Three and the Golden Ratio

Divide 1 by 7.

1/7 => 0.|142857|
2/7 => 0.|285714|
3/7 => 0.|428571|
4/7 => 0.|571428|
5/7 => 0.|714285|
6/7 => 0.|857142|

Six fractions. The same six digits every time, started from a different place. 142857, rotated. In decimal, all six fractions belong to one repeating orbit.

Now divide by 12. Here are the first six rows.

 1/12 => 0.08|333|
 2/12 => 0.1|666|
 3/12 => 0.25
 4/12 => 0.|333|
 5/12 => 0.41|666|
 6/12 => 0.5

Some terminate. Others repeat immediately or after a nonrepeating prefix. I call the latter mixed cycles.

The bars enclose a block that repeats indefinitely. I repeat whole copies of the shortest pattern until dividing their digit sum by one less than the base leaves no remainder. Here the base is ten, so I divide by nine. One 3 gives three; two give six; three give nine. The rule therefore gives |333| in this example. The shortest repeating pattern is still a single 3.

How much of this fractional field stays in step with $1/12$? Advance all eleven rows past the first two decimal places. Four repeat 3, four repeat 6, and three have terminated. Literal agreement with the reference tail accounts for four rows out of eleven.

The alignment score $\alpha$ used here also gives the terminating rows full credit. That is a convention in the definition. Their trailing zeros do not match the reference’s 3s. At denominator 12, it gives $\alpha(12)=7/11$. If we compared only the eight nonterminating rows, the matching proportion would instead be $4/8=1/2$.

For denominators $3m$, where $m$ is built from powers of 2 and 5, the same count gives a formula.

The prime factors of ten are what let us clear $m$ by advancing to a common decimal depth. The fractions $k/(3m)$ then split into three classes mod 3. Multiples of 3 terminate. The class $k \equiv 1$ carries the same one-digit repetend as $1/(3m)$. The class $k \equiv 2$ carries its nines complement, the digit that sums with the first to make 9. The first class contains $m-1$ rows and each of the other two contains $m$. Counting the terminating and matching classes gives $2m-1$ out of $3m-1$.

As $m$ grows, $\alpha$ climbs toward $2/3$.

The dots mark admissible denominators $3m$. The red line is $1/\varphi \approx 0.618$, the reciprocal of the golden ratio. The first dot above it is at $m=4$, giving denominator 12 and score $7/11$. The score increases thereafter toward $2/3$.

The golden threshold

Choose $1/\varphi$ as the threshold and solve $\frac{2m-1}{3m-1} \geq \frac{1}{\varphi}$. Using the golden identity $\varphi^2=\varphi+1$, the inequality becomes $m\geq\varphi^2$. The count supplies the curve; the choice of threshold brings the golden ratio into the calculation.

Since $\varphi^2 \approx 2.618$, the integer threshold is $m \geq 3$. But 3 is not a product of 2’s and 5’s, so it does not belong to the family. The first admissible value is $m = 4$, which gives $n = 12$.

The equality at $m=\varphi^2$ belongs to the real curve obtained by extending the formula beyond integer inputs. There is no denominator ensemble at that irrational scale. The actual decimal-twelve score is $7/11$, not $1/\varphi$.

To replace 3 with another prime $p$, we must also choose a base $b$ with $b\equiv1\pmod p$, and let $m$ divide a power of that base. These conditions give one-digit tails and the same residue count. The score is $(2m-1)/(pm-1)$, approaching $2/p$.

For an odd prime’s curve to reach $1/\varphi$, its limit must exceed the threshold, so $p<2\varphi\approx3.236$. Only 3 qualifies. The separate case $p=2$ has score 1 throughout. At $p=5$ the limit is $0.4$; at $p=7$ it is about $0.286$. These are comparisons across admissible base families, not claims about every prime in decimal.

Among the odd primes shown, one dot lies above the line. The cutoff $2\varphi$ sits between 3 and 5. For $p=3$, every base $b\equiv1\pmod3$ gives the same score formula. Base ten is one instance.

The cubic

Any threshold between $2/5$ and $2/3$ separates the same odd primes. The interval is wide, so separation alone does not distinguish the golden threshold. There is a more specific relation to examine.

At $p=3$, the real scale where the curve equals $1/\varphi$ is $\varphi^2$, the reciprocal square of the threshold. Ask when that relation holds for other primes. Write $\tau$ for a threshold in $(0,1)$ and require its crossing scale to equal $\tau^{-2}$. This is an additional compatibility condition we choose to study. Substitution gives a cubic in $\tau$, one for each prime.

Write $f_p$ for this cubic. For every prime $p\geq3$, it has a unique root between 0 and 1. The rational root theorem says the only rational-root candidates are $\pm1$ and $\pm2$. Evaluate.

$f_p(1) = 2 - p$. Zero at $p = 2$.

$f_p(2) = 6 - 2p$. Zero at $p = 3$.

$f_p(-2) = 2p - 10$. Zero at $p = 5$.

$f_p(-1) = p$. Never zero at a prime.

So the cubic factors over $\mathbb{Q}$ for $p \in \{2, 3, 5\}$ and is irreducible for every prime past 5.

At $p = 3$, the factorization is $(\tau - 2)(\tau^2 + \tau - 1)$. The quadratic factor is the minimal polynomial of $1/\varphi$. At $p = 5$, it is $(\tau + 2)(\tau^2 - 3\tau + 1)$, whose root is $1/\varphi^2$. Both quadratics have discriminant 5. Both roots live in $\mathbb{Q}(\sqrt{5})$, the golden field.

For $p\geq7$, the threshold roots are cubic irrationals. They do not belong to any quadratic field, so these later members of the family cannot be expressed in the golden field.

The gold curve, $p=3$, vanishes at $\tau=1/\varphi$. The reason is visible in polynomial division. Divide $\tau^3-\tau^2-p\tau+2$ by $\tau^2+\tau-1$. The quotient is $\tau-2$ and the remainder is $(3-p)\tau$.

That remainder is zero when $p = 3$ and nowhere else.

Of the two self-referential thresholds in $\mathbb{Q}(\sqrt{5})$, only $1/\varphi$ lies in the separating interval $(2/5, 2/3)$. The other, $1/\varphi^2 \approx 0.382$, falls below $2/5$. Under the self-referential criterion, $1/\varphi$ is the unique distinguished value in the interval where separation occurs.

The underlying arithmetic is classical. Midy’s theorem concerns sums of pieces within a repetend, and Leavitt extended it in 1967. Hardy and Wright treat repeating expansions in An Introduction to the Theory of Numbers; Koshy’s Fibonacci and Lucas Numbers with Applications develops the golden-ratio algebra. The research note gives the references and proofs. The claim here is confined to the stated score and reciprocal-square condition.

The decimal-twelve field remains an ordinary collection of eleven fractions, with an alignment score of $7/11$. Its associated curve meets $1/\varphi$ at $\varphi^2$, the reciprocal square of the threshold. Requiring that same relation throughout the prime-indexed family produces a classification, and the remainder $(3-p)\tau$ shows exactly where three enters it.