= Solution
There are three free-particle rest-energy terms and five nuclear corrections. In the paper's <natural units>, their meanings are:
* $Zm_p$: the rest <energy> of $Z$ free <protons>.
* $Zm_e$: the rest <energy> of the atomic <Electrons>; including this term makes the result an atomic, rather than bare nuclear, <mass>. Small atomic binding energies are neglected.
* $(A-Z)m_n$: the rest <energy> of the free <neutrons>.
* $a_1A$, with $a_1<0$: the attractive volume term. The short-range <nuclear force> mainly binds each <nucleon> to nearby neighbours, so the leading attraction is extensive.
* $a_2A^{2/3}>0$: the surface correction. Surface <nucleons> have fewer neighbours and are less bound; surface area grows as $A^{2/3}$ at approximately constant nuclear density.
* $a_3Z^2/A^{1/3}>0$: the <nuclear Coulomb energy> cost of repelling <protons>, scaling as $Z^2/R$ with nuclear radius $R\propto A^{1/3}$.
* $a_4(A-2Z)^2/A>0$: the <nuclear symmetry energy>, penalizing unequal neutron and proton populations. The <Pauli exclusion principle> forces an excess species into higher single-particle states; strong-interaction effects also contribute.
* $\delta(A)/A^{1/2}$: <nuclear pairing>. It lowers the mass for even-even <nuclei>, raises it for odd-odd <nuclei>, and vanishes for odd $A$ in the stated model.
The sign convention is significant: the quantity labeled $E_{\mathrm{bind}}$ in the supplied mass formula is added to the free-particle masses and is normally negative for a bound nucleus. It is the negative of the conventional positive <nuclear binding energy> $B$:
$$
\boxed{M=Zm_p+Zm_e+(A-Z)m_n+E_{\mathrm{bind}},\qquad B=-E_{\mathrm{bind}}>0.}
$$
The coefficients must therefore retain their printed signs; using positive $B$ while still adding it to the masses would be inconsistent.
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