Start from a just lattice. Fill the gaps between its tones with a recurrence relation. The result has one melodic structure: an MOS, which includes equal temperaments. Work in two directions. In direction 1, seed a recurrence with the lattice and let it converge to a generator. In direction 2, fit an MOS to the lattice directly. The MOS is the answer, and no recurrence is needed.
The lattice supplies the harmony. Its tones become the seeds in station 3 and the targets in station 4.
The page reduces each tone to one octave and sorts them. It removes duplicates.
| # | ratio | cents | harmonic seed |
|---|
Form: H[n] = a·H[n−i] + b·H[n−j]. The ratio H[n]/H[n−1] converges to the positive root G. The generator is log2(G) in octaves.
The lift multiplies the characteristic polynomial by (xk − ρk). The k extra modes decay by ρ/G per step. The transient of the base recurrence decays by |λ₂|/G per step, where λ₂ is its second largest root. The slower of the two rates controls the transient. MERU-002 measured a ladder of ρ values. A larger ρ makes the transient longer. The mean count of 2-cent triads during the transient decreases as ρ increases. A long transient does not keep the harmony.
How to read the chart: the axis is the note count N. A dot at N means that N consecutive generators, reduced to one octave, make a scale with exactly two step sizes, L and s. Those scales are the MOS of this generator. The page finds them by the scale tree. It walks the Stern–Brocot tree toward g. Each fraction p/q on the path is an MOS with q notes. The fraction p/q is also the equal temperament that the MOS approximates: q-ET, with p steps per generator. The table gives each MOS, its fraction, that ET's generator in cents, the two step sizes in cents, and their ratio. The generator and its complement 1 − g have the same MOS sizes, because the chain is the same scale read from the other end. Erv drew this walk by hand as the "zig-zag pattern" on each page of meruthree.pdf. For 0.238186 his page says "24 places", and the table shows the 25-note MOS with s = 2.3 cents.
| N (MOS size) | p/q | ET generator ¢ | L ¢ | s ¢ | L : s |
|---|
The first terms are the lattice tones. Each new term is a sum of earlier terms. The new terms converge to the generator chain. Move the step slider to see harmony become melody.
2 cents is below the 12-ET error of 3/2 (1.96 cents) and far below its error of 5/4 (13.7 cents).
Search generators and note counts. Keep the chains that contain every lattice tone within the tolerance. Select a row to see the scale against the lattice.
What the fit searches, in Wilson's terms. A scale here is a chain: N consecutive generators, reduced to one octave. The octave is the period, and it does not change. Three things vary. The generator g runs over 0 to 600 cents, and its complement covers the rest, because g and 1 − g make the same scale. The note count N runs from 2 to the limit you set. In the Brun model N is the level, and the "only MOS" switch keeps the N values that are levels on the scale tree. The window start is the murchana. The chain contains the lattice's 1/1 at one of its N degrees. The search tries all N positions, which is the same as trying every mode of the chain. For each N the search is exact. For a tolerance ε, each tone and each degree k allow g to lie in an interval of half-width ε/|k|. A sweep over the interval endpoints finds the first g where some window of N degrees holds every tone. A bisection on ε then finds the smallest ε that works. A final refinement moves g to minimize the largest tone error. The result for each N is the generator, the murchana, the largest error and the mean error. Equal temperaments are a separate row for each N with g = 1200/N, where the murchana does not matter.
What the fit does not do. It does not change the period. It does not place the lattice's 1/1 between two degrees. It fits tones and not triads. CPSFIT-002 in the research lane fits the triads. It uses one generator, so it is rank 2 with the octave. A lattice that needs two generators, such as a 7-limit CPS at a small N, gets a large error here. The page reports the 1/1 position for each fit so you can read the murchana directly.
This generator's chains against the lattice appear in the first table. The free search below it finds the best generator for each N.
Press Run. The search is exact: for each N it finds the generator that minimizes the largest lattice error (interval sweep, bisection on the tolerance). Large lattices at large N take a few seconds.
No fit selected.
Move the slider to rotate the mode. Each step drops one tone from one end of the chain and adds one at the other end. The step pattern below the slider rotates with it, and the lattice errors show which modes hold the lattice.
| N | type | level | murchana (1/1 at degree) | max err | mean err | captured |
|---|
Every N is listed with the generator that makes the largest lattice error as small as possible at that N. The max error column is that smallest possible error, so it is often above the tolerance. A row above the tolerance says that no chain of N notes holds the lattice any closer. Rows within the tolerance are highlighted and come first. The mean error is the average over the lattice tones at the same generator.
| N | type | level | murchana (1/1 at degree) | generator | cents | max err | mean err | captured |
|---|
A recurrence H[n] = a·H[n−i] + b·H[n−j] with positive a and b has exactly one positive root G. The ratio H[n]/H[n−1] converges to G. The generator is log2(G), measured in octaves.
The ratio converges only when G is strictly larger than every other root. For positive a and b, this is true only when i and j have no common factor. Erv's Fig 12 lists only such pairs. For j ≤ 10, it lists 31 pairs, and 31 is the number of coprime pairs.
When i and j share a factor d, the sequence splits into d interleaved chains. The ratio H[n]/H[n−1] does not converge. Example: H[n] = H[n−2] + H[n−4] is two Fibonacci sequences. The consecutive ratio alternates between two values that sum to the Fibonacci generator.
The seeds do not change the generator. Any seeds that are not orthogonal to the dominant mode converge to the same G. The recurrence sets the melody. The seeds set the harmony. Erv seeded his own recurrences with harmonic material. Example: 192, 336, 588, 1029, a chain of 7/4, into 4·H[n−4] + H[n−1].
A generator g and its complement 1 − g make the same MOS. The chain ratio is different: 2^g or 2^(1−g). Erv's 0.238186 recurrence, 4·H[n−3] − 2·H[n−2] = H[n], diverges forward because a complex root pair has modulus 1.84 and the real root is 1.18. Erv wrote "converges left". The forward form is H[n] = H[n−1] + 2·H[n−3]. It converges to 1.69562, which is 2 divided by 1.17951.
Sums, differences, and every k-th term of a solution are solutions of a recurrence with the same limit. Erv recorded this: every difference-tone series of a sequence approaches the limit chain. This closure is the basis for seeding. The Lucas identity F[n] = 3·F[n−2] − F[n−4] is one case of it.
To use more seeds than the order j, raise the order. Multiply the characteristic polynomial by (x^k − ρ^k). This adds k seeds. If ρ is smaller than G, the extra modes decay by ρ/G per step. If ρ equals G, the extra modes never decay, and the sequence keeps parallel chains. Station 2 does this with the lift control.
Recurrences converge at very different speeds. The speed is the ratio of the second largest root to G. Fibonacci shrinks its transient by 0.38 per step, so 5 steps reduce it 100 times. Erv's 4·H[n−8] + 8·H[n−9] shrinks by 0.992 per step and needs about 600 steps. A slow recurrence keeps the seed harmony for a long time. This is a second control, next to ρ.
Some generators have no exact recurrence. The generator √2 − 1 = 0.414214 is a scale-tree generator: 1/2, 2/5, 5/12, 12/29, 29/70. The number 2^(√2−1) is transcendental, so no integer recurrence reaches it. The closest two-term recurrence up to order 12 is 8·H[n−9] + 7·H[n−10] at +0.04 cents. The pure fourth 4/3 is 0.99 cents above it.
The exact recurrences for the five favourite generators: 0.238186 is H[n−1] + 2·H[n−3] read as the complement. 0.328173 is H[n−1] + H[n−7], Fig 12 entry 12. 0.415227 is 4·H[n−8] + 8·H[n−9], 0.23 cents above 4/3. 0.418935 is H[n−1] + 32·H[n−4], the octave-shifted form of 2L⁴ = L³ + 4. The meta-meantone fifth 695.63 cents is 2·H[n−3] + 2·H[n−4], from the condition that 1, G⁴/4, G is a 4:5:6 proportional triad.
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