Angular size of the largest region in causal contact at the moment of recombination

I want to calculate the angular size of the largest region in causal contact at the moment of recombination (z = 1100) to our universe, so we have: $H_0 = 68$ (km/s)/Mpc, $Omega_{m,0} = 0.31$, $Omega_{Lambda,0} = 0.69$ and $Omega_{r,0}=9times 10^{-5}$.
I know that:

$$ theta = frac{d_{hor}(t_{ls})}{d_A}$$
where $d_{hor}(t_{ls})$ is the physical size of the largest region at the recombination and $d_A$ is the angular diameter distance.

My attempt:
To find the distance $d_{A}$ at z = 1100, I can calculate the horizon distance today than use that $$d_A = frac{d_{hor}(t_0)}{1+z} $$
To calculate this we note that:

$$d_{hor}(t_0) = cint_{t_e}^{t_o}frac{dt}{a(t)}$$
But you can write this equation in terms of z and (making limits from $infty$ to $0$) to make calculations easier. Using that $$1 + z = a(t)^{-1}$$ we have:

$$frac{dz}{dt} = frac{dz} {da} frac{da} {dt}$$
$$frac{dz}{dt} = -frac{1}{a^2} dot{a}$$
$$dz = -frac{H}{a}dt = -H(1+z)dt$$
$$dt = -frac{dz}{H(1+z)}$$

Since at $t_e$ corresponds to $z = infty$ and $t_0 = 0$ we have

$$d_{hor}(t_0) = cint_{t_e}^{t_o}frac{dt}{a(t)} = -cint_{infty}^{0}frac{dz}{H}$$

Also we know using Friedmann equation that $$H = H_0sqrt{Omega_r(1+z)^4 + Omega_m(1+z)^3 + Omega_{Lambda}}$$

So:

$$ d_{hor}(t_0) = -frac{c}{H_0}int_{infty}^{0}frac{dz}{sqrt{Omega_r(1+z)^4 + Omega_m(1+z)^3 + Omega_{Lambda}}}$$
Using the above given data, I can do this integral numerically and find that:

$$d_{hor} approx 14121; Mpc$$
and
$$ d_A = 14121/1101 approx 12.943 ; Mpc$$
Until this point I think I’m right. Now the hard and dubious part:

To calculate $d_{hor}(t_{ls})$ I again tought about using the formula:
$$ d_{hor}(t_ls) = -frac{c}{H_0}int_{1100}^{0}frac{dz}{sqrt{Omega_r(1+z)^4 + Omega_m(1+z)^3 + Omega_{Lambda}}}$$
Since at $t_e$ corresponds to $z = 1100$ and $t_0 = 0$, but the result (13842 Mpc) here is very different from the one I expected (0.251 Mpc). Can someone give-me a help to calculate $d_{hor}(t_{ls})$?