Ternary scale theorems: Difference between revisions

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# the stack has only copies of '''Q''' and '''R'''; or

# the stack has one '''T''' and does not contain any '''R''' (since it's more than {{nowrap|''f'' − 1}} generators away).

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

These give exactly three distinct sizes for every interval class. Hence ''s'' is SV3. In this case a window stacking argument shows that the second type of ''fk''-step {{nowrap|((''f'' − 1)''Q'' + ''T'')}} alternates with the first type {{nowrap|((''f'' − 1)''Q'' + ''R'')}}, and ''fQ'' occurs only once, so ''s'' has generator sequence {{nowrap|GS((''f'' − 1)''Q'' + ''T'', (''f'' − 1)''Q'' + ''R'')}}. Since ''n'' is odd, ''s'' is odd-regular.

In this case ~~''s'' has two chains~~ of '''Q'~~'', one with~~ {{~~floor~~|''n''~~/2}} notes and one offset by {{nowrap|'~~''Q'''~~(~~''f'' ~~− 1~~)~~R~~}} with {{~~ceil~~|''n''~~/2}} notes. Every instance of Q must be a~~ ''k''~~-step, since by ℤ-linear independence {{nowrap|~~'''Q'~~'' {{=~~}} ~~α'''a''' + β'~~''b''~~' + γ'''c'''}} is the~~ only ~~way to write~~ '''Q'~~'' in the basis~~ {{nowrap|('''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:''' {{nowrap|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 ~~{{nowrap|~~Σ {{=}} '''~~QRQR~~'~~''…'''QRT'''}}~~, and ''s'' ~~is well-formed with respect to the~~ generator {{nowrap|'''Q''~~' + '~~''R'''}}, thus ''s'' ~~satisfies the generator~~-~~offset property. By Proposition 1, ''s'' is SV3~~.

In this case we have Σ = ''QRQR…QRT'', and ''s'' has generator sequence {{nowrap|GS(''Q'', ''R'').}} We thus have that ''s'' is odd-regular.

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

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 # the stack has only copies of '''Q''' and '''R'''; or
 # the stack has one '''T''' and does not contain any '''R''' (since it's more than {{nowrap|''f'' − 1}} generators away).
-These give exactly three distinct sizes for every interval class. Hence ''s'' is SV3.
+These give exactly three distinct sizes for every interval class. Hence ''s'' is SV3. In this case a window stacking argument shows that the second type of ''fk''-step {{nowrap|((''f'' &minus; 1)''Q'' + ''T'')}} alternates with the first type {{nowrap|((''f'' &minus; 1)''Q'' + ''R'')}}, and ''fQ'' occurs only once, so ''s'' has generator sequence {{nowrap|GS((''f'' &minus; 1)''Q'' + ''T'', (''f'' &minus; 1)''Q'' + ''R'')}}. Since ''n'' is odd, ''s'' is odd-regular.
-In this case ''s'' has two chains of '''Q''', one with {{floor|''n''/2}} notes and one offset by {{nowrap|'''Q'''<sup>(''f'' − 1)</sup>R}} with {{ceil|''n''/2}} notes. Every instance of Q must be a ''k''-step, since by ℤ-linear independence {{nowrap|'''Q''' {{=}} α'''a''' + β'''b''' + γ'''c'''}} is the only way to write '''Q''' in the basis {{nowrap|('''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:''' {{nowrap|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&nbsp;{{ceil|''n''/2}}β′, and Λ<sub>3</sub> is {{ceil|''n''/2}}γ&nbsp;{{floor|''n''/2}}γ′. There is a unique mode of {{ceil|''n''/2}}γ&nbsp;{{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&nbsp;{{ceil|''n''/2}}β′, and Λ<sub>3</sub> is {{ceil|''n''/2}}γ&nbsp;{{floor|''n''/2}}γ′. There is a unique mode of {{ceil|''n''/2}}γ&nbsp;{{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 {{nowrap|Σ {{=}} '''QRQR'''…'''QRT'''}}, and ''s'' is well-formed with respect to the generator {{nowrap|'''Q''' + '''R'''}}, thus ''s'' satisfies the generator-offset property. By Proposition 1, ''s'' is SV3.
+In this case we have Σ = ''QRQR…QRT'', and ''s'' has generator sequence {{nowrap|GS(''Q'', ''R'').}} We thus have that ''s'' is odd-regular.
 '''Case 3:''' {{nowrap|3 &le; μ &le; {{floor|''n''/2}}}}.