 Solution

A point $z_0$ is regular singular when $(z-z_0)p$ and $(z-z_0{)}^{2}q$ are analytic there. The Papperitz symbol records the three singular points and their two local exponents. Here the finite points are $0,1$ and the third is $\infty$. The indicial equations give exponents $(0,-1)$ at $0$ and $(0,0)$ at $1$. Comparing with the hypergeometric equation gives $c=2$, $a+b=2$, $ab=1$, hence $a=b=1$ and exponents $(1,1)$ at infinity. Thus