<s>[INST] Why does solving the spherical Bessel equation using Frobenius series produce two quadratic equations for the exponents at the singularity? The spherical Bessel equation is:
    $$x^2y'' + 2xy' + (x^2 - \frac{5}{16})y = 0$$
If I seek a Frobenius series solution, I will have:
    \begin{align*}
        &\quad y = \sum_{n = 0}^{\infty} a_nx^{n + r} \\
        &\implies y' = \sum_{n = 0}^{\infty} (n + r)a_nx^{n + r - 1} \\
        &\implies y'' = \sum_{n = 0}^{\infty} (n + r)(n + r - 1)a_nx^{n + r - 2}
    \end{align*}
Substituting into the ODE of interest:
    \begin{align*}
        &\quad x^2y'' + 2xy' + (x^2 - \frac{5}{16})y = 0 \\
        &\equiv \sum_{n = 0}^{\infty} (n + r)(n + r - 1)a_nx^{n + r} + \sum_{n = 0}^{\infty} 2(n + r)a_nx^{n + r} + \sum_{n = 0}^{\infty} a_nx^{n + r + 2} + \sum_{n = 0}^{\infty} \frac{-5}{16}a_nx^{n + r} = 0 \\
        &\equiv \sum_{n = 0}^{\infty} [(n + r)(n + r - 1) + 2(n + r) - \frac{5}{16}]a_nx^{n + r} + \sum_{n = 2}^{\infty} a_{n - 2}x^{n + r} = 0 \\
        &\equiv [r(r-1) + 2r - (5/16)]a_0 + [(r+1)(r) + 2(r+1) - (5/16)]a_1 + \\
        &\quad \sum_{n = 2}^{\infty} ([(n + r)(n + r - 1) + 2(n + r) - \frac{5}{16}]a_n + a_{n-2})x^{n + r} = 0 \\
        &\implies [r(r-1) + 2r - (5/16)] = 0 \wedge  \\
        &\quad [(r+1)(r) + 2(r+1) - (5/16)] = 0 \wedge \\
        &\quad [(n + r)(n + r - 1) + 2(n + r) - \frac{5}{16}]a_n + a_{n-2} = 0
    \end{align*}
The first conjunct is the standard indicial equation specifying $r$. The second conjunct is yet another quadratic that $r$ has to satisfy. Is there a mistake? Or should $a_1 = 0$?
 [/INST] The main indicial is the the first equation. You get two roots from the first equation so you should set $a_1=0$. If there is a common root between the first and second equations, then you can consider that root and set both $a_1$ and $a_0$ nonzero.
</s>