= Solution
The small <baryon-to-photon ratio> suppresses nuclear equilibrium abundances through the factor $\eta^{A-1}$. Cooling eventually makes the binding-energy exponential favourable, but the <deuterium bottleneck> delays substantial nuclear burning until well after <cosmological weak freeze-out>. In ordinary <Big Bang nucleosynthesis>, nearly every <neutron> surviving that delay ends up in helium-4. This makes the final <primordial helium mass fraction> chiefly a <neutron> budget rather than an unrestricted equilibrium abundance.
Let $r_F$ be the <neutron-to-proton ratio> at freeze-out, let $\Delta t$ be the delay until effective nuclear burning, and let $\tau_n$ be the <neutron> mean lifetime. The initial fraction of <baryons> which are <neutrons> is $x_{n,F}=r_F/(1+r_F)$. Because <neutron> decay converts a <neutron> into a <proton>, total <baryon number> remains constant, while the <neutron> fraction becomes $x_{n,B}=x_{n,F}e^{-\Delta t/\tau_n}$. Each helium nucleus uses two <neutrons> and four <baryons>. Hence <neutron-limited helium synthesis> gives
$$
\boxed{Y_4\simeq\frac{2r_F}{1+r_F}e^{-\Delta t/\tau_n},\qquad
\tau_n=\frac{\tau_{1/2}(n)}{\ln2}.}
$$
Equivalently, $Y_4=2r_B/(1+r_B)$ with $r_B=x_{n,B}/(1-x_{n,B})$. This accounts for the extra <protons> produced by <neutron> decay; multiplying $r_F$ alone by the survival factor would miss their contribution. The following responses refer to ordinary neutron-limited burning, without exotic particle injection or a change to the nuclear network.
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