TE error cutoffs and ETs

I was talking in another place about the function to return
all equal temperaments with a given number of steps to the
octave within a given Tenney-Euclidean (TE) error. That
depends on having a measure that always increases when you
consider a new prime. With a fairly uninteresting piece of
algebra, I've proved that this is the case. Here it is:

http://x31eq.com/increase.pdf

It's the variance of the weighted mapping multiplied by the
number of primes you're considering. There are hand waving
proofs that this does always increase. The variance is the
mean-squared deviation relative to the mean, and the mean
is the point where the mean-squared deviation is
minimized. Adding a point to the sum must increase it.
Moving the point the deviations are being measured from
must increase it *before* that extra point gets added. But
now there's no need to wave any hands because it's all in
algebra.

Graham