Suppose that we are given a first-order system of ordinary differential equations $$\frac{\mathrm{d} f(x)}{\mathrm{d}x} = A(x) f(x),$$ where $A(x)$ is an $N \times N$ matrix which has an expansion of the form $A(x) = \sum_{n=n_0}^\infty A_n x^n$ where $n_0 \in \mathbb{Z}$ and $A_0 \neq 0_{N \times N}$. How are regular singular points defined for ODE systems like this?
If $n_0=0$ then $x=0$ is an ordinary point and we will find power series solutions for $f(x)$. For $n_0=-1$, $x=0$ is a regular singular point and one finds generalised power series solutions for $f(x)$. But it seems to me that having $n_0=-1$ (i.e. only simple poles in $A(x)$) is sufficient, but not necessary. In general, having double poles can lead to irregular singular points, but at least for systems of differential equations it seems that one can have double poles in $x=0$ and still find that it appears to be a regular singular point.
Example: Consider the system with $$A(x) = \begin{pmatrix} \frac{1}{x} & 1 \\ 1 & \frac{1}{x} \end{pmatrix}.$$ This matrix has simple poles, i.e., $n_0 = -1$, and I find the following solutions $$f(x) = C_1 x \mathrm{e}^{-x} \begin{pmatrix} 1 \\ -1 \end{pmatrix} + C_2 x \mathrm{e}^x \begin{pmatrix} 1 \\ 1 \end{pmatrix}.$$ $x=0$ appears to be a regular singular point.
Now consider the transformation $\tilde{f}(x) = T(x) f(x)$ with $$T(x) = \begin{pmatrix} 1 & 0 \\ \frac{1}{x} & 1 \end{pmatrix}.$$ This leads to the system $$\frac{\mathrm{d} \tilde{f}(x)}{\mathrm{d} x} = B(x) \tilde{f}(x), \text{ where } B(x) = T(x)A(x)T^{-1}(x) + \frac{\mathrm{d} T(x)}{\mathrm{d}x} T^{-1}(x) = \begin{pmatrix} 0 & 1 \\ 1-\frac{2}{x^2} & \frac{2}{x} \end{pmatrix}.$$ Obviously, this system is solved by $$\tilde{f}(x) = T(x) f(x) = C_1 \mathrm{e}^{-x} \begin{pmatrix} x \\ 1-x \end{pmatrix} + C_2 \mathrm{e}^{x} \begin{pmatrix} x \\ 1+x \end{pmatrix}.$$ This solution still falls into the category of generalised power series. Moreover, if one derives scalar second-order differential equations for the components of $\tilde{f}(x) = (\tilde{f}_1(x),\tilde{f}_2(x))^T$, one finds that they fulfill the equations $$\tilde{f}_1''(x) = \frac{2}{x} \tilde{f}_1'(x) + \left(1-\frac{2}{x^2}\right) \tilde{f}_1(x), \\ \tilde{f}_2''(x) = \frac{2 x}{x^2-2} \tilde{f}_2'(x) + \left(1 + \frac{4}{2-x^2}\right) \tilde{f}_2(x).$$ The ODE for $\tilde{f}_1(x)$ has $x=0$ as a regular singular point and the ODE for $\tilde{f}_2(x)$ has $x=0$ even as an ordinary point. This is not surprising, given that the power series expansion of $\tilde{f}_2(x)$ starts with a constant term.
I bring up this example just to illustrate that a first order system whose matrix $B(x)$ has a double pole, seems to have $x=0$ as a regular singular point / ordinary point.
This brings me back to the original question: How are regular singular points defined for ODE systems and how does one characterise whether a ODE system with higher poles in its matrix $A(x)$ admits (generalised) power series solutions?