$$V_{2}(r) =
    \displaystyle\frac{2\alpha^{2}}
    {\biggl(r + r_{0} + \displaystyle\frac{1}{C\alpha}\biggr)^{2}}.
\label{eq.3.1.5}$$ We see, that this potential is finite in the whole region of its definition at any values of the parameters $C>0$ and $r_{0} \ge 0$. Thus, we have obtained the reflectionless potential of the inverse power type with a shift to the left, which is defined on the whole positive semiaxis of $r$ (including $r=0$ and $r_{0}=0$).

In accordance with [@Maydanyuk.2005.APNYA] (see p. 452–455, sec. 5.1.2), one can construct a hierarchy of the inverse power potentials, and a general solution of the potential with arbitrary number $n$ can be written down so: $$\begin{array}{ll}
  V_{n}(r) =
    \displaystyle\frac{\gamma_{n} \alpha^{2}}{\bar{r}^{2}}, &
  \gamma_{n \pm 1} = 1 + \gamma_{n} \pm \sqrt{4\gamma_{n}+1}.
\end{array}
\label{eq.3.1.6}$$ If to require, that the first potential $V_{1}(r)$ in this hierarchy must be constant (i. e. at $\gamma_{1}=0$ and $n=1$), then all hierarchy of the inverse power potentials (\[eq.3.1.6\]) becomes the *hierarchy of the reflectionless inverse power potentials*, and the solution (\[eq.3.1.5\]) becomes the general solution for the reflectionless inverse power potential. Note, that *when the hierarchy of the inverse power potentials becomes reflectionless, then the coefficients $\gamma_{n}$ become integer numbers*. We write its first values: $$\gamma_{n} = 0, 2, 6, 12, 20, 30, 42...
\label{eq.3.1.7}$$

Now, if to calculate $\beta_{n}$ for given $\gamma_{n}$ with number $n$ from (\[eq.3.1.7\]) from the following condition: $$\beta_{n} (\beta_{n}-\alpha) = \gamma_{n} \alpha^{2},
\label{eq.3.1.8}$$ then the first potential $V_{1}(r)$ from (\[eq.3.1.4\]) becomes reflectionless inverse power potential (at $\beta =\beta_{n}$). The second potential $V_{2}(r)$ from (\[eq.3.1.4\]) is finite in the whole region of its definition (including $r=0$) and should be reflectionless also, however it is not inverse power potential. So, substituting the coefficients $\gamm