Partition Puzzle 5: Feldman Analysis of Z-rels mod 31

Yesterday David Feldman left a comment on a previous E&EN blog entry. I felt it ought to be brought forward as a post of its own so it doesn't get overlooked. He has agreed to let me post it here as a Preliminary Report.

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[Jon Wild wrote:]

"There is a quintuplet of Z-related hexachords in 31-tone equal
temperament whose interval vectors consist solely of 1s--they are all interval hexachords, expressing 15 interval classes in just six pitches. This set of 5 hexachords can be packed into the 31-tone aggregate leaving only one pitch uncovered. The odds that this is by chance are astronomical-there is some other principle at work, related to the observations Stephen has started to make, but I don't know what that principle is."

Here is a description of [this] example which may remove some of the mystery.

Modulo 31, the five sets can be written as[1]         

{3^0 , 3^10, 3^20, 3^3,  3^13, 3^23}         

{3^2 , 3^12, 3^22, 3^5,  3^15, 3^25}         

{3^4 , 3^14, 3^24, 3^7,  3^17, 3^27}         

{3^6 , 3^16, 3^26, 3^9,  3^19, 3^29}      

{3^8 , 3^18, 3^28, 3^11, 3^21, 3^31}


In the language of abstract algebra,
              {3^0 , 3^10, 3^20}
constitutes a subgroup of the multiplicative group of the field with 31elements and
              {3^3,  3^13, 3^23}
one of it's cosets.
 Moreover
              {3^0 , 3^6, 3^12, 3^18, 3^24}
constitutes a disjoint subgroup with 5 elements.

The five sets come from the first by multiplication, as one see quickly if one keeps in mind the congruence of 3^30 with 1 modulo 31.  Multiplication permutes intervals,but preserves all interval hexachords.

One could still argue that {3^0 , 3^10, 3^20, 3^3,  3^13, 3^23} giving an all-interval hexachord seems miraculous, but one surely gets all the intervals if no interval occurs twice. So write
            A = {3^0 , 3^10, 3^20}  B = {3^3 ,  3^13 , 3^23}
Obviously distinct pairs in A give distinct intervals (or the interval would have to be preserved by multiplication by 3^10); similarly distinct pairs in B; similarly a pair in A compared with a pair in B similarly two pairs each crossing A to B. 
Thus the only not obvious case would be a pair in A, say, and a pair crossing from A to B.  One can also rule out some of these cases a priori, and reduce the total number by symmetry, but in any case one doesn't have so many cases that the non-occurrence of an equality seems miraculous anymore.

– David Victor Feldman

9 April 2014 

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[1] For a direct relationship to these sets in more recognizable integer notation, it may be helpful to use the following distribution of 3^n as an overlay:

1	25	5	27	24	11
9	8	14	26	30	6
19	10	2	17	22	23
16	28	18	29	12	21
20	4	7	13	15	3

If X is any one of these five sets,

V(X) = [6111111111111111]
V(X, complX) = [0 a a a a a a a a a a a a a a a]  (a=10)

remembering that complX always includes 0, the odd man out.