= Solution
Put $x=\sqrt{s}R>0$. Orthogonality of the <Legendre polynomials> lets us match each degree separately. Continuity of the field and its radial derivative gives
$$
B_\ell j_\ell(x)=\frac{A_\ell}{R^{\ell+1}},\qquad \sqrt{s}B_\ell j_\ell'(x)=-\frac{(\ell+1)A_\ell}{R^{\ell+2}}.
$$
Eliminate $A_\ell$ by multiplying the first equation by $(\ell+1)/R$ and adding it to the second. A nonzero mode therefore requires
$$
\boxed{xj_\ell'(x)+(\ell+1)j_\ell(x)=0.}
$$
This argument avoids dividing by a potentially zero boundary value. Apply the supplied recurrence with $\nu=\ell-1$:
$$
xj_\ell'(x)+(\ell+1)j_\ell(x)=xj_{\ell-1}(x).
$$
Consequently <spherical Helmholtz matching to a decaying harmonic exterior> gives
$$
\boxed{j_{\ell-1}(\sqrt{s}R)=0,}
$$
where $j_{-1}(x)=\cos x/x$ for the $\ell=0$ sector.
\b[The original PDF has an index error here:] it prints $f_\ell$ rather than $f_{\ell-1}$ in the resulting eigenvalue condition. This cannot be repaired by silently choosing another normalization. If $j_\ell(x)$ itself were zero at a positive matching root, the boundary equation would also give $j_\ell'(x)=0$, contradicting uniqueness for this nonzero solution of a regular differential equation at that point.
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