Wow had no idea about lancularity, thanks!

I attached a scalar to the lancularity input on the advanced perlin, and can now better understand the affects. However, I'm not quite sure this answers the question since I'm still not sure whether each octave is scaled down by the same percentage.

Also, the lancularity input requires a scalar, not an interger, so how does that decimal correspond to the scaling?

It seems to me that a lancularity of 1.0 means the octave is scaled down by 50%, so basically, you can use the scalar input to choose between 0 and 50% decrease in scale for each octave.