Octave-Equivalent Epimericity

Let us define Octave-Equivalent Epimericity (OEE) of a ratio Q as follows:

OEE=min[D(S)], where S is the set of all ratios Q*2^n, and where D(S) denotes a function which takes the elements A/B in S and maps them to |A-B|.

Examples:
OEE=1 --- 3/2, 4/3, 5/4, 8/5, 6/5, 5/3, ...
OEE=2 --- 7/5, 10/7, 9/7, 14/9, ...
OEE=3 --- 11/8, 16/11, 13/10, 20/13, ...

Theorem #1: If we have two ratios 1/1 and (N+1)/N, there are a finite number of non-octave-equivalent ratios we can add to this scale without any intervals exceeding OEE=1.

Proof: Find the largest interval (N+1+m)/(N+m) smaller than (N+1)/N such that the OEE of all intervals in the scale is 1. This means the interval between (N+1+m)/(N+m) and (N+1)/N is the smallest interval with OEE=1. The closed interval formed by these upper and lower bounds contain a finite number of intervals with OEE=1, since the only acclamation points are at 1/1 and 2/1. To test for ratios larger than (N+1)/N, simply take the octave inversion and repeat the same process above.

(Thank you to Keenan Pepper for the above proof!)

Theorem #2: There are no scales with more than 2 notes such that all the intervals all have OEE=2.

Proof: Start by constructing a general scale, with notes 1/1 and (N+2)/N and (N+2+k)/(N+k), for odd N and non-zero even k. The interval spanned between the latter two notes is (N+2)*(N+k)/[N*(N+2+k)]. The difference between the numerator and denominator is 2*k. Now suppose this fraction is not in lowest terms, but that GCD(num, den) = Z. This means that OEE=2*k/Z. This can never be 2, since k is always even and Z is always odd.

There is one more case we are forgetting: suppose our scale was formed by the notes 1/1, 2N/(N+2) and (k+2)/k. The interval between these latter two notes is 2*k*N/[(k+2)*(N+2)]. The difference between the numerator and the denominator is k*N-2*k-2*n-4. Now suppose this fraction is not in lowest terms, but that GCD(num, den) = Z. This means that OEE=(k*N-2*k-2*n-4)/Z. Assume WLOG that Z divides k evenly, such that k/Z = X. Therefore OEE=X*N-2*X-(2*N+4)/Z. Since Z is odd, and it must divide 2*N+4 evenly, we know that X=1, and therefore k=Z. Therefore, OEE=N-2-2*(N+2)/k. This can never be 2 given our initial conditions, hence OEE != 2.

(Note: There are a few lemmas I used in the above proof without stating or proving them. One of these is that a fraction (N+2)/N with OEE=2 will always be smaller than 600 cents. Without this, the above proof is not possible.)

Ryan Avella