In this section, we study the powers under umbral composition applied to associated sequences. Throughout this section, let \(r_{n}(x)=\sum_{k=0}^{n} r_{n,k}x^{k}\sim(1,f(t))\) and \(r_{n}^{(m)}(x)=\sum_{k=0}^{n}r_{n,k}^{(m)}x^{k}\sim(1,f^{m}(t))\).
Generalized falling factorial polynomials
As a first interesting case, let us consider the generalized falling factorial polynomials \((x|\lambda)_{n}=x(x-\lambda)\cdots(x-(n-1)\lambda)\), for \(n\geq1\), and \((x|\lambda)_{0}=1\) (see [4]). So
$$r_{n}(x)=(x|\lambda)_{n}=\sum_{k=0}^{n} \lambda^{n-k}S_{1}(n,k)x^{k}\sim\bigl(1,f(t)\bigr) $$
with \(f(t)=\frac{e^{\lambda t}-1}{\lambda}\). Then, by (3), we obtain
$$\begin{aligned} r_{n,k}^{(m)}&=\sum_{\ell_{1},\ldots,\ell_{m-1}=0}^{n} \lambda^{n-\ell _{1}}S_{1}(n,\ell_{1})\lambda^{\ell_{1}-\ell_{2}}S_{1}( \ell_{1},\ell_{2}) \cdots\lambda^{\ell_{m-1}-k}S_{1}( \ell_{m-1},k) \\ &=\lambda^{n-k}\sum_{\ell_{1},\ldots,\ell_{m-1}=0}^{n}S_{1}(n, \ell_{1})S_{1}(\ell _{1},\ell_{2}) \cdots S_{1}(\ell_{m-1},k). \end{aligned}$$
(4)
To proceed, we recall the transfer formula (see [2]): for \(p_{n}(x)\sim(1,f(t))\) and \(q_{n}(x)\sim(1,\ell(t))\), we have \(q_{n}(x)=x(f(t)/\ell(t))^{n}x^{-1}p_{n}(x)\), for all \(n\geq1\). As \(x^{n}\sim(1,t)\), we have, for \(n\geq1\),
$$\begin{aligned} r_{n}(x)&=x \biggl(\frac{\lambda t}{e^{\lambda t}-1} \biggr)^{n}x^{-1}x^{n}=x \sum_{k\geq0}\beta_{k}^{(n)} \frac{\lambda^{k}}{k!}t^{k}x^{n-1}\\ &=x\sum_{k=0}^{n-1}\binom{n-1}{k} \lambda^{k}\beta_{k}^{(n)}x^{n-1-k}=x\sum _{k=0}^{n-1}\binom{n-1}{k} \lambda^{n-1-k}\beta_{n-1-k}^{(n)}x^{k} \\ &=\sum_{k=1}^{n}\binom{n-1}{k-1} \lambda^{n-k}\beta_{n-k}^{(n)}x^{k}. \end{aligned}$$
Thus, for \(n\geq1\), \(0\leq m\leq n-1\), we observe that
$$\bigl(f(t)\bigr)^{m}x^{-1}r_{n}(x)=\bigl(f(t) \bigr)^{m}\bigl(t/f(t)\bigr)^{n}x^{n-1}=\bigl(t/f(t) \bigr)^{n-m}t^{m}x^{n-1}. $$
Therefore,
$$\bigl(f(t)\bigr)^{m}x^{-1}r_{n}(x)=\sum _{\ell=0}^{n-1-m}\frac{(n-1)_{\ell+m}}{\ell !}\lambda^{\ell}B_{\ell}^{(n-m)}x^{n-1-\ell-m}. $$
Now, we get, for \(n\geq1\), \(r_{n}^{(2)}(x)=x(f(t)/f^{2}(t))^{n}x^{-1}r_{n}(x)\), which gives
$$\begin{aligned} r_{n}^{(2)}(x)&=x\sum_{k_{2}=0}^{n-1} B_{k_{2}}^{(n)}\frac{\lambda ^{k_{2}}}{k_{2}!}\bigl(f(t)\bigr)^{k_{2}}x^{-1}r_{n}(x) \\ &=\sum_{k_{2}=0}^{n-1}\sum _{k_{1}=0}^{n-1-k_{2}}\binom {n-1}{k_{1},k_{2},n-1-k_{1}-k_{2}}\lambda ^{k_{1}+k_{2}}B_{k_{2}}^{(n)}B_{k_{1}}^{(n-k_{2})}x^{n-k_{1}-k_{2}} \\ &=\sum_{k=1}^{n} \biggl(\sum _{k_{1}+k_{2}=n-k}\binom{n-1}{k_{1},k_{2},k-1}\lambda ^{n-k}B_{k_{2}}^{(n)}B_{k_{1}}^{(n-k_{1})} \biggr)x^{k}. \end{aligned}$$
(5)
By induction on m, we obtain the following result.
Theorem 3.1
For all
\(m,n\geq1\),
$$\begin{aligned} r_{n}^{(m)}(x)=\sum_{k=1}^{n} \Biggl(\sum_{k_{1}+\cdots+k_{m}=n-k}\binom {n-1}{k_{1},\ldots,k_{m},k-1} \lambda^{n-k} \prod_{j=1}^{m}B_{k_{j}}^{(n-\sum_{i=j+1}^{m}k_{i})} \Biggr)x^{k}. \end{aligned}$$
Note that by combining the two expressions (see (4) and Theorem 3.1) for \(r_{n}^{(m)}(x)\), we obtain the same result as obtained in [8], Theorem 4:
Corollary 3.2
For all
\(1\leq k\leq n\)
and
\(m\geq1\),
$$\begin{aligned} &\sum_{\ell_{1},\ldots,\ell_{m-1}=0}^{n}S_{1}(n, \ell_{1})S_{1}(\ell_{1},\ell_{2}) \cdots S_{1}(\ell_{m-1},k) \\ & =\sum_{k_{1}+\cdots+k_{m}=n-k}\binom{n-1}{k_{1},\ldots,k_{m},k-1} B_{k_{m}}^{(n)}B_{k_{m-1}}^{(n-k_{m})} \cdots B_{k_{1}}^{(n-k_{m}-\cdots-k_{2})}. \end{aligned}$$
Note that for \(m=1\), the above corollary reduces to \(S_{1}(n,k)=\binom {n-1}{k-1}B_{n-k}^{(n)}\).
Degenerate Bell polynomials
Now, let us take the associated sequence \(r_{n}(x)\) to \(f(t)=\frac{(1+t)^{\lambda}-1}{\lambda}\). So \(r_{n}(x)=\sum_{k=0}^{n}S_{2}(n, k|\lambda)x^{k}\sim(1,f(t))\) and \(r_{n}^{(m)}(x)=\sum_{k=0}^{n}r_{n,k}^{(m)}x^{k}\sim(1,f^{m}(t))\). Here, \(S_{2}(n,k|\lambda)\) are the degenerate Stirling numbers of the second kind obtained by \(\frac{1}{k!}((1+\lambda t)^{1/\lambda}-1)^{k} =\sum_{n\geq k}S_{2}(n,k|\lambda)\frac{t^{n}}{n!}\) (see [9]). Indeed, as \(\bar{f}(t)=(1+\lambda t)^{1/\lambda}-1\), we get
$$\begin{aligned} \sum_{n\geq0}r_{n}(x)\frac{t^{n}}{n!}&=e^{x((1+\lambda t)^{1/\lambda}-1)} =\sum_{k\geq0}\bigl((1+\lambda t)^{1/\lambda}-1 \bigr)^{k}\frac{x^{k}}{k!} \\ &=\sum_{n\geq0} \Biggl(\sum _{k=0}^{n}S_{2}(n,k|\lambda)x^{k} \Biggr)\frac{t^{n}}{n!}. \end{aligned}$$
Thus, \(r_{n}(x)=\sum_{k=0}^{n}S_{2}(n,k|\lambda)x^{k}\). As \(\lambda\rightarrow0 \), \(r_{n}(x)\) tends to the Bell polynomial \(Bel_{n}(x)=\sum_{k=0}^{n}S_{2}(n,k)x^{k}\). Hence they may be called the degenerate Bell polynomials. From (3), we obtain
$$\begin{aligned} r_{n,k}^{(m)}&=\sum_{\ell_{1},\ldots,\ell_{m-1}=0}^{n}S_{2}(n, \ell_{1}|\lambda )S_{2}(\ell_{1}, \ell_{2}|\lambda) \cdots S_{2}(\ell_{m-1},k| \lambda). \end{aligned}$$
(6)
On the other hand, from the transfer formula, we get, for \(n\geq1\), \(r_{n}(x)=x(\lambda t/((1+t)^{\lambda}-1))^{n}x^{-1}x^{n}\).
Recall that the Korobov polynomials
\(K_{n,(r)}(\lambda,x)\) of order r (see [10]) are given by generating function \((\lambda t/((1+t)^{\lambda}-1))^{r}(1+t)^{n}=\sum_{n\geq0}K_{n,(r)}(\lambda ,x)\frac{t^{n}}{n!}\). For \(x=0\), \(K_{n,(r)}(\lambda)=K_{n,(r)}(\lambda ,0)\) are called the Korobov numbers of order r. Note that \(K_{n,(r)}(\lambda,x)\) should be distinguished from \(K_{n}^{(r)}(\lambda ,x)\), which denotes the rth power under umbral composition of \(K_{n}(\lambda,x)\). Thus,
$$\begin{aligned} r_{n}(x)&=x\sum_{k\geq0}K_{k,(n)}( \lambda)\frac{t^{k}}{k!}x^{n-1} =\sum_{k=1}^{n} \binom{n-1}{k-1}K_{n-k,(n)}(\lambda)x^{k}. \end{aligned}$$
To proceed further, we observe the following: for \(0\leq m\leq n-1\),
$$\begin{aligned} f(t)^{m}x^{-1}r_{n}(x)&=f(t)^{m} \bigl(t/f(t)\bigr)^{n}x^{n-1}=\bigl(t/f(t)\bigr)^{n-m}t^{m}x^{n-1} \\ &=(n-1)_{m}\bigl(\lambda t/\bigl((1+t)^{\lambda}-1\bigr) \bigr)^{n-m}x^{n-1-m} \\ &=(n-1)_{m}\sum_{\ell\geq0}K_{\ell,(n-m)}( \lambda)\frac{t^{\ell}}{\ell !}x^{n-1-m} \\ &=\sum_{\ell=0}^{n-1-m}\frac{(n-1)_{\ell+m}}{\ell!}K_{\ell ,(n-m)}( \lambda)x^{n-1-\ell-m}. \end{aligned}$$
Thus, by induction on m and (3), we can state the following formula.
Theorem 3.3
For all
\(m, n\geq1\),
$$r_{n}^{(m)}(x)=\sum_{k=1}^{n} \Biggl(\sum_{k_{1}+\cdots+k_{m}=n-k}\binom {n-1}{k_{1},\ldots,k_{m},k-1}\prod _{j=1}^{m}K_{k_{j},(n-\sum _{i=j+1}^{m}k_{i})}(\lambda) \Biggr)x^{k}. $$
Combining the two expressions for \(r_{n}^{(m)}(x)\) (see (6) and Theorem 3.3), we obtain the following corollary.
Corollary 3.4
For all
\(1\leq k\leq n\)
and
\(m\geq1\),
$$\begin{aligned} &\sum_{\ell_{1},\ldots,\ell_{m-1}=0}^{n}S_{2}(n, \ell_{1}|\lambda)S_{2}(\ell_{1},\ell _{2}|\lambda)\cdots S_{2}(\ell_{m-1},k|\lambda)\\ &\quad =\sum_{k_{1}+\cdots+k_{m}=n-k}\binom{n-1}{k_{1},\ldots ,k_{m},k-1}\prod _{j=1}^{m}K_{k_{j},(n-\sum_{i=j+1}^{m}k_{i})}(\lambda). \end{aligned}$$
Note that the above corollary with \(m=1\) shows that \(S_{2}(n,\ell _{1}|\lambda)=\binom{n-1}{k-1}K_{n-k,(n)}(\lambda)\).
Degenerate falling factorial polynomials
As for third example, let us consider the associated sequence \(r_{n}(x)\) to \(f(t)=(1+\lambda t)^{1/\lambda}-1\). So, \(r_{n}(x)=\sum_{k=0}^{n} S_{1}(n,k|\lambda)x^{k}\sim(1,f(t))\) and \(r_{n}^{(m)}(x)=\sum_{k=0}^{n} r_{n,k}^{(m)}x^{k}\sim(1,f^{m}(t))\). Here, \(S_{1}(n,k|\lambda)\) are the degenerate Stirling numbers of the first kind (see [4, 9]) given by \(\frac{1}{k!}((1+t)^{\lambda}-1)^{k}/\lambda^{k}=\sum_{n\geq k}S_{1}(n,k|\lambda )\frac{t^{n}}{n!}\). \(r_{n}(x)\) may be called the degenerate falling factorial polynomials, since, as \(\lambda\rightarrow0\), \(r_{n}(x)\) tends to the falling factorial polynomial \((x)_{n}=\sum_{k=0}^{n} S_{1}(n,k)x^{k}\). From (3), we obtain
$$\begin{aligned} &r_{n,k}^{(m)}=\sum _{\ell_{1},\ldots,\ell_{m-1}=0}^{n}S_{1}(n,\ell_{1}| \lambda )S_{1}(\ell_{1},\ell_{2}|\lambda)\cdots S_{1}(\ell_{m-1},k|\lambda). \end{aligned}$$
(7)
We recall that the degenerate Bernoulli polynomials \(\beta _{n,(r)}(\lambda,x)\) of order r are defined by the generating function
$$t^{r}/\bigl((1+\lambda t)^{1/\lambda}-1\bigr)^{r}(1+ \lambda t)^{x/\lambda}=\sum_{n\geq 0} \beta_{n,(r)}(\lambda,x)\frac{t^{n}}{n!}. $$
For \(x=0\), \(\beta_{n,(r)}(\lambda)=\beta_{n,(r)}(\lambda,0)\) are called the degenerate Bernoulli numbers of order r. Here, \(\beta _{n,(r)}(\lambda,x)\) should not be confused with \(\beta ^{(r)}_{n}(\lambda,x)\), which denotes the rth power under umbral composition of \(\beta_{n}(\lambda,x)\). So, by these definitions, for \(n\geq1\), we have
$$\begin{aligned} r_{n}(x)&=x\frac{t^{n}}{((1+\lambda t)^{1/\lambda}-1)^{n}}x^{n-1}=x\sum _{k\geq 0}\beta_{k,(n)}(\lambda)\frac{t^{k}}{k!}x^{n-1} \\ &=\sum_{k=1}^{n}\binom{n-1}{k-1} \beta_{n-k,(n)}(\lambda)x^{k}. \end{aligned}$$
Thus, for \(0\leq m\leq n-1\), we have
$$\begin{aligned} f(t)^{m}x^{-1}r_{n}(x)&=\bigl(t/f(t) \bigr)^{n-m}t^{m}x^{n-1} \\ &=(n-1)_{m}\frac{t^{n-m}}{((1+\lambda t)^{1/\lambda}-1)^{n-m}}x^{n-1-m} \\ &=\sum_{\ell=0}^{n-1-m}\frac{(n-1)_{\ell+m}}{\ell!} \beta_{\ell ,(n-m)}(\lambda)x^{n-1-\ell-m}. \end{aligned}$$
By using similar arguments to (5), we obtain
$$\begin{aligned} r_{n}^{(2)}(x)&=x\bigl(f(t)/f^{2}(t) \bigr)^{n}x^{-1}r_{n}(x) \\ &=\sum_{k=1}^{n} \biggl(\sum _{k_{+}k_{2}=n-k}\binom{n-1}{k_{1},k_{2},k-1}\beta _{k_{2},(n)}(\lambda) \beta_{k_{1},(n-k_{2})}(\lambda) \biggr)x^{k}. \end{aligned}$$
Hence, by induction on m, we derive the following result.
Theorem 3.5
For all
\(m, n\geq1\),
$$r_{n}^{(m)}(x)=\sum_{k=1}^{n} \Biggl(\sum_{k_{1}+\cdots+k_{m}=n-k}\binom {n-1}{k_{1},\ldots,k_{m},k-1}\prod _{j=1}^{m}\beta_{k_{j},(n-\sum _{i=j+1}^{m}k_{i})}(\lambda) \Biggr)x^{k}. $$
Combining the two expressions for \(r_{n}^{(m)}(x)\) (see (7) and Theorem 3.5), we obtain the following corollary.
Corollary 3.6
For all
\(1\leq k\leq n\)
and
\(m\geq1\),
$$\begin{aligned} &\sum_{\ell_{1},\ldots,\ell_{m-1}=0}^{n}S_{1}(n, \ell_{1}|\lambda)S_{1}(\ell_{1},\ell _{2}|\lambda)\cdots S_{1}(\ell_{m-1},k|\lambda)\\ &\quad =\sum_{k_{1}+\cdots+k_{m}=n-k}\binom{n-1}{k_{1},\ldots ,k_{m},k-1}\prod _{j=1}^{m}\beta_{k_{j},(n-\sum_{i=j+1}^{m}k_{i})}(\lambda). \end{aligned}$$
Note that the above corollary with \(m=1\) shows that \(S_{1}(n,k|\lambda )=\binom{n-1}{k-1}\beta_{n-k,(n)}(\lambda)\).
Generalized Bell polynomials
One more example is Bell polynomials, also called the exponential polynomials, which are given by the generating function \(e^{x(e^{t}-1)}=\sum_{n\geq0}Bel_{n}(x)\frac {t^{n}}{n!}\) (see [4]). Let \(r_{n}(x)=\lambda^{n}Bel_{n}(x/\lambda) =\sum_{k=0}^{n}\lambda^{n-k}S_{n}(n,k)x^{k}\sim(1,f(t))\) with \(f(t)=\frac {1}{\lambda}\log(1+\lambda t)\). Thus, \(\bar{f}(t)=\frac{e^{\lambda t}-1}{\lambda}\), which implies
$$\begin{aligned} \sum_{n\geq0}r_{n}(x)\frac{t^{n}}{n!}&=e^{\frac{x}{\lambda}(e^{\lambda t}-1)}= \sum_{n\geq0}\lambda^{n}Bel_{n}(x/ \lambda)\frac{t^{n}}{n!} \\ &=\sum_{n\geq0} \Biggl(\sum _{k=0}^{n}\lambda^{n-k}S_{2}(n,k)x^{k} \Biggr)\frac{t^{n}}{n!}. \end{aligned}$$
Thus, from (3), we get
$$\begin{aligned} r_{n,k}^{(m)}=\lambda^{n-k}\sum _{\ell_{1},\ldots,\ell_{m-1}=0}^{n} S_{2}(n, \ell_{1})S_{2}(\ell_{1},\ell_{2})\cdots S_{2}(\ell_{m-1},k). \end{aligned}$$
(8)
To proceed we recall that the Bernoulli polynomials of the second kind
\(b_{n,(r)}(x)\) of order r are given by the generating function
$$\bigl(t/\log(1+t)\bigr)^{r}(1+t)^{x}=\sum _{n\geq0}b_{n,(r)}(x)\frac{t^{n}}{n!} $$
(see [4]). For \(x=0\), \(b_{n,(r)}=b_{n,(r)}(0)\) are the Bernoulli numbers of the second kind of order r. Here, \(b_{n,(r)}(x)\) should be distinguished from \(b_{n}^{(r)}(x)\), which denotes the rth power under umbral composition of \(b_{n}(x)\). So, for \(n\geq1\),
$$\begin{aligned} r_{n}(x)&=x \biggl(\frac{\lambda t}{\log(1+\lambda t)} \biggr)^{n}x^{n-1} =x\sum_{k\geq0}b_{k,(n)}\frac{\lambda^{k}t^{k}}{k!}x^{n-1} \\ &=\sum_{k=1}^{n}\binom{n-1}{k-1} \lambda^{n-k}b_{n-k,(n)}x^{k}. \end{aligned}$$
For \(0\leq m\leq n-1\), we observe that
$$f(t)^{m}x^{-1}r_{n}(x)=\bigl(t/f(t) \bigr)^{n-m}t^{m}x^{n-1}=\sum _{\ell=0}^{n-1-m}\frac {(n-1)_{\ell+m}}{\ell!}\lambda^{\ell}b_{\ell,(n-m)}x^{n-1-\ell-m}. $$
Thus, by induction on m (similar to Theorem 3.1), one can obtain the following formula.
Theorem 3.7
For all
\(m, n\geq1\),
$$r_{n}^{(m)}(x)=\sum_{k=1}^{n} \Biggl(\sum_{k_{1}+\cdots+k_{m}=n-k}\lambda ^{n-k} \binom{n-1}{k_{1},\ldots,k_{m},k-1}\prod_{j=1}^{m}b_{k_{j},(n-\sum _{i=j+1}^{m}k_{i})} \Biggr)x^{k}. $$
Combining the two expressions for \(r_{n}^{(m)}(x)\) (see (8) and Theorem 3.7), we obtain the following corollary.
Corollary 3.8
For all
\(1\leq k\leq n\)
and
\(m\geq1\),
$$\begin{aligned} &\sum_{\ell_{1},\ldots,\ell_{m-1}=0}^{n}S_{2}(n, \ell_{1})S_{2}(\ell_{1},\ell _{2})\cdots S_{2}(\ell_{m-1},k)\\ &\quad =\sum_{k_{1}+\cdots+k_{m}=n-k}\binom{n-1}{k_{1},\ldots ,k_{m},k-1}\prod _{j=1}^{m}b_{k_{j},(n-\sum_{i=j+1}^{m}k_{i})}. \end{aligned}$$
Note that the above corollary with \(m=1\) shows that \(S_{2}(n,k)=\binom {n-1}{k-1}b_{n-k,(n)}\).