The Coherence Decomposition

When the alignment is high, why is it high?

Take the fractions of seven. Write one repeating block from each row.

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

Read down any column. No digit appears twice. The six rows are rotations of the same block, but at their actual positions they never agree with one another.

The reference alignment is still $1/6$. We compare every row with $1/7$, including $1/7$ itself. That self-match contributes the entire score. The other five rows contribute zero.

Now use $77 = 7\times11$. Just two rows already look different.

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

Their first digits agree. So do their fourth digits. They match in two of six positions, then repeat that agreement indefinitely. Elsewhere in the table there are more such pairs.

Across all $2{,}850$ distinct unordered pairs, the average agreement is about $0.088421$. The alignment with $1/77$ is about $0.100877$. The reference average exceeds the pairwise average by only $0.012456$.

Seven and seventy-seven both fall in the low tier of The Three-Tier Theorem. At seven, the reference score survives entirely on its self-match. At seventy-seven, much of the reference score is matched by an average that never singles out $1/77$ at all. A low total can conceal quite different arrangements of agreement.

Two pieces, one sum

Call the average over distinct pairs $\sigma(n)$, the pairwise alignment. It measures the internal coherence of the fractional field. Here coherence means agreement of digits at the same positions. The tails stay on one common clock.

Subtract that average from the reference alignment $\alpha(n)$. The difference is the focused alignment, $F(n)$.

$$\alpha(n)=F(n)+\sigma(n).$$

Denominator	Reference $\alpha$	Pairwise $\sigma$	Focused $F$
$7$	$1/6$	$0$	$1/6$
$77$	$23/228$	$42/475$	$71/5700$

The identity follows from the definition of $F$. By itself it explains nothing. The question is whether the counts on the two sides have a useful relation.

Nor is $F$ a quantity that we can assign to $1/n$ in isolation. It compares two averages over different sets of comparisons. It can be signed. For the digit-partitioning family, however, the count gives it a particularly simple meaning.

Eleven rows, fifty-five pairs

Return to the fractions of twelve. After two decimal places, the eleven rows separate into three groups.

Numerators	Tail after two places	Rows
$3,6,9$	Terminated	$3$
$1,4,7,10$	$333\ldots$	$4$
$2,5,8,11$	$666\ldots$	$4$

For reference alignment, the four rows that match $1/12$ score one. The three terminating rows also score one, by the alignment convention. Seven out of eleven gives $\alpha(12)=7/11$.

For pairwise alignment, compare distinct rows. Two terminating rows score one, a terminating row paired with a repeating row scores zero, and two repeating rows score their proportion of matching digits. In this table, pairs agree exactly when they belong to the same group.

The group of three supplies three pairs. Each group of four supplies six. That gives fifteen agreeing pairs out of fifty-five.

$$\sigma(12)=\frac{3+6+6}{55}=\frac3{11}.$$

This is the equality I want to isolate. The two ways of counting use different denominators and count different objects. After reduction, they give the same fraction. Subtracting leaves

$$F(12)=\frac7{11}-\frac3{11}=\frac4{11},$$

the size of the reference’s group divided by the number of rows.

The same calculation works for $n=pm$ whenever $p$ is digit-partitioning and every prime factor of $m$ divides the base. There are $m-1$ terminating rows and $p-1$ repeating groups, each containing $m$ rows. Within a group, every pair agrees. Between groups, none does.

Counting pairs within those groups gives

$$\sigma(pm)=\frac{m-1}{pm-1},\qquad F(pm)=\frac{m}{pm-1}.$$

Thus the pairwise average equals the terminating contribution to $\alpha$, and the focused residual equals the normalized size of the reference class. Neither the lengths of the repeating blocks nor the number of distinct rotations enters this count. The digit function’s ability to keep distinct remainders apart is enough.

There is also a useful endpoint. If every fraction terminates, as at $n=10$, both averages are one and $F=0$. The reference has no excess over the pairwise average. The pairwise statistic starts at $n=3$ because the table for $n=2$ has only one row and no distinct pair.

The first shared digits

Put the pair scores in a square grid. Each fraction labels a row and a column. A square records how much those two fractions agree. Every diagonal square is one because a fraction matches itself. The mean of the squares away from the diagonal is $\sigma$.

At eleven, only the diagonal is lit. At thirteen, six distinct pairs acquire a score of $1/3$. One of them is familiar.

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

The third and sixth digits agree. Those two matches produce a pair of squares away from the diagonal, one for each order of the comparison. The other five pairs appear the same way.

This gives an exact characterization. For a prime $p\ge3$ not dividing the base,

$$\sigma_b(p)=0\quad\Longleftrightarrow\quad p\text{ is digit-partitioning in base }b.$$

If the digit function assigns different digits to different nonzero remainders, two distinct rows cannot agree in any column. Conversely, if two remainders emit the same digit, the corresponding rows already have a match in the first column. Their pair score is positive, so the pairwise average is positive too.

In decimal, three, seven and eleven all have zero pairwise background. Five is different because its fractions terminate. Every prime above eleven has some agreement away from the diagonal. The boundary $p\le b+1$ can therefore be read directly from the pairwise statistic.

The golden gap

For a fixed digit-partitioning prime $p$, let the supported factor $m$ grow. The focused part approaches $1/p$. The pairwise part approaches the same number. Their sum approaches $2/p$.

At $p=3$, neither component reaches the golden threshold $1/\varphi\approx0.618$. Each approaches $1/3$.

But $1/3+1/3=2/3$. Above.

The crossing is already visible in a finite table. At twelve, the two parts are $4/11$ and $3/11$. Neither reaches the threshold; their sum $7/11$ does. At $120$, the two parts are $40/119$ and $39/119$. They sit closer together, while the sum sits closer to $2/3$.

The limiting gap contains the prescribed threshold precisely when

$$\frac1p<\frac1\varphi<\frac2p.$$

Equivalently, $\varphi<p<2\varphi$. Since $2\varphi\approx3.236$, only the primes two and three fit. The prime-two case is available in odd bases and gives reference alignment identically one. Among odd primes, three is the sole survivor.

This selects a prime for the chosen threshold. It does not uniquely select the golden ratio. The threshold $3/5$, for example, also lies between $1/3$ and $2/3$ and in no corresponding interval for a larger odd prime. The contribution here is the exact decomposition that exposes the gap.

The average and the arrangement

Counting equal symbols has a long history. Lempel and Greenberger studied Hamming correlation in families of finite sequences. Kak and Chatterjee studied distances and correlations in reciprocal digit sequences, including cyclic shifts of maximum-length sequences. Armstrong and Armstrong described repetend multiplication and its group structure.

The calculation here uses every distinct unordered pair in a denominator’s table, then compares that average with the score of one marked reference. For digit-partitioning prime cores it gives the closed forms above without requiring the base to generate every nonzero residue. The same statistic also detects exactly where digit separation fails.

Seven and seventy-seven now have more to tell us. Both have low reference alignment, but only seven has no agreement at all between distinct rows. At twelve, the pair count reduces to the same fraction as the terminating contribution, although most agreeing pairs are nonterminating. The decomposition makes those differences explicit.

At thirteen, $\sigma=1/33$ tells us the average agreement. The six pairs tell us where it occurs. At twelve, a different arrangement gives the exact split $7/11=4/11+3/11$. The decomposition lets us keep the reference score while seeing how much agreement is already shared across the field.