K*N subgroups: Difference between revisions

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For any [[~~Harmonic_Limit~~|prime limit]] p, EDO N and positive integer k, the p-limit k*N subgroup is the largest [[~~Just_intonation_subgroups|~~just intonation subgroup]] of the p-limit on which N-edo and k*N-edo approximate intervals to the same values using the mapping supplied by the [[~~Patent_val|~~patent val]] for k*N-edo. This also means they temper out the same commas.

For any [[Harmonic_limit|prime limit]] ''p'', EDO ''N'', and positive integer ''k'', the ''p''-limit ''k''*''N'' subgroup is the largest [[just intonation subgroup]] of the ''p''-limit on which ''N''-edo and ''k''*''N''-edo approximate intervals to the same values using the mapping supplied by the [[patent val]] for ''k''*''N''-edo. This also means they temper out the same commas.

A procedure for finding the k*N subgroup which seems to suffice is to take the product m of the odd primes less than or equal to p, and then find the [[~~Euler_genera|~~Euler genus]] ~~Euler~~(m^i) for integers 1, 2, and so forth. Here ~~Euler~~(d) for an odd integer d is the set of all divisors of d reduced to an octave, and including 2. For each such genus, select those intervals such that the k*N patent val maps the interval to a number divisible by k, and then find the corresponding [[~~Normal_lists~~|normal interval list]]. When two successive values i and i+1 lead to the same normal list, add to that a basis for the commas of the k*N patent val, and return the normal interval list for that.

A procedure for finding the ''k''*''N'' subgroup which seems to suffice is to take the product ''m'' of the odd primes less than or equal to ''p'', and then find the [[Euler genus]] Eu(''m''''i'') for integers 1, 2, and so forth. Here Eu(''d'') for an odd integer ''d'' is the set of all divisors of ''d'' reduced to an octave, and including 2. For each such genus, select those intervals such that the ''k''*''N'' patent val maps the interval to a number divisible by ''k'', and then find the corresponding [[normal lists|normal interval list]]. When two successive values ''i'' and {{nowrap|''i'' + 1}} lead to the same normal list, add to that a basis for the commas of the ''k''*''N'' patent val, and return the normal interval list for that.

For example, to find the 7-limit 2*6 subgroup we first find m = 3*5*7 = 105. The subgroup of ~~Euler~~(105) consisting of those intervals mapped to an even number by ~~<~~12 19 28 34| is 2.5.7. The subgroup of ~~Euler~~(105^2) is 2.9.5.7, not the same. However, the subgroup of ~~Euler~~(105^3) is again 2.9.5.7. If to this we add 81/80, we still obtain 2.9.5.7. It is clear this is maximal, as 3 is mapped by 12et to an odd number (19), and the rest of the values are prime.

For example, to find the 7-limit 2*6 subgroup we first find {{nowrap|''m'' {{=}} 3 × 5 × 7 {{=}} 105}}. The subgroup of Eu(105) consisting of those intervals mapped to an even number by {{vector|12 19 28 34}} is 2.5.7. The subgroup of Eu(1052) is 2.9.5.7, not the same. However, the subgroup of Eu(1053) is again 2.9.5.7. If to this we add 81/80, we still obtain 2.9.5.7. It is clear this is maximal, as 3 is mapped by 12et to an odd number (19), and the rest of the values are prime.

[[Category:~~algorithm~~]]

[[Category:~~math~~]]

[[Category:Algorithms]]

[[Category:theory]]

[[Category:Math]]

[[Category:Regular temperament theory]]

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-For any [[Harmonic_Limit|prime limit]] p, EDO N and positive integer k, the p-limit k*N subgroup is the largest [[Just_intonation_subgroups|just intonation subgroup]] of the p-limit on which N-edo and k*N-edo approximate intervals to the same values using the mapping supplied by the [[Patent_val|patent val]] for k*N-edo. This also means they temper out the same commas.
+{{todo|explain its xenharmonic value}}
+For any [[Harmonic_limit|prime limit]] ''p'', EDO ''N'', and positive integer ''k'', the ''p''-limit ''k''*''N'' subgroup is the largest [[just intonation subgroup]] of the ''p''-limit on which ''N''-edo and ''k''*''N''-edo approximate intervals to the same values using the mapping supplied by the [[patent val]] for ''k''*''N''-edo. This also means they temper out the same commas.
-A procedure for finding the k*N subgroup which seems to suffice is to take the product m of the odd primes less than or equal to p, and then find the [[Euler_genera|Euler genus]] Euler(m^i) for integers 1, 2, and so forth. Here Euler(d) for an odd integer d is the set of all divisors of d reduced to an octave, and including 2. For each such genus, select those intervals such that the k*N patent val maps the interval to a number divisible by k, and then find the corresponding [[Normal_lists|normal interval list]]. When two successive values i and i+1 lead to the same normal list, add to that a basis for the commas of the k*N patent val, and return the normal interval list for that.
+A procedure for finding the ''k''*''N'' subgroup which seems to suffice is to take the product ''m'' of the odd primes less than or equal to ''p'', and then find the [[Euler genus]] Eu(''m''<sup>''i''</sup>) for integers 1, 2, and so forth. Here Eu(''d'') for an odd integer ''d'' is the set of all divisors of ''d'' reduced to an octave, and including 2. For each such genus, select those intervals such that the ''k''*''N'' patent val maps the interval to a number divisible by ''k'', and then find the corresponding [[normal lists|normal interval list]]. When two successive values ''i'' and {{nowrap|''i'' + 1}} lead to the same normal list, add to that a basis for the commas of the ''k''*''N'' patent val, and return the normal interval list for that.
-For example, to find the 7-limit 2*6 subgroup we first find m = 3*5*7 = 105. The subgroup of Euler(105) consisting of those intervals mapped to an even number by &lt;12 19 28 34| is 2.5.7. The subgroup of Euler(105^2) is 2.9.5.7, not the same. However, the subgroup of Euler(105^3) is again 2.9.5.7. If to this we add 81/80, we still obtain 2.9.5.7. It is clear this is maximal, as 3 is mapped by 12et to an odd number (19), and the rest of the values are prime.
+For example, to find the 7-limit 2*6 subgroup we first find {{nowrap|''m'' {{=}} 3 × 5 × 7 {{=}} 105}}. The subgroup of Eu(105) consisting of those intervals mapped to an even number by {{vector|12 19 28 34}} is 2.5.7. The subgroup of Eu(105<sup>2</sup>) is 2.9.5.7, not the same. However, the subgroup of Eu(105<sup>3</sup>) is again 2.9.5.7. If to this we add 81/80, we still obtain 2.9.5.7. It is clear this is maximal, as 3 is mapped by 12et to an odd number (19), and the rest of the values are prime.
-[[Category:algorithm]]
-[[Category:math]]
+[[Category:Algorithms]]
-[[Category:theory]]
+[[Category:Math]]
+[[Category:Regular temperament theory]]