The Three-Tier Theorem

The alignment formula from The Alignment Limit works for prime-times-smooth denominators. It already showed that the prime 3 sits above the golden line at $1/\varphi \approx 0.618$ and the other primes do not. But most integers are not of that form. To classify every integer, I needed a different argument.

Every positive integer falls into one of exactly three tiers. Above the golden line, on it, or below it. No integer sits ambiguously near the threshold.

Tier 1: the smooth numbers

Some integers produce only terminating fractions. Divide anything by $8$, or $25$, or $200$, and the decimal stops. No repeating block, no cycle. These are the smooth numbers, integers built entirely from the prime factors of the base. In base $10$, smooth means a product of $2$s and $5$s.

./nfield field 8
  1/8 => 0.125
  2/8 => 0.25
  3/8 => 0.375
  4/8 => 0.5
  5/8 => 0.625
  6/8 => 0.75
  7/8 => 0.875

Every decimal stops. Nothing repeats, so nothing can mismatch. The alignment is $1$.

Tier 2: the golden family

Tier 2 is the golden family. These are the integers of the form $n = 3m$, where $m$ is smooth and at least $4$. They have repeating fractions, genuine disagreements between decimal patterns, and yet their alignment stays above $1/\varphi$.

The smallest example is $n = 12$. Its field has three flavors of fraction at once. Terminating, mixed, and pure repeating. The pipes mark the repeating block where there is one.

./nfield field 12
  1/12 => 0.08|3|
  2/12 => 0.1|6|
  3/12 => 0.25
  4/12 => 0.|3|
  5/12 => 0.41|6|
  6/12 => 0.5
  7/12 => 0.58|3|
  8/12 => 0.|6|
  9/12 => 0.75
 10/12 => 0.8|3|
 11/12 => 0.91|6|

Four of the eleven fractions repeat the digit $3$ inside their pipes, four repeat $6$, three terminate. The alignment of $12$ is $0.636$.

./nfield align 12       # 0.636
./nfield align 24       # 0.652
./nfield align 60       # 0.661
./nfield align 120      # 0.665

Each value is above $0.618$. Each one climbs slowly toward $2/3$.

The golden family is small. It is 12, 15, 24, 30, 48, 60, and so on. The integers of the form three times a smooth number, with smooth factor at least 4.

Tier 3: everything else

Every prime from $5$ onward. Every composite whose rough part is anything other than $3$. Every number outside the smooth family and the golden family.

./nfield align 7        # 0.167
./nfield align 13       # 0.111
./nfield align 77       # 0.088
./nfield align 91       # 0.100

Every integer from 2 to 500, plotted by alignment. White at the top, the smooth numbers. Gold above the line, the golden family. Blue below, everything else. Between the gold and the blue, nothing.

The mirror bound

The proof rests on a simple symmetry of repeating decimals. Every non-terminating fraction $k/n$ has a partner, $(n - k)/n$. The repeating digits of $k/n$ and its partner sum to $9$ at every position. If $k/n$ has the digit $3$ at some position, then $(n-k)/n$ has $6$ there. If $k/n$ has $1$, the partner has $8$. They are mirror images in the digit system. Mathematicians call this the nines complement.

Within each pair, at most one fraction can match the digit of $1/n$ at any position. The other is forced to differ.

This bounds the average alignment over the non-terminating fractions by $1/2$. Add back the terminating fractions, which always contribute alignment $1$, and the total alignment of $n$ satisfies

where $t$ is the rough part of $n$ (the prime factors not in the base).

Now a fact specific to base 10. The rough part of any integer is always coprime to 10. The integers 4, 5, and 6 all share a factor with the base, so they cannot appear as rough parts. The smallest rough part strictly larger than 3 is therefore $t = 7$. At $t = 7$:

That is below $1/\varphi \approx 0.618$. Every rough part larger than 7 gives a smaller bound still.

Three cases remain. For $t = 1$, the integer is smooth: Tier 1. For $t = 3$ with $m \ge 4$, the alignment formula gives

which puts these in Tier 2. For $t = 3$ with $m < 4$, direct computation gives $\alpha(3) = 1/2$ and $\alpha(6) = 3/5$, both below $1/\varphi$: Tier 3.

The whole argument pivots on the gap between 3 and 7 in the rough part.

The curve is the mirror bound $(t+1)/(2t)$ as a function of the rough part $t$. At $t = 1$, the bound is 1. At $t = 3$, it is $2/3$, above the golden line. At $t = 7$, it drops to $4/7$, below the line, and it never comes back. The gap between 3 and 7 is empty because 4, 5, and 6 all share a factor with the base. No admissible rough part exists in between.

The rough part and the smooth part

Every positive integer splits uniquely into two pieces. The smooth part absorbs the prime factors the base contains. The rough part is everything else. In base $10$, the integer $60 = 20 \times 3$ has smooth part $20$ and rough part $3$. The integer $77 = 1 \times 77$ has smooth part $1$ and rough part $77$. The integer $8$ has smooth part $8$ and rough part $1$.

The smooth part controls the resolution of the field. It tells you how finely the field is sampled. The rough part controls the structure. It tells you which prime geometry governs the repeating digits.

The three tiers are determined entirely by the rough part. A quantity defined by comparing decimal expansions ends up depending only on one arithmetic datum. The part of the denominator the base does not already own.

A note from 2026

April 2026

The mirror bound is the part of this paper that carried forward. Every fraction $k/n$ is paired with its complement $(n-k)/n$, and the digits of those two fractions add to $b-1$ position by position. In The Collision Periodic Table the same involution appears as a reflection identity on residue classes rather than on decimal strings. Different setting, same symmetry.

The rough part also turned out to carry more weight than it first seemed to. Here it controls the three-tier classification. In The Collision Invariant and The Collision Periodic Table, the arithmetic slicing becomes finer, but the basic idea holds: structure is carried by the part of the modulus the base does not already absorb.

.:.

Try it yourself

Start with the smooth numbers. No rough prime, no repeating block, perfect alignment.

./nfield align 8        # 1.000
./nfield align 25       # 1.000
./nfield align 200      # 1.000

Now the golden family. Rough part is 3 in every case.

./nfield align 12       # 0.636   (= 3 × 4)
./nfield align 60       # 0.661   (= 3 × 20)
./nfield align 120      # 0.664   (= 3 × 40)

All above $0.618$. All climbing toward $2/3$. Now try anything with a different rough prime.

./nfield align 7        # 0.167
./nfield align 13       # 0.111
./nfield align 77       # 0.088   (= 7 × 11)
./nfield align 91       # 0.100   (= 7 × 13)
./nfield align 1001     # 0.099   (= 7 × 11 × 13)

Divide out the factors of 2 and 5 from any integer. If 3 is all that remains, the alignment is above the line. If anything else remains, it is below.

Code: github.com/alexspetty/nfield
Paper: The Three-Tier Theorem

Alexander S. Petty
October 2020 (updated April 2026)
.:.