= Solution
First retain the exact initial normalization. The <scale factor> obeys $a/a_i=(t/t_i)^{2/3}$, so the <cosmological redshift> is
$$
1+z=(1+z_i)\left[\frac{t_i}{B(\theta-\sin\theta)-T}\right]^{2/3}.
$$
Also $\rho=3M/(4\pi r^3)$ and $\rho_m=1/(6\pi Gt^2)$. Using $A^3=GMB^2$ gives
$$
1+\delta=\frac92\frac{[\theta-\sin\theta-T/B]^2}{(1-\cos\theta)^3}.
$$
These equations exactly accommodate the finite-time initial conditions in <2/b>.
Now neglect $T$ and retain leading order in $\delta_i\ll1$. Since $t_i/B=4\delta_i^{3/2}/[3(1+\delta_i)]$, the <cosmological redshift> relation reduces to
$$
1+z\simeq\left(\frac43\right)^{2/3}\frac{\delta_i(1+z_i)}{(\theta-\sin\theta)^{2/3}}.
$$
Define $\delta_0=\frac35\delta_i(1+z_i)$. Then
$$
\boxed{1+z\simeq\frac53\left(\frac43\right)^{2/3}
\frac{\delta_0}{(\theta-\sin\theta)^{2/3}},\qquad
\delta\simeq\frac92\frac{(\theta-\sin\theta)^2}{(1-\cos\theta)^3}-1.}
$$
The second equality is exact for the ideal common-bang growing-mode <spherical-collapse model>; it is approximate for the paper's finite-time initial Hubble-velocity data.
The factor $3/5$ has a physical origin. Linear mass conservation gives $\dot\delta_i=0$ when the initial local expansion equals the background <Hubble flow>. The general linear dust solution is $\delta_L=C_g(t/t_i)^{2/3}+C_d(t/t_i)^{-1}$. Imposing $\delta_L(t_i)=\delta_i$ and $\dot\delta_L(t_i)=0$ yields
$$
\delta_L(t)=\frac35\delta_i\left(\frac t{t_i}\right)^{2/3}
+\frac25\delta_i\left(\frac t{t_i}\right)^{-1}.
$$
Thus $\delta_0$ is the present extrapolation of the \i[growing component], after neglecting the decaying transient. It is not the exact present value of the full linear solution for finite $t_i$.
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