The Coherence Decomposition

Write one repeating block from each of the six proper sevenths.

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

The rows contain the same six digits in the same cyclic order. Read down any column, though, and no digit appears twice. These fractions agree about the cycle and disagree at every position in it.

The reference alignment is $1/6$. We compare all six rows with $1/7$, including $1/7$ itself. Its self-match supplies the entire score. The other five rows contribute nothing.

Now multiply the denominator by eleven.

1/77 = 0.|012987|
2/77 = 0.|025974|

The first digits agree, and so do the fourth. That gives two matches in six positions, repeated indefinitely. Other pairs in the table have agreements of their own.

The reference alignment at seventy-seven is about $0.100877$. Average over all $2{,}850$ distinct pairs instead, with no row appointed as the reference, and the result is about $0.088421$. The two averages are close. At seven, the corresponding pair average is zero.

Both denominators belong to the low tier of The Three-Tier Theorem. The classification puts them on the same side of the threshold, but it leaves this difference untouched. I want to separate the agreement with one marked row from the agreement already present among the rows themselves.

A table without a reference

Call the average over distinct unordered pairs $\sigma(n)$, the pairwise alignment. Each pair is counted once, and no row is paired with itself. This is the coherence in the title, agreement of digits at corresponding positions. We keep the tails on one common clock, without rotating individual rows to improve their matches.

Subtract this average from the reference alignment $\alpha(n)$. I call the difference the focused alignment, $F(n)$.

That equation follows from the definition. The work begins when we count the two averages separately and find an exact relation between them. Nor does $F$ count just the pairs involving $1/n$. It is the difference between a row average and a pair average, taken over different sets of comparisons.

The exact scores at seven and seventy-seven

Quantity	Seven	Seventy-seven
Reference $\alpha$	$1/6$	$23/228$
Pairwise $\sigma$	$0$	$42/475$
Difference $F$	$1/6$	$71/5700$

These pair averages exclude self-matches. The reference averages include them. Pairwise alignment is defined for $n\geq3$, since the table at two has only one row.

Eleven rows, fifty-five pairs

The twelfths make the count small enough to follow in full. Move past two decimal places in every row. The fractions with numerators $3,6,9$ have terminated. Those with numerators $1,4,7,10$ repeat 3, and those with numerators $2,5,8,11$ repeat 6.

For reference alignment, the four repeating threes match $1/12$. The three terminating rows also receive full credit under the convention used throughout the alignment series. That gives seven credited rows out of eleven.

Pairwise alignment needs its own termination rule. Two terminating rows score one; a terminating row paired with a repeating row scores zero. Two repeating rows score the fraction of positions where their digits agree. This is different from automatically crediting every terminating row in the reference count.

At twelve, a pair therefore scores one exactly when both rows belong to the same group. The three terminating rows supply three pairs. Each group of four repeating rows supplies six. There are fifteen agreeing pairs among the fifty-five possible pairs.

Three elevenths again. We reached it once by counting terminating rows and now by counting agreements between pairs. Yet twelve of the fifteen agreeing pairs are nonterminating. The equality is between the two totals, not between the objects being counted.

Subtracting this pair average from the reference score leaves

Four elevenths is exactly the share of rows in the reference’s group. At twelve the two counts recover the two pieces of the earlier alignment formula. The question is whether that cancellation survives when the table grows.

The count that survives

Take $n=pm$, where $p$ is a digit-partitioning prime and every prime factor of $m$ divides the base. In decimal, $m$ can use twos and fives, while $p$ can be three, seven, or eleven. In base $b$, the conditions are $p\nmid b$ and $p\leq b+1$.

Digit-Partitioning Primes and the Alignment Formula establishes what this buys us. Once all rows have passed a common clearing depth, each nonzero remainder modulo $p$ has $m$ rows behind it. Rows with the same remainder agree forever. Different remainders remain different as the division advances, and digit partitioning ensures they emit different digits.

There are $p-1$ repeating groups of $m$ rows, together with one terminating group of $m-1$ rows. We can count pairs within those groups exactly as we did at twelve. Dividing by the total number of pairs gives

This is the central result. An average over pairs reduces to the same fraction as the terminating-row contribution to reference alignment, for every denominator in this family. The remaining part is the reference group’s share of the rows.

Follow the pair count through the cancellation

A group of $r$ rows contains $\binom r2=r(r-1)/2$ distinct pairs. The number of agreeing pairs is

$$\begin{aligned} \Sigma&=\binom{m-1}{2}+(p-1)\binom m2\\ &=\frac{(m-1)(pm-2)}2. \end{aligned}$$

There are $\binom{pm-1}{2}=(pm-1)(pm-2)/2$ pairs altogether. Dividing cancels the factor $(pm-2)/2$ and leaves $(m-1)/(pm-1)$. We require $pm\geq3$, so this denominator counts at least one pair.

Nothing in the calculation requires the base to visit every nonzero remainder in a single cycle. The rows may run through several cycles of any allowed length. The count needs only equal-sized remainder classes and a digit function that keeps those classes apart.

The entirely terminating tables have a simpler outcome. At ten, for example, every row receives reference credit and every pair consists of two terminating rows. Both averages are one, so $F=0$.

The first shared digits

Put the pair scores in a square grid, with the fractions labeling both axes. The diagonal records self-matches and is left out of $\sigma$. The rest of the grid shows the comparisons we are averaging, with each unordered pair appearing twice by symmetry.

At eleven, every square away from the diagonal is empty. At thirteen, six distinct pairs acquire a score of $1/3$. One of them is the pair we met in the digit-partitioning article.

 1/13 = 0.|076923|
11/13 = 0.|846153|

The third and sixth digits match. Each match returns every six places, giving a pair score of $2/6$.

There is an exact test behind this change in the picture. A prime $p\geq3$ that does not divide the base has zero pairwise alignment exactly when it is digit-partitioning. Using the earlier characterization,

When every nonzero remainder emits a different digit, distinct rows cannot agree in any column. Conversely, if two remainders emit the same digit, their fractions match in the first column. That match returns with the period, making their pair score positive. An average of nonnegative scores can be zero only when every one of them is zero.

So the pair statistic detects precisely the failure of digit separation. In decimal, three, seven, and eleven have $\sigma=0$. Every prime above eleven has a positive pair average. Five is outside this test because its decimal fractions terminate.

From a few rows to thousands

At thirteen, we can point to every matching pair. Increase the denominator and those isolated squares give way to bands, crossings, and fine-grained patterns. The rule that produces a square has not changed. It still asks how often two fractions show the same digit in the same position.

The gallery runs from ten rows to $5{,}002$, with twelve decimal denominators along the way. The largest grid contains just over twenty-five million cells. Four further views keep the denominator at $601$ and change the base, so we can compare changes in the digit system without changing the number of rows.

The colors here have a different job from the simple yes-or-no highlights above. Nine color bands separate lower scores in teal from higher scores in gold, with a neutral band around each panel’s pair average. A nonlinear scale expands the small differences near that average. Zero agreement is left blank and the self-match diagonal is gray. Each plate has its own numerical key. A color in one plate need not represent the same score in another.

Both axes remain in numerator order. The broad views are reduced for the page; open an individual plate and use 100% or zoom to inspect its cells. These are exact finite tables. The patterns give us more to examine, without turning their visual resemblance into a theorem.

Small grids, from eleven to fifty-three

More rows, from seventy-seven to 401

Dense grids, from 601 to 5003

One denominator in four other bases

Two thirds, one third at a time

Let the supported factor $m$ grow while the digit-partitioning prime $p$ stays fixed. The focused part and the pairwise part both approach $1/p$. Together they approach $2/p$.

For the three-core family, each approaches one third. The golden threshold, $1/\varphi\approx0.618$, lies above either component but below their sum of two thirds. Here $\varphi=(1+\sqrt5)/2$.

We already see the crossing at twelve. Neither $4/11$ nor $3/11$ reaches the threshold, but their sum $7/11$ does. At $120$, the components are $40/119$ and $39/119$, closer to equal thirds. The combined score is closer to two thirds.

The limiting gap contains the prescribed threshold when

or equivalently $\varphi<p<2\varphi$. Since $2\varphi\approx3.236$, only the primes two and three fit. Two is available in odd bases and gives reference alignment identically one. Among odd primes, three is the only case.

The components stay below the threshold at finite scales too. For an odd prime, $F_b(pm)\leq1/(p-1)\leq1/2$, and $\sigma_b(pm)<1/p$. At $p\geq5$, even their sum is below $2/p\leq2/5$. No larger odd digit-partitioning prime can make the crossing.

The threshold is part of the question. Choosing $3/5$ would also separate one third from two thirds and exclude every larger odd prime. The golden ratio does not own this interval. What the decomposition supplies is an exact account of the two quantities on either side of it.

Back to seventy-seven

Seventy-seven has neither the equal-digit classes of twelve nor the empty off-diagonal of seven. Its rows include different periods. The fraction $7/77=1/11$ repeats 09, while $11/77=1/7$ repeats 142857. Comparing six positions accommodates both periods without changing either row’s starting point.

The prime-core formulas do not apply here. Both seven and eleven remain in the denominator after removing the factors of the base, and distinct remainder classes can share digits. We count those partial matches on the common clock. In this particular table, $756$ of the $2{,}850$ distinct pairs score $1/3$; the rest score zero. Thus

Counting equal symbols has an established history. Lempel and Greenberger study Hamming correlation in finite sequence families. Kak and Chatterjee study distances and correlations in reciprocal digit sequences, including shifts of maximum-length sequences. Armstrong and Armstrong describe repetend multiplication and its group structure.

My construction compares the average over every distinct pair in a denominator’s table with the score of one marked reference. For digit-partitioning prime cores, the pair count gives the exact decomposition above without a single-cycle assumption. For a prime denominator, a zero pair average characterizes digit partitioning itself. Those are the results developed in the companion paper.

Pair-count diagrams and larger grids

The first diagram follows the three groups at twelve. The second extends the pairwise grids to denominators 31, 53, and 151, where one average conceals increasingly detailed arrangements of matches. The third places the finite three-core scores beside the limiting intervals for the decimal digit-partitioning primes.

Return to the squares for seven and seventy-seven. Their reference scores place them in the same low tier. Erase the self-matches at seven and nothing remains. Do the same at seventy-seven and there are still $756$ pairs to account for.