Ternary scale theorems: Difference between revisions
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==== Statement (2) ==== | ==== Statement (2) ==== | ||
In case 2, let (2, 1) − (1, 1) = g<sub>1</sub>, (1, 2) − (2, 1) = g<sub>2</sub> be the two alternants. Let g<sub>3</sub> be the leftover 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>. | In case 2, let (2, 1) − (1, 1) = g<sub>1</sub>, (1, 2) − (2, 1) = g<sub>2</sub> be the two alternants. Let g<sub>3</sub> be the leftover 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, ''k'' must be 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> | ||