**Case 1**

Regarding the first, we define the operator [7]:

$$\begin{aligned} \hbar^r \psi\_0 &= \frac{1}{r}, \quad r > 0, \\ \hbar^0 \psi\_0 &= 1, \end{aligned} \tag{46}$$

thus casting the first of Equation (45) in the form (see Equation (32) in Reference [7]):

$$
\mu\_1 \mathfrak{h}\_n = 1 - (1 - \mathfrak{k})^n \mathfrak{\psi}\_{0\prime} \tag{47}
$$

which can be exploited to once more derive the Gosper generating function. By multiplying both sides of Equation (47) by *tnn*!, and then by summing up on the index *n*, we obtain:

$$\sum\_{n=0}^{\infty} \frac{t^n}{n!} h\_n = e^t \left(1 - e^{-\hat{\mathbf{x}}t}\right). \tag{48}$$

Keeping the m-th derivative with respect to both sides of Equation (48) yields:

$$\sum\_{n=0}^{\infty} \frac{t^n}{n!} \,\_1h\_{n+m} = e^t \sum\_{r=0}^m \binom{m}{r} \partial\_t^r \phi(t),\tag{49}$$

where

$$\begin{aligned} \boldsymbol{\phi}(t) &= 1 - e^{-\boldsymbol{\kappa}t} \boldsymbol{\psi}\_{0}, \\ \boldsymbol{\partial}'\_{t} \boldsymbol{\phi}(t) &= -(-\boldsymbol{\kappa})' e^{-\boldsymbol{\kappa}t} \boldsymbol{\psi}\_{0} = -\sum\_{s=0}^{\infty} \frac{(-1)^{s+r} \boldsymbol{\kappa}^{s+r}}{s!} t^{s} \boldsymbol{\psi}\_{0} = \begin{cases} -\sum\_{s=0}^{\infty} \frac{(-1)^{s+r}}{s!(s+r)} t^{s}, & r > 0, \\ -\sum\_{r=1}^{\infty} \frac{-t^{r}}{r!r}, & r = 0. \end{cases} \end{aligned} \tag{50}$$
