When silence becomes 5

One period, n = 0 … 24. The usual remainder H gives 0 at the 9 silences; Lluis's grouping gives 5 there (purple) and leaves every other value unchanged. Nothing is lost: 5 and 0 are the same class mod 5. What changes is that 5 can be used: as a ratio, a harmonic, a note.

The paper reads H as the ratio H : (H − 1), the consecutive ratios of the tetractys: 1:1, 2:1, 3:2, 4:3, with 0 as silence. Read the silence as 5 and the same rule gives 5 : 4, the just major third, the next superparticular ratio after the tetractys. It completes the major triad (1, 5/4, 3/2) and is the arithmetic mean of 1:1 and 3:2. Each period then sounds 9 major thirds and 4 of each of the other intervals.

n = 0 … 50. Each column sounds two notes in just intonation: the fundamental and the interval above it. Purple: the major thirds that were silences.

Lluis's second reading divides each value by the one before it: i(n)/i(n − 1). With the usual remainder this breaks at every silence (division by 0, or a ratio of 0); with the silence as 5 every step is a ratio. And because the steps multiply in a chain, the note at step n is just i(n) times a fundamental: the cycle plays only the harmonics 1, 2, 3, 4, 5 of one note (in C: C, C, G, C, E going up), like a natural trumpet or a horn call, and each step is a leap between two of them.

proved The 25 steps of a period multiply to exactly 1: the melody comes home every 25 steps. (They are i(1)/i(0) · i(2)/i(1) · … · i(25)/i(24) = i(25)/i(0) = 5/5.) Click a row or a point to hear that step.

The notes of the palindrome read the same backwards, so the horn call is its own retrograde. In its leaps this shows as a mirror with a twist: because i(50 − n) = i(n), the leap at step 51 − n is the reciprocal of the leap at step n, step(51 − n) = i(n − 1)/i(n). The second half makes the leaps of the first half in reverse order and the other way up: every rise of the first half is a fall of the same size in the second. proved

And the mirror repeats, mirror inside mirror, like two mirrors facing each other (Lluis's observation: the infinite mirror effect). Inside the first half the leaps mirror the same way about step 13, the single unison leap where the fifth sounds twice (3, 3 at n = 12 | 13): step(26 − k) = 1/step(k). Step 12 goes down, step 14 goes up; step 10 goes up, step 16 goes down; and so on. So the first 12 leaps, taken together, go down (×3/5, from harmonic 5 to harmonic 3) and the next 12 go up by the same amount (×5/3). The second half does the same about step 38. Then the whole block of 24 around step 13 is mirrored again by the block around step 38, across the big axis: step 12 (down) meets step 39 (up), and the inner part 14–37 mirrors itself. proved

The leaps, up (green) or down (red), height in octaves (log₂ of the step). Each shaded column or dashed line is a mirror: across it every bar meets a bar of the same size and opposite sign. Yellow arcs cross the small mirrors (steps 13, 38, 63, 88), blue arcs the mirrors between halves (25|26, 75|76), red arcs the mirror between blocks of 50 (50|51).

Why, and how far it goes. The notes have two mirrors: i(−n) = i(n), because 𝔓(−n) = 𝔓(n), and i(25 − n) = i(n), because 𝔓(25 − n) − 𝔓(n) = 125 − 10n, a multiple of 5. Two mirrors 12½ apart make a repeat of 25 and an endless row of mirrors, one every 12½ notes: on the silences (n = 0, 25, 50, …) and between the repeated fifths (12½, 37½, …). For the leaps that puts a mirror at 13, 25½, 38, 50½, 63, … for ever. So the block of 25 mirrors the next 25, the block of 50 the next 50, the block of 100 the next 100, ad infinitum, as Lluis says (checked by computer to 1000 notes). proved It is an infinite mirror rather than a fractal: a fractal would also repeat at smaller and smaller sizes, and here the mirrors repeat at every larger size but none is smaller than 12½ (the first 12 leaps are not a mirror of themselves). So far no smaller one is known. interpretation

The same reading works with any modulus m: i_m(n) = 1 + MOD(𝔓 − 1; m), and play the harmonics i_m(n) of one fundamental. Lluis tried other moduli in his spreadsheet; here they are side by side. Choose one and listen: the steps that are not consonances of just intonation are marked in red.

Every modulus makes a palindrome. 𝔓(−n) = 𝔓(n), and 𝔓(n + 5m) − 𝔓(n) = 2mn + 5m², a multiple of m, so the notes repeat after at most 5m steps. A pattern that repeats and reads the same from 0 backwards also reads the same backwards from half its period: the infinite mirror of section 4 is there for every m. proved So the palindrome is not what makes 5 special (see also Other primes).

The sound is. Every ratio of two whole numbers from 1 to 6 is a consonance (Zarlino's senario again), and 7 is the first number that brings a dissonance. So the moduli 2 to 6 sound only consonances, and from 7 on dissonances come in: 8/7, 7/4, 7/5 with m = 7, then 9/8, 16/9, 11/8 and more. (m = 8 stays consonant only because it uses just the harmonics 1, 4, 5 and 8.) proved

Among the consonant ones, only 5 and 6 use all seven consonances, with opposite characters. Mod 6 is open, made mostly of fourths and fifths (14 of its 30 steps). Mod 5 is sweet: thirds and sixths lead (14 of its 25 steps), the fourth and the fifth sound once each, the fifth only at the inner wall of the palindrome, and it has the most silences, 9 in 25 notes. Mod 3 has no thirds at all; 2 and 4 are only octaves. computed

And 5 is the number of the formula itself: 𝔓 is built by dividing by 5, and H = 𝔓 mod 5 is the consonance indicator of the papers. Of the moduli that stay fully consonant, 5 is the largest prime. That is why we take 5 as the one. interpretation

Mod 10 repeats after exactly 50 notes: the 50-note palindrome is the natural cycle of base 10, played twice over by mod 5. computed In Lluis's view of numbers this is structural: base 10 has nine symbols and the 0, and 10 is one full cycle, of which mod 5 is the half.