Based on the shape of the curve shown in this post and the possible relationship shown here, I am willing to venture a conjecture:

$$\frac{\sigma (n)}{e^{\gamma} n \log \log n}<1-\frac{0.242692}{\ln(n)}$$

This conjecture holds for the largest Colossally Abundant Number I could find, the $143215^{th}$ with over 800k digits computed by Schwabhäuser. From his statistics:

$$ \frac{X(n_{143215})}{e^\gamma}=0.99995934<1-\frac{0.242692}{\ln(n_{143215})}=0.999999873$$

Can this bound or something similar be proven to hold for all Superabundant Numbers (and therefore all numbers)?


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