Interval-count complexity

To get to a note-class in what I called shell n^2 from the unison in
7-limit, you need at minimum n steps because a straight line path of
1-step intervals takes you only out to a distance n. Hence, for
something in shell n, at a distance of sqrt(n), you need at minimum
ceil(sqrt(n)) steps. For 81/80 this is ceil(sqrt(13))=4 steps, and for
2401/2400 it is ceil(sqrt(11))=4 steps also. In fact, both can be
reached in four steps in only one way, up to commuitivity; we have

81/80 = (6/5)(3/2)^3 (1/4)
2401/2400 = (7/6)(7/5)^2(7/4) (1/4)

Any comma pump for these commas will be related to some arragement of
these steps. You could say from this that 81/80 and 2401/2400 are
equally complex; however 2401/2400 involves all septimal intervals,
and 81/80 all 5-limit intervals, and weighting the intervals will
make it less complex. 4375/4374 is in shell 35, so
we need at least six steps to get to it; but in fact seven are
required and again this is unique

4375/4374 = (7/6)(4/3)^2(5/3)^4 (1/16)

Weighting will bring down the relative complexity of this compared to
2401/2400, since we have six 5-limit steps and only one septimal step.