Otonality and utonality: Difference between revisions

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Assume a chord is ambitonal. Then its largest integer, max(chord), is equal to the largest integer of its inverse, which is LCM(chord)/min(chord). Therefore min(chord)*max(chord) = LCM(chord). Conversely, if a set of integers has gcd 1 and also satisfies this, then it is an ambitonal chord.

Thus, for any given odd number N (where N is not prime), all ambitonal chords with LCM N can easily be found by considering subsets of the factors of N. If a subset has at least three factors, has a GCD of 1, an LCM of N, and also satisfies min(subset)*max(subset) = N, then it is an ambitonal chord. These conditions are satisfied by any subset which includes 1 and N. There are usually other valid subsets as well.

Thus, for any given odd number N (where N is not prime), all ambitonal chords with LCM N can easily be found by considering subsets of the factors of N. If a subset has at least three factors (as mentioned above, the statement always holds for two or fewer), has a GCD of 1, an LCM of N, and also satisfies min(subset)*max(subset) = N, then it is an ambitonal chord. These conditions are satisfied by any subset which includes 1 and N. There are usually other valid subsets as well.

For N = 15, the factors are 1, 3, 5 and 15, and the ambitonal chords are {1, 3, 5, 15}, {1, 3, 15} and {1, 5, 15}. These [[octave-reduce]] to {1/1, 3/2, 5/4, 15/8} = maj7 chord, {1/1, 3/2, 15/8} = maj7no3 chord, and {1/1, 5/4, 15/8} = maj7no5 chord.

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 Assume a chord is ambitonal. Then its largest integer, max(chord), is equal to the largest integer of its inverse, which is LCM(chord)/min(chord). Therefore min(chord)*max(chord) = LCM(chord). Conversely, if a set of integers has gcd 1 and also satisfies this, then it is an ambitonal chord.
-Thus, for any given odd number N (where N is not prime), all ambitonal chords with LCM N can easily be found by considering subsets of the factors of N. If a subset has at least three factors, has a GCD of 1, an LCM of N, and also satisfies min(subset)*max(subset) = N, then it is an ambitonal chord. These conditions are satisfied by any subset which includes 1 and N. There are usually other valid subsets as well.
+Thus, for any given odd number N (where N is not prime), all ambitonal chords with LCM N can easily be found by considering subsets of the factors of N. If a subset has at least three factors (as mentioned above, the statement always holds for two or fewer), has a GCD of 1, an LCM of N, and also satisfies min(subset)*max(subset) = N, then it is an ambitonal chord. These conditions are satisfied by any subset which includes 1 and N. There are usually other valid subsets as well.
 For N = 15, the factors are 1, 3, 5 and 15, and the ambitonal chords are {1, 3, 5, 15}, {1, 3, 15} and {1, 5, 15}. These [[octave-reduce]] to {1/1, 3/2, 5/4, 15/8} = maj7 chord, {1/1, 3/2, 15/8} = maj7no3 chord, and {1/1, 5/4, 15/8} = maj7no5 chord.