Talk:K*N subgroups

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Termination condition

When two successive values i and i+1 lead to the same normal list

This may cause termination too early. For example... 5-limit 3*24 subgroup of 24edo, I want a result "2.3.125 subgroup", but no additional base at i=2, cause terminating the algorithm and result in "2.3 subgroup". Simply it should be do up to i=k. --Dummy index (talk) 13:29, 13 June 2024 (UTC)

Hmm, subgroup may be extended by this → add to that a basis for the commas of the k*N patent val

Commas are all unison, 0 is divisible by k, not intrusive. Ah, before that, 5-limit 24edo is enfactored... Sorry. --Dummy index (talk) 14:55, 13 June 2024 (UTC)

New example, 5-limit 9*19 subgroup of 19edo. Because 3 and 5 are equally flattened on 19edo, patent val of 171edo maps these to 1\171 sharpened than intervals of 19edo, respectively. i=1 and i=2 result in intervals {1/1}(*2^n), its subgroup is 2-limit. Add commas (such as unimodular basis) then we gain only 2.3^171.3^8*5 subgroup. In this case we need at least i=5, {1/1, 3^5*5^4, 3^4*5^5} then we gain 2.3^9.3^8*5 subgroup. And of course, in 3-limit, we need i=9 to gain 2.3^9 subgroup. --Dummy index (talk) 14:07, 23 June 2024 (UTC)