In this section, we derive several identities between the degenerate Lah–Bell polynomials and some other polynomials arising from degenerate Sheffer sequences.
Kim–Kim introduced the degenerate Lah–Bell polynomials given by
$$ \begin{aligned} e_{\lambda }^{x} \biggl( \frac{t}{1-t} \biggr) = \sum_{n=0}^{\infty } \mathbf{B}_{n,\lambda }^{L}(x) \frac{t^{n}}{n!}\quad (n,k\ge 0) \text{ (see [10])}. \end{aligned} $$
(26)
When \(x=1\), \(\mathbf{B}_{n,\lambda }^{L} : = \mathbf{B}_{n,\lambda }^{L}(1)\) are called the nth degenerate Lah–Bell numbers.
When \(\lambda \rightarrow 0\), \(\lim_{\lambda \rightarrow 0}\mathbf{B}_{n,\lambda }^{L} = \mathbf{B}_{n}^{L}\) are the nth Lah–Bell numbers.
For \(n \in \mathbb{N}\cup \{0\}\) and \(P(x)= \sum_{k=0}^{n} Z_{k} \mathbf{B}_{k,\lambda }^{L} (x) \in \mathbb{P}_{n}\),
by using (23), we observe that
$$ \begin{aligned} \biggl\langle \biggl(\frac{t}{1+t} \biggr)^{k} \Big| P(x) \biggr\rangle _{\lambda }&=\sum_{l=0}^{n} Z_{l} \biggl\langle \biggl(\frac{t}{1+t} \biggr)^{k} \Big| \mathbf{B}_{l,\lambda }^{L}(x) \biggr\rangle _{\lambda }= \sum _{l=0}^{n} Z_{l} l! \delta _{k,l} = k!Z_{k}. \end{aligned} $$
(27)
From (27), we have
$$ \begin{aligned} Z_{k}= \frac{1}{k!} \biggl\langle \biggl(\frac{t}{1+t} \biggr)^{k} \Big| P(x) \biggr\rangle _{\lambda }. \end{aligned} $$
(28)
Thus, we have
$$ P(x)= \sum_{k=0}^{n} Z_{k} \mathbf{B}_{k,\lambda }^{L}(x) \quad \text{where } Z_{k} = \frac{1}{k!} \biggl\langle \biggl(\frac{t}{1+t} \biggr)^{k} \Big| P(x) \biggr\rangle _{\lambda }. $$
(29)
Theorem 1
For \(n \in \mathbb{N}\cup \{0\}\), we have
$$ \mathbf{B}_{n,\lambda }^{L}(x) = \sum _{k=0}^{n} L(n,k) (x)_{k,\lambda } = \sum _{k=0}^{n} \Biggl(\frac{1}{k!}\sum _{l=0}^{k} \binom{k}{l} (-1)^{k-l} \langle l\rangle _{n} \Biggr) (x)_{k,\lambda }. $$
(30)
As the inversion formula of (30), we have
$$ (x)_{n,\lambda } = \sum_{k=0}^{n}(-1)^{n-k} L(n,k) \mathbf{B}_{k, \lambda }^{L}(x)= \sum _{k=0}^{n} \Biggl( \frac{1}{k!} \sum _{l=0}^{k} \binom{k}{l} (-1)^{l+n}\langle l\rangle _{n} \Biggr) \mathbf{B}_{k,\lambda }^{L}(x). $$
(31)
Proof
From (5), (24) and (26), we consider the following two Sheffer sequences:
$$ \begin{aligned} \mathbf{B}_{n,\lambda }^{L}(x) \sim \biggl(1, \frac{t}{1+t} \biggr)_{\lambda }\quad \text{and} \quad (x)_{n,\lambda } \sim (1, t )_{\lambda }. \end{aligned} $$
(32)
From (6), (25) and (32), we have
$$ \begin{aligned} \mathbf{B}_{n,\lambda }^{L}(x) = \sum_{k=0}^{n} \mu _{n,k} (x)_{k,\lambda }, \end{aligned} $$
(33)
where
$$ \begin{aligned} & \mu _{n,k} = \frac{1}{k!} \biggl\langle \biggl( \frac{t}{1-t} \biggr)^{k} \Big| (x)_{n,\lambda } \biggr\rangle _{\lambda }= L(n,k), \\ & \text{or} \quad \mu _{n,k} =\frac{1}{k!}\sum _{l=0}^{k} \binom{k}{l} (-1)^{k-l} \biggl\langle \biggl(\frac{1}{1-t} \biggr)^{l} \Big| (x)_{n,\lambda } \biggr\rangle _{\lambda }=\frac{1}{k!}\sum_{l=0}^{k} \binom{k}{l} (-1)^{k-l} \langle l\rangle _{n}. \end{aligned} $$
(34)
Therefore, we have the identity (30).
To find the inversion formula of (30), let \(P(x)=(x)_{n,\lambda }\). From (29),
$$ \begin{aligned} (x)_{n,\lambda } = \sum _{k=0}^{n} Z_{k} \mathbf{B}_{k, \lambda }^{L}(x), \end{aligned} $$
(35)
where
$$ \begin{aligned} & Z_{k}=\frac{1}{k!} \biggl\langle \biggl( \frac{t}{1+t} \biggr)^{k} \Big| (x)_{n,\lambda } \biggr\rangle _{\lambda }=(-1)^{n-k}L(n,k), \\ &\text{or} \quad Z_{k} =\frac{1}{k!} \sum _{l=0}^{k} \binom{k}{l} (-1)^{l} \biggl\langle \biggl(\frac{1}{1+t} \biggr)^{l} \Big| (x)_{n,\lambda } \biggr\rangle _{\lambda }=\frac{1}{k!} \sum_{l=0}^{k} \binom{k}{l} (-1)^{l+n} \langle l\rangle _{n}. \end{aligned} $$
(36)
Therefore, from (35) and (36), we have the identity (31). □
Theorem 2
For \(n \in \mathbb{N}\cup \{0\}\) and \(r \in \mathbb{N}\), we have
$$ \mathbf{B}_{n,\lambda }^{L}(x)=\sum _{k=0}^{n} \Biggl(\sum_{l=k}^{n} \sum_{m=0}^{n-l}\binom{n}{l} \frac{(1)_{m+1,\lambda }}{(m+1)} L(l,k)L(n-l,m) \Biggr)\beta _{k,\lambda }(x). $$
(37)
As the inversion formula of (37), we have
$$ \begin{aligned} \beta _{n,\lambda }(x) &= \sum _{k=0}^{n} \Biggl( \frac{1}{k!} \sum _{l=0}^{k} \sum_{m=0}^{n} \binom{k}{l} \binom{l+n-m-1}{n-m} \binom{n}{m}(-1)^{l+n-m}(n-m)! \beta _{m,\lambda } \Biggr) \mathbf{B}_{k,\lambda }^{L}(x) \\ & = \sum_{k=0}^{n} \Biggl(\sum _{m=0}^{n} \binom{n}{m} (-1)^{n-m-k}L(n-m,k)\beta _{m,\lambda } \Biggr) \mathbf{B}_{k, \lambda }^{L}(x), \end{aligned} $$
(38)
where \(\beta _{n,\lambda }(x)\) are the degenerate Bernoulli polynomials.
Proof
From (7), (24) and (26), we consider the following two degenerate Sheffer sequences:
$$ \begin{aligned} \mathbf{B}_{n,\lambda }^{L}(x) \sim \biggl(1, \frac{t}{1+t} \biggr)_{\lambda }\quad \text{and} \quad \beta _{n,\lambda }(x) \sim \biggl(\frac{e_{\lambda }(t)-1}{t}, t \biggr)_{\lambda }. \end{aligned} $$
(39)
From (2), (5) and (25), we have
$$ \begin{aligned} \mathbf{B}_{n,\lambda }^{L}(x)= \sum_{k=0}^{n} \mu _{n,k} \beta _{k,\lambda }(x), \end{aligned} $$
(40)
where
$$ \begin{aligned} \mu _{n,k} &= \frac{1}{k!} \biggl\langle \biggl( \frac{e_{\lambda }(\frac{t}{1-t})-1}{\frac{t}{1-t}} \biggr) \biggl( \frac{t}{1-t} \biggr)^{k} \Big| (x)_{n,\lambda } \biggr\rangle _{\lambda } \\ &= \biggl\langle \biggl(\frac{e_{\lambda }(\frac{t}{1-t})-1}{\frac{t}{1-t}} \biggr) \Big| \biggl(\frac{1}{k!} \biggl( \frac{t}{1-t} \biggr)^{k} \biggr)_{\lambda }(x)_{n,\lambda } \biggr\rangle _{\lambda } \\ &= \sum_{l=k}^{n} \binom{n}{l} L(l,k) \biggl\langle \biggl( \frac{e_{\lambda }(\frac{t}{1-t})-1}{\frac{t}{1-t}} \biggr) \Big| (x)_{n-l, \lambda } \biggr\rangle _{\lambda } \\ &=\sum_{l=k}^{n} \binom{n}{l} L(l,k) \biggl\langle \sum_{m=1}^{\infty } \frac{(1)_{m,\lambda }}{m!} \biggl( \frac{t}{1-t} \biggr)^{m-1} \Big| (x)_{n-l, \lambda } \biggr\rangle _{\lambda } \\ &=\sum_{l=k}^{n} \binom{n}{l} L(l,k) \biggl\langle \sum_{m=0}^{\infty } \frac{(1)_{m+1,\lambda }}{(m+1)!} \biggl( \frac{t}{1-t} \biggr)^{m} \Big| (x)_{n-l, \lambda } \biggr\rangle _{\lambda } \\ &=\sum_{l=k}^{n} \binom{n}{l} L(l,k) \sum_{m=0}^{n-l} \frac{(1)_{m+1,\lambda }}{(m+1)} \biggl\langle \frac{1}{m!} \biggl( \frac{t}{1-t} \biggr)^{m} \Big| (x)_{n-l,\lambda } \biggr\rangle _{\lambda } \\ &=\sum_{l=k}^{n} \binom{n}{l} L(l,k) \sum_{m=0}^{n-l} \frac{(1)_{m+1,\lambda }}{(m+1)}L(n-l,m). \end{aligned} $$
(41)
To find the inversion formula of (37), let \(P(x)=\beta _{n,\lambda }(x)\). From (29), we have
$$ \begin{aligned} \beta _{n,\lambda }(x) = \sum _{k=0}^{n} Z_{k} \mathbf{B}_{k, \lambda }^{L}(x)\quad (n\geq 0). \end{aligned} $$
where
$$ \begin{aligned} Z_{k}&=\frac{1}{k!} \biggl\langle \biggl( \frac{t}{1+t} \biggr)^{k} \Big| \beta _{n,\lambda }(x) \biggr\rangle _{\lambda } \\ &=\frac{1}{k!} \sum_{l=0}^{k} \binom{k}{l} (-1)^{l} \biggl\langle \biggl( \frac{1}{1+t} \biggr)^{l} \Big| \beta _{n,\lambda }(x) \biggr\rangle _{\lambda } \\ &=\frac{1}{k!} \sum_{l=0}^{k} \binom{k}{l} (-1)^{l} \sum_{\nu =0}^{n} \binom{l+\nu -1}{\nu }(-1)^{\nu } \biggl\langle t^{\nu } \Big| \sum _{m=0}^{n} \binom{n}{m}\beta _{m,\lambda } (x)_{n-m,\lambda } \biggr\rangle _{\lambda } \\ &=\frac{1}{k!} \sum_{l=0}^{k} \binom{k}{l} (-1)^{l} \sum_{\nu =0}^{n} \binom{l+\nu -1}{\nu }(-1)^{\nu }\sum_{m=0}^{n} \binom{n}{m} \beta _{m, \lambda } \bigl\langle t^{\nu } | (x)_{n-m,\lambda } \bigr\rangle _{\lambda } \\ &=\frac{1}{k!} \sum_{l=0}^{k} \binom{k}{l} \binom{l+n-m-1}{n-m}(-1)^{l+n-m} \sum _{m=0}^{n}\binom{n}{m} \beta _{m,\lambda }(n-m)!. \end{aligned} $$
(42)
Stated differently, we get
$$\begin{aligned} Z_{k}&=\frac{1}{k!} \biggl\langle \biggl( \frac{t}{1+t} \biggr)^{k} \Big| \beta _{n,\lambda }(x) \biggr\rangle _{\lambda } \\ &= \biggl\langle \frac{1}{k!} \biggl(\frac{t}{1+t} \biggr)^{k} \Big| \sum _{m=0}^{n} \binom{n}{m}\beta _{m,\lambda } (x)_{n-m,\lambda } \biggr\rangle _{\lambda } \\ &=\sum_{m=0}^{n} \binom{n}{m}\beta _{m,\lambda } \biggl\langle \frac{1}{k!} \biggl(\frac{t}{1+t} \biggr)^{k} \Big| (x)_{n-m,\lambda } \biggr\rangle _{\lambda }=\sum _{m=0}^{n} \binom{n}{m}\beta _{m,\lambda } (-1)^{n-m-k}L(n-m,k) . \end{aligned}$$
(43)
Therefore, from (42) and (43) we have the identity (38). □
Theorem 3
For \(n \in \mathbb{N}\cup \{0\}\), we have
$$ \begin{aligned} &\mathbf{B}_{n,\lambda }^{L}(x)\\ &\quad = \frac{1}{(1-u)^{r}}\sum_{k=0}^{n} \Biggl(\sum _{l=k}^{n}\sum_{j=0}^{r} \sum_{m=0}^{n-l}\binom{n}{l} \binom{r}{j} (-u)^{r-j}(j)_{m,\lambda } L(l,k)L(n-l,m) \Biggr)h^{(r)}_{k, \lambda }(x|u). \end{aligned} $$
(44)
As the inversion formula of (44), we have
$$ \begin{aligned} &h^{(r)}_{n,\lambda }(x|u)\\ &\quad = \sum_{k=0}^{n} \Biggl( \frac{1}{k!} \sum_{l=0}^{k} \sum _{m=0}^{n}\binom{k}{l} \binom{n}{m} \binom{l+n-m-1}{n-m}(-1)^{l+n-m} (n-m)! h^{(r)}_{m,\lambda } (u) \Biggr) \mathbf{B}^{L}_{k,\lambda }(x) \\ & \quad = \sum_{k=0}^{n} \Biggl(\sum _{m=0}^{n} \binom{n}{m} (-1)^{n-m-k}L(n-m,k)h^{(r)}_{m,\lambda }(u) \Biggr) \mathbf{B}_{k,\lambda }^{L}(x), \end{aligned} $$
(45)
where \(h^{(r)}_{n,\lambda }(x|u)\) are the degenerate Frobenius–Euler polynomials of order r.
Proof
From (9), (24) and (26), we consider the following two degenerate Sheffer sequences:
$$ \begin{aligned} \mathbf{B}_{n,\lambda }^{L}(x) \sim \biggl(1, \frac{t}{1+t} \biggr)_{\lambda }\quad \text{and} \quad h^{(r)}_{n,\lambda }(x|u) \sim \biggl( \biggl(\frac{e_{\lambda }(t)-u}{1-u} \biggr)^{r}, t \biggr)_{\lambda }. \end{aligned} $$
(46)
By using (2), (5), (25) and (46), we have
$$ \begin{aligned} \mathbf{B}_{n,\lambda }^{L}(x)= \sum_{k=0}^{n} \mu _{n,k} h^{(r)}_{k, \lambda }(x|u), \end{aligned} $$
(47)
Here
$$\begin{aligned} \mu _{n,k} &= \frac{1}{k!} \biggl\langle \biggl( \frac{(e_{\lambda }(\frac{t}{1-t})-u)}{1-u} \biggr)^{r} \biggl( \frac{t}{1-t} \biggr)^{k} \Big| (x)_{n,\lambda } \biggr\rangle _{\lambda } \\ &= \biggl\langle \biggl( \frac{(e_{\lambda }(\frac{t}{1-t})-u)}{1-u} \biggr)^{r} \Big| \biggl( \frac{1}{k!} \biggl(\frac{t}{1-t} \biggr)^{k} \biggr)_{\lambda }(x)_{n,\lambda } \biggr\rangle _{\lambda } \\ &= \frac{1}{(1-u)^{r}}\sum_{l=k}^{n} \binom{n}{l} L(l,k) \biggl\langle \biggl(e_{\lambda } \biggl(\frac{t}{1-t} \biggr)-u \biggr)^{r} \Big| (x)_{n-l, \lambda } \biggr\rangle _{\lambda } \\ \\ &= \frac{1}{(1-u)^{r}}\sum_{l=k}^{n} \binom{n}{l} L(l,k)\sum_{j=0}^{r} \binom{r}{j}(-u)^{r-j} \biggl\langle e^{j}_{\lambda } \biggl( \frac{t}{1-t} \biggr) \Big| (x)_{n-l,\lambda } \biggr\rangle _{\lambda } \\ &= \frac{1}{(1-u)^{r}}\sum_{l=k}^{n} \binom{n}{l} L(l,k)\sum_{j=0}^{r} \binom{r}{j}(-u)^{r-j} \sum_{m=0}^{n-l}(j)_{m,\lambda } \biggl\langle \frac{1}{m!} \biggl(\frac{t}{1-t} \biggr)^{m} \Big| (x)_{n-l,\lambda } \biggr\rangle _{\lambda } \\ &= \frac{1}{(1-u)^{r}}\sum_{l=k}^{n} \binom{n}{l} L(l,k)\sum_{j=0}^{r} \binom{r}{j}(-u)^{r-j} \sum_{m=0}^{n-l}(j)_{m,\lambda }L(n-l,m). \end{aligned}$$
(48)
Therefore, from (47) and (48), we get the identity (44).
To find the inversion formula of (44), by (29), we have
$$ \begin{aligned} h^{(r)}_{n,\lambda }(x|u) = \sum _{k=0}^{n} Z_{k} \mathbf{B}_{k,\lambda }^{L}(x). \end{aligned} $$
In the same way as (42) and (43), we have
$$ \begin{aligned} Z_{k}&=\frac{1}{k!} \biggl\langle \biggl( \frac{t}{1+t} \biggr)^{k} \Big| h^{(r)}_{n,\lambda } (x|u) \biggr\rangle _{\lambda } \\ &=\frac{1}{k!} \sum_{l=0}^{k} \binom{k}{l} (-1)^{l} \biggl\langle \biggl( \frac{1}{1+t} \biggr)^{l} \Big| h^{(r)}_{n,\lambda } (x|u) \biggr\rangle _{\lambda } \\ &=\frac{1}{k!} \sum_{l=0}^{k} \binom{k}{l} (-1)^{l} \sum_{\nu =0}^{n} \binom{l+\nu -1}{\nu }(-1)^{\nu } \biggl\langle t^{\nu } \Big| \sum _{m=0}^{n} \binom{n}{m}h^{(r)}_{m,\lambda } (u) (x)_{n-m,\lambda } \biggr\rangle _{\lambda } \\ &=\frac{1}{k!} \sum_{l=0}^{k} \binom{k}{l} (-1)^{l} \sum_{\nu =0}^{n} \binom{l+\nu -1}{\nu }(-1)^{\nu }\sum_{m=0}^{n} \binom{n}{m} h^{(r)}_{m, \lambda } (u) \bigl\langle t^{\nu } | (x)_{n-m,\lambda } \rangle_{\lambda } \\ &=\frac{1}{k!} \sum_{l=0}^{k} \binom{k}{l} \binom{l+n-m-1}{n-m}(-1)^{l+n-m} \sum _{m=0}^{n}\binom{n}{m} h^{(r)}_{m,\lambda } (u) (n-m)!. \end{aligned} $$
(49)
In another way, we can get
$$ \begin{aligned} Z_{k}&=\frac{1}{k!} \biggl\langle \biggl(\frac{t}{1+t} \biggr)^{k} \Big| h^{(r)}_{n,\lambda } (x|u) \biggr\rangle _{\lambda }=\sum_{m=0}^{n} \binom{n}{m} h^{(r)}_{m,\lambda }(u) (-1)^{n-m-k}L(n-m,k) . \end{aligned} $$
(50)
Therefore, from (49) and (50), we have the identity (45). □
When \(u=-1\) in Theorem 3, we have the following corollary.
Corollary 4
For \(n \in \mathbb{N}\cup \{0\}\) and \(r \in \mathbb{N}\), we have
$$ \mathbf{B}_{k,\lambda }^{L}(x) =\frac{1}{2^{r}} \sum_{k=0}^{n} \Biggl( \sum _{l=k}^{n}\sum_{j=0}^{r} \sum_{m=0}^{n-l}\binom{n}{l} \binom{r}{j}(j)_{m,\lambda } L(l,k)L(n-l,m) \Biggr)E^{(r)}_{k,\lambda }(x). $$
(51)
By the inversion formula of (51), we have
$$ E^{(r)}_{n,\lambda }(x) = \sum_{k=0}^{n} \Biggl(\frac{1}{k!} \sum_{l=0}^{k} \sum_{m=0}^{n}\binom{k}{l} \binom{n}{m}\binom{l+n-m-1}{n-m}(-1)^{l+n-m} (n-m)! E^{(r)}_{m,\lambda } \Biggr) \mathbf{B}_{k,\lambda }^{L}(x), $$
where \(E^{(r)}_{n,\lambda }(x)\) are the degenerate Euler polynomials of order r.
Theorem 5
For \(n \in \mathbb{N}\cup \{0\}\) and \(r \in \mathbb{N}\), we have
$$ \begin{aligned} \mathbf{B}_{n,\lambda }^{L}(x)= \sum_{k=0}^{n} \Biggl( \sum _{l=k}^{n}\sum_{m=0}^{n} \sum_{j=0}^{n-m} \binom{n}{m} \frac{(1)_{j+1,\lambda }}{j+1}S_{2,\lambda }(l, k) L(m,l)L(n-m,j) \Biggr) D_{k, \lambda }(x). \end{aligned} $$
(52)
As the inversion formula of (52), we have
$$ \begin{aligned} D_{n,\lambda }(x)=\sum _{k=0}^{n} \Biggl(\sum_{m=0}^{n} \sum_{j=0}^{n-m}\binom{n}{m}(-1)^{n-m-k}S_{1,\lambda }(n-m,j)L(n-m,k) D_{m, \lambda } \Biggr) \mathbf{B}^{L}_{k,\lambda }(x), \end{aligned} $$
(53)
where \(D_{n,\lambda }(x)\) are the degenerate Daehee polynomials.
Proof
From (10) and (24), we consider the following two degenerate Sheffer sequences:
$$ \begin{aligned} \mathbf{B}_{n,\lambda }^{L}(x) \sim \biggl(1 , \frac{t}{1+t} \biggr)_{\lambda }\quad \text{and} \quad D_{n,\lambda }(x) \sim \biggl(\frac{e_{\lambda }(t)-1}{t}, e_{\lambda }(t)-1 \biggr)_{\lambda }. \end{aligned} $$
(54)
From (2), (16), (25) and (54), we have
$$ \begin{aligned} \mathbf{B}_{n,\lambda }^{L}(x)= \sum_{k=0}^{n} \mu _{n,k} D_{k, \lambda }(x), \end{aligned} $$
(55)
where
$$ \begin{aligned} \mu _{n,k} &= \frac{1}{k!} \biggl\langle \frac{e_{\lambda }(\frac{t}{1-t})-1}{\frac{t}{1-t}} \biggl(e_{\lambda } \biggl( \frac{t}{1-t} \biggr)-1 \biggr)^{k} \Big| (x)_{n,\lambda } \biggr\rangle _{\lambda } \\ &= \biggl\langle \frac{1-t}{t} \biggl( e_{\lambda } \biggl(\frac{t}{1-t} \biggr)-1 \biggr)\frac{1}{k!} \biggl(e_{\lambda } \biggl( \frac{t}{1-t} \biggr)-1 \biggr)^{k} \Big| (x)_{n,\lambda } \biggr\rangle _{\lambda } \\ &= \sum_{l=k}^{n} S_{2,\lambda }(l, k) \biggl\langle \frac{1-t}{t} \biggl( e_{\lambda } \biggl(\frac{t}{1-t} \biggr)-1 \biggr) \Big| \biggl(\frac{1}{l!} \biggl(\frac{t}{1-t} \biggr)^{l} \biggr)_{\lambda }(x)_{n,\lambda } \biggr\rangle _{\lambda } \\ &=\sum_{l=k}^{n} S_{2,\lambda }(l, k) \sum_{m=0}^{n} \binom{n}{m} L(m,l) \biggl\langle \frac{1-t}{t} \biggl( e_{\lambda } \biggl(\frac{t}{1-t} \biggr)-1 \biggr) \Big| (x)_{n-m,\lambda } \biggr\rangle _{\lambda } \\ &=\sum_{l=k}^{n} S_{2,\lambda }(l, k) \sum_{m=0}^{n} \binom{n}{m} L(m,l) \biggl\langle \sum _{j=1}^{\infty }(1)_{j,\lambda } \frac{1}{j!} \biggl( \frac{t}{1-t} \biggr)^{j-1} \Big| (x)_{n-m,\lambda } \biggr\rangle _{\lambda } \\ &=\sum_{l=k}^{n} S_{2,\lambda }(l, k) \sum_{m=0}^{n} \binom{n}{m} L(m,l) \biggl\langle \sum _{j=0}^{\infty }(1)_{j+1,\lambda } \frac{1}{(j+1)!} \biggl( \frac{t}{1-t} \biggr)^{j} \Big| (x)_{n-m,\lambda } \biggr\rangle _{\lambda } \\ &=\sum_{l=k}^{n} S_{2,\lambda }(l, k) \sum_{m=0}^{n} \binom{n}{m} L(m,l) \sum _{j=0}^{n-m} \frac{(1)_{j+1,\lambda }}{j+1} \biggl\langle \frac{1}{j!} \biggl(\frac{t}{1-t} \biggr)^{j} \Big| (x)_{n-m,\lambda } \biggr\rangle _{\lambda } \\ &=\sum_{l=k}^{n} S_{2,\lambda }(l, k) \sum_{m=0}^{n} \binom{n}{m} L(m,l) \sum _{j=0}^{n-m} \frac{(1)_{j+1,\lambda }}{j+1}L(n-m,j). \end{aligned} $$
(56)
Therefore, from (55) and (56), we get the identity (52).
To find the inversion formula of (52), from (29), we have
$$ \begin{aligned} D_{n,\lambda }(x) = \sum _{k=0}^{n} Z_{k} B_{k,\lambda }^{L}(x). \end{aligned} $$
(57)
By using \((1+t)^{x} = \sum_{n=0}^{\infty }(x)_{n} \frac{t^{n}}{n!}\) and the first equation of (15), we have
$$ \begin{aligned}Z_{k}&= \biggl\langle \frac{1}{k!} \biggl( \frac{t}{1+t} \biggr)^{k} \Big| D_{n,\lambda }(x) \biggr\rangle _{\lambda } \\ &= \biggl\langle \frac{1}{k!} \biggl( \frac{t}{1+t} \biggr)^{k} \Big| \sum _{m=0}^{n} \binom{n}{m}D_{m,\lambda } (x)_{n-m} \biggr\rangle _{\lambda } \\ &= \sum_{m=0}^{n}\binom{n}{m} D_{m,\lambda } \biggl\langle \frac{1}{k!} \biggl( \frac{t}{1+t} \biggr)^{k} \Big| \sum_{j=0}^{n-m}S_{1,\lambda }(n-m,j) (x)_{n-m, \lambda } \biggr\rangle _{\lambda } \\ &= \sum_{m=0}^{n}\binom{n}{m} D_{m,\lambda }\sum_{j=0}^{n-m}S_{1, \lambda }(n-m,j) \biggl\langle \frac{1}{k!} \biggl( \frac{t}{1+t} \biggr)^{k} \Big| (x)_{n-m,\lambda } \biggr\rangle _{\lambda } \\ &= \sum_{m=0}^{n}\binom{n}{m} D_{m,\lambda }\sum_{j=0}^{n-m}S_{1, \lambda }(n-m,j) (-1)^{n-m-k}L(n-m,k). \end{aligned} $$
(58)
Therefore, from (57) and (58), we have the identity (53). □
Theorem 6
For \(n \in \mathbb{N}\cup \{0\}\), we have
$$ \mathbf{B}_{n,\lambda }^{L}(x) = \sum _{k=0}^{n} \Biggl(\sum_{l=k}^{n} S_{1, \lambda }(l,k)L(n,k) \Biggr)\mathrm{Bel}_{k,\lambda }(x). $$
(59)
As the inversion formula of (59), we have
$$ \mathrm{Bel}_{n,\lambda }(x) = \sum _{k=0}^{n} \Biggl(\sum_{l=0}^{n} (-1)^{l-k} S_{2, \lambda }(n,l) L(l,k) \Biggr) \mathbf{B}_{k,\lambda }^{L}(x), $$
(60)
where \(\mathrm{Bel}_{n,\lambda }(x)\) are the degenerate Bell polynomials.
Proof
From (12), (24) and (26), we consider two degenerate Sheffer sequences as follows:
$$ \begin{aligned} \mathbf{B}_{n,\lambda }^{L}(x) \sim \biggl(1, \frac{t}{1+t} \biggr)_{\lambda }\quad \text{and} \quad \mathrm{Bel}_{n,\lambda } (x) \sim \bigl(1, \log _{\lambda }(1+t) \bigr)_{\lambda }. \end{aligned} $$
(61)
By using (2), (15), (25) and (61), we have
$$ \begin{aligned} \mathbf{B}_{n,\lambda }^{L}(x) = \sum_{k=0}^{n} \mu _{n,k} \mathrm{Bel}_{k,\lambda }(x), \end{aligned} $$
(62)
where
$$ \begin{aligned} \mu _{n,k}&= \frac{1}{k!} \biggl\langle \biggl(\log _{\lambda } \biggl(1+\frac{t}{1-t} \biggr) \biggr)^{k} \Big| (x)_{n,\lambda } \biggr\rangle _{\lambda } \\ &= \sum_{l=k}^{n} S_{1,\lambda }(l,k) \biggl\langle \frac{1}{l!} \biggl( \frac{t}{1-t} \biggr)^{l} \Big| (x)_{n,\lambda } \biggr\rangle _{\lambda }= \sum_{l=k}^{n} S_{1,\lambda }(l,k)L(n,l). \end{aligned} $$
(63)
Therefore from (62) and (63), we get the identity (59).
To find inversion formula of (59), from (29), we have
$$ \begin{aligned} \mathrm{Bel}_{n,\lambda }(x) = \sum _{k=0}^{n} Z_{k} B_{k,\lambda }^{L}(x). \end{aligned} $$
(64)
From (5) and (16), we observe that
$$ \begin{aligned} \sum_{n=0}^{\infty } \mathrm{Bel}_{n,\lambda }(x) \frac{t^{n}}{n!}= e_{\lambda }^{x} \bigl(e_{\lambda }(t)-1\bigr)=\sum_{n=0}^{\infty } \Biggl(\sum_{l=0}^{n} S_{2,\lambda }(n,l) (x)_{l,\lambda } \Biggr)\frac{t^{n}}{n!}. \end{aligned} $$
Thus, by using \(\mathrm{Bel}_{n,\lambda }(x) = \sum_{l=0}^{n} S_{2,\lambda }(n,l)(x)_{l, \lambda }\) and (6), we have
$$ \begin{aligned} Z_{k}&=\frac{1}{k!} \biggl\langle \biggl( \frac{t}{1+t} \biggr)^{k} \Big| \mathrm{Bel}_{n,\lambda }(x) \biggr\rangle _{\lambda } \\ &= \biggl\langle \frac{1}{k!} \biggl( \frac{t}{1+t} \biggr)^{k} \Big| \sum _{l=0}^{n} S_{2,\lambda }(n,l) (x)_{l,\lambda } \biggr\rangle _{\lambda } \\ &= \sum_{l=0}^{n} S_{2,\lambda }(n,l) \biggl\langle \frac{1}{k!} \biggl( \frac{t}{1+t} \biggr)^{k} \Big| (x)_{l,\lambda } \biggr\rangle _{\lambda } \\ &= \sum_{l=0}^{n} S_{2,\lambda }(n,l) (-1)^{l-k} L(l,k) . \end{aligned} $$
(65)
Therefore, from (64) and (65), we have the identity (60). □
Theorem 7
For \(n \in \mathbb{N}\cup \{0\}\), we have
$$ \mathbf{B}_{n,\lambda }^{L}(x) = \sum _{k=0}^{n} \Biggl( \sum_{l=k}^{n} S_{2,\lambda }(l,k)L(n,l) \Biggr) (x)_{n} \quad (n\geq 0). $$
(66)
Proof
Since \(e_{\lambda }^{x}(\log (1+t))=(1+t)^{x} = \sum_{n=0}^{\infty }(x)_{n} \frac{t^{n}}{n!}\), we have \((x)_{n} \sim (1,e_{\lambda }(t)-1)_{\lambda }\).
Therefore, we consider the two degenerate Sheffer sequences as follows:
$$ \begin{aligned} \mathbf{B}_{n,\lambda }^{L}(x) \sim \biggl(1, \frac{t}{1+t} \biggr)_{\lambda }\quad \text{and} \quad (x)_{n} \sim \bigl(1, e_{\lambda }(t)-1\bigr)_{\lambda }. \end{aligned} $$
(67)
Thus, from (2), (16) and (67), we have
$$ \begin{aligned} \mathbf{B}_{n,\lambda }^{L}(x) = \sum_{k=0}^{n} \mu _{n,k}(x)_{k}\quad (n\geq 0), \end{aligned} $$
(68)
where
$$ \begin{aligned} \mu _{n,k}&= \frac{1}{k!} \biggl\langle \biggl(e_{\lambda } \biggl( \frac{t}{1-t} \biggr)-1 \biggr)^{k} \Big| (x)_{n,\lambda } \biggr\rangle _{\lambda } \\ &= \sum_{l=k}^{n} S_{2,\lambda }(l,k) \biggl\langle \frac{1}{l!} \biggl( \frac{t}{1-t} \biggr)^{l} \Big| (x)_{n,\lambda } \biggr\rangle _{\lambda }= \sum_{l=k}^{n} S_{2,\lambda }(l,k)L(n,l). \end{aligned} $$
(69)
Therefore, from (68) and (69), we have the identity (66). □
Theorem 8
For \(n \in \mathbb{N}\cup \{0\}\) and \(s \in \mathbb{N}\), we have
$$ \mathcal{E}_{n,\lambda }^{(s)}(x) = \sum _{k=0}^{n} \Biggl( \sum_{l=k}^{n} \binom{n}{l}(-1)^{l-k} L(l,k)\mathcal{E}_{n-l,\lambda }^{(s)} \Biggr) \mathbf{B}_{k,\lambda }^{L}(x), $$
(70)
where \(\mathcal{E}_{n,\lambda }^{(s)}(x)\) are type 2 degenerate poly-Euler polynomials.
Proof
From (18), (24) and (26), we consider the following two degenerate Sheffer sequences as follows:
$$ \begin{aligned} \mathbf{B}_{n,\lambda }^{L}(x) \sim \biggl(1, \frac{t}{1+t} \biggr)_{\lambda }\quad \text{and} \quad \mathcal{E}_{n, \lambda }^{(s)} (x)\sim \biggl( \frac{t(e_{\lambda }(t)+1)}{\mathrm{Ei}_{s}(\log (1+2t))}, t \biggr)_{\lambda }. \end{aligned} $$
(71)
By using (2) and (25), we have
$$ \begin{aligned} \mathcal{E}_{n,\lambda }^{(s)}(x) = \sum_{k=0}^{n} \mu _{n,k} \mathbf{B}_{k\lambda }^{L}(x), \end{aligned} $$
(72)
where
$$ \begin{aligned} \mu _{n,k}&= \frac{1}{k!} \biggl\langle \frac{\mathrm{Ei}_{s}(\log (1+2t))}{t(e_{\lambda }(t)+1)} \biggl(\frac{t}{1+t} \biggr)^{k} \Big| (x)_{n,\lambda } \biggr\rangle _{\lambda } \\ &= \biggl\langle \frac{\mathrm{Ei}_{s}(\log (1+2t))}{t(e_{\lambda }(t)+1)} \Big| \biggl(\frac{1}{k!} \biggl(\frac{t}{1+t} \biggr)^{k} \biggr)_{\lambda }(x)_{n, \lambda } \biggr\rangle _{\lambda } \\ &= \sum_{l=k}^{n} \binom{n}{l} (-1)^{l-k}L(l,k) \biggl\langle \frac{\mathrm{Ei}_{s}(\log (1+2t))}{t(e_{\lambda }(t)+1)} \Big| (x)_{n-l, \lambda } \biggr\rangle _{\lambda }\\ &=\sum_{l=k}^{n} \binom{n}{l}(-1)^{l-k} L(l,k) \mathcal{E}_{n-l,\lambda }^{(s)}. \end{aligned} $$
(73)
Therefore, from (72) and (73), we get the identity (70). □