Primes and the Major Scale

I have been a musician for most of my life. I have also spent most of my life hearing that the golden ratio explains music. The explanations I encountered failed on contact with the mathematics, but I kept looking. After years of working on the question, I found a connection through repeating decimals. I call it the alignment-deficit lattice.

The route runs through repeating decimals and the primes in the base. For certain denominators, the proportion of aligned fractions approaches a limit. The distance left over is a small, exact fraction. Compare two of those distances and familiar musical ratios begin to appear.

Decimal gives us the octave and the just major third, but it leaves out the perfect fifth. To reach that interval, we need a base with a factor of 3.

The deficit below two thirds

Among the eleven proper twelfths, four fractions eventually share the repeating tail of $1/12$. Three others terminate and receive a whole credit under the alignment convention. That gives seven credits among eleven fractions, or an alignment of $7/11$.

This is one of the counts behind the middle band of decimal alignment. For a denominator $3m$, where $m$ is built entirely from 2 and 5, the formula is

Each repeating tail in this family consists of a single digit. Once the nonrepeating beginnings are cleared, a row either agrees with the reference in every position or misses it in every position. Whole-tail agreement gives the same count as averaging the individual digit matches here.

As $m$ grows, the alignment approaches $2/3$ from below. Call the distance still to go the deficit.

At $m=4$, the deficit is $1/33$. Doubling $m$ to 8 brings it down to $1/69$, and doubling again gives $1/141$. Each step leaves a little less than half the previous deficit.

Dividing the first deficit by the second gives $69/33=23/11$, a little more than 2. The next pair gives $47/23$. As the denominators grow, these ratios approach exactly 2, the frequency ratio of an octave.

Used as a frequency ratio, each finite value gives an interval slightly wider than an octave. The next doubling brings it closer, and the formula lets us calculate the remaining error.

The fifth that decimal cannot supply

The same calculation works in other bases. Choose a prime $p\geq3$ dividing $b-1$, and let $m$ use only prime factors of the base $b$. The denominator $pm$ again gives one-digit repeating tails after its beginning is cleared. Its alignment approaches $2/p$, with deficit

For two allowed integers $u$ and $v$, scale both by a common factor $t$ made from the base’s primes. The ratio of their deficits is

The deficit at the smaller integer goes on top because deficits decrease as the integers grow. The two subtractions by 1 keep the finite ratio slightly away from $v/u$; their effect vanishes as $t$ increases.

In base 10, the allowed integers are products of 2 and 5. Their ratios give octaves and the just major third, $5/4$. For example, the points 40 and 50 have ratio $5/4$.

A fifth above 40 would require 60, whose factor of 3 puts it outside this grid.

No multiplication or division using only 2 and 5 gives exactly $3/2$. There is a 3 in the denominator $3m$, but it determines the alignment family. The allowed factors of $m$ come from the base, and decimal supplies only 2 and 5.

In base 12, both 2 and 3 are available. We use $p=11$, since 11 divides $12-1$, and place the allowed integers on a grid with one axis for each prime. A step to the right multiplies by 2; a step up multiplies by 3.

Starting at 24, one step left and one step up brings us to 36. We have divided by 2 and multiplied by 3, giving the fifth $3/2$.

These are the same prime-exponent coordinates used in Pythagorean tuning. The alignment calculation assigns a deficit to every integer point, and its ratio formula tells us how close two finite deficits come to the corresponding musical interval.

Decimal can still approximate a fifth by choosing different pairs of points. It has no fixed pair whose limiting ratio is exactly $3/2$, whereas in base 12 the pair 24 and 36 does the job.

Twelve fifths and seven octaves

Twelve pure fifths take us almost, but not quite, to the same pitch as seven octaves. The discrepancy is small enough to disappear in a rough calculation and large enough to cause centuries of trouble in tuning.

Twelve fifths multiply the frequency by $(3/2)^{12}$. Seven octaves multiply it by $2^7$. The ratio between the endpoints is

This is the Pythagorean comma, about 23.46 cents. A semitone in twelve-tone equal temperament is 100 cents, so the gap is almost a quarter of one. Pure fifths make a spiral, not a closed circle. Equal temperament narrows each fifth by one twelfth of the comma, measured in cents, so that twelve fifths reach seven octaves. Benson follows the arithmetic of the mismatch.

The two integers in the reduced fraction, $3^{12}$ and $2^{19}$, are allowed points in base 12. Put their deficits into the ratio formula.

This finite ratio is slightly larger than the comma. The two already agree to six significant figures, but the exact fractions retain the difference.

Scaling both integer points makes the deficit ratio converge to the classical comma. The mismatch between pure fifths and octaves survives intact. Equal temperament adjusts the fifths to close the circle; increasing the deficit scale reproduces the pure intervals and their failure to close.

Eight notes from nine points

Base $30=2\times3\times5$ adds the third axis. With $p=29$, the deficit now lives on the integer grid used in five-limit just intonation. The name means that no prime larger than 5 enters an interval ratio.

The just major scale has ratios

Their least common denominator is 24. Multiplying by 24 gives the integers 24, 27, 30, 32, 36, 40, 45 and 48. Each uses only 2, 3 and 5. This is a classical representation of the scale, also set out in Kurenniemi’s account of musical lattices.

The integer $25=5^2$ also lies between 24 and 48 and uses only the allowed primes.

Its interval, $25/24$, belongs to the lattice but falls outside the familiar eight-note selection. Recovering that selection still requires a musical choice. The arithmetic supplies all nine points.

Dividing consecutive just intervals gives the steps $9/8$, $10/9$, $16/15$, $9/8$, $10/9$, $9/8$ and $16/15$. The whole tones come in two sizes, $9/8$ and $10/9$, with $16/15$ for the diatonic semitone.

We get the familiar whole, whole, half, whole, whole, whole, half pattern, although the whole steps are unequal in this tuning.

The fifth’s extra fraction

The finite column in the figure comes directly from the deficits. Taking the deficit at 24 over the deficit at each selected point gives $(29m-1)/695$.

For the fifth, $m=36$. Its finite ratio is

The extra $1/1390$ makes this fifth a little wide. The octave is wider too, and has the largest relative error among the eight notes, just under 0.072 percent.

Now multiply both fifth endpoints, 24 and 36, by the same allowed factor $t$. The exact excess becomes

Each tenfold increase in $t$ removes roughly another decimal place of error. The same convergence holds for every interval in the scale.

An octave-wide plot would hide most of the change, so the lower panel measures relative error in parts per million. Following Sol down the three curves shows the fifth approaching $3/2$. The starting Do has no error because its deficit is divided by itself.

A seventh in base 210

The factors of the base determine the grid. Bases 6, 12 and 24 all supply 2 and 3, so they have the same limiting Pythagorean intervals. Bases 30, 60 and 120 supply 2, 3 and 5. Their finite deficit values can differ because the chosen prime $p$ also enters the formula.

Base $210=2\times3\times5\times7$ retains those intervals and admits a fourth prime. With $p=11$, which divides 209, deficit ratios at $4t$ and $7t$ approach the harmonic seventh, $7/4$. The pair $7t$ and $8t$ gives the septimal whole tone, $8/7$, in the limit. Both intervals require the new factor of 7.

The tuning lattice is classical, and any positive function behaving like a constant divided by $m$ would recover its limiting ratios. The alignment-deficit coordinate comes specifically from counting repeating tails, with an exact finite correction. To my knowledge, this coordinate has not been previously described.

At the fifth, that count gives $1043/695$. Subtract the musical ratio $3/2$ and exactly $1/1390$ remains. We can account for that fraction and make it as small as we please by scaling the two integer points together.

I wanted a connection that would survive contact with the mathematics. Here, even the part that is still out of tune has an exact value.

The one-digit count in any supported base

Let $p\geq3$ be prime with $p\mid b-1$, and let $m$ be a positive integer whose prime factors all divide $b$. Choose a clearing length $D$ with $m\mid b^D$. Among the $pm-1$ proper fractions $k/(pm)$, there are $m-1$ terminating rows. Each receives a whole credit by convention.

For the other rows, multiplication by $b$ preserves the remainder modulo $p$, since $b\equiv1\pmod p$. Their cleared tails repeat a single digit. The $m$ rows with $k\equiv1\pmod p$ share the tail of $1/(pm)$. The remaining $(p-2)m$ rows do not. Thus

$$\alpha_b(pm)=\frac{(m-1)+m}{pm-1},$$ $$\delta_p(m)=\frac2p-\alpha_b(pm) =\frac{p-2}{p(pm-1)}.$$

For example, take base 12, $p=11$ and $m=2$. The denominator is 22. After one digit, $1/22$ repeats the digit 6. Rows 1 and 12 match that tail; row 11 terminates. The alignment is $3/21=1/7$, and its deficit from $2/11$ is $3/77$.

The exact difference from a limiting ratio is

$$\frac{ptv-1}{ptu-1}-\frac vu =\frac{v-u}{u(ptu-1)}.$$

It is positive when $v>u$. The finite interval approaches its target from above. The limit is taken along allowed factors $t$, such as successive powers of any prime dividing the base.

Exact scale values and the fifth’s convergence

The last column is $\delta_{29}(24)/\delta_{29}(m)$. The just interval in the middle column is its limit when both points are scaled together.

Note	$m$	Just interval	Finite ratio
Do	24	$1$	$1$
Re	27	$9/8$	$782/695$
Mi	30	$5/4$	$869/695$
Fa	32	$4/3$	$927/695$
Sol	36	$3/2$	$1043/695$
La	40	$5/3$	$1159/695$
Ti	45	$15/8$	$1304/695$
Do	48	$2$	$1391/695$

For a selected point $m$ between 24 and 48, the relative error after scaling by $t$ is $(m-24)/(m(696t-1))$. Its maximum occurs at 48 and equals $1/[2(696t-1)]$. At $t=1$, the bound is $1/1390$. The fifth’s absolute error is also $1/1390$ at this scale, but its relative error is $1/2085$.

Here is the fifth at six starting points. The displayed decimal values are rounded.

Starting point $m_0$	Deficit ratio at $3m_0/2$
24	1.500719
240	1.500072
2400	1.500007
24000	1.5000007
240000	1.50000007
2400000	1.500000007

Earlier illustrations

These illustrations from the earlier version are retained for comparison. The paired figures above separate the shrinking deficits, prime grid, comma, scale selection, and finite errors.

Further reading

The prime-based description of tuning has a long history. Euler’s work, Helmholtz’s On the Sensations of Tone, and Barbour’s Tuning and Temperament provide historical background. The following accounts develop the arithmetic used here.

• Benson, Music: A Mathematical Offering, including the Pythagorean comma and temperaments.
• Kurenniemi, Chords, Scales, and Divisor Lattices, including the prime-exponent lattice and the integer representation of the just major scale.
• Archibald, Mathematics and Music (1924), an earlier survey of the subject.

The alignment-deficit calculation supplies a long-division-derived coordinate on an established lattice. It is not a claim that the classical intervals or the major-scale integer sequence are new.