Ternary scale theorems: Difference between revisions

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We also know that ''s'' is of the form ''w''('''x''', '''y''', '''z''')''w''('''x''', '''z''', '''y'''). Hence the count of '''x''' + '''y''' = the count of '''x''' + '''z''' in all of ''s''.

Write ''s''1 for the scale word made from stacked 2-steps from the 0-degree, and let ''s''2 be the scale word of stacked 2-steps from the ~~(''n''/2)~~-degree. We know that they differ only by interchanging '''y''' and '''z''', hence that they have the same period. Hence both ''s''1 and ''s''2 are primitive.

Write ''s''1 for the scale word made from stacked 2-steps from the 0-degree, and let ''s''2 be the scale word of stacked 2-steps from the 1-degree. We know that they differ only by interchanging '''y''' and '''z''', hence that they have the same period. Hence both ''s''1 and ''s''2 are primitive.

We prove that ''s''1 and ''s''2 are MOS substitution scales with a filling MOS of period 2. The number the 2-step (1) occurs must be the same in both ''s''1 and ''s''2. The word of stacked 2-steps of the template MOS (which is of the form ''w''('''x''', '''X''', '''X''')''w''('''x''', '''X''', '''X''')), which is itself a MOS word, consists of letters (1) '''x''' + '''X''' and (2) 2'''X''' if more '''X''''s than '''x''''s, 2'''x''' if more '''x''''s than '''X''''s. The word of stacked 2-steps from the n/2-degree is also this same MOS word.

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 We also know that ''s'' is of the form ''w''('''x''', '''y''', '''z''')''w''('''x''', '''z''', '''y'''). Hence the count of '''x''' + '''y''' = the count of '''x''' + '''z''' in all of ''s''.
-Write ''s''<sub>1</sub> for the scale word made from stacked 2-steps from the 0-degree, and let ''s''<sub>2</sub> be the scale word of stacked 2-steps from the (''n''/2)-degree. We know that they differ only by interchanging '''y''' and '''z''', hence that they have the same period. Hence both ''s''<sub>1</sub> and ''s''<sub>2</sub> are primitive.
+Write ''s''<sub>1</sub> for the scale word made from stacked 2-steps from the 0-degree, and let ''s''<sub>2</sub> be the scale word of stacked 2-steps from the 1-degree. We know that they differ only by interchanging '''y''' and '''z''', hence that they have the same period. Hence both ''s''<sub>1</sub> and ''s''<sub>2</sub> are primitive.
 We prove that ''s''<sub>1</sub> and ''s''<sub>2</sub> are MOS substitution scales with a filling MOS of period 2. The number the 2-step (1) occurs must be the same in both ''s''<sub>1</sub> and ''s''<sub>2</sub>. The word of stacked 2-steps of the template MOS (which is of the form ''w''('''x''', '''X''', '''X''')''w''('''x''', '''X''', '''X''')), which is itself a MOS word, consists of letters (1) '''x''' + '''X''' and (2) 2'''X''' if more '''X''''s than '''x''''s, 2'''x''' if more '''x''''s than '''X''''s. The word of stacked 2-steps from the n/2-degree is also this same MOS word.