\geq 1$ for all $w$, and thus, in this case, $$\hat{\psi}_{t \backslash g}(n)  \geq {\displaystyle}\sum_{w = 1}^{\left\lfloor \frac{n}{g} \right\rfloor} g = g \left\lfloor \frac{n}{g} \right\rfloor.\qedhere$$

Positivity for self-conjugate core partitions.
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We have the following self-conjugate analogues to Theorem \[infinitelymanystcores\] and Theorem \[f\].

If $s,t,g>1$, then there are infinitely many self-conjugate $(s,t)$-core partitions that are not $g$-core partitions.

Recall that there are infinitely many self-conjugate $g$-cores for $g>1$. For any self-conjugate $g$-core ${\gamma}$, consider the (completely and uniquely defined) partition ${\lambda}$ with ${\mathcal Cor}_g({{\lambda}})={\gamma}$ and ${\mathcal Quo}_g({\lambda})=( (1), \, \emptyset, \, \ldots , \, \emptyset , \, (1))$. Then note that $(1)$ is an $(s',t')$-core since $s'>1$ and $t'>1$. Thus ${\lambda}$ is a self-conjugate $(s'g,t'g)$-core that is not a $g$-core. Allowing ${\gamma}$ to vary completes the proof.

Fix integers $g=8$ or $g\geq10$ and $t'>1$, and let $t=t'g$. Then the number $\hat{\psi}_{t^*\backslash g}(n)$ of self-conjugate $t$-cores of $n$ which are not $g$-cores satisfies

1.  $\hat{\psi}_{t^*\backslash g}(n) \geq {\displaystyle}\sum_{w=1}^{\left\lfloor \frac{n}{2g} \right\rfloor} Q_{t'}\left(\frac{g}{2},w\right)$ if $g$ is even, and

2.  $\hat{\psi}_{t^*\backslash g}(n) \geq {\displaystyle}\sum_{2w_1+w_2=1}^{\left\lfloor \frac{n}{g} \right\rfloor} Q_{t'}\left(\frac{g-1}{2},w_1\right)f^*_{t'}(w_2)$ if $g$ is odd.

In particular, if $g$ is even and $n \geq 2g$, or if $g$ is odd and $n \geq g$, then $\hat{\psi}_{t^*\backslash g}(n) >0$ 