The correlation is from the off-diagonal terms of the residual interaction, which is the TBME of the shell model. Consider following Hamiltonian of these nuclei

$H(Z,N-1) = H(Z,N-2) + h_n + R_n$

$H(Z,N) = H(Z,N-2) + h_{n_1} + h_{n_2} + R_{n_1} + R_{n_2} + R_{nn}$

the correlation energy between the two neutrons is the term $R_{nn}$. Be aware that this is a residual interaction, not the nucleon-nucleon interaction $V_{ij}$.

$V_{ij}= \begin{pmatrix} V_{11} & V_{12} & V_{1C} \\ V_{21} & V_{22} & V_{2C} \\ V_{C1} & V_{C2} & V_{CC} \end{pmatrix}$

Thus,

$R_{n} = \begin{pmatrix} V_{11} - U_1 & V_{1C} \\ V_{C1} & V_{CC}-U_C \end{pmatrix}$,

where $U_i$ is mean field.

The separation energies is proportional to the terms

$S_n(Z,N-1) \sim h_n + V_n$

$S_{2n}(Z,N)\sim h_{n_1} + h_{n_2} + V_{n_1} + V_{n_2} + V_{nn}$

Thus, the neutron-neutron correlation energy is

$\Delta_{pn}(N,Z) = 2*S_n(Z,N-1) - S_{2n}(N,Z)$

For 18O, $S_n(^{17}O) = 4.1431$ MeV, and $S_{2n}(^{18}O) = 12.1885$ MeV, thus, $\Delta_{2n}(^{18}O) = 3.9023$ MeV.

In the shell model calculation, the single particle energy of the 1d5/2 and 2s1/2 neutron are -4.143 MeV and -3.27 MeV respectively. The residual interaction is

$V = \begin{pmatrix} -1.79 & -0.83 \\ -0.83 & -2.53\end{pmatrix}$

The eigenenergy of the ground state of 18O is -10.539 on top of 16O.

The non-correlated binding energy of 18O is  2*-4.143 = -8.286 MeV.

Therefore, the theoretical correlated energy is 2.258 MeV.