If, on the other hand, $n \geq 2g$, then $$\hat{\psi}_{t^*\backslash g}(n) \geq \sum_{w = 1}^{\left\lfloor \frac{n}{2g} \right\rfloor} Q_{t'}\left(\frac{g}{2},w\right) \geq \sum_{w = 1}^{\left\lfloor \frac{n}{2g} \right\rfloor} \frac{g}{2} f_{t'}(w).$$ Since $t'>1$, we have $f_{t'}(1)=1$ (and $Q_{t'}\left(\frac{g}{2},1\right)= \frac{g}{2}$), so that $$\hat{\psi}_{t^*\backslash g}(n) \geq \frac{g}{2} f_{t'}(1)=\frac{g}{2},$$ as claimed. In particular, there exists a self-conjugate $t$-core of $n$ which is not a $g$-core.

If, furthermore, $t'\geq 4$, then, by Theorem \[Ono\], $f_{t'}(w) \geq 1$ for all $w$, and thus, in this case, $$\hat{\psi}_{t^*\backslash g}(n) \geq   \sum_{w = 1}^{\left\lfloor \frac{n}{2g} \right\rfloor} \frac{g}{2} = \frac{g}{2} \left\lfloor \frac{n}{2g} \right\rfloor,$$ as claimed.

Suppose now that $g$ is odd. For any $w_1 \geq 1$, we have $Q_{t'}\left(\frac{g-1}{2},w_1\right) \geq \frac{g-1}{2} f_{t'}(w_1)$ while, for $w_1=0$, we have $Q_{t'}\left(\frac{g-1}{2},w_1\right)=1$. Separating the sum according to $w_1=0$ and $w_1 \geq 1$, we thus obtain $$\begin{array}{rcl} \hat{\psi}_{t^*\backslash g}(n) & \geq & {\displaystyle}\sum_{w_2 = 1}^{\left\lfloor \frac{n}{g} \right\rfloor} Q_{t'}\left(\frac{g-1}{2},0\right)f^*_{t'}(w_2) +   {\displaystyle}\sum_{{\genfrac{}{}{0pt}{}{w_1 \geq 1}{2w_1+w_2 = 1}}}^{\left\lfloor \frac{n}{g} \right\rfloor}\frac{g-1}{2} f_{t'}(w_1) f^*_{t'}(w_2)\\
 & = & {\displaystyle}\sum_{w_2 = 1}^{\left\lfloor \frac{n}{g} \right\rfloor} f^*_{t'}(w_2) +       {\displaystyle}\sum_{{\genfrac{}{}{0pt}{}{w_1 \geq 1}{2w_1+w_2 = 1}}}^{\left\lfloor \frac{n}{g} \right\rfloor}\frac{g-1}{2} f_{t'}(w_1) f^*_{t'}(w_2) .\end{array}$$

Since $t'>1$, we have $f^*_{t'}(1)=1$ (corresponding to the self-conjugate $t'$-core $(1)$). In particular, if $n \geq g$, then $\hat{\psi}_{t^*\backslash g}(n) \geq 1$, and there exists a self-conjugate $t$-core of $n$ 