Digit-Partitioning Primes and the Alignment Formula

Digit-Partitioning Primes and the Alignment Formula
The same rotating field from the first paper is no longer continuous. It has been cut into contiguous regions by the digit function, each interval carrying its own trajectory. No two paths overlap. The coherence observed before is now explained: the field separates cleanly into independent channels, and alignment becomes a consequence of perfect partition.

The previous result

In the companion paper, we proved that the repetend alignment of n = 3m equals (2m − 1)/(3m − 1), and that the golden ratio selects p = 3 uniquely through a self-referential algebraic condition.

A natural question follows: does the same formula hold for primes other than 3?

The answer

Yes. The formula

$$\alpha(pm) = \frac{2m - 1}{pm - 1}$$
holds for every prime p satisfying a single condition: pb + 1, where b is the positional base. We call such primes digit-partitioning.

In base 10, the digit-partitioning primes are exactly 3, 7, and 11.

Why it works

The proof rests on one object: the digit function
$$\delta(r) = \left\lfloor \frac{br}{p} \right\rfloor$$
This is the machine inside long division. Given a remainder r, it produces the next digit. When pb + 1, the digit function is injective: different remainders always produce different digits.

Injectivity means that at every position of the repetend, no two fractions k/p and 1/p can share the same digit (unless k = 1). Every non-matching fraction contributes zero alignment. The counting argument from the first paper carries through unchanged, with 3 replaced by any digit-partitioning prime p.

The boundary case

When p = b + 1, the digit function simplifies to

$$\delta(r) = r - 1$$
A bijection from {1, ..., b} to {0, ..., b − 1}. Every remainder maps to a unique digit. Every digit is used exactly once. No room to spare.

In base 10, this gives p = 11. The five cosets of ⟨10⟩ in (ℤ/11ℤ)* produce the repetend pairs:

  • {1, 10} → 09
  • {2, 9} → 18
  • {3, 8} → 27
  • {4, 7} → 36
  • {5, 6} → 45

These are the nines-complement pairs. They partition all ten digits with nothing left over.

Three mechanisms, one condition

In base 10, the three digit-partitioning primes arise through different structural mechanisms:

  • p = 3: single-digit repetends (ord₃(10) = 1)
  • p = 7: cyclic number 142857, all digits distinct (ord₇(10) = 6 = p − 1)
  • p = 11: complement-pair partition (10 ≡ −1 mod 11)

These initially appeared to require separate treatment. The characterization pb + 1 unifies them. All three are consequences of the injectivity of the digit function.

The golden ratio, universally

Since the formula (2m − 1)/(pm − 1) holds for all digit-partitioning primes, the golden threshold analysis applies uniformly. The alignment limit as m → ∞ is 2/p, which exceeds 1/φ only when

$$p < 2\varphi \approx 3.236$$
Among primes, only p = 3 satisfies this. Across all bases and all digit-partitioning primes, the golden ratio selects p = 3 uniquely. The self-referential characterization, the fourth-power identity τ⁴ = 2 − 3τ, and the factorization (τ² + τ − 1)(τ − 2) = 0 extend without modification.

The digit-partitioning framework reveals that this selection is not an artifact of base 10. It is a universal feature of the alignment formula.

The paper

The full paper, with proofs of the characterization theorem, the universal alignment formula, and the enumeration across bases:

Digit-Partitioning Primes and the Alignment Formula (PDF)

Both papers can be verified computationally using nfield:

./nfield verify    # paper 1
./nfield verify2   # paper 2

nfield on GitHub

.:.