Ternary scale theorems: Difference between revisions

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* Bolded Latin variables refer to step vectors (linear combinations of step sizes).

* Indices for all words are 1-indexed.

** If ''s'' is a circular word and ''i'' < 1 or ''i'' > len(''s''), we first replace ''i'' with ''i'' % len(''s'') + 1 before using it as an argument in ''s''[-].

* The notation ''s''('''X'''1, ..., '''X'''''r'') is used for an ''r''-ary scale word with variables '''X'''1, ..., '''X'''''r'' possibly standing in for any sizes. If ''s''('''X''', '''Y''') = '''XXY''' then ''s''('''A''', '''B''') = '''AAB'''.

* We leave the distinction between linear words (words in the ordinary sense) and circular words up to context. We usually also elide the distinction between subwords and the dyad sizes that subtend them.

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* An ''odd-step'' is a ''k''-step where ''k'' is odd; an ''even-step'' is defined similarly.

* Given a linear or circular word ''s'' with a step size '''X''', define ''E'''''X'''(''s'') as the scale word resulting from deleting all instances of '''X''' from ''s''.

* By a ''subword'', ''substring'', or ''slice'' of a word ''s'', denoted ''s''[''i'' : ''j''] (''j'' > ''i''), we mean ''s''[''i''] ''s''[''i'' + 1] ... ''s''[''j'' − 1].

* Given a MOS a'''X''' b'''Y''', a ''chunk'' of '''X''''s is a maximal (possibly length 0) substring made of '''X''''s, bounded by '''Y''''s. We do not include the boundary '''Y''''s.

* ''Length'' is another term for a scale's size. The length of a scale ''s'' is denoted len(''s'').

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== Theorem 1 (Properties of SGA scales) ==

Let ''s'' be a ternary scale word in '''L''', '''M''', and '''s''' of length ''n'', and suppose ''s'' is SGA. Then:

# The length of ''s'' is odd, or ''s'' is equivalent to ('''xy''')''r'''''xz''' for some integer ''r'' ≥ 1.

# If ''n'' is odd, ''s'' is of the form ''a'''''x''' ''b'''''y''' ''b'''''z''' for some permutation ('''x''', '''y''', '''z''') of ('''L''', '''M''', '''s''').

# If ''n'' is odd, ''s'' is abstractly SV3 (i.e. SV3 for almost all tunings).

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O-O-...-O ((n-1)/2 notes).

Label the notes (1, ''j'') and (2, ''j''), 1 ≤ ''j'' ≤ (number of notes in the chain), for notes in the upper and lower chain respectively.

==== Statement (1) ====

In case 1, let '''g'''1 = (2, 1) − (1, 1), '''g'''2 = (1, 2) − (2, 1), and '''g'''3 = (1, 1) − (''n''/2, 2) = ((−''n''/2 − 1)*'''g'''1 − ''n''/2*'''g'''2) mod '''e'''. We assume that '''g'''1, '''g'''2 and '''e''' are ℤ-linearly independent. We have the chain '''g'''1 '''g'''2 '''g'''1 '''g'''2 ... '''g'''1 '''g'''3 which visits every note in ''s''.

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Since ''s'' is generator-offset it is well-formed with respect to the aggregate generator '''g''' = ('''g'''2 + '''g'''1). Since '''g'''1 and '''g'''2 subtend the same number of steps by the SGA assumption, each is an odd-step. All multiples of the aggregate generator '''g''' must be even-steps, and those dyads that are "offset" by '''g'''1 must be odd-steps. Letting ''M'' be the subset consisting of all even-numbered notes (which are generated by '''g''') and considering ''M'' as a scale by dividing degree indices in ''M'' by two, ''M'' is well-formed with respect to '''g''', thus ''M'' (and its offset) must be a MOS subset. Hence ('''g'''3 + '''g'''1), the imperfect generator of the MOS generated by '''g''', subtends the same number of steps as '''g'''. Thus '''g'''2 and '''g'''3 subtend the same number of steps, a fact we need in order to be able to substitute one instance of '''g'''2 with '''g'''3 in the next part.

Let ''r'' be odd and ''r'' ≥ 3. Consider the following abstract sizes for the dyad class of ''k''-steps reached by stacking ''r'' generators:

# from '''g'''1 '''g'''2 ... '''g'''1, we get a1 = (''r'' − 1)/2 * '''g''' + '''g'''1 = ceil(''r''/2) '''g'''1 + floor(''r''/2) '''g'''2

# from '''g'''1 '''g'''2 ... '''g'''1, we get a1 = (''r'' − 1)/2 * '''g''' + '''g'''1 = {{ceil|''r''/2}} '''g'''1 + {{floor|''r''/2}} '''g'''2

# from '''g'''2 '''g'''1 ... '''g'''2, we get a2 = (''r'' − 1)/2 * '''g''' + '''g'''2 = floor(''r''/2) '''g'''1 + ceil(''r''/2) '''g'''2

# from '''g'''2 '''g'''1 ... '''g'''2, we get a2 = (''r'' − 1)/2 * '''g''' + '''g'''2 = {{floor|''r''/2}} '''g'''1 + {{ceil|''r''/2}} '''g'''2

# from '''g'''2 (...even # of gens...) '''g'''1 '''g'''3 '''g'''1 (...even # of gens...) '''g'''2, we get a3 = (''r'' − 1)/2 '''g'''1 + (''r'' − 1)/2 '''g'''2 + '''g'''3 ≡ (''r'' − ''n''/2 − 3/2)'''g'''1 + (''r'' − ''n''/2 − 1/2)'''g'''2 mod e.

# from '''g'''1 (...odd # of gens...) '''g'''1 '''g'''3 '''g'''1 (...odd # of gens...) '''g'''1, we get a4 = (''r'' + 1)/2 '''g'''1 + (''r'' − 3)/2 '''g'''2 + '''g'''3 ≡ (''r'' − ''n''/2 − 1/2)'''g'''1 + (''r'' − ''n''/2 − 3/2)'''g'''2 mod e.

Since ''n'' > 0, these are all distinct by ℤ-linear independence; hence there are at least 4 sizes for ''k''-steps. A 1-step must be reached by stacking an odd number of generators, thus by applying this argument to 1-steps, we see that there must be at least 4 step sizes in some tuning, a contradiction. Thus '''g'''1 and '''g'''2 must themselves be step sizes. Thus we see that an even-length SGA ternary scale must be of the form (xy)''r''xz. (Note that (xy)''r''xz is not SV3, since it has only two kinds of 2-steps, '''xy''' and '''xz'''.) This proves (1).

==== Statement (2) ====

In case 2, let ''n'' ≥ 3 and let (2, 1) − (1, 1) = '''g'''1, (1, 2) − (2, 1) = '''g'''2 be the two alternants. Let '''g'''3 be the closing generator after stacking alternating '''g'''1 and '''g'''2. Then the generator circle is ('''g'''1 '''g'''2)floor(''n''/2) '''g'''3. If a step is formed by stacking ''k'' generators, we may assume that ''k'' is odd, and the combinations of alternants corresponding to a step come in exactly 3 sizes:

In case 2, let ''n'' ≥ 3 and let (2, 1) − (1, 1) = '''g'''1, (1, 2) − (2, 1) = '''g'''2 be the two alternants. Let '''g'''3 be the closing generator after stacking alternating '''g'''1 and '''g'''2. Then the generator circle is ('''g'''1 '''g'''2){{floor|''n''/2}} '''g'''3. If a step is formed by stacking ''k'' generators, we may assume that ''k'' is odd, and the combinations of alternants corresponding to a step come in exactly 3 sizes:

# ceil(''k''/2)'''g'''1 + floor(''k''/2)'''g'''2

# {{ceil|''k''/2}}'''g'''1 + {{floor|''k''/2}}'''g'''2

# floor(''k''/2)'''g'''1 + ceil(''k''/2)'''g'''2

# {{floor|''k''/2}}'''g'''1 + {{ceil|''k''/2}}'''g'''2

# floor(''k''/2)'''g'''1 + floor(''k''/2) '''g'''2 + '''g'''3

# {{floor|''k''/2}}'''g'''1 + {{floor|''k''/2}} '''g'''2 + '''g'''3

(since the scale size is odd, we can always ensure this by taking octave complements of all the generators). By counting the length-''k'' subwords of the (linear) word ('''g'''1 '''g'''2)floor(''n''/2), we see that the first two sizes must both occur (''n'' − ''k'')/2 times. This proves (2).

(since the scale size is odd, we can always ensure this by taking octave complements of all the generators). By counting the length-''k'' subwords of the (linear) word ('''g'''1 '''g'''2){{floor|''n''/2}}, we see that the first two sizes must both occur (''n'' − ''k'')/2 times. This proves (2).

==== Statement (3) ====

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x x x ... x

x x x ... x x

and use the vectors (-1, 2) and (ceil(n/2), 1) as the Fokker block chromas. A rank-3 Fokker block has the property that tempering out by each of the chromas gives two MOSes. These correspond to two of the temperings '''X''' = '''Y''', '''Y''' = '''Z''' and '''X''' = '''Z'''. The third tempering follows by symmetry (by taking the other chirality).

and use the vectors (-1, 2) and ({{ceil|n/2}}, 1) as the Fokker block chromas. A rank-3 Fokker block has the property that tempering out by each of the chromas gives two MOSes. These correspond to two of the temperings '''X''' = '''Y''', '''Y''' = '''Z''' and '''X''' = '''Z'''. The third tempering follows by symmetry (by taking the other chirality).

==== Statement (5) ====

By part (2), we have that ''s'' has step signature ''a'''''X''' ''b'''''Y''' ''b'''''Z''', ''a'' odd. By part (4), we have that ''T''('''X''', '''W''') = ''s''('''X''', '''W''', '''W''') is a MOS scale ''a'''''X'''2''b'''''W'''. If ''b'' = 1, there's nothing to prove, so assume ''b'' > 1.

Consider the two generators in the GS of ''s'', which are detemperings of the generator ''i'''''X''' + ''j'''''W''' of ''T''('''X''', '''W'''), where gcd(''j'', 2''k'') = 1. Assume, possibly after inverting the generator, that the imperfect generator of ''T'' has ''j'' + 1 '''W'''s and the perfect generator has ''j'' '''W'''s.

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Assume that the generator '''g''' is a ''k''-step and ''k'' is even. (If ''k'' is not even, invert the generator.) On some note ''p'' we have a chain of (''n'' + 1)/2 notes and on ''p′'' = ''p'' + offset we'll have (''n'' − 1)/2) notes.

Assume 1 < gcd(''k'', ''n'') < ''n'' and ''n'' ≥ 5. Since ''n'' is odd, ''d'' = gcd(''k'', ''n'') is an odd number at least 3, and by well-formedness with respect to the generator, there must be a circle of ''n''/''d'' < floor(''n''/2) notes formed by '''g''', contrary to the assumption of GO. Thus, gcd(''k'', ''n'') = 1.

Assume 1 < gcd(''k'', ''n'') < ''n'' and ''n'' ≥ 5. Since ''n'' is odd, ''d'' = gcd(''k'', ''n'') is an odd number at least 3, and by well-formedness with respect to the generator, there must be a circle of ''n''/''d'' < {{floor|''n''/2}} notes formed by '''g''', contrary to the assumption of GO. Thus, gcd(''k'', ''n'') = 1.

Since ''n'' is odd, ''rk'' ≡ ''k''/2 mod ''n'' iff ''r'' ≡ (''n'' + 1)/2 mod ''n''. (Note that both 2 and ''k'' are coprime with ''n'', hence multiplicatively invertible mod ''n''.) This proves that the offset, which must be reached after (''n'' + 1)/2 ''k''-steps, is a ''k''/2-step, as desired. (As [''k''] is a generator of ℤ/''n'', stacking (''n'' − 1)-many ''k''-steps must visit every note exactly once. Thus if the offset wasn't reached in (''n'' + 1)/2 steps, the two generator chains wouldn't have the assumed lengths.) {{qed}}

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=== Proof ===

==== If the generator of a projection of ''s'' is a ''k''-step, the word of stacked ''k''-steps in ''s'' is pairwise well-formed ====

Suppose ''s'' has ''n'' notes (after dealing with small cases, we may assume ''n'' ≥ 7) and ''s'' projects to primitive MOSes ''s''1 (via identifying '''b''' with '''c'''), ''s''2 (via identifying '''a''' with '''c'''), and ''s''3 (via identifying '''a''' with '''b'''). Suppose ''s''1's generator is a ''k''-step, which comes in two sizes: '''P''', the perfect ''k''-step, and '''I''', the imperfect ''k''-step. By stacking ''n''-many ''k''-steps, we get two words of length ''n'' of ''k''-steps of ''s''2 and ''s''3, respectively. These binary words, which we call Σ2 and Σ3, must be MOSes, since ''m''-steps in the new words correspond to ''mk''-steps in the MOS words ''s''1 and ''s''2, which come in at most two sizes. Since ''s''1 is a primitive MOS, gcd(''k'', ''n'') = 1. Hence when 0 < ''m'' < ''n'', ''mk'' is ''not'' divisible by ''n'' and ''mk''-steps come in ''exactly'' two sizes; hence both Σ2 and Σ3 are primitive MOSes.

index: 1 2 3 4 ... ''n''

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==== Case analysis ====

Since, by our assumption, Λ3 has two γ in a row, Λ3 must have more γ than γ′, so μ − 1 < ''n''/2. Since Λ3 is a MOS, μ − 1 ≥ 1. So we have 2 ≤ μ ≤ ceil(''n''/2).

Since, by our assumption, Λ3 has two γ in a row, Λ3 must have more γ than γ′, so μ − 1 < ''n''/2. Since Λ3 is a MOS, μ − 1 ≥ 1. So we have 2 ≤ μ ≤ {{ceil|''n''/2}}.

We have three cases to consider:

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'''Case 1''': μ = 2, i.e. Λ2 is the MOS (''n'' − 2)β 2β′.

For Λ2 to be a MOS, the first, and only, occurrence of '''R''' must be at either ''f'' = floor(''n''/2) or ceil(''n''/2). We may assume that it is at ''f''; otherwise reverse the chain and reindex the words to start at 2''f''.

For Λ2 to be a MOS, the first, and only, occurrence of '''R''' must be at either ''f'' = {{floor|''n''/2}} or {{ceil|''n''/2}}. We may assume that it is at ''f''; otherwise reverse the chain and reindex the words to start at 2''f''.

1 … ''f'' … 2''f'' ''n''

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These give exactly three distinct sizes for every dyad class. Hence ''s'' is SV3.

In this case ''s'' has two chains of '''Q''', one with floor(''n''/2) notes and one offset by '''Q'''(''f''−1)R with ceil(''n''/2) notes. Every instance of Q must be a ''k''-step, since by ℤ-linear independence '''Q''' = α'''a''' + β'''b''' + γ'''c''' is the only way to write '''Q''' in the basis ('''a''', '''b''', '''c'''); so ''s'' is well-formed with respect to '''Q'''. Thus ''s'' also satisfies the generator-offset property with generator '''Q'''.

In this case ''s'' has two chains of '''Q''', one with {{floor|''n''/2}} notes and one offset by '''Q'''(''f''−1)R with {{ceil|''n''/2}} notes. Every instance of Q must be a ''k''-step, since by ℤ-linear independence '''Q''' = α'''a''' + β'''b''' + γ'''c''' is the only way to write '''Q''' in the basis ('''a''', '''b''', '''c'''); so ''s'' is well-formed with respect to '''Q'''. Thus ''s'' also satisfies the generator-offset property with generator '''Q'''.

'''Case 2:''' μ ≥ ceil(''n''/2), i.e. Λ2 has fewer β than β′.

'''Case 2:''' μ ≥ {{ceil|''n''/2}}, i.e. Λ2 has fewer β than β′.

Since Λ3 has more β than β′, Λ2 is floor(''n''/2)β ceil(''n''/2)β′, and Λ3 is ceil(''n''/2)γ floor(''n''/2)γ′. There is a unique mode of ceil(''n''/2)γ floor(''n''/2)γ′ that both begins and ends with γ, namely γγ′γγ′…γγ′γ. Thus Λ2 is ββ′ββ′…ββ′β′. It is now easy to see that if the number of ''k''-steps stacked is odd, then there are two sizes that do not contain '''T''' and one size that contains '''T'''; if the number of ''k''-steps stacked is even, then there is one size that does not contain '''T''' and two sizes that contain T. Hence ''s'' is SV3.

Since Λ3 has more β than β′, Λ2 is {{floor|''n''/2}}β {{ceil|''n''/2}}β′, and Λ3 is {{ceil|''n''/2}}γ {{floor|''n''/2}}γ′. There is a unique mode of {{ceil|''n''/2}}γ {{floor|''n''/2}}γ′ that both begins and ends with γ, namely γγ′γγ′…γγ′γ. Thus Λ2 is ββ′ββ′…ββ′β′. It is now easy to see that if the number of ''k''-steps stacked is odd, then there are two sizes that do not contain '''T''' and one size that contains '''T'''; if the number of ''k''-steps stacked is even, then there is one size that does not contain '''T''' and two sizes that contain T. Hence ''s'' is SV3.

In this case we have Σ = '''QRQR'''...'''QRT''', and ''s'' is well-formed with respect to the generator '''Q''' + '''R''', thus ''s'' satisfies the generator-offset property. By Proposition 1, ''s'' is SV3.

'''Case 3:''' 3 ≤ μ ≤ floor(''n''/2).

'''Case 3:''' 3 ≤ μ ≤ {{floor|''n''/2}}.

Λ2 has a chunk of β (after the first β′) of size ''x'' where ''x'' = floor(''n''/μ~~) ≥~~ floor(''n''/floor(''n''/2)) = 2 or ''x'' = ceil(''n''/μ) = floor(''n''/μ) + 1. Hence Λ3 has a chunk of γ of size ''x''. Λ3 also has a chunk that contains Λ3[''n'' : 2] as a subword. This chunk must be of size ''y'', where

Λ2 has a chunk of β (after the first β′) of size ''x'' where ''x'' = {{floor|''n''/μ}} ≥ {{floor|''n''/floor(''n''/2}}) = 2 or ''x'' = {{ceil|''n''/μ}} = {{floor|''n''/μ}} + 1. Hence Λ3 has a chunk of γ of size ''x''. Λ3 also has a chunk that contains Λ3[''n'' : 2] as a subword. This chunk must be of size ''y'', where

<math>2 \lfloor\frac{n}{\mu}\rfloor - 1 = 2 \big(\lfloor \frac{n}{\mu} \rfloor - 1\big) + 1 \leq y \leq 2 \big(\lfloor\frac{n}{\mu}\rfloor + 1 \big) + 1 = 2\lfloor\frac{n}{\mu}\rfloor + 3.</math>

(The lower bound is reached if Λ3 has chunks of sizes floor(''n''/μ) − 1 and floor(''n''/μ), and the upper bound is reached if Λ3 has chunks of sizes floor(''n''/μ) and ceil(''n''/μ).)

(The lower bound is reached if Λ3 has chunks of sizes {{floor|''n''/μ}} − 1 and {{floor|''n''/μ}}, and the upper bound is reached if Λ3 has chunks of sizes {{floor|''n''/μ}} and {{ceil|''n''/μ}}.)

The difference between the chunk sizes of Λ3 is ''y'' − ''x'', which must be 1 since Λ3 is pairwise well-formed. We thus have the following subcases: (In the following, chunk of Λ2 means chunk of β, and chunk of Λ3 means chunk of γ.)

'''Case 3.1:''' (''x'', ''y'') = (floor(''n''/μ), 2*floor(''n''/μ) − 1).

'''Case 3.1:''' (''x'', ''y'') = ({{floor|''n''/μ}}, 2*{{floor|''n''/μ}} − 1).

Since ''y'' − ''x'' = floor(''n''/μ) − 1, we have ''x'' = floor(''n''/μ) = 2 and ''y'' = 3. The chunk in Λ3 whose size was defined to be ''y'' is made from two consecutive chunks in Λ2 of size 1. (So Λ2 has chunks of size 1 and 2, and Λ3 has chunks of size 2 and 3.) Since chunk sizes of a MOS themselves form a MOS, Λ2 has more chunks of size 1 than it has chunks of size 2.

Since ''y'' − ''x'' = {{floor|''n''/μ}} − 1, we have ''x'' = {{floor|''n''/μ}} = 2 and ''y'' = 3. The chunk in Λ3 whose size was defined to be ''y'' is made from two consecutive chunks in Λ2 of size 1. (So Λ2 has chunks of size 1 and 2, and Λ3 has chunks of size 2 and 3.) Since chunk sizes of a MOS themselves form a MOS, Λ2 has more chunks of size 1 than it has chunks of size 2.

Λ2 has only two chunks of size 1, Λ2[''n'' − 1] and Λ2[1], since otherwise Λ3 would have a chunk of size 1 within Λ3[1 : ''n'']. Thus Λ2 has exactly one chunk of size 2. Thus Λ2 = ββ′βββ′ββ′ and Λ3 = γγ′γγγ′γγ. Thus we have:

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(This also implies ''s'' is SV3.)

'''Case 3.2''': (''x'', ''y'') = (floor(''n''/μ) + 1, 2*floor(''n''/μ) − 1) is impossible: Here (''x'', ''y'') = (4, 5). But then Λ2 has a chunk of size < 3 because of the β' at index ''n'', contradicting that ''x'' is one of the chunk sizes of Λ2.

'''Case 3.2''': (''x'', ''y'') = ({{floor|''n''/μ}} + 1, 2*{{floor|''n''/μ}} − 1) is impossible: Here (''x'', ''y'') = (4, 5). But then Λ2 has a chunk of size < 3 because of the β' at index ''n'', contradicting that ''x'' is one of the chunk sizes of Λ2.

'''Case 3.3''': (''x'', ''y'') = (floor(''n''/μ) + 1, 2*floor(''n''/μ)) is impossible: Here (''x'', ''y'') = (3, 4). But then Λ2 has a chunk of size 1 because of the β' at index ''n'', and another chunk of size 0 or 2, contradicting that ''x'' is one of the chunk sizes of Λ2.

'''Case 3.3''': (''x'', ''y'') = ({{floor|''n''/μ}} + 1, 2*{{floor|''n''/μ}}) is impossible: Here (''x'', ''y'') = (3, 4). But then Λ2 has a chunk of size 1 because of the β' at index ''n'', and another chunk of size 0 or 2, contradicting that ''x'' is one of the chunk sizes of Λ2.

The remaining cases are all impossible because they imply ''y'' − ''x'' ≥ 2:

* '''Case 3.4''': (''x'', ''y'') = (floor(''n''/μ) + 1, 2*floor(''n''/μ) + 1)

* '''Case 3.4''': (''x'', ''y'') = ({{floor|''n''/μ}} + 1, 2*{{floor|''n''/μ}} + 1)

* '''Case 3.5''': (''x'', ''y'') = (floor(''n''/μ) + 1, 2*floor(''n''/μ) + 2)

* '''Case 3.5''': (''x'', ''y'') = ({{floor|''n''/μ}} + 1, 2*{{floor|''n''/μ}} + 2)

* '''Case 3.6''': (''x'', ''y'') = (floor(''n''/μ) + 1, 2*floor(''n''/μ) + 3)

* '''Case 3.6''': (''x'', ''y'') = ({{floor|''n''/μ}} + 1, 2*{{floor|''n''/μ}} + 3)

* '''Case 3.7''': (''x'', ''y'') = (floor(''n''/μ), 2*floor(''n''/μ))

* '''Case 3.7''': (''x'', ''y'') = ({{floor|''n''/μ}}, 2*{{floor|''n''/μ}})

* '''Case 3.8''': (''x'', ''y'') = (floor(''n''/μ), 2*floor(''n''/μ) + 1)

* '''Case 3.8''': (''x'', ''y'') = ({{floor|''n''/μ}}, 2*{{floor|''n''/μ}} + 1)

* '''Case 3.9''': (''x'', ''y'') = (floor(''n''/μ), 2*floor(''n''/μ) + 2)

* '''Case 3.9''': (''x'', ''y'') = ({{floor|''n''/μ}}, 2*{{floor|''n''/μ}} + 2)

* '''Case 3.10''': (''x'', ''y'') = (floor(''n''/μ), 2*floor(''n''/μ) + 3)

* '''Case 3.10''': (''x'', ''y'') = ({{floor|''n''/μ}}, 2*{{floor|''n''/μ}} + 3)

== Theorem 5 (PWF scales are balanced) ==

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=== Proof ===

The result of substituting '''Y''' with '''X''' (let us call this map ''p'') is the MOS ''M'' = 2''a'''''X'''2''c'''''Z''', which has exactly 2 periods since gcd(''a'', ''c'') = 1. ''M'' thus consists of two generator chains separated by the period of ''M'', which has ''a'' + ''c'' = len(''s'') steps. It thus suffices for there to exist ''k'', 0 < ''k'' < ''a'' + ''c'', such that every perfect ''k''-step generator has the same preimage in ''s'', which will be our desired generator. Suppose that the perfect ''k''-step of ''M'' is ''i'''''W''' + ''j'''''Z''' where 0 < ''i'' < ''a''. Since ''a'' is odd, possibly after taking the period-complement we may assume that ''i'' is even. Hence each subword ''w'' of ''s'' such that its projection ''p''(''w'') subtends a perfect ''k''-step satisfies |''w''|'''X''' = |''w''|'''Y''' (= ''i''/2). It plainly follows that every such ''w'' satisfies |''w''|'''X''' = |''w''|'''Y''' = ''i''/2 and |''w''|'''Z''' = ''j''.

It remains to show that ''s'' is balanced...

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# A primitive [[balanced]] MV3 scale ''s'' satisfies one of the following:

## '''sporadic balanced''': ''s'' is equivalent to '''XYXZXYX''', the ternary [[Fraenkel word]], with step signature 4'''X'''2'''Y'''1'''Z'''.

## '''odd-regular''': len(''s'') is odd, and ''s'' is equivalent to a word constructed from taking the brightest mode of the MOS ''c'''''X'''''b'''''Z''' with ''c'' even and gcd(''c'', ''b'') = 1, and replacing every other '''X''' with '''Y'''. We assume '''X''' > '''Z''' when constructing the MOS. In particular, ''s'' has [[step signature]] ''a'''''X'''''a'''''Y'''''b'''''Z''' where ''b'' is odd (with ''a'' = ''c''/2).

## '''even-regular''': len(''s'') is even, and ''s'' is equivalent to a word constructed from taking the brightest mode of the MOS 2''a'''''X'''2''c'''''Z''' with ''a'' odd and gcd(''a'', ''c'') = 1, and replacing every other '''X''' with '''Y'''. In particular, ''s'' has [[step signature]] ''a'''''X'''''a'''''Y'''''b'''''Z''' with ''a'' odd and ''b'' even.

# All primitive balanced ternary scales are MV3.

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(a) Let ''s'' be a ternary balanced word; then for any given letter '''y''' the number of '''y'''s in a subword of any given length ''L'' varies by at most 1. Thus the same is true when we count all non-'''y''' letters in any subword of length ''L''; thus when we equate '''x''' and '''z''', the count of the resulting letter in any subword of length ''L'' differs by 1. Being a binary balanced word is one characterization of the MOS property.

(b) Assume that the projection ''p'''''YZ'''(''s'') identifying '''Y''' and '''Z''' of a primitive PMOS scale ''s'' with signature ''ar'''''X''' ''b'''''Y''' ''c'''''Z''' is an ''r''-period MOS, ''r'' > 1, with step signature ''ar'''''X''' ''dr'''''W'''. We claim that neither ''b'' nor ''c'' is divisible by ''r''. Since ''p'''''XY'''(''s'') is the MOS (''ar'' + ''b'')'''W''' ''c'''''Z''' and ''p'''''XZ'''(''s'') is the MOS (''ar'' + ''c'')'''W''' ''b'''''Y''', if either ''b'' and ''c'' is divisible by ''r'', then the distributions of two of the letters have ''r'' periods. Then the distribution of the third letter also has ''r'' periods, meaning that ''s'' itself has ''r'' periods, a contradiction. It suffices to show that ''r'' = 2. If ''r'' > 2, then...

For 7.1.2: Suppose ''s'' is balanced and has at least three sizes for ''k''-steps, ''a''''i'''''X''' + ''b''''i'''''Y''' + ''c''''i'''''Z''' = (''a''''i'', ''b''''i'', ''c''''i'') for ''i'' = 1, 2, 3. We may assume (''a''2, ''b''2, ''c''2) = (''a''1, ''b''1 + 1, ''c''1 - 1). Then either (''a''3, ''b''3, ''c''3) = (''a''1 + 1, ''b''1, ''c''1 - 1) or (''a''3, ''b''3, ''c''3) = (''a''1 - 1, ''b''1 + 1, ''c''1). In both cases, by balancedness applied to subwords of length ''k'', the three vectors represent the only possible interval sizes.

For 7.1.3: The ternary Fraenkel word may be verified as SV3 by inspection, and we have already shown in Theorem 1 that odd-regular balanced scales are SV3. To show that even-regular balanced scales are ''not'' SV3, observe that (''a''+''c'')-steps come in only 2 sizes in such a scale ''s'': floor(''a''/2)'''X''' + ceil(''a''/2)'''Y''' + ''c'''''Z''' and ceil(''a''/2)'''X''' + floor(''a''/2)'''Y''' + ''c'''''Z''', since the underlying MOS 2''a'''''X'''2''c'''''Y''' only has the (''a''+''c'')-step ''a'''''X''' + ''c'''''Z'''. The construction replaces the '''X'''s in these subwords with alternating '''X'''s and '''Y'''s; either of '''X''' or '''Y''' may occur first, corresponding to the two possible sizes, since ''a'' is odd and thus the (''a''+''c'')-step subword ''s''[''k'' : ''k''+''a''+''c''] becomes the subword ''s''[''k''+''a''+''c'' : ''k''+2''a''+2''c''] via interchanging '''X''' and '''Y'''.

For 7.1.3: The ternary Fraenkel word may be verified as SV3 by inspection, and we have already shown in Theorem 1 that odd-regular balanced scales are SV3. To show that even-regular balanced scales are ''not'' SV3, observe that (''a''+''c'')-steps come in only 2 sizes in such a scale ''s'': {{floor|''a''/2}}'''X''' + {{ceil|''a''/2}}'''Y''' + ''c'''''Z''' and {{ceil|''a''/2}}'''X''' + {{floor|''a''/2}}'''Y''' + ''c'''''Z''', since the underlying MOS 2''a'''''X'''2''c'''''Y''' only has the (''a''+''c'')-step ''a'''''X''' + ''c'''''Z'''. The construction replaces the '''X'''s in these subwords with alternating '''X'''s and '''Y'''s; either of '''X''' or '''Y''' may occur first, corresponding to the two possible sizes, since ''a'' is odd and thus the (''a''+''c'')-step subword ''s''[''k'' : ''k''+''a''+''c''] becomes the subword ''s''[''k''+''a''+''c'' : ''k''+2''a''+2''c''] via interchanging '''X''' and '''Y'''.

Claim 7.1.4 can be verified by noting that such scales are PWF and using Theorem 4.

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# Given any arbitrary MOS scale with at least three notes per period, is there *always* a MV3 generator-offset scale which can be derived as a "detempering" of that scale? Or is this only true for some MOS's? For instance, the MOS LLsLLLs has the MV3 generator-offset scale LmsLmLs as a detempering. Does a similar MV3 detempering exist for every possible DE scale with at least three notes per period, or at least for strict MOS's with one period per octave (e.g. well-formed scales)?

# The scale tree is a great way to analyze MOS scales. For any generator, we can compute the various MOS's it forms if we simply look at the scale tree, and indeed MOS "words" like LLsLLLs can be identified with regions on the scale tree (in this situation the interval between 4/7 and 3/5). A similar "scale plane" should exist for generator-offset-MV3 scales, where given some word representing a generator-offset-MV3 scale, we can look at the set of points on the generator plane which generates it; these seem to often be triangles, with the lines corresponding to MOS's and the vertices corresponding to EDOs (though is this always true?). What is the big picture of this scale plane? Can we use Viggo Brun's algorithm for this, generalizing the theory of continued fractions? Is there some simple formula we can use to predict, given some generator-offset-MV3 scale, which region on the scale plane it corresponds to? Can we plot simple generator-size-proportions as points in this space? And so on.

# In the theory of MOS, there is a second [[MOS Scale Family Tree|scale tree]] that is less frequently talked about, which Erv Wilson calls the "Rabbit Sequence" ([http://www.anaphoria.com/RabbitSequence.pdf Erv Wilson's original version], [https://mikebattagliamusic.com/MOSTree/MOSTreeab.html interactive version 1], [https://mikebattagliamusic.com/MOSTree/MOSTreeLs.html interactive version 2]). This is a tree for which each MOS word has two children, depending on if the MOS is "soft" (with L/s < 2) or "hard" (with L/s > 2). For instance, LsLss has the two children LLsLLLs and ssLsssL. Does a similar scale plane exist for these generator-offset-MV3 scales?

[[Category:Math]]

@@ Line 5: / Line 5: @@
 * Bolded Latin variables refer to step vectors (linear combinations of step sizes).
 * Indices for all words are 1-indexed.
-** If ''s'' is a circular word and ''i'' < 1 or ''i'' > len(''s''), we first replace ''i'' with ''i'' % len(''s'') + 1 before using it as an argument in ''s''[-].
+** If ''s'' is a circular word and ''i'' &lt; 1 or ''i'' &gt; len(''s''), we first replace ''i'' with ''i'' % len(''s'') + 1 before using it as an argument in ''s''[-].
 * The notation ''s''('''X'''<sub>1</sub>, ..., '''X'''<sub>''r''</sub>) is used for an ''r''-ary scale word with variables '''X'''<sub>1</sub>, ..., '''X'''<sub>''r''</sub> possibly standing in for any sizes. If ''s''('''X''', '''Y''') = '''XXY''' then ''s''('''A''', '''B''') = '''AAB'''.
 * We leave the distinction between linear words (words in the ordinary sense) and circular words up to context. We usually also elide the distinction between subwords and the dyad sizes that subtend them.
@@ Line 22: / Line 22: @@
 * An ''odd-step'' is a ''k''-step where ''k'' is odd; an ''even-step'' is defined similarly.
 * Given a linear or circular word ''s'' with a step size '''X''', define ''E''<sub>'''X'''</sub>(''s'') as the scale word resulting from deleting all instances of '''X''' from ''s''.
-* By a ''subword'', ''substring'', or ''slice'' of a word ''s'', denoted ''s''[''i'' : ''j''] (''j'' > ''i''), we mean ''s''[''i''] ''s''[''i'' + 1] ... ''s''[''j'' &minus; 1].
+* By a ''subword'', ''substring'', or ''slice'' of a word ''s'', denoted ''s''[''i'' : ''j''] (''j'' &gt; ''i''), we mean ''s''[''i''] ''s''[''i'' + 1] ... ''s''[''j'' &minus; 1].
 * Given a MOS a'''X''' b'''Y''', a ''chunk'' of '''X''''s is a maximal (possibly length 0) substring made of '''X''''s, bounded by '''Y''''s. We do not include the boundary '''Y''''s.
 * ''Length'' is another term for a scale's size. The length of a scale ''s'' is denoted len(''s'').
@@ Line 30: / Line 30: @@
 == Theorem 1 (Properties of SGA scales) ==
 Let ''s'' be a ternary scale word in '''L''', '''M''', and '''s''' of length ''n'', and suppose ''s'' is SGA. Then:
-# The length of ''s'' is odd, or ''s'' is equivalent to ('''xy''')<sup>''r''</sup>'''xz''' for some integer ''r'' ≥ 1.
+# The length of ''s'' is odd, or ''s'' is equivalent to ('''xy''')<sup>''r''</sup>'''xz''' for some integer ''r'' &ge; 1.
 # If ''n'' is odd, ''s'' is of the form ''a'''''x''' ''b'''''y''' ''b'''''z''' for some permutation ('''x''', '''y''', '''z''') of ('''L''', '''M''', '''s''').
 # If ''n'' is odd, ''s'' is abstractly SV3 (i.e. SV3 for almost all tunings).
@@ Line 51: / Line 51: @@
   O-O-...-O ((n-1)/2 notes).
-Label the notes (1, ''j'') and (2, ''j''), 1 ≤ ''j'' ≤ (number of notes in the chain), for notes in the upper and lower chain respectively.
+Label the notes (1, ''j'') and (2, ''j''), 1 &le; ''j'' &le; (number of notes in the chain), for notes in the upper and lower chain respectively.
 ==== Statement (1) ====
 In case 1, let '''g'''<sub>1</sub> = (2, 1) &minus; (1, 1), '''g'''<sub>2</sub> = (1, 2) &minus; (2, 1), and '''g'''<sub>3</sub> = (1, 1) &minus; (''n''/2, 2) = ((&minus;''n''/2 &minus; 1)*'''g'''<sub>1</sub> &minus; ''n''/2*'''g'''<sub>2</sub>) mod '''e'''. We assume that '''g'''<sub>1</sub>, '''g'''<sub>2</sub> and '''e''' are ℤ-linearly independent. We have the chain '''g'''<sub>1</sub> '''g'''<sub>2</sub> '''g'''<sub>1</sub> '''g'''<sub>2</sub> ... '''g'''<sub>1</sub> '''g'''<sub>3</sub> which visits every note in ''s''.
@@ Line 57: / Line 57: @@
 Since ''s'' is generator-offset it is well-formed with respect to the aggregate generator '''g''' = ('''g'''<sub>2</sub> + '''g'''<sub>1</sub>). Since '''g'''<sub>1</sub> and '''g'''<sub>2</sub> subtend the same number of steps by the SGA assumption, each is an odd-step. All multiples of the aggregate generator '''g''' must be even-steps, and those dyads that are "offset" by '''g'''<sub>1</sub> must be odd-steps. Letting ''M'' be the subset consisting of all even-numbered notes (which are generated by '''g''') and considering ''M'' as a scale by dividing degree indices in ''M'' by two, ''M'' is well-formed with respect to '''g''', thus ''M'' (and its offset) must be a MOS subset. Hence ('''g'''<sub>3</sub> + '''g'''<sub>1</sub>), the imperfect generator of the MOS generated by '''g''', subtends the same number of steps as '''g'''. Thus '''g'''<sub>2</sub> and '''g'''<sub>3</sub> subtend the same number of steps, a fact we need in order to be able to substitute one instance of '''g'''<sub>2</sub> with '''g'''<sub>3</sub> in the next part.
-Let ''r'' be odd and ''r'' ≥ 3. Consider the following abstract sizes for the dyad class of ''k''-steps reached by stacking ''r'' generators:
+Let ''r'' be odd and ''r'' &ge; 3. Consider the following abstract sizes for the dyad class of ''k''-steps reached by stacking ''r'' generators:
-# from '''g'''<sub>1</sub> '''g'''<sub>2</sub> ... '''g'''<sub>1</sub>, we get a<sub>1</sub> = (''r'' &minus; 1)/2 * '''g''' + '''g'''<sub>1</sub> = ceil(''r''/2) '''g'''<sub>1</sub> + floor(''r''/2) '''g'''<sub>2</sub>
+# from '''g'''<sub>1</sub> '''g'''<sub>2</sub> ... '''g'''<sub>1</sub>, we get a<sub>1</sub> = (''r'' &minus; 1)/2 * '''g''' + '''g'''<sub>1</sub> = {{ceil|''r''/2}} '''g'''<sub>1</sub> + {{floor|''r''/2}} '''g'''<sub>2</sub>
-# from '''g'''<sub>2</sub> '''g'''<sub>1</sub> ... '''g'''<sub>2</sub>, we get a<sub>2</sub> = (''r'' &minus; 1)/2 * '''g''' + '''g'''<sub>2</sub> = floor(''r''/2) '''g'''<sub>1</sub> + ceil(''r''/2) '''g'''<sub>2</sub>
+# from '''g'''<sub>2</sub> '''g'''<sub>1</sub> ... '''g'''<sub>2</sub>, we get a<sub>2</sub> = (''r'' &minus; 1)/2 * '''g''' + '''g'''<sub>2</sub> = {{floor|''r''/2}} '''g'''<sub>1</sub> + {{ceil|''r''/2}} '''g'''<sub>2</sub>
 # from '''g'''<sub>2</sub> (...even # of gens...) '''g'''<sub>1</sub> '''g'''<sub>3</sub> '''g'''<sub>1</sub> (...even # of gens...) '''g'''<sub>2</sub>, we get a<sub>3</sub> = (''r'' &minus; 1)/2 '''g'''<sub>1</sub> + (''r'' &minus; 1)/2 '''g'''<sub>2</sub> + '''g'''<sub>3</sub> ≡ (''r'' &minus; ''n''/2 &minus; 3/2)'''g'''<sub>1</sub> + (''r'' &minus; ''n''/2 &minus; 1/2)'''g'''<sub>2</sub> mod e.
 # from '''g'''<sub>1</sub> (...odd # of gens...) '''g'''<sub>1</sub> '''g'''<sub>3</sub> '''g'''<sub>1</sub> (...odd # of gens...) '''g'''<sub>1</sub>, we get a<sub>4</sub> = (''r'' + 1)/2 '''g'''<sub>1</sub> + (''r'' &minus; 3)/2 '''g'''<sub>2</sub> + '''g'''<sub>3</sub> ≡ (''r'' &minus; ''n''/2 &minus; 1/2)'''g'''<sub>1</sub> + (''r'' &minus; ''n''/2 &minus; 3/2)'''g'''<sub>2</sub> mod e.
-Since ''n'' > 0, these are all distinct by ℤ-linear independence; hence there are at least 4 sizes for ''k''-steps. A 1-step must be reached by stacking an odd number of generators, thus by applying this argument to 1-steps, we see that there must be at least 4 step sizes in some tuning, a contradiction. Thus '''g'''<sub>1</sub> and '''g'''<sub>2</sub> must themselves be step sizes. Thus we see that an even-length SGA ternary scale must be of the form (xy)<sup>''r''</sup>xz. (Note that (xy)<sup>''r''</sup>xz is not SV3, since it has only two kinds of 2-steps, '''xy''' and '''xz'''.) This proves (1).
+Since ''n'' &gt; 0, these are all distinct by ℤ-linear independence; hence there are at least 4 sizes for ''k''-steps. A 1-step must be reached by stacking an odd number of generators, thus by applying this argument to 1-steps, we see that there must be at least 4 step sizes in some tuning, a contradiction. Thus '''g'''<sub>1</sub> and '''g'''<sub>2</sub> must themselves be step sizes. Thus we see that an even-length SGA ternary scale must be of the form (xy)<sup>''r''</sup>xz. (Note that (xy)<sup>''r''</sup>xz is not SV3, since it has only two kinds of 2-steps, '''xy''' and '''xz'''.) This proves (1).
 ==== Statement (2) ====
-In case 2, let ''n'' ≥ 3 and let (2, 1) &minus; (1, 1) = '''g'''<sub>1</sub>, (1, 2) &minus; (2, 1) = '''g'''<sub>2</sub> be the two alternants. Let '''g'''<sub>3</sub> be the closing generator after stacking alternating '''g'''<sub>1</sub> and '''g'''<sub>2</sub>. Then the generator circle is ('''g'''<sub>1</sub> '''g'''<sub>2</sub>)<sup>floor(''n''/2)</sup> '''g'''<sub>3</sub>. If a step is formed by stacking ''k'' generators, we may assume that ''k'' is odd, and the combinations of alternants corresponding to a step come in exactly 3 sizes:
+In case 2, let ''n'' &ge; 3 and let (2, 1) &minus; (1, 1) = '''g'''<sub>1</sub>, (1, 2) &minus; (2, 1) = '''g'''<sub>2</sub> be the two alternants. Let '''g'''<sub>3</sub> be the closing generator after stacking alternating '''g'''<sub>1</sub> and '''g'''<sub>2</sub>. Then the generator circle is ('''g'''<sub>1</sub> '''g'''<sub>2</sub>)<sup>{{floor|''n''/2}}</sup> '''g'''<sub>3</sub>. If a step is formed by stacking ''k'' generators, we may assume that ''k'' is odd, and the combinations of alternants corresponding to a step come in exactly 3 sizes:
-# ceil(''k''/2)'''g'''<sub>1</sub> + floor(''k''/2)'''g'''<sub>2</sub>
+# {{ceil|''k''/2}}'''g'''<sub>1</sub> + {{floor|''k''/2}}'''g'''<sub>2</sub>
-# floor(''k''/2)'''g'''<sub>1</sub> + ceil(''k''/2)'''g'''<sub>2</sub>
+# {{floor|''k''/2}}'''g'''<sub>1</sub> + {{ceil|''k''/2}}'''g'''<sub>2</sub>
-# floor(''k''/2)'''g'''<sub>1</sub> + floor(''k''/2) '''g'''<sub>2</sub> + '''g'''<sub>3</sub>
+# {{floor|''k''/2}}'''g'''<sub>1</sub> + {{floor|''k''/2}} '''g'''<sub>2</sub> + '''g'''<sub>3</sub>
-(since the scale size is odd, we can always ensure this by taking octave complements of all the generators). By counting the length-''k'' subwords of the (linear) word ('''g'''<sub>1</sub> '''g'''<sub>2</sub>)<sup>floor(''n''/2)</sup>, we see that the first two sizes must both occur (''n'' &minus; ''k'')/2 times. This proves (2).
+(since the scale size is odd, we can always ensure this by taking octave complements of all the generators). By counting the length-''k'' subwords of the (linear) word ('''g'''<sub>1</sub> '''g'''<sub>2</sub>)<sup>{{floor|''n''/2}}</sup>, we see that the first two sizes must both occur (''n'' &minus; ''k'')/2 times. This proves (2).
 ==== Statement (3) ====
@@ Line 79: / Line 79: @@
   x x x ... x
   x x x ... x x
-and use the vectors (-1, 2) and (ceil(n/2), 1) as the Fokker block chromas. A rank-3 Fokker block has the property that tempering out by each of the chromas gives two MOSes. These correspond to two of the temperings '''X''' = '''Y''', '''Y''' = '''Z''' and '''X''' = '''Z'''. The third tempering follows by symmetry (by taking the other chirality).
+and use the vectors (-1, 2) and ({{ceil|n/2}}, 1) as the Fokker block chromas. A rank-3 Fokker block has the property that tempering out by each of the chromas gives two MOSes. These correspond to two of the temperings '''X''' = '''Y''', '''Y''' = '''Z''' and '''X''' = '''Z'''. The third tempering follows by symmetry (by taking the other chirality).
 ==== Statement (5) ====
-By part (2), we have that ''s'' has step signature ''a'''''X''' ''b'''''Y''' ''b'''''Z''', ''a'' odd. By part (4), we have that ''T''('''X''', '''W''') = ''s''('''X''', '''W''', '''W''') is a MOS scale ''a'''''X'''2''b'''''W'''. If ''b'' = 1, there's nothing to prove, so assume ''b'' > 1.
+By part (2), we have that ''s'' has step signature ''a'''''X''' ''b'''''Y''' ''b'''''Z''', ''a'' odd. By part (4), we have that ''T''('''X''', '''W''') = ''s''('''X''', '''W''', '''W''') is a MOS scale ''a'''''X'''2''b'''''W'''. If ''b'' = 1, there's nothing to prove, so assume ''b'' &gt; 1.
 Consider the two generators in the GS of ''s'', which are detemperings of the generator ''i'''''X''' + ''j'''''W''' of ''T''('''X''', '''W'''), where gcd(''j'', 2''k'') = 1. Assume, possibly after inverting the generator, that the imperfect generator of ''T'' has ''j'' + 1 '''W'''s and the perfect generator has ''j'' '''W'''s.
@@ Line 106: / Line 106: @@
 Assume that the generator '''g''' is a ''k''-step and ''k'' is even. (If ''k'' is not even, invert the generator.) On some note ''p'' we have a chain of (''n'' + 1)/2 notes and on ''p′'' = ''p'' + offset we'll have (''n'' &minus; 1)/2) notes.
-Assume 1 < gcd(''k'', ''n'') < ''n'' and ''n'' &ge; 5. Since ''n'' is odd, ''d'' = gcd(''k'', ''n'') is an odd number at least 3, and by well-formedness with respect to the generator, there must be a circle of ''n''/''d'' < floor(''n''/2) notes formed by '''g''', contrary to the assumption of GO. Thus, gcd(''k'', ''n'') = 1.
+Assume 1 &lt; gcd(''k'', ''n'') &lt; ''n'' and ''n'' &ge; 5. Since ''n'' is odd, ''d'' = gcd(''k'', ''n'') is an odd number at least 3, and by well-formedness with respect to the generator, there must be a circle of ''n''/''d'' &lt; {{floor|''n''/2}} notes formed by '''g''', contrary to the assumption of GO. Thus, gcd(''k'', ''n'') = 1.
 Since ''n'' is odd, ''rk'' ≡ ''k''/2 mod ''n'' iff ''r'' ≡ (''n'' + 1)/2 mod ''n''. (Note that both 2 and ''k'' are coprime with ''n'', hence multiplicatively invertible mod ''n''.) This proves that the offset, which must be reached after (''n'' + 1)/2 ''k''-steps, is a ''k''/2-step, as desired. (As [''k''] is a generator of ℤ/''n'', stacking (''n'' &minus; 1)-many ''k''-steps must visit every note exactly once. Thus if the offset wasn't reached in (''n'' + 1)/2 steps, the two generator chains wouldn't have the assumed lengths.) {{qed}}
@@ Line 123: / Line 123: @@
 === Proof ===
 ==== If the generator of a projection of ''s'' is a ''k''-step, the word of stacked ''k''-steps in ''s'' is pairwise well-formed ====
-Suppose ''s'' has ''n'' notes (after dealing with small cases, we may assume ''n'' ≥ 7) and ''s'' projects to primitive MOSes ''s''<sub>1</sub> (via identifying '''b''' with '''c'''), ''s''<sub>2</sub> (via identifying '''a''' with '''c'''), and ''s''<sub>3</sub> (via identifying '''a''' with '''b'''). Suppose ''s''<sub>1</sub>'s generator is a ''k''-step, which comes in two sizes: '''P''', the perfect ''k''-step, and '''I''', the imperfect ''k''-step. By stacking ''n''-many ''k''-steps, we get two words of length ''n'' of ''k''-steps of ''s''<sub>2</sub> and ''s''<sub>3</sub>, respectively. These binary words, which we call Σ<sub>2</sub> and Σ<sub>3</sub>, must be MOSes, since ''m''-steps in the new words correspond to ''mk''-steps in the MOS words ''s''<sub>1</sub> and ''s''<sub>2</sub>, which come in at most two sizes. Since ''s''<sub>1</sub> is a primitive MOS, gcd(''k'', ''n'') = 1. Hence when 0 < ''m'' < ''n'', ''mk'' is ''not'' divisible by ''n'' and ''mk''-steps come in ''exactly'' two sizes; hence both Σ<sub>2</sub> and Σ<sub>3</sub> are primitive MOSes.
+Suppose ''s'' has ''n'' notes (after dealing with small cases, we may assume ''n'' &ge; 7) and ''s'' projects to primitive MOSes ''s''<sub>1</sub> (via identifying '''b''' with '''c'''), ''s''<sub>2</sub> (via identifying '''a''' with '''c'''), and ''s''<sub>3</sub> (via identifying '''a''' with '''b'''). Suppose ''s''<sub>1</sub>'s generator is a ''k''-step, which comes in two sizes: '''P''', the perfect ''k''-step, and '''I''', the imperfect ''k''-step. By stacking ''n''-many ''k''-steps, we get two words of length ''n'' of ''k''-steps of ''s''<sub>2</sub> and ''s''<sub>3</sub>, respectively. These binary words, which we call Σ<sub>2</sub> and Σ<sub>3</sub>, must be MOSes, since ''m''-steps in the new words correspond to ''mk''-steps in the MOS words ''s''<sub>1</sub> and ''s''<sub>2</sub>, which come in at most two sizes. Since ''s''<sub>1</sub> is a primitive MOS, gcd(''k'', ''n'') = 1. Hence when 0 &lt; ''m'' &lt; ''n'', ''mk'' is ''not'' divisible by ''n'' and ''mk''-steps come in ''exactly'' two sizes; hence both Σ<sub>2</sub> and Σ<sub>3</sub> are primitive MOSes.
   index: 1 2 3 4 ...   ''n''
@@ Line 153: / Line 153: @@
 ==== Case analysis ====
-Since, by our assumption, Λ<sub>3</sub> has two γ in a row, Λ<sub>3</sub> must have more γ than γ′, so μ &minus; 1 < ''n''/2. Since Λ<sub>3</sub> is a MOS, μ &minus; 1 ≥ 1. So we have 2 ≤ μ ≤ ceil(''n''/2).
+Since, by our assumption, Λ<sub>3</sub> has two γ in a row, Λ<sub>3</sub> must have more γ than γ′, so μ &minus; 1 &lt; ''n''/2. Since Λ<sub>3</sub> is a MOS, μ &minus; 1 &ge; 1. So we have 2 &le; μ &le; {{ceil|''n''/2}}.
 We have three cases to consider:
@@ Line 159: / Line 159: @@
 '''Case 1''': μ = 2, i.e. Λ<sub>2</sub> is the MOS (''n'' &minus; 2)β 2β′.
-For Λ<sub>2</sub> to be a MOS, the first, and only, occurrence of '''R''' must be at either ''f'' = floor(''n''/2) or ceil(''n''/2). We may assume that it is at ''f''; otherwise reverse the chain and reindex the words to start at 2''f''.
+For Λ<sub>2</sub> to be a MOS, the first, and only, occurrence of '''R''' must be at either ''f'' = {{floor|''n''/2}} or {{ceil|''n''/2}}. We may assume that it is at ''f''; otherwise reverse the chain and reindex the words to start at 2''f''.
 …   ''f''    … 2''f'' ''n''
@@ Line 172: / Line 172: @@
 These give exactly three distinct sizes for every dyad class. Hence ''s'' is SV3.
-In this case ''s'' has two chains of '''Q''', one with floor(''n''/2) notes and one offset by '''Q'''<sup>(''f''&minus;1)</sup>R with ceil(''n''/2) notes. Every instance of Q must be a ''k''-step, since by ℤ-linear independence '''Q''' = α'''a''' + β'''b''' + γ'''c''' is the only way to write '''Q''' in the basis ('''a''', '''b''', '''c'''); so ''s'' is well-formed with respect to '''Q'''. Thus ''s'' also satisfies the generator-offset property with generator '''Q'''.
+In this case ''s'' has two chains of '''Q''', one with {{floor|''n''/2}} notes and one offset by '''Q'''<sup>(''f''&minus;1)</sup>R with {{ceil|''n''/2}} notes. Every instance of Q must be a ''k''-step, since by ℤ-linear independence '''Q''' = α'''a''' + β'''b''' + γ'''c''' is the only way to write '''Q''' in the basis ('''a''', '''b''', '''c'''); so ''s'' is well-formed with respect to '''Q'''. Thus ''s'' also satisfies the generator-offset property with generator '''Q'''.
-'''Case 2:''' μ ≥ ceil(''n''/2), i.e. Λ<sub>2</sub> has fewer β than β′.
+'''Case 2:''' μ &ge; {{ceil|''n''/2}}, i.e. Λ<sub>2</sub> has fewer β than β′.
-Since Λ<sub>3</sub> has more β than β′, Λ<sub>2</sub> is floor(''n''/2)β ceil(''n''/2)β′, and Λ<sub>3</sub> is ceil(''n''/2)γ floor(''n''/2)γ′. There is a unique mode of ceil(''n''/2)γ floor(''n''/2)γ′ that both begins and ends with γ, namely γγ′γγ′…γγ′γ. Thus Λ<sub>2</sub> is ββ′ββ′…ββ′β′. It is now easy to see that if the number of ''k''-steps stacked is odd, then there are two sizes that do not contain '''T''' and one size that contains '''T'''; if the number of ''k''-steps stacked is even, then there is one size that does not contain '''T''' and two sizes that contain T. Hence ''s'' is SV3.
+Since Λ<sub>3</sub> has more β than β′, Λ<sub>2</sub> is {{floor|''n''/2}}β {{ceil|''n''/2}}β′, and Λ<sub>3</sub> is {{ceil|''n''/2}}γ {{floor|''n''/2}}γ′. There is a unique mode of {{ceil|''n''/2}}γ {{floor|''n''/2}}γ′ that both begins and ends with γ, namely γγ′γγ′…γγ′γ. Thus Λ<sub>2</sub> is ββ′ββ′…ββ′β′. It is now easy to see that if the number of ''k''-steps stacked is odd, then there are two sizes that do not contain '''T''' and one size that contains '''T'''; if the number of ''k''-steps stacked is even, then there is one size that does not contain '''T''' and two sizes that contain T. Hence ''s'' is SV3.
 In this case we have Σ = '''QRQR'''...'''QRT''', and ''s'' is well-formed with respect to the generator '''Q''' + '''R''', thus ''s'' satisfies the generator-offset property. By Proposition 1, ''s'' is SV3.
-'''Case 3:''' 3 ≤ μ ≤ floor(''n''/2).
+'''Case 3:''' 3 &le; μ &le; {{floor|''n''/2}}.
-Λ<sub>2</sub> has a chunk of β (after the first β′) of size ''x'' where ''x'' = floor(''n''/μ) ≥ floor(''n''/floor(''n''/2)) = 2 or ''x'' = ceil(''n''/μ) = floor(''n''/μ) + 1. Hence Λ<sub>3</sub> has a chunk of γ of size ''x''. Λ<sub>3</sub> also has a chunk that contains Λ<sub>3</sub>[''n'' : 2] as a subword. This chunk must be of size ''y'', where
+Λ<sub>2</sub> has a chunk of β (after the first β′) of size ''x'' where ''x'' = {{floor|''n''/μ}} &ge; {{floor|''n''/floor(''n''/2}}) = 2 or ''x'' = {{ceil|''n''/μ}} = {{floor|''n''/μ}} + 1. Hence Λ<sub>3</sub> has a chunk of γ of size ''x''. Λ<sub>3</sub> also has a chunk that contains Λ<sub>3</sub>[''n'' : 2] as a subword. This chunk must be of size ''y'', where
 <math>2 \lfloor\frac{n}{\mu}\rfloor - 1 = 2 \big(\lfloor \frac{n}{\mu} \rfloor - 1\big) + 1 \leq y \leq 2 \big(\lfloor\frac{n}{\mu}\rfloor + 1 \big) + 1 = 2\lfloor\frac{n}{\mu}\rfloor + 3.</math>
-(The lower bound is reached if Λ<sub>3</sub> has chunks of sizes floor(''n''/μ) &minus; 1 and floor(''n''/μ), and the upper bound is reached if Λ<sub>3</sub> has chunks of sizes floor(''n''/μ) and ceil(''n''/μ).)
+(The lower bound is reached if Λ<sub>3</sub> has chunks of sizes {{floor|''n''/μ}} &minus; 1 and {{floor|''n''/μ}}, and the upper bound is reached if Λ<sub>3</sub> has chunks of sizes {{floor|''n''/μ}} and {{ceil|''n''/μ}}.)
 The difference between the chunk sizes of Λ<sub>3</sub> is ''y'' &minus; ''x'', which must be 1 since Λ<sub>3</sub> is pairwise well-formed. We thus have the following subcases: (In the following, chunk of Λ<sub>2</sub> means chunk of β, and chunk of Λ<sub>3</sub> means chunk of γ.)
-'''Case 3.1:''' (''x'', ''y'') = (floor(''n''/μ), 2*floor(''n''/μ) &minus; 1).
+'''Case 3.1:''' (''x'', ''y'') = ({{floor|''n''/μ}}, 2*{{floor|''n''/μ}} &minus; 1).
-Since ''y'' &minus; ''x'' = floor(''n''/μ) &minus; 1, we have ''x'' = floor(''n''/μ) = 2 and ''y'' = 3. The chunk in Λ<sub>3</sub> whose size was defined to be ''y'' is made from two consecutive chunks in Λ<sub>2</sub> of size 1. (So Λ<sub>2</sub> has chunks of size 1 and 2, and Λ<sub>3</sub> has chunks of size 2 and 3.) Since chunk sizes of a MOS themselves form a MOS, Λ<sub>2</sub> has more chunks of size 1 than it has chunks of size 2.
+Since ''y'' &minus; ''x'' = {{floor|''n''/μ}} &minus; 1, we have ''x'' = {{floor|''n''/μ}} = 2 and ''y'' = 3. The chunk in Λ<sub>3</sub> whose size was defined to be ''y'' is made from two consecutive chunks in Λ<sub>2</sub> of size 1. (So Λ<sub>2</sub> has chunks of size 1 and 2, and Λ<sub>3</sub> has chunks of size 2 and 3.) Since chunk sizes of a MOS themselves form a MOS, Λ<sub>2</sub> has more chunks of size 1 than it has chunks of size 2.
 Λ<sub>2</sub> has only two chunks of size 1, Λ<sub>2</sub>[''n'' &minus; 1] and Λ<sub>2</sub>[1], since otherwise Λ<sub>3</sub> would have a chunk of size 1 within Λ<sub>3</sub>[1 : ''n'']. Thus Λ<sub>2</sub> has exactly one chunk of size 2. Thus Λ<sub>2</sub> = ββ′βββ′ββ′ and Λ<sub>3</sub> = γγ′γγγ′γγ. Thus we have:
@@ Line 210: / Line 210: @@
 (This also implies ''s'' is SV3.)
-'''Case 3.2''': (''x'', ''y'') = (floor(''n''/μ) + 1, 2*floor(''n''/μ) &minus; 1) is impossible: Here (''x'', ''y'') = (4, 5). But then Λ<sub>2</sub> has a chunk of size < 3 because of the β' at index ''n'', contradicting that ''x'' is one of the chunk sizes of Λ<sub>2</sub>.
+'''Case 3.2''': (''x'', ''y'') = ({{floor|''n''/μ}} + 1, 2*{{floor|''n''/μ}} &minus; 1) is impossible: Here (''x'', ''y'') = (4, 5). But then Λ<sub>2</sub> has a chunk of size &lt; 3 because of the β' at index ''n'', contradicting that ''x'' is one of the chunk sizes of Λ<sub>2</sub>.
-'''Case 3.3''': (''x'', ''y'') = (floor(''n''/μ) + 1, 2*floor(''n''/μ)) is impossible: Here (''x'', ''y'') = (3, 4). But then Λ<sub>2</sub> has a chunk of size 1 because of the β' at index ''n'', and another chunk of size 0 or 2, contradicting that ''x'' is one of the chunk sizes of Λ<sub>2</sub>.
+'''Case 3.3''': (''x'', ''y'') = ({{floor|''n''/μ}} + 1, 2*{{floor|''n''/μ}}) is impossible: Here (''x'', ''y'') = (3, 4). But then Λ<sub>2</sub> has a chunk of size 1 because of the β' at index ''n'', and another chunk of size 0 or 2, contradicting that ''x'' is one of the chunk sizes of Λ<sub>2</sub>.
-The remaining cases are all impossible because they imply ''y'' &minus; ''x'' ≥ 2:
+The remaining cases are all impossible because they imply ''y'' &minus; ''x'' &ge; 2:
-* '''Case 3.4''': (''x'', ''y'') = (floor(''n''/μ) + 1, 2*floor(''n''/μ) + 1)
+* '''Case 3.4''': (''x'', ''y'') = ({{floor|''n''/μ}} + 1, 2*{{floor|''n''/μ}} + 1)
-* '''Case 3.5''': (''x'', ''y'') = (floor(''n''/μ) + 1, 2*floor(''n''/μ) + 2)
+* '''Case 3.5''': (''x'', ''y'') = ({{floor|''n''/μ}} + 1, 2*{{floor|''n''/μ}} + 2)
-* '''Case 3.6''': (''x'', ''y'') = (floor(''n''/μ) + 1, 2*floor(''n''/μ) + 3)
+* '''Case 3.6''': (''x'', ''y'') = ({{floor|''n''/μ}} + 1, 2*{{floor|''n''/μ}} + 3)
-* '''Case 3.7''': (''x'', ''y'') = (floor(''n''/μ), 2*floor(''n''/μ))
+* '''Case 3.7''': (''x'', ''y'') = ({{floor|''n''/μ}}, 2*{{floor|''n''/μ}})
-* '''Case 3.8''': (''x'', ''y'') = (floor(''n''/μ), 2*floor(''n''/μ) + 1)
+* '''Case 3.8''': (''x'', ''y'') = ({{floor|''n''/μ}}, 2*{{floor|''n''/μ}} + 1)
-* '''Case 3.9''': (''x'', ''y'') = (floor(''n''/μ), 2*floor(''n''/μ) + 2)
+* '''Case 3.9''': (''x'', ''y'') = ({{floor|''n''/μ}}, 2*{{floor|''n''/μ}} + 2)
-* '''Case 3.10''': (''x'', ''y'') = (floor(''n''/μ), 2*floor(''n''/μ) + 3)
+* '''Case 3.10''': (''x'', ''y'') = ({{floor|''n''/μ}}, 2*{{floor|''n''/μ}} + 3)
 {{qed}}
 == Theorem 5 (PWF scales are balanced) ==
@@ Line 256: / Line 256: @@
 === Proof ===
-The result of substituting '''Y''' with '''X''' (let us call this map ''p'') is the MOS ''M'' =  2''a'''''X'''2''c'''''Z''', which has exactly 2 periods since gcd(''a'', ''c'') = 1. ''M'' thus consists of two generator chains separated by the period of ''M'', which has ''a'' + ''c'' = len(''s'') steps. It thus suffices for there to exist ''k'', 0 < ''k'' < ''a'' + ''c'', such that every perfect ''k''-step generator has the same preimage in ''s'', which will be our desired generator. Suppose that the perfect ''k''-step of ''M'' is ''i'''''W''' + ''j'''''Z''' where 0 < ''i'' < ''a''. Since ''a'' is odd, possibly after taking the period-complement we may assume that ''i'' is even. Hence each subword ''w'' of ''s'' such that its projection ''p''(''w'') subtends a perfect ''k''-step satisfies |''w''|<sub>'''X'''</sub> = |''w''|<sub>'''Y'''</sub> (= ''i''/2). It plainly follows that every such ''w'' satisfies |''w''|<sub>'''X'''</sub> = |''w''|<sub>'''Y'''</sub> = ''i''/2 and |''w''|<sub>'''Z'''</sub> = ''j''.
+The result of substituting '''Y''' with '''X''' (let us call this map ''p'') is the MOS ''M'' =  2''a'''''X'''2''c'''''Z''', which has exactly 2 periods since gcd(''a'', ''c'') = 1. ''M'' thus consists of two generator chains separated by the period of ''M'', which has ''a'' + ''c'' = len(''s'') steps. It thus suffices for there to exist ''k'', 0 &lt; ''k'' &lt; ''a'' + ''c'', such that every perfect ''k''-step generator has the same preimage in ''s'', which will be our desired generator. Suppose that the perfect ''k''-step of ''M'' is ''i'''''W''' + ''j'''''Z''' where 0 &lt; ''i'' &lt; ''a''. Since ''a'' is odd, possibly after taking the period-complement we may assume that ''i'' is even. Hence each subword ''w'' of ''s'' such that its projection ''p''(''w'') subtends a perfect ''k''-step satisfies |''w''|<sub>'''X'''</sub> = |''w''|<sub>'''Y'''</sub> (= ''i''/2). It plainly follows that every such ''w'' satisfies |''w''|<sub>'''X'''</sub> = |''w''|<sub>'''Y'''</sub> = ''i''/2 and |''w''|<sub>'''Z'''</sub> = ''j''.
 It remains to show that ''s'' is balanced... <!--{{qed}}-->
@@ Line 265: / Line 265: @@
 # A primitive [[balanced]] MV3 scale ''s'' satisfies one of the following:
 ## '''sporadic balanced''': ''s'' is equivalent to '''XYXZXYX''', the ternary [[Fraenkel word]], with step signature 4'''X'''2'''Y'''1'''Z'''.
-## '''odd-regular''': len(''s'') is odd, and ''s'' is equivalent to a word constructed from taking the brightest mode of the MOS ''c'''''X'''''b'''''Z''' with ''c'' even and gcd(''c'', ''b'') = 1, and replacing every other '''X''' with '''Y'''. We assume '''X''' > '''Z''' when constructing the MOS. In particular, ''s'' has [[step signature]] ''a'''''X'''''a'''''Y'''''b'''''Z''' where ''b'' is odd (with ''a'' = ''c''/2).
+## '''odd-regular''': len(''s'') is odd, and ''s'' is equivalent to a word constructed from taking the brightest mode of the MOS ''c'''''X'''''b'''''Z''' with ''c'' even and gcd(''c'', ''b'') = 1, and replacing every other '''X''' with '''Y'''. We assume '''X''' &gt; '''Z''' when constructing the MOS. In particular, ''s'' has [[step signature]] ''a'''''X'''''a'''''Y'''''b'''''Z''' where ''b'' is odd (with ''a'' = ''c''/2).
 ## '''even-regular''': len(''s'') is even, and ''s'' is equivalent to a word constructed from taking the brightest mode of the MOS 2''a'''''X'''2''c'''''Z''' with ''a'' odd and gcd(''a'', ''c'') = 1, and replacing every other '''X''' with '''Y'''. In particular,  ''s'' has [[step signature]] ''a'''''X'''''a'''''Y'''''b'''''Z''' with ''a'' odd and ''b'' even.
 # All primitive balanced ternary scales are MV3.
@@ Line 278: / Line 278: @@
 (a) Let ''s'' be a ternary balanced word; then for any given letter '''y''' the number of '''y'''s in a subword of any given length ''L'' varies by at most 1. Thus the same is true when we count all non-'''y''' letters in any subword of length ''L''; thus when we equate '''x''' and '''z''', the count of the resulting letter in any subword of length ''L'' differs by 1. Being a binary balanced word is one characterization of the MOS property.
-(b) Assume that the projection ''p''<sub>'''YZ'''</sub>(''s'') identifying '''Y''' and '''Z''' of a primitive PMOS scale ''s'' with signature ''ar'''''X''' ''b'''''Y''' ''c'''''Z'''  is an ''r''-period MOS, ''r'' > 1, with step signature ''ar'''''X''' ''dr'''''W'''. We claim that neither ''b'' nor ''c'' is divisible by ''r''. Since ''p''<sub>'''XY'''</sub>(''s'') is the MOS (''ar'' + ''b'')'''W''' ''c'''''Z''' and  ''p''<sub>'''XZ'''</sub>(''s'') is the MOS (''ar'' + ''c'')'''W''' ''b'''''Y''', if either ''b'' and ''c'' is divisible by ''r'', then the distributions of two of the letters have ''r'' periods. Then the distribution of the third letter also has ''r'' periods, meaning that ''s'' itself has ''r'' periods, a contradiction. It suffices to show that ''r'' = 2. If ''r'' > 2, then...
+(b) Assume that the projection ''p''<sub>'''YZ'''</sub>(''s'') identifying '''Y''' and '''Z''' of a primitive PMOS scale ''s'' with signature ''ar'''''X''' ''b'''''Y''' ''c'''''Z'''  is an ''r''-period MOS, ''r'' &gt; 1, with step signature ''ar'''''X''' ''dr'''''W'''. We claim that neither ''b'' nor ''c'' is divisible by ''r''. Since ''p''<sub>'''XY'''</sub>(''s'') is the MOS (''ar'' + ''b'')'''W''' ''c'''''Z''' and  ''p''<sub>'''XZ'''</sub>(''s'') is the MOS (''ar'' + ''c'')'''W''' ''b'''''Y''', if either ''b'' and ''c'' is divisible by ''r'', then the distributions of two of the letters have ''r'' periods. Then the distribution of the third letter also has ''r'' periods, meaning that ''s'' itself has ''r'' periods, a contradiction. It suffices to show that ''r'' = 2. If ''r'' &gt; 2, then...
 For 7.1.2: Suppose ''s'' is balanced and has at least three sizes for ''k''-steps, ''a''<sub>''i''</sub>'''X''' + ''b''<sub>''i''</sub>'''Y''' + ''c''<sub>''i''</sub>'''Z''' = (''a''<sub>''i''</sub>, ''b''<sub>''i''</sub>, ''c''<sub>''i''</sub>) for ''i'' = 1, 2, 3. We may assume (''a''<sub>2</sub>, ''b''<sub>2</sub>, ''c''<sub>2</sub>) = (''a''<sub>1</sub>, ''b''<sub>1</sub> + 1, ''c''<sub>1</sub> - 1). Then either (''a''<sub>3</sub>, ''b''<sub>3</sub>, ''c''<sub>3</sub>) = (''a''<sub>1</sub> + 1, ''b''<sub>1</sub>, ''c''<sub>1</sub> - 1) or (''a''<sub>3</sub>, ''b''<sub>3</sub>, ''c''<sub>3</sub>) =  (''a''<sub>1</sub> - 1, ''b''<sub>1</sub> + 1, ''c''<sub>1</sub>). In both cases, by balancedness applied to subwords of length ''k'', the three vectors represent the only possible interval sizes.
-For 7.1.3: The ternary Fraenkel word may be verified as SV3 by inspection, and we have already shown in Theorem 1 that odd-regular balanced scales are SV3. To show that even-regular balanced scales are ''not'' SV3, observe that (''a''+''c'')-steps come in only 2 sizes in such a scale ''s'': floor(''a''/2)'''X''' + ceil(''a''/2)'''Y''' + ''c'''''Z''' and ceil(''a''/2)'''X''' + floor(''a''/2)'''Y''' + ''c'''''Z''', since the underlying MOS 2''a'''''X'''2''c'''''Y''' only has the (''a''+''c'')-step ''a'''''X''' + ''c'''''Z'''. The construction replaces the '''X'''s in these subwords with alternating '''X'''s and '''Y'''s; either of '''X''' or '''Y''' may occur first, corresponding to the two possible sizes, since ''a'' is odd and thus the (''a''+''c'')-step subword ''s''[''k'' : ''k''+''a''+''c''] becomes the subword ''s''[''k''+''a''+''c'' : ''k''+2''a''+2''c''] via interchanging '''X''' and '''Y'''.
+For 7.1.3: The ternary Fraenkel word may be verified as SV3 by inspection, and we have already shown in Theorem 1 that odd-regular balanced scales are SV3. To show that even-regular balanced scales are ''not'' SV3, observe that (''a''+''c'')-steps come in only 2 sizes in such a scale ''s'': {{floor|''a''/2}}'''X''' + {{ceil|''a''/2}}'''Y''' + ''c'''''Z''' and {{ceil|''a''/2}}'''X''' + {{floor|''a''/2}}'''Y''' + ''c'''''Z''', since the underlying MOS 2''a'''''X'''2''c'''''Y''' only has the (''a''+''c'')-step ''a'''''X''' + ''c'''''Z'''. The construction replaces the '''X'''s in these subwords with alternating '''X'''s and '''Y'''s; either of '''X''' or '''Y''' may occur first, corresponding to the two possible sizes, since ''a'' is odd and thus the (''a''+''c'')-step subword ''s''[''k'' : ''k''+''a''+''c''] becomes the subword ''s''[''k''+''a''+''c'' : ''k''+2''a''+2''c''] via interchanging '''X''' and '''Y'''.
 Claim 7.1.4 can be verified by noting that such scales are PWF and using Theorem 4. <!--{{Qed}}-->
@@ Line 314: / Line 314: @@
 # Given any arbitrary MOS scale with at least three notes per period, is there *always* a MV3 generator-offset scale which can be derived as a "detempering" of that scale? Or is this only true for some MOS's? For instance, the MOS LLsLLLs has the MV3 generator-offset scale LmsLmLs as a detempering. Does a similar MV3 detempering exist for every possible DE scale with at least three notes per period, or at least for strict MOS's with one period per octave (e.g. well-formed scales)?
 # The scale tree is a great way to analyze MOS scales. For any generator, we can compute the various MOS's it forms if we simply look at the scale tree, and indeed MOS "words" like LLsLLLs can be identified with regions on the scale tree (in this situation the interval between 4/7 and 3/5). A similar "scale plane" should exist for generator-offset-MV3 scales, where given some word representing a generator-offset-MV3 scale, we can look at the set of points on the generator plane which generates it; these seem to often be triangles, with the lines corresponding to MOS's and the vertices corresponding to EDOs (though is this always true?). What is the big picture of this scale plane? Can we use Viggo Brun's algorithm for this, generalizing the theory of continued fractions? Is there some simple formula we can use to predict, given some generator-offset-MV3 scale, which region on the scale plane it corresponds to? Can we plot simple generator-size-proportions as points in this space? And so on.
-# In the theory of MOS, there is a second [[MOS Scale Family Tree|scale tree]] that is less frequently talked about, which Erv Wilson calls the "Rabbit Sequence" ([http://www.anaphoria.com/RabbitSequence.pdf Erv Wilson's original version], [https://mikebattagliamusic.com/MOSTree/MOSTreeab.html interactive version 1], [https://mikebattagliamusic.com/MOSTree/MOSTreeLs.html interactive version 2]). This is a tree for which each MOS word has two children, depending on if the MOS is "soft" (with L/s < 2) or "hard" (with L/s > 2). For instance, LsLss has the two children LLsLLLs and ssLsssL. Does a similar scale plane exist for these generator-offset-MV3 scales?
+# In the theory of MOS, there is a second [[MOS Scale Family Tree|scale tree]] that is less frequently talked about, which Erv Wilson calls the "Rabbit Sequence" ([http://www.anaphoria.com/RabbitSequence.pdf Erv Wilson's original version], [https://mikebattagliamusic.com/MOSTree/MOSTreeab.html interactive version 1], [https://mikebattagliamusic.com/MOSTree/MOSTreeLs.html interactive version 2]). This is a tree for which each MOS word has two children, depending on if the MOS is "soft" (with L/s &lt; 2) or "hard" (with L/s &gt; 2). For instance, LsLss has the two children LLsLLLs and ssLsssL. Does a similar scale plane exist for these generator-offset-MV3 scales?
 [[Category:Math]]

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