MOS substitution

A procedure for obtaining a ternary scale with arbitrary aLbmcs scale signature, intended to take advantage of extra symmetry when a, c or b, c is not a coprime pair and generalizing the congruence substitution procedure of building balanced words to obtain more non-balanced but still more "even" scales.

Take for example d = gcd(a,c), let a' = a/d and c' = c/d. Consider the MOS word (a+c)Xbm, which we call the template MOS. The most even arrangement of a'-many L steps and c'-many s steps is the MOS a'Lc's, so this method prescribes following the latter MOS, called the filling MOS, to fill in the X's. Fixing a choice of which X in (a + c)Xbm you start from, of which there are gcd(a + c, b) choices, you have to choose a mode of a'Lc's, of which there are lcm(a, c) choices. When a' = c' = 1, we obtain a balanced (equivalently MV3) scale; when in addition b is odd, the scale is also SV3 and chiral, and we recover the two chiralities from the two modes of a'Lc's. Of course, one may do this using aL(b + c)X and (b/(b,c))m (c/(b,c))s instead.

For 5L2m4s, we obtain mLsLsLmLsLs, sLsLmLsLsLm, and sLmLsLsLmLs. The first two are a chiral pair that are billiard scales, and the last is achiral, but it is not deletion-MOS. All three scales admit short generator sequences of 2-steps, respectively GS(L+s, L+s, L+m), GS(L+s, L+m, L+s), and GS(L+m, L+s, L+s), notably representing all 3 possible permutations of (L+s, L+m, L+s).

Open question: Is the length of the shortest guided generator sequence related to the length of the filler MOS? That wouldn't be true for 5L2m3s whose filler MOS is 2m3s and whose shortest generator sequence is GS(s, L, s, m). It could hold when the scale pattern has the divisibilities that this procedure is intended to take advantage of.