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[Jon Wild wrote:]

"There is a quintuplet of Z-related hexachords in 31-tone equal

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}

{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.

{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

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:

If X is any one of these five sets,

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

V(X) = [6111111111111111]remembering that complX always includes 0, the odd man out.

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