While answering another question, I made an error and wound up solving for the wrong ODE. However, it occurred to me that the wrong ODE that I solved for would be susceptible to applying the Frobenius method. Having only recently (re)encountered the Frobenius method, I decided to take the opportunity to better acquaint myself with it, but I've encountered similar recurrences that I'm unable to deal with.
Consider the ODE $$(x-1)^2y''-2(x-1)y'+\ln(x)y=0,\tag{1}$$ which has a regular singular point at $x=1.$ Suppose for some $r\in\Bbb R$ that $$y=\sum_{n=0}^\infty a_n(x-1)^{n+r}\tag{2}$$ with $a_0\ne 0.$
Differentiation gives us \begin{eqnarray}(x-1)^2y'' &=& (x-1)^2\sum_{n=0}^\infty a_n\frac{d^2}{dx^2}\bigl[(x-1)^{n+r}\bigr]\\ &=& (x-1)^2\sum_{n=0}^\infty (n+r)(n+r-1)a_n(x-1)^{n+r-2}\\ &=& \sum_{n=0}^\infty (n+r)(n+r-1)a_n(x-1)^{n+r}\end{eqnarray} and \begin{eqnarray}2(x-1)y' &=& 2(x-1)\sum_{n=0}^\infty\frac{d}{dx}\left[a_n(x-1)^{n+r}\right]\\ &=& 2(x-1)\sum_{n=0}^\infty (n+r)a_n(x-1)^{n+r-1}\\ &=& \sum_{n=0}^\infty 2(n+r)a_n(x-1)^{n+r},\end{eqnarray} so that $$(x-1)^2y''-2(x-1)y'=\sum_{n=0}^\infty (n+r)(n+r-3)a_n(x-1)^{n+r}\tag{3}$$
Now, we see that for any $n\ge 0$ we have $$a_n(x-1)^{n+r}\ln(x)=a_n(x-1)^{n+r}\sum_{m=1}^\infty\frac{(-1)^{m-1}}m(x-1)^m=\sum_{m=1}^\infty\frac{(-1)^{m-1}}ma_n(x-1)^{m+n+r},$$ allowing us to rewrite
\begin{eqnarray}\ln(x)y &=& \sum_{n=0}^\infty\sum_{m=1}^\infty\frac{(-1)^{m-1}}ma_n(x-1)^{m+n+r}\\ &=& \sum_{j=0}^\infty\left(\sum_{k=0}^{j-1}\frac{(-1)^{(1+k)-1}}{1+k}a_{j-1-k}\right)(x-1)^{(1+k)+(j-1-k)+r}\\ &=& \sum_{j=0}^\infty\left(\sum_{k=0}^{j-1}\frac{(-1)^ka_{j-1-k}}{1+k}\right)(x-1)^{j+r}\\ &=& \sum_{n=0}^\infty\left(\sum_{k=0}^{n-1}\frac{(-1)^ka_{n-1-k}}{1+k}\right)(x-1)^{n+r},\end{eqnarray}
and reindexing by $i=n-1-k$, we get $$\ln(x)y=\sum_{n=0}^\infty\left(\sum_{i=0}^{n-1}\frac{(-1)^{n-1-i}a_i}{n-i}\right)(x-1)^{j+r}.\tag{4}$$
Now, using $(3)$ and $(4)$ lets us rewrite $(1)$ as $$\sum_{n=0}^\infty\left[(n+r)(n+r-3)a_n+\sum_{i=0}^{n-1}\frac{(-1)^{n-1-i}a_i}{n-i}\right](x-1)^{n+r}=0,$$ or $$\sum_{n=0}^\infty\left[(n+r)(n+r-3)a_n-\sum_{i=0}^{n-1}\frac{(-1)^{n-i}a_i}{n-i}\right](x-1)^{n+r}=0.\tag{5}$$
The $n=0$ term is $$r(r-3)a_0(x-1)^r,$$ which gives us the indicial equation $r(r-3)=0,$ so that $r=0$ or $r=3.$
Case 1: Taking $r=3,$ we suppose that $$y_1=(x-1)^3\sum_{n=0}^\infty a_n(x-1)^n$$ is a solution to $(1)$ and rewrite $(5)$ as $$\sum_{n=0}^\infty\left[(n+3)na_n-\sum_{i=0}^{n-1}\frac{(-1)^{n-i}a_i}{n-i}\right](x-1)^{n+3}=0,$$ or $$\sum_{n=1}^\infty\left[(n+3)na_n-\sum_{i=0}^{n-1}\frac{(-1)^{n-i}a_i}{n-i}\right](x-1)^{n+3}=0.\tag{6}$$ This gives us the recurrence $$a_n=\frac{1}{(n+3)n}\sum_{i=0}^{n-1}\frac{(-1)^{n-i}a_i}{n-i}$$ for all $n\ge 1,$ but while I've been able to find the first bunch of coefficients, I am no closer to finding a closed form. It's possible that none such form exists, and the recurrence means that $y_1$ is some special function with which I'm unfamiliar, but either way, I'm stuck in that case.
Case 2: Taking $r=0,$ we suppose that $$y_2=\sum_{n=0}^\infty b_n(x-1)^n$$ is a solution to $(1)$ and rewrite $(5)$ as $$\sum_{n=0}^\infty\left[n(n-3)b_n-\sum_{i=0}^{n-1}\frac{(-1)^{n-i}b_i}{n-i}\right](x-1)^n=0,$$ or $$\sum_{n=1}^\infty\left[n(n-3)b_n-\sum_{i=0}^{n-1}\frac{(-1)^{n-i}b_i}{n-i}\right](x-1)^n=0.\tag{7}$$ This gives us the condition $$n(n-3)b_n=\sum_{i=0}^{n-1}\frac{(-1)^{n-i}b_i}{n-i}\tag{8}$$ for all $n\ge 1,$ and in particular, the recurrence $$b_n=\frac{1}{n(n-3)}\sum_{i=0}^{n-1}\frac{(-1)^{n-i}b_i}{n-i}$$ for $n>3.$ Applying $(8)$ in the $n=1,2,3$ cases readily gives $b_0=b_1=b_2=0,$ but again, I'm stuck on how to tackle the recurrence.
Case 2 (Alternative approach): Since the roots of the indicial equation are separated by an integer, then if I understand correctly, $$y_2=cy_1\ln(x-1)+\sum_{n=0}^\infty c_n(x-1)^n$$ should be another solution to $(1),$ but I think this still leads me to the second recurrence if I'm to determine $c$ and the constants $c_n$ for $n\ne 3.$ Of course, I'm getting this by trying to extrapolate from a blurb on Wikipedia, so it's quite possible that I'm not on target with this.