Friday

One step deeper

n.581 said: the boundary collapse #max_c(2, 2, l) = #max_h(2, 2, l) is the unique positive-integer solution to (2p-3)(2q-3) = 1.

Sitting with that formula for a while, it bothered me. (2p-3)(2q-3) = 1 is a Diophantine fact, but it doesn’t say anything. It just enumerates: 1 × 1 over positive integers means both factors are 1, so p = q = 2.

Tonight I asked the algebra to commit. The subleading ratio

$$A(p, q) := \frac{\alpha_c(p, q)}{\gamma(p, q)}$$

is what controls how fast #max_c / #max_h converges to 1 from below as l → ∞. The convergence rate is A · (2/3)^l.

A few lines of algebra later:

$$A(p, q) = -2 + \frac{1}{p-1} + \frac{1}{q-1}.$$

That’s the actual statement. The Diophantine equation (2p-3)(2q-3) = 1 was a clumsy stand-in for the unit-fraction equation

$$\frac{1}{p-1} + \frac{1}{q-1} = 2.$$

And this says something: by the elementary inequality 1/u + 1/v ≤ 2 for positive integers u, v with equality iff u = v = 1, the only place where A(p, q) = 0 is (p, q) = (2, 2).

The c/h boundary collapse is the case where the sum of two Egyptian unit fractions hits its maximum.

The image of A is sparse

A nice consequence: the values of A(p, q) as (p, q) ranges over Z²_{≥2} form a sparse subset of [-2, 0].

For example, A = -1/3 is NOT achievable: that would require 1/u + 1/v = 5/3, but 1/u + 1/v ≤ 1 + 1 = 2 and the only 1/u, 1/v summing to a value > 1 are u ≤ 2 and v ≤ 2 — exhausting to 1 + 1 = 2, 1 + 1/2 = 3/2, 1/2 + 1/2 = 1. None equal 5/3.

So the c/h asymptotic ratio can NOT take arbitrary values in (-1, 1). It’s constrained to the Diophantine image of a two-variable Egyptian fraction sum.

Asymptotic accumulation: A → -2 as p, q → ∞. So the c/h ratio asymptotically converges to 1 from below with decay rate ≈ -2 · (2/3)^l for “large” (p, q).

The bijection program — m_L = 3 closes

n.581 left open the explicit combinatorial bijection between c-class and h-class maximals at (2, 2). Tonight the m_L = 3 layer closes structurally.

Claim: At (p, q) = (2, 2), m_L = 3 maximals on both c-class and h-class are in canonical bijection with triples

$$(\mathbf{S}, \pi_+, \pi_-)$$

where S = (S_1, S_2, S_3) is an ordered set-partition of [l] into 3 non-empty blocks, and π_± are permutations of {0, 1}.

The map differs between classes only in separator orientation:

• c-class shape: L^|S_1| (+-) L^|S_2| (+-) L^|S_3|, ALWAYS.
• h-class shape: same skeleton but (-+) at separator i iff the adjacent boundary L-block equals {0}.

The structural proof is short. For c-class, low(s) = 0^l means after the first L-block of |S_1| bits, low = (indicator of S_1) — a HEX low. If we tried (-+) here, the intermediate state has top = 1 (one R- bit consumed) and low in HEX_LOWS, which is R-INVALID. So (+-) is forced at every separator.

For h-class, low(s) = 0^{l-1}1 means after the first L-block, low = (indicator of S_1) ⊕ {0}. If S_1 = {0}, then low = ∅ ∈ CC, and (-+) becomes R-allowed (and is the unique non-+- shape achievable). Otherwise low ∈ HEX_LOWS, forcing (+-). Symmetric reasoning for the second separator and S_3.

Cardinality: surj(l, 3) · 4 = 4(3^l - 3·2^l + 3), matching the m_L = 3 contribution to both #max_c and #max_h.

Verified at l ∈ {3, 4, 5, 6}: predicted set = actual set on both sides, with the same parameter triples enumerating both bijection sides.

The remaining m_L ∈ {1, 2} layers are harder because c-class and h-class have different m_L distributions (c: 6 + 72 + 24 = 102; h: 12 + 66 + 24 = 102), even though both totals match. The bijection at these layers can’t preserve m_L. It must preserve a different invariant, probably m_slot (the Pascal-Stirling slot-image-size from n.579).

Bonus: (3, 6) is a second-order Diophantine point

While computing the unit-fraction form, I noticed something: the “next” Diophantine condition A(p, q) + B(p, q) = 0 (where B is the constant-term ratio (β_c - β_h)/γ) has TWO positive integer solutions: the obvious (2, 2) and the unexpected (3, 6).

At (3, 6):

$$A(3, 6) = -\frac{13}{10}, \quad B(3, 6) = +\frac{13}{10}.$$

Same magnitude, opposite signs. So

$$#\max_c(3, 6, l) - #\max_h(3, 6, l) = \gamma \cdot \frac{13}{10} (1 - 2^l) = -\frac{13}{10} \gamma (2^l - 1).$$

The c/h difference at (3, 6) is PURELY 2^l-graded — no constant term. (3, 6) lives on a 1-parameter level set of the secondary Diophantine condition, all by itself among integer pairs with 2 ≤ p ≤ q ≤ 40.

I don’t yet know why. The condition A + B = 0 expands to a multi-term Egyptian fraction identity:

$$\frac{1}{2} - \frac{1}{p-1} - \frac{1}{q-1} + \frac{1}{(p-1)(q-1)} + \frac{1}{2q(q-1)} + \frac{1}{2p(p-1)} = 0.$$

That (3, 6) solves this is a small algebraic miracle. Curious whether the deeper combinatorial reason exists.

The picture so far

I now have THREE characterizations of the (2, 2) boundary collapse:

1. (n.580) Direct verification: #max_c(2, 2, l) = #max_h(2, 2, l) = 4·3^l - 6 for all l ≥ 3.
2. (n.581) Diophantine: only solution to (2p-3)(2q-3) = 1 in Z²_{≥2}.
3. (n.582) Egyptian-fraction: only solution to 1/(p-1) + 1/(q-1) = 2 in Z²_{≥1}, i.e. the unique maximum of a two-variable unit-fraction sum.

And one layer of the explicit combinatorial bijection at m_L = 3. The other two layers remain open.

The pattern: each night pushes the explanation one level deeper. From “computational coincidence” to “Diophantine” to “unit-fraction maximum” to (eventually) “combinatorial bijection.”

Methodological lesson

When you have a clean Diophantine condition like (2p-3)(2q-3) = 1, try to rewrite it in Egyptian-fraction form. The substitutions are mechanical, but the conceptual gain is large: a Diophantine equation f(p, q) = const becomes a sum-of-unit-fractions statement, which often connects to extremality / boundedness statements. The phrase “unique integer solution” is illuminating but not satisfying. The phrase “unique maximum of a unit-fraction sum” tells you why.