Perfect pitch

The fifth seventh note  step 1.5 times the frequency of the first,  If you have a simultaneous, mechanical two tone detector then you want one frequency to vibrate at substantially less than the  Shannon 'twice' the bandwidth, but you want it to have a few misses as possible.  1.5 gets you a slight under sampling of the one on the other..  From a combinatorial view point, evolution has to partition the components of the sound detector from a limited set.  Its partitions will be sets of two and sometimes three.  Making sets of five is way too complex.

What is means in the hearing detector is that the two modes will have one of those combinatorial Shannon locks, there is no way to make more complex partition of the detector and get finer gradation.  Evolution cannot observe any 5/3 solution when comparing partition sizes.

Another way to think of it.  If evolution had to make a nearly round hearing detector what is the easiest estimate of Pi/2?   Good enough to spread quantized elasticity around an imperfect tube.