The Three-Tier Theorem

Compute the alignment of every integer from 2 to 500 and plot the results.

For each denominator, write out all its proper fractions. Compare their repeating digits with those of the first row, keeping every comparison on the same long-division clock after the prefixes have cleared. Each row receives its proportion of matching positions. Terminating rows receive score 1. The average is the alignment.

The white points sit at 1. The gold points approach $2/3$. The blue points stay below them, with a single highest point at denominator 6. Between blue and gold is an empty strip.

Its edges are exact.

$$\alpha_{10}(6)=\frac35, \qquad \alpha_{10}(12)=\frac7{11}.$$

No integer denominator has alignment strictly between those values. Six is the only denominator that attains the lower edge. Twelve is the only one that attains the upper edge.

The plot stops at 500. The gap does not.

The Alignment Limit for All Primes gave an exact formula when the part of the denominator left after clearing the base factors is prime. Here that remaining part can be composite. A pairing of the fractions will let us count them together.

Tier 1. The fractions stop

Divide by 8.

1/8 = 0.125
2/8 = 0.250
3/8 = 0.375
4/8 = 0.500
5/8 = 0.625
6/8 = 0.750
7/8 = 0.875

Every row terminates. Under the alignment convention, each receives score 1, so the average is 1.

The same happens at 25, 200, and every denominator built entirely from twos and fives. These are the integers supported on base ten. Dividing out all their factors of two and five leaves 1.

They form the white tier.

Tier 2. Three remains

At denominator 12, dividing out the factors of two leaves 3. The fractions now include both terminating rows and repeating tails.

 1/12 = 0.083333...
 2/12 = 0.166666...
 3/12 = 0.250000...
 4/12 = 0.333333...
 5/12 = 0.416666...
 6/12 = 0.500000...
 7/12 = 0.583333...
 8/12 = 0.666666...
 9/12 = 0.750000...
10/12 = 0.833333...
11/12 = 0.916666...

Start every comparison after two decimal places. Four rows then repeat the reference digit 3. Four repeat 6. Three terminate.

The four matching rows and three terminating rows give seven credits among eleven rows.

$$\alpha_{10}(12)=\frac{4+3}{11}=\frac7{11}.$$

The count extends to every denominator $n=3m$ whose factor $m$ contains only twos and fives. There are $m$ matching rows, $m$ opposing rows, and $m-1$ terminating rows. Therefore

$$\alpha_{10}(3m)=\frac{2m-1}{3m-1}.$$

The family begins below the gap. At $m=1$, denominator 3 has alignment $1/2$. At $m=2$, denominator 6 has alignment $3/5$. The next permitted factor is $m=4$, giving denominator 12 and alignment $7/11$.

The missing factor is 3. It cannot belong to the supported part in base ten.

From $m=4$ onward, the values rise toward $2/3$.

Denominator	Supported factor	Exact alignment	Decimal
12	4	$7/11$	0.6364
24	8	$15/23$	0.6522
60	20	$39/59$	0.6610
120	40	$79/119$	0.6639

These are the gold points. More generally, the gold tier consists of $3m$ with $m$ supported on ten and $m\ge4$.

Factoring makes the rule easy to use. Remove every factor of two and five from 96, 120, or 75. Each leaves 3, and each has a supported factor at least 4. All three belong to this tier.

Tier 3. Everything else

Remove the twos and fives from 28 and 77. The remainders are 7 and 77. Remove them from 45 and 91. The remainders are 9 and 91.

The remainder after this removal is the rough part. It need not be prime. Writing the denominator as

$$n=tm,$$

we use $m$ for its supported part and $t$ for its rough part.

Every denominator outside the first two tiers has alignment at most $3/5$. The two small three-core cases, 3 and 6, belong here too. Six sits at the top. Every other rough part in decimal is at least 7 and has alignment strictly below $4/7$.

A long repeating block can have many partial agreements with the reference. To bound their total, we will pair each row with its complement.

One match between two digits

Take a fraction and subtract it from one. At denominator 21, for example,

$$\frac2{21}+\frac{19}{21}=1.$$

Their repeating blocks are

 2/21 = 0.095238095238...
19/21 = 0.904761904761...

Read down the columns. Every pair of digits adds to nine.

Now compare both rows with $1/21$. At any one position, the two partner digits are different. If one matches the reference, the other cannot. They may both miss, but they cannot both match.

The same pairing works for every nonterminating fraction $k/n$ and its partner $(n-k)/n$. Shorter periods cause no problem. Repeat them across the reference period and keep the decimal positions synchronized.

There are exactly $m-1$ terminating rows. The other $(t-1)m$ rows form $(t-1)m/2$ complement pairs. Each pair contributes at most one credit on average. The whole field therefore satisfies

$$\alpha_{10}(n) \le \frac{m-1+(t-1)m/2}{tm-1} = \frac{m(t+1)/2-1}{tm-1}.$$

The numerator records both counts. Terminating rows contribute individually. Repeating rows contribute in pairs.

For $t>1$, this finite bound is strictly below

$$\frac{t+1}{2t}.$$

The strictness is useful. A ceiling approached as denominators grow need not be reached by any finite field.

The integers the base removes

In decimal, the possible rough parts begin

$$1,\ 3,\ 7,\ 9,\ 11,\ 13,\ldots$$

Two and five have already been removed, together with every factor they contribute. Between 3 and 7, the integers 4, 5, and 6 are unavailable because

$$\gcd(4,10)=2,\qquad \gcd(5,10)=5,\qquad \gcd(6,10)=2.$$

Three small greatest common divisors close that stretch of the list. The complement bound supplies the consequence. At every rough part $t\ge7$,

$$\alpha_{10}(n)<\frac{t+1}{2t}\le\frac47.$$

Those denominators stay below the gap. The core-three family is settled by its exact formula, including the small cases 3 and 6.

The rough part selects the family. The supported part decides where the small members of the family of three fall. Both are needed for the classification.

The same gap in other bases

The pairing used nine because we were writing decimals. In base $b$, the partner digits add to $b-1$.

If $b$ is even, that sum is odd. No integer digit can equal its own complement, so at most one member of a pair can match the reference. The same bound follows.

An even base also makes every rough part odd. Outside 1 and 3, the first possible rough part is at least 5. Thus the uniform ceiling is $3/5$. Decimal notation improves it to $4/7$ for those other rough parts because five divides ten.

Provided three does not divide the base, the three-core formula still applies. The supported factor 3 is unavailable and the supported factor 4 is available. The same two denominators, 6 and 12, give the same sharp edges in every even base not divisible by three.

In an odd base, the middle digit can be its own complement. In base three, for example, $1/2=0.111\ldots$. Its partner is itself, and both match the reference at every position. That is the precise step where this proof stops applying.

The interval left empty

The reciprocal golden ratio lies between the edges.

$$\frac35 < \frac1\varphi < \frac7{11}.$$

It therefore separates the blue tier from the other two. Any threshold strictly inside this interval gives the same classification. The connection to the golden ratio comes from the three-core profile studied in Why the Golden Ratio Selects the Prime Three. The gap is forced by the integer counts.

Complementary repetends have a substantial history. Shrader-Frechette surveys their patterns. Kak and Chatterjee study complement structure and digit-sequence correlations, while Lewittes and Armstrong and Armstrong develop related period and repetend arithmetic. Here the pairing is applied across the complete denominator field, including composite rough parts. It yields the finite bound, the exact empty interval, and the classification of every denominator in the stated bases.

Return to the plot. The field at 6 ends one tier. The field at 12 begins another. The fractions between those denominators have not been omitted, and going beyond 500 cannot fill the strip. The repeating rows have partners, the terminating rows have an exact count, and the factors permitted by the base leave no place for an intermediate alignment.