Ternary scale theorems: Difference between revisions

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For 5.1.1: We showed previously that the Fraenkel, odd-regular, and even-regular circular words are balanced.

We will first prove that a balanced circular word is primitive iff ~~rhe~~ gcd of the step signature is 1. Proof sketch: let ''d'' be the gcd of the step signature. (''n''/''d'')-step multisets come in 1 size, namely the equave divided by ''d'', because if some letter count differs, then we get 3 values for this letter count for (''n''/''d'')-step multisets by the discrete IVT.

We will first prove that a balanced circular word is primitive iff the gcd of the step signature is 1. Proof sketch: let ''d'' be the gcd of the step signature. (''n''/''d'')-step multisets come in 1 size, namely the equave divided by ''d'', because if some letter count differs, then we get 3 values for this letter count for (''n''/''d'')-step multisets by the discrete IVT.

It remains to show that (a) ternary balanced words are pairwise-MOS (b) if ''a'' > ''b'' > ''c'', then ''s'' is equivalent to the Fraenkel word (c) assuming ''a'' != ''b'' = ''c'' any ''s'' that is not odd-regular or even-regular is not balanced.

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 For 5.1.1: We showed previously that the Fraenkel, odd-regular, and even-regular circular words are balanced.
-We will first prove that a balanced circular word is primitive iff rhe gcd of the step signature is 1. Proof sketch: let ''d'' be the gcd of the step signature. (''n''/''d'')-step multisets come in 1 size, namely the equave divided by ''d'', because if some letter count differs, then we get 3 values for this letter count for (''n''/''d'')-step multisets by the discrete IVT.
+We will first prove that a balanced circular word is primitive iff the gcd of the step signature is 1. Proof sketch: let ''d'' be the gcd of the step signature. (''n''/''d'')-step multisets come in 1 size, namely the equave divided by ''d'', because if some letter count differs, then we get 3 values for this letter count for (''n''/''d'')-step multisets by the discrete IVT.
 It remains to show that (a) ternary balanced words are pairwise-MOS (b) if ''a'' > ''b'' > ''c'', then ''s'' is equivalent to the Fraenkel word (c) assuming ''a'' != ''b'' = ''c'' any ''s'' that is not odd-regular or even-regular is not balanced.