty }\frac{x^{l} (e^{\frac{t}{2}}-e^{-\frac{t}{2}})}{(l-1)!l^{k}} \\ &=1+ \sum_{l=1}^{\infty }\frac{x^{l}}{l^{k-1}} \frac{(e^{\frac{t}{2}}-e^{-\frac{t}{2}})^{l}}{l!} \\ &= 1+\sum_{l=1}^{\infty }\frac{x^{l}}{l^{k-1}} \sum _{n=l}^{\infty }T(n,l) \frac{t^{n}}{n!} \\ &= 1+\sum_{n=1}^{\infty } \Biggl(\sum _{l=1}^{n} \frac{1}{l^{k-1}}T(n,l) x^{l} \Biggr) \frac{t^{n}}{n!}. \end{aligned} $$
(28)
Combining (23) with (28), we get
$$ B_{n}^{(c,k)}(x) = \sum _{l=1}^{n} \frac{1}{l^{k-1}}T(n,l)x^{l} \quad (n \geq 1). $$
□
In Theorem 3, when \(k=1\), we note that
$$ \begin{aligned} B_{n}^{(c)}(x)=\sum _{l=0}^{n} T(n,l)x^{l} . \end{aligned} $$
Theorem 4
For \(k \in \mathbb{Z}\) and \(n\geq 1\), we have
$$ \begin{aligned} B_{n}^{(c,k)}(x) = \sum _{m=1}^{n} \sum_{l=1}^{m} \binom{n}{m} \biggl(-\frac{1}{2} \biggr)^{n-m}l^{n-m-k+1} S_{2}(m,l)x^{l}. \end{aligned} $$
Proof
From (13) and (23), we observe that
$$ \begin{aligned} 1+J_{k}\bigl(x \bigl(e^{\frac{t}{2}}-e^{-\frac{t}{2}}\bigr)\bigr) &=1+ \sum _{l=1}^{\infty }\frac{x^{l}e^{-\frac{t}{2}l}(e^{t}-1)^{l}}{(l-1)!l^{k}} \\ &= 1+\sum_{l=1}^{\infty }\frac{x^{l}e^{-\frac{t}{2}l}}{l^{k-1}} \sum _{m=l}^{\infty }S_{2}(m,l) \frac{t^{m}}{m!} \\ &= 1+\sum_{m=1}^{\infty }\sum _{l=1}^{m} \frac{1}{l^{k-1}} S_{2}(m,l)x^{l} \frac{t^{m}}{m!} \sum_{j=0}^{\infty } \biggl(- \frac{l}{2} \biggr)^{j} \frac{t^{j}}{j!} \\ &= 1+\sum_{n=1}^{\infty } \Biggl(\sum _{m=1}^{n} \sum_{l=1}^{m} \binom{n}{m}\frac{1}{l^{k-1}} \biggl(-\frac{1}{2}l \biggr)^{n-m} S_{2}(m,l)x^{l} \Biggr) \frac{t^{n}}{n!}. \end{aligned} $$
(29)
Combining (23) with (29), we have the desired result. □
In Theorem 4, when \(k=1\), we note that
$$ \begin{aligned} B_{n}^{(c)}(x)= \sum _{m=1}^{n} \sum_{l=1}^{m} \binom{n}{m} \biggl(-\frac{1}{2}l \biggr)^{n-m} S_{2}(m,l)x^{l}. \end{aligned} $$
Theorem 5
For \(k \in \mathbb{Z}\) and \(n\geq 1\), we have
$$ \begin{aligned} B_{n}^{(c,k)}(x) = \sum _{l=1}^{n} \binom{n}{l} \biggl(- \frac{1}{2}l \biggr)^{n-l}\mathit{bel}_{n-l}^{(k)}(x), \end{aligned} $$
where \(\mathit{bel}_{n}^{(k)}(x)\) are the poly-Bell polynomials.
Proof
From (19) and (23), we get
$$ \begin{aligned} 1+ J_{k}\bigl(x \bigl(e^{\frac{t}{2}}-e^{-\frac{t}{2}}\bigr)\bigr) &= 1+ J_{k} \bigl(xe^{- \frac{t}{2}}\bigl(e^{t}-1\bigr)\bigr)= 1+ \sum _{l=1}^{\infty }\frac{x^{l} e^{-\frac{t}{2}}(e^{t}-1)^{l}}{(l-1)!l^{k}} \\ &= 1+ \sum_{l=1}^{\infty }\frac{(x(e^{t}-1))^{l}}{(l-1)!l^{k}}\sum _{m=0}^{\infty } \biggl(-\frac{1}{2}l \biggr)^{m}\frac{t^{m}}{m!} \\ &=1+ \sum_{l=1}^{\infty }\mathit{bel}_{l}^{(k)}(x) \frac{t^{l}}{l!} \sum_{m=0}^{\infty } \biggl(- \frac{1}{2}l \biggr)^{m}\frac{t^{m}}{m!} \\ &=1+ \sum_{n=1}^{\infty }\sum _{l=1}^{n} \binom{n}{l} \biggl(- \frac{1}{2}l \biggr)^{n-l}\mathit{bel}_{n-l}^{(k)}(x) \frac{t^{n}}{n!} . \end{aligned} $$
(30)
Combining (23) with (30), we have the desired result. □
Theorem 6
For \(k \in \mathbb{Z}\) and \(n\geq 1\), we have
$$ \begin{aligned} \sum_{j=1}^{n} 2^{j} S_{1}(n,j)B_{j}^{(c,k)}(x) = \sum _{m=1}^{n} \sum_{l=1}^{m} \binom{n}{m}\frac{2^{l} x^{l}}{l^{k-1}} S_{1}(m,l)b_{n-m}^{*(l)}, \end{aligned} $$
where \(b_{n}^{*(l)}\) are the order l type 2 Bernoulli polynomials of the second kind.
Proof
By replacing t with \(2\log (1+t)\) in (23), from (11) and (17), the left-hand side of (23) is
$$ \begin{aligned} 1+ J_{k}(x \bigl((1+t)-(1+t)^{-1}\bigr) &= 1+ \sum_{l=1}^{\infty } \frac{x^{l} ((1+t)-(1+t)^{-1})^{l}}{(l-1)! l^{k}} \\ &= 1+ \sum_{l=1}^{\infty }\frac{2^{l} x^{l}}{l^{k-1}} \biggl( \frac{(1+t)-(1+t)^{-1}}{2\log (1+t)} \biggr)^{l} \frac{\log (1+t)^{l}}{l!} \\ &= 1+ \sum_{l=1}^{\infty }\frac{2^{l} x^{l}}{l^{k-1}} S_{1}(m,l) \frac{t^{m}}{m!}\sum_{j=0}^{\infty }b_{j}^{*(l)} \frac{t^{j}}{j!} \\ &= 1+ \sum_{m=1}^{\infty }\sum _{l=1}^{m} \frac{2^{l} x^{l}}{l^{k-1}} S_{1}(m,l) \frac{t^{m}}{m!} \sum_{j=0}^{\infty }b_{j}^{*(l)} \frac{t^{j}}{j!} \\ &=1+\sum_{n=1}^{\infty }\sum _{m=1}^{n}\sum_{l=1}^{m} \binom{n}{m} \frac{2^{l} x^{l}}{l^{k-1}} S_{1}(m,l) b_{n-m}^{*(l)} \frac{t^{n}}{n!}. \end{aligned} $$
(31)
On the other hand, by (11), the right-hand side of (23) is
$$ \begin{aligned} \sum_{m=0}^{\infty }B_{m}^{(c,k)}(x) \frac{(2\log (1+t))^{m}}{m!} &= \sum_{m=0}^{\infty }2^{m} B_{m}^{(c,k)}(x) \sum_{n=m}^{\infty }S_{1}(n,m) \frac{t^{n}}{n!} \\ &=\sum_{n=0}^{\infty }\sum _{m=0}^{n} 2^{m} S_{1}(n,m)B_{m}^{(c,k)}(x) \frac{t^{n}}{n!} \\ &=1+ \sum_{n=1}^{\infty }\sum _{m=0}^{n} 2^{m} S_{1}(n,m)B_{m}^{(c,k)}(x). \end{aligned} $$
(32)
Since \(S_{1}(n,0)=0\) for \(n\geq 1\), by comparing with the coefficients of (31) and (32), we have the desired result. □
Theorem 7
For \(k \in \mathbb{Z}\) and \(n\geq 1\), we have
$$ \begin{aligned} \sum_{j=0}^{n} \binom{n}{j}b_{j} B_{n-j+1}^{(c,k)}(x)= \sum _{m=0}^{n} \binom{n}{m}E_{n-m} B_{m+1}^{(c,k)}(x), \end{aligned} $$
where \(b_{n}\) are the ordinary Bernoulli numbers and \(E_{n}\) are the ordinary Euler numbers.
Proof
Differentiating with respect to t in (23), the left-hand side of (23) is
$$ \begin{aligned} \frac{\partial }{\partial t} \bigl(1+J_{k} \bigl(x\bigl(e^{\frac{t}{2}}-e^{- \frac{t}{2}}\bigr)\bigr)\bigr) &= \sum _{n=1}^{\infty }\frac{x^{n} (e^{\frac{t}{2}}-e^{-\frac{t}{2}})^{n-1}}{(n-1)! n^{k-1}} \frac{1}{2} \bigl(e^{\frac{t}{2}}+e^{-\frac{t}{2}}\bigr) \\ &= \frac{e^{\frac{t}{2}}+e^{-\frac{t}{2}}}{2(e^{\frac{t}{2}}-e^{-\frac{t}{2}})} \sum_{n=1}^{\infty } \frac{x^{n}(e^{\frac{t}{2}}-e^{-\frac{t}{2}})^{n}}{(n-1)!n^{k-1}} \\ &= \frac{e^{\frac{t}{2}}+e^{-\frac{t}{2}}}{2(e^{\frac{t}{2}}-e^{-\frac{t}{2}})} J_{k-1}\bigl(x\bigl(e^{\frac{t}{2}}-e^{-\frac{t}{2}} \bigr)\bigr) \\ &= \frac{e^{t}+1}{2(e^{t}-1)} \sum_{n=1}^{\infty }B_{n}^{(c,k)}(x) \frac{t^{n}}{n!}. \end{aligned} $$
(33)
On the other hand, the right-hand side of (23) is
$$ \begin{aligned} \frac{\partial }{\partial t} \Biggl(\sum _{n=0}^{\infty }B_{n}^{(c,k)}(x) \frac{t^{n}}{n!} \Biggr) &= \sum_{n=0}^{\infty }B_{n+1}^{(c,k)}(x) \frac{t^{n}}{n!}. \end{aligned} $$
(34)
Combining (33) with (34), we get
$$ \begin{aligned} \frac{1}{e^{t}-1} \sum _{n=1}^{\infty }B_{n}^{(c,k)}(x) \frac{t^{n}}{n!} =\frac{2}{e^{t}+1}\sum_{n=0}^{\infty }B_{n+1}^{(c,k)}(x) \frac{t^{n}}{n!}. \end{aligned} $$
(35)
From (16) and (35), we have
$$ \begin{aligned} \sum_{j=0}^{\infty }b_{j} \frac{t^{j}}{j!}\sum_{m=0}^{\infty } \frac{1}{m+1}B_{m+1}^{(c,k)}(x)\frac{t^{m}}{m!}=\sum _{i=0}^{\infty }E_{i}\frac{t^{i}}{i!} \sum_{m=0}^{\infty }B_{m+1}^{(c,k)}(x) \frac{t^{m}}{m!}. \end{aligned} $$
(36)
By (36), we have
$$ \begin{aligned} \sum_{n=0}^{\infty } \sum_{j=0}^{n} \binom{n}{j}b_{j} B_{n-j+1}^{(c,k)}(x) \frac{t^{n}}{n!}=\sum _{n=0}^{\infty }\sum_{m=0}^{n} \binom{n}{m}E_{n-m} B_{m+1}^{(c,k)}(x) \frac{t^{n}}{n!}. \end{aligned} $$
(37)
By comparing with the coefficients of both sides of (37), we have the desired result. □