Ternary scale theorems: Difference between revisions

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

# from g1 ... g1, we get a1 = (''r'' − 1)/2*g~~0~~ + g1 = (''r'' ~~+ 1~~/2) g1 + (''r'' ~~− 1~~/2) g2

# from g1 ... g1, we get a1 = (''r'' − 1)/2*g + g1 = ceil(''r''/2) g1 + floor(''r''/2) g2

# from g2 ... g2, we get a2 = (''r'' − 1)/2*g~~0~~ + g2 = (''r'' ~~− 1~~/2) g1 + (''r'' ~~+ 1~~/2) g2

# from g2 ... g2, we get a2 = (''r'' − 1)/2*g + g2 = floor(''r''/2) g1 + ceil(''r''/2) g2

# from g2 (...even # of gens...) g1 g3 g1 (...even # of gens...) g2, we get a3 = (''r'' − 1)/2 g1 + (''r'' − 1)/2 g2 + g3 ≡ (''r'' − ''n''/2 − 3/2)g1 + (''r'' − ''n''/2 − 1/2)g2 mod e.

# from g1 (...odd # of gens...) g1 g3 g1 (...odd # of gens...) g1, we get a4 = (''r'' + 1)/2 g1 + (''r'' − 3)/2 g2 + g3 ≡ (''r'' − ''n''/2 − 1/2)g1 + (''r'' − ''n''/2 − 3/2)g2 mod e.

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 Let ''r'' be odd and ''r'' ≥ 3. Consider the following abstract sizes for the interval class (''k''-steps) reached by stacking ''r'' generators:
-# from g<sub>1</sub> ... g<sub>1</sub>, we get a<sub>1</sub> = (''r'' &minus; 1)/2*g<sub>0</sub> + g<sub>1</sub> = (''r'' + 1/2) g<sub>1</sub> + (''r'' &minus; 1/2) g<sub>2</sub>
+# from g<sub>1</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>2</sub>, we get a<sub>2</sub> = (''r'' &minus; 1)/2*g<sub>0</sub> + g<sub>2</sub> = (''r'' &minus; 1/2) g<sub>1</sub> + (''r'' + 1/2) g<sub>2</sub>
+# from g<sub>2</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.