How does the author derive these relations?

$color{brown}{textbf{Preliminary note.}}$

Equation $$(r^2-r)f''(r) + f'(r) -l(l+1)f(r)= 0tag1$$ has exact solution $$f(r) = c_1 r^2operatorname{_2F_1}(1-l, 2+l; 3; r) + c_2 operatorname {G_{2,,2}^{2,,0}} left(rbig|_{0,,2}^{1-l,,2+l}right),$$ where $operatorname{_2F_1}(a,b;c;z)$ is the Gauss hypergeometric function and $G$ is the Meijer G-function,

To solve equation $(1)$ can be used Frobenius Metod.

$color{brown}{textbf{Substitution.}}$

The author of the paper uses Frobenius method after the substitution $$f(r) = left(dfrac{r^2}{r-1},e^{r-1}right)^{-rho} g(r),tag2$$ which reduces the polynomials degree.

The given linear homogenius ODE is $$(r^2-r)f''_{rr} + f'_r-left(dfrac{rho^2r^3}{r-1}+l^2+l-dfrac3rright)f(r) = 0.tag3$$

Substiution $(2)$ leads to the equation $$dfrac{r-1}r (r^2g''_{rr}) + left(dfrac{4rho+1}{r^2}-2rhoright) (r^2 g'_r) - left(4rho^2+l^2+l + dfrac{4rho^2+4rho-3}rright)g =0.tag4$$

$color{brown}{textbf{Frobenius method.}}$

Let $$t=dfrac{r-1}r,quad dfrac1r = 1-t,quad g(r)=sumlimits_{n=0}^infty a_n t^n,tag5$$ then begin{align} &dfrac{4rho+1}{r^2}-2rho = (4rho+1)(1-t)^2-2rho = 2rho+1 - (8rho+2)t + (4rho+1)t^2,\[4pt] &4rho^2+l^2+l + dfrac{4rho^2+4rho-3}r =l^2+l+8rho^2+4rho-3 -(4rho^2+4rho-3)t,\[4pt] &t'_r = dfrac1{r^2} = (1-t)^2,\[4pt] &r^2g'_r = dfrac{g'_r}{t'_r} = g'_t = sumlimits_{n=1}^infty n a_n t^{n-1},\[4pt] &g'_r = (1-t)^2g'_t = sumlimits_{n=1}^infty n a_n,t^{n-1} -2sumlimits_{n=1}^infty n a_n,t^n +sumlimits_{n=1}^infty na_n,t^{n+1},\[4pt] &r^2g''_{rr} = (g'_r)'_t = sumlimits_{n=1}^infty n(n-1)a_n,t^{n-2} -2sumlimits_{n=1}^infty n^2a_n,t^{n-1} +sumlimits_{n=1}^infty(n+1)n a_n,t^n, end{align}

and from $(4)$ should begin{align} &sumlimits_{n=1}^infty n(n-1)a_n,t^{n-1} -2sumlimits_{n=1}^infty n^2a_n,t^{n} +sumlimits_{n=1}^infty(n+1)n a_n,t^{n+1}\ &+(2rho+1) sumlimits_{n=1}^infty n a_n t^{n-1} -(8rho+2)sumlimits_{n=1}^infty n a_n t^{n} + (4rho+1)sumlimits_{n=1}^infty n a_n t^{n+1}\ & - (l^2+l+8rho^2+4rho-3)sumlimits_{n=0}^infty a_n t^n +(4rho^2+4rho-3)sumlimits_{n=0}^infty a_n t^{n+1} =0,\[4pt] &sumlimits_{n=1}^infty(n+1)na_{n+1},t^n -2sumlimits_{n=1}^infty n^2a_n,t^n +sumlimits_{n=1}^infty n(n-1)a_{n-1},t^{n}\ &+(2rho+1)a_1+(2rho+1)sumlimits_{n=1}^infty(n+1)a_{n+1}t^{n}\ &-(8rho+2)sumlimits_{n=1}^infty na_{n}t^{n} + (4rho+1)sumlimits_{n=1}^infty (n-1) a_{n-1} t^{n}\ &- (l^2+l+8rho^2+4rho-3)a_0 - (l^2+l+8rho^2-4rho-3)sumlimits_{n=1}^infty a_{n} t^{n}\ & +(4rho^2+4rho-3)sumlimits_{n=1}^infty a_{n-1} t^{n} = 0,\[4pt] &(2rho+1)a_1- (l^2+l+8rho^2+4rho-3)a_0\[4pt] &+sumlimits_{n=1}^infty big((n+1)n+(2rho+1)(n+1)big)a_{n+1},t^n\[4pt] &+sumlimits_{n=1}^infty big(-2n^2-(8rho+2)n-(l^2+l+8rho^2+4rho-3)big)a_n,t^n\[4pt] &+sumlimits_{n=1}^infty big(n(n-1)+ (4rho+1)(n-1)+(4rho^2+4rho-3)big)a_{n-1},t^{n}=0,\[4pt] end{align} begin{cases} (2rho+1)a_1- (l^2+l+8rho^2+4rho-3)a_0 = 0,\[4pt] big(n^2 + (2rho+2)n+2rho+1big)a_{n+1}\[4pt] -big(color{red}{mathbf{2}}n^2 +(8rho+2) n+(l^2+l+8rho^2+4rho-3)big)a_n\[4pt] +big(n^2+4rho n + 4rho^2-4big)a_{n-1}=0,\[4pt] end{cases} with the single difference in the coefficients from the pointed.