  If $N = {p^k}{m^2}$ is an odd perfect number with special prime factor $p$,
then it is proved that ${p^k} < (2/3){m^2}$. Numerical results on the abundancy
indices $\frac{\sigma(p^k)}{p^k}$ and $\frac{\sigma(m^2)}{m^2}$, and the ratios
$\frac{\sigma(p^k)}{m^2}$ and $\frac{\sigma(m^2)}{p^k}$, are used. It is also
showed that $m^2 > \frac{\sqrt{6}}{2}({10}^{150})$.
