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Solidification
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The semi-empirical mass formula's pairing term is attractive only for even-even nuclei, see for example, Miriam Hein et al. 2022$2022$ Eur. J. Phys. 43 035801$43$ $035801$. That is, it is pairing between protons or pairing between neutrons. This is the isospin T=1$T=1$ channel, where the S=0$S=0$ interaction is most attractive. Your statement that S=1$S=1$ is most attractive is true for T=0$T=0$ i.e. the deuteron channel, but this is not the pairing contained in the semi-empirialempirical mass formula term.

The semi-empirical mass formula's pairing term is attractive only for even-even nuclei, see for example, Miriam Hein et al. 2022 Eur. J. Phys. 43 035801. That is, it is pairing between protons or pairing between neutrons. This is the isospin T=1 channel, where the S=0 interaction is most attractive. Your statement that S=1 is most attractive is true for T=0 i.e. the deuteron channel, but this is not the pairing contained in the semi-empirial mass formula term.

The semi-empirical mass formula's pairing term is attractive only for even-even nuclei, see for example, Miriam Hein et al. $2022$ Eur. J. Phys. $43$ $035801$. That is, it is pairing between protons or pairing between neutrons. This is the isospin $T=1$ channel, where the $S=0$ interaction is most attractive. Your statement that $S=1$ is most attractive is true for $T=0$ i.e. the deuteron channel, but this is not the pairing contained in the semi-empirical mass formula term.

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user200143
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The semi-empirical mass formula's pairing term is attractive only for even-even nuclei, see for example, Miriam Hein et al. 2022 Eur. J. Phys. 43 035801. That is, it is pairing between protons or pairing between neutrons. This is the isospin T=1 channel, where the S=0 interaction is most attractive. Your statement that S=1 is most attractive is true for T=0 i.e. the deuteron channel, but this is not the pairing contained in the semi-empirial mass formula term.