What is the relationship between binding energy / mass number as a function of mass number?

Specifically, in this figure, which shows the binding energy per nucleon of β-stable nuclei, what is the main contribution to the increasing value of B/A at low values of A? And what is the main cause of the reduction in B/A beyond the maximum value?

You are looking for a formula that expresses the binding energy, with the mass number.

The semi-empirical mass formula states that the binding energy will take the following form:$$$E_B = a_V A - a_{S} A^{2/3} - a_C \frac{Z(Z-1)}{A^{1/3}} - a_A \frac{(A - 2Z)^2} A + \delta(A,Z)$$$

A=Z(proton)+N(neutron)

Now in the function the constituents mean:

$a_V$ volume term: the bigger the volume because the more nucleons, the more interactions due to strong force and the stronger binding energy

$a_S$ surface term: the nucleons on the surface interact with less other nucleons (then the ones at the center) so this decreases the effect of the volume term

$a_C$ Coulomb term: is the EM repulsion between protons

$a_A$ Asymmetry term: because of the Pauli exclusion principle, one type of nucleus, neutron occupies different quantum states then the other type, proton. Now as one type occupies higher energy levels, the new same type of nulceons can only go to even higher energy levels. If there are too many neutrons (compared to the protons), then the total energy will be more then it needs to be. You can think of the neutrons and protons as being in different pools.

This will decrease the binding energy.

$\delta$ pairing term: is the spin coupling, it is energetically favorable (lower energy) for protons to couple with protons with opposite spin, and the same for neutrons.

If you want to find the most strongly bound nucleus, you need to use :

$$$N/Z \approx 1 + \frac{a_C}{2a_A} A^{2/3}.$$$

If you put Z back into Eb, you will get the binding energy as a function of atomic weight, and the max valuo for that is like your table shows A=63.

The growth of B/A value pertains to increasing binding due to pairing attractive forces of nucleons.

The 4He is a nucleus where the binding has good value(7.1 Mev per nucleon) and it has come up from deuteron which has B/A of about 1.1 MeV.

The four nuclei which are spread about the maximum has even-even neutrons and protons and the possibility of alpha clustering is also there. They are tightly bound.

As A increases the p-p repulsion, as well as surface deformities, play a larger role.

As nuclei are having oscillation as well as vibrations along with larger surface effects leads to smaller B values.

For larger nuclei, the saturation of nuclear forces plays a role in the reduction of B.

see the reference below-

The most tightly bound of the nuclei is 62Ni, a case made convincingly by M. P. Fewell in an article in the American Journal of Physics. Though the championship of nuclear binding energy is often attributed to 56Fe, it actually comes in a close third. The four most tightly bound nuclides are listed in the table below with a tabulation of the binding energy B divided by the mass number A. The curve adapted from Fewell shows those nuclides that are close to the peak.

Nuclide

B/A (keV/A)

62Ni 8794.60

58Fe 8792.23

56Fe 8790.36

60Ni 8780.79

Data from Wapstra and Bos.