= Solution
For the <Frobenius solution>, differentiate term by term:
$$
y'=\sum_{n\geq0}(n+\sigma)a_nx^{n+\sigma-1},\qquad y''=\sum_{n\geq0}(n+\sigma)(n+\sigma-1)a_nx^{n+\sigma-2}.
$$
The unshifted terms in the equation have coefficient $[4(n+\sigma)(n+\sigma-1)-4(n+\sigma)+3]a_n$. The extra factors of $x$ give the shifted contribution $[4(n+\sigma)(n+\sigma-1)+1]a_nx^{n+\sigma+1}$. Equating the coefficient of $x^{k+\sigma}$ for $k\geq1$ gives the <linear recurrence relation>
$$
\boxed{[4(k+\sigma)^2-8(k+\sigma)+3]a_k+[4(k+\sigma-1)(k+\sigma-2)+1]a_{k-1}=0.}
$$
Factoring both coefficients is especially useful:
$$
\boxed{(2k+2\sigma-3)\big[(2k+2\sigma-1)a_k+(2k+2\sigma-3)a_{k-1}\big]=0.}
$$
The separate lowest-power equation is $(4\sigma^2-8\sigma+3)a_0=0$. Do not cancel $2k+2\sigma-3$ when it vanishes: that is the possible <undetermined coefficient at Frobenius resonance>.
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