The Coherence Decomposition

The Three-Tier Theorem classified integers by alignment. But it left a natural question. When the alignment is high, why is it high?

The alignment compares every fraction of $n$ to $1/n$. That mixes together two different effects. Sometimes $1/n$ is genuinely unusual inside its own field. Sometimes the whole field is already internally coherent, and $1/n$ benefits from that shared structure. The alignment number does not separate those two contributions.

Two pieces, one sum

They come apart cleanly. Define a second number, the pairwise alignment, written $\sigma(n)$, as the average digit-match proportion between all pairs of fractions, not just pairs that involve $1/n$. This measures how much the fractions agree with each other on average, regardless of which one you pick as the reference.

Now define a third number, the focused alignment, by

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

This is the excess agreement that $1/n$ has over and above what the field already has with itself.

The two new numbers add up to the original one:

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

One piece is focused on $1/n$. The other is spread across the field. The alignment is the sum.

The paper proves explicit closed forms for both pieces when $p$ is a digit-partitioning prime in the base and $n = pm$ for $m$ a smooth integer. The focused part is

$$F(pm) = \frac{m}{pm-1}$$

and the pairwise part is

$$\sigma(pm) = \frac{m-1}{pm-1}.$$
As $m$ grows, both numbers approach the same limit $1/p$, and the total alignment approaches $2/p$. At infinite resolution, the total coherence splits evenly between the focused part and the pairwise part.

The golden gap

At $p = 3$, the focused part approaches $1/3 \approx 0.333$. The pairwise part approaches the same number. Neither one reaches the golden threshold at $0.618$. Not separately. Not even close.

But $1/3 + 1/3 = 2/3 \approx 0.667$. Above.

That is the golden gap. The threshold sits above each component but below their sum. The two pieces need each other to clear the line, and at $p = 3$ they have each other. At no other prime.

The gold bar is the focused part. The blue bar is the pairwise part. At $p = 3$, the two stack together and cross the golden line. At $p = 5$, the sum is $0.4$. At $p = 7$, it is $0.286$. The bars shrink as the prime grows.

The golden ratio selected 3 in Why the Golden Ratio Selects the Prime Three by sitting between $2/3$ and $2/5$, in the gap between consecutive primes. The decomposition gives another view of the same fact. At $p = 3$, the threshold sits between the components and their sum. For every larger prime, even the total falls below the line.

The grid

Compare every fraction of an integer to every other fraction, digit by digit. Record the proportion of matches for each pair. Lay them in a grid. At $p = 11$, the grid tells the whole story.

$ ./nfield field 11
   1/11 => 0.|09|      6/11 => 0.|54|
   2/11 => 0.|18|      7/11 => 0.|63|
   3/11 => 0.|27|      8/11 => 0.|72|
   4/11 => 0.|36|      9/11 => 0.|81|
   5/11 => 0.|45|     10/11 => 0.|90|

Ten fractions. Ten distinct two-digit blocks. No two share a digit at any position. So the grid has a one everywhere a fraction meets itself and a zero everywhere it meets a stranger.

        1/11 2/11 3/11 4/11 5/11 6/11 7/11 8/11 9/11 10/11
 1/11 [  [1]   0    0    0    0    0    0    0    0    0  ]
 2/11 [   0   [1]   0    0    0    0    0    0    0    0  ]
 3/11 [   0    0   [1]   0    0    0    0    0    0    0  ]
 4/11 [   0    0    0   [1]   0    0    0    0    0    0  ]
 5/11 [   0    0    0    0   [1]   0    0    0    0    0  ]
 6/11 [   0    0    0    0    0   [1]   0    0    0    0  ]
 7/11 [   0    0    0    0    0    0   [1]   0    0    0  ]
 8/11 [   0    0    0    0    0    0    0   [1]   0    0  ]
 9/11 [   0    0    0    0    0    0    0    0   [1]   0  ]
10/11 [   0    0    0    0    0    0    0    0    0   [1] ]

The pairwise part $\sigma$ is zero. No fraction agrees with any other. The cross-alignment matrix of $11$ is the identity.

Now take $p = 13$. Twelve fractions, six-digit repeating blocks.

$ ./nfield field 13
   1/13 => 0.|076923|      7/13 => 0.|538461|
   2/13 => 0.|153846|      8/13 => 0.|615384|
   3/13 => 0.|230769|      9/13 => 0.|692307|
   4/13 => 0.|307692|     10/13 => 0.|769230|
   5/13 => 0.|384615|     11/13 => 0.|846153|
   6/13 => 0.|461538|     12/13 => 0.|923076|

Build the grid the same way. Dots are zeros.

          1    2    3    4    5    6    7    8    9   10   11   12
   1  [  [1]   .    .    .    .    .    .    .    .    .  0.3    . ]
   2  [   .   [1]   .    .    .    .    .    .    .    .    .  0.3 ]
   3  [   .    .   [1]   .    .    .  0.3    .    .    .    .    . ]
   4  [   .    .    .   [1] 0.3    .    .    .    .    .    .    . ]
   5  [   .    .    .  0.3  [1]   .    .    .    .    .    .    . ]
   6  [   .    .    .    .    .   [1]   .    .    .  0.3    .    . ]
   7  [   .    .  0.3    .    .    .   [1]   .    .    .    .    . ]
   8  [   .    .    .    .    .    .    .   [1] 0.3    .    .    . ]
   9  [   .    .    .    .    .    .    .  0.3  [1]   .    .    . ]
  10  [   .    .    .    .    .  0.3    .    .    .   [1]   .    . ]
  11  [ 0.3    .    .    .    .    .    .    .    .    .   [1]   . ]
  12  [   .  0.3    .    .    .    .    .    .    .    .    .   [1]]

Six pairs of fractions share digits at some positions. The $0.3$ entries are small, but they are there. That is the pairwise part. Average any row including the diagonal and you get $\alpha$. Average the off-diagonal entries and you get $\sigma$.

At $p = 11$, there are no off-diagonal matches. At $p = 13$, they begin to appear.

Scale up.

Twenty primes, from 3 to 151. At $p = 3$, the grid is nearly full. At $p = 5$, completely full. At $p = 7$ and $p = 11$, the grid goes dark: white diagonal, black everywhere else. These are the digit-partitioning primes, with $\sigma = 0$. Then at $p = 13$, the first off-diagonal entries appear and the filling begins.

More off-diagonal agreement does not mean higher alignment. The pairwise part can grow while the focused part shrinks.

The pinch

The golden gap has a simple algebraic form. The condition for $1/\varphi$ to sit inside the gap is

$$\frac{1}{p} < \frac{1}{\varphi} < \frac{2}{p}.$$

The left inequality says $p > \varphi \approx 1.618$. Every prime satisfies that. The right one says $p < 2\varphi \approx 3.236$. Only two primes satisfy that, $p = 2$ and $p = 3$. The prime 2 is degenerate, so among odd primes only $p = 3$ remains.

Each bar is the interval from $1/p$ to $2/p$ for one prime. The golden line at $1/\varphi$ cuts through only one of them. Only the bar for $p = 3$ reaches across the line, with $1/3$ on one side and $2/3$ on the other.

The golden ratio is not in this picture for aesthetic reasons. It is the unique number whose reciprocal fits between $1/p$ and $2/p$ at exactly one odd prime.

The threshold is crossed not by one dominant piece, but by two smaller pieces whose sum succeeds only at a single prime.

A note from 2026

April 2026

This paper is easier to appreciate after The Cross-Alignment Matrix. The grid that appears here as a computational tool becomes the real object there. Seen from that angle, the split into focused and pairwise parts is already asking a structural question: how much of the alignment belongs to $1/n$ itself, and how much belongs to the field around it?

That way of splitting a quantity into a distinguished part and a background part shows up again in The Collision Transform, where the global mean is removed before the Fourier analysis begins. The language is different, but the instinct is the same.

The exceptional role of the prime $3$ also survives the transition. Here $3$ is the only prime for which the two pieces can add up across the golden line. In The Collision Spectrum, the same prime becomes exceptional again from a different direction. The agreement is too clean to feel accidental.

One point remains especially satisfying. The pairwise term $\sigma$ vanishes exactly at the digit-partitioning primes. The dark grids at $p=7$ and $p=11$ are not just visually simple; they are the places where the background coherence drops all the way to zero.

.:.

Try it yourself

Pull the alignment apart and see what each piece is doing.

$ ./nfield decompose 12
  alpha  = 0.636  (total)
  sigma  = 0.273  (background)
  F      = 0.364  (focused)
  1/phi in gap = yes

$ ./nfield decompose 120
  alpha  = 0.664  (total)
  sigma  = 0.328  (background)
  F      = 0.336  (focused)
  1/phi in gap = yes

Watch what happens as the smooth part grows. The focused number $F$ drops from $0.364$ toward the limit $1/3$. The pairwise number $\sigma$ rises from $0.273$ toward the same limit. The two components move toward each other while their sum stays above $1/\varphi$.

Now try $7$. No collisions, no shared digits.

$ ./nfield decompose 7
  alpha  = 0.167  (total)
  sigma  = 0.000  (background)
  F      = 0.167  (focused)
  1/phi in gap = no

Zero pairwise background. The fractions of $7$ are completely independent. The matrix is the identity. The only coherence is focused, and $1/7$ alone cannot carry the total anywhere near $0.618$.

Now $77 = 7 \times 11$. Two primes that individually had zero pairwise part.

$ ./nfield decompose 77
  alpha  = 0.088  (total)
  sigma  = 0.065  (background)
  F      = 0.022  (focused)
  1/phi in gap = no

Both components are present but both are tiny. Together they reach $0.088$.

The decomposition is not just bookkeeping. It tells you why an integer fails to make the cut. At digit-partitioning primes like $7$ and $11$, the pairwise part vanishes. At composite Tier 3 numbers like $77$, both components are present but small. At Tier 2, both are needed. At Tier 1, the alignment is trivially perfect.

Code: github.com/alexspetty/nfield
Paper: The Coherence Decomposition

Alexander S. Petty
January 2021 (updated April 2026)
.:.