Unfortunately, erratua appear in the statement corresponding Theorems 2.6 and 2.10 in the original paper. In the statemets of the theorems, the identities must be corrected as follows. After this modification, the amended proofs are also provided.
Theorem 2.6
For \(n,i \in \mathbb {N}\), with \(i \le n\), and \(x \in [0,1]\), we have
$$\begin{aligned} \begin{aligned} (x)_{i,\lambda }=\frac{1}{(1-i\lambda )_{n-i,\lambda }}\sum _{k=i}^{n} \frac{{k \atopwithdelims ()i}}{{n \atopwithdelims ()i}}B_{k,n}(x|\lambda ). \end{aligned} \end{aligned}$$
Theorem 2.10
For \(n,i \in \mathbb {N}\), with \(i \le n\), and \(x \in [0,1]\), we have
$$\begin{aligned} \begin{aligned} \sum _{l=0}^{i} (x)_{l}S_{2,\lambda }(i,l)=\frac{1}{(1-i\lambda )_{n-i,\lambda }}\sum _{k=i}^{n} \frac{{k \atopwithdelims ()i}}{{n \atopwithdelims ()i}}B_{k,n}(x|\lambda ), \end{aligned} \end{aligned}$$
where \((x)_{l}=x(x-1)(x-2)\cdots (x-l+1)\).
Proof of Theorem 2.6
$$\begin{aligned} \sum _{k=1}^{n} \frac{{k \atopwithdelims ()1}}{{n \atopwithdelims ()1}} B_{k,n}(x|\lambda )&=\sum _{k=1}^{n}\frac{k}{n}{n \atopwithdelims ()k}(x)_{k,\lambda }(1-x)_{n-k,\lambda }\\&=\sum _{k=1}^{n} {n-1 \atopwithdelims ()k-1}(x)_{k,\lambda }(1-x)_{n-k,\lambda }\\&=\sum _{k=0}^{n-1} {n-1 \atopwithdelims ()k}(x)_{k+1,\lambda }(1-x)_{n-k-1,\lambda }\\&=(x)_{1,\lambda }\sum _{k=0}^{n-1}{n-1 \atopwithdelims ()k}(x-\lambda )_{k,\lambda }(1-x)_{n-k-1,\lambda }\\&=(x)_{1,\lambda }(1-\lambda )_{n-1,\lambda }. \end{aligned}$$
Now, we observe that
$$\begin{aligned} \begin{aligned} \sum _{k=2}^{n} \frac{{k \atopwithdelims ()2}}{{n \atopwithdelims ()2}}B_{k,n}(x|\lambda )&=\sum _{k=2}^{n} \frac{k(k-1)}{n(n-1)}{n \atopwithdelims ()k} (x)_{k,\lambda }(1-x)_{n-k,\lambda }\\&=\sum _{k=2}^{n} {n-2 \atopwithdelims ()k-2}(x)_{k,\lambda }(1-x)_{n-k,\lambda }\\&=\sum _{k=0}^{n-2}{n-2 \atopwithdelims ()k}(x)_{k+2,\lambda }(1-x)_{n-k-2,\lambda }\\&=(x)_{2,\lambda }\sum _{k=0}^{n-2}{n-2 \atopwithdelims ()k}(x-2\lambda )_{k,\lambda }(1-x)_{n-k-2,\lambda }\\&=(x)_{2,\lambda } (1-2\lambda )_{n-2,\lambda }. \end{aligned} \end{aligned}$$
Similarly, we have
$$\begin{aligned} \begin{aligned} \sum _{k=i}^{n} \frac{{k \atopwithdelims ()i}}{{n \atopwithdelims ()i}}B_{k,n}(x|\lambda )=(x)_{i,\lambda }(1-i\lambda )_{n-i,\lambda }. \end{aligned} \end{aligned}$$
\(\square \)
Proof of Theorem 2.10
For \(n \ge 0\), the degenerate Stirling number of the second kind is given by
$$\begin{aligned} \begin{aligned} (x)_{n,\lambda }=\sum _{k=0}^{n} S_{2,\lambda }(n,k)(x)_{k}. \end{aligned} \end{aligned}$$
Thus, by Theorem 2.6, we get
$$\begin{aligned} \begin{aligned} \sum _{l=0}^{i} (x)_{l} S_{2,\lambda }(i,l)=\frac{1}{(1-i\lambda )_{n-i,\lambda }}\sum _{k=i}^{n} \frac{{k \atopwithdelims ()i}}{{n \atopwithdelims ()i}}B_{k,n}(x|\lambda ). \end{aligned} \end{aligned}$$
\(\square \)
