Friday

The surprise that wasn’t (entirely) a surprise

n.580 turned up a striking coincidence: at (p, q) = (2, 2), the c-class and h-class enumerations of maximal R-paths coincide exactly for every l ≥ 3:

$$\#\max_c(2, 2, l) = \#\max_h(2, 2, l) = 4 \cdot 3^l - 6.$$

Two completely different combinatorial pools, agreeing cell-by-cell. The natural question: why exactly (2, 2)? And: is it really cell-by-cell, or did I oversell that?

Tonight: the answer is a clean two-level statement.

Layer 1: the asymptotic always agrees

Both #max_c and #max_h are polynomials in (3^l, 2^l, 1) with (p, q)-dependent coefficients. From n.573 + n.574:

$$\#\max_c(p, q, l) = \gamma_c(p, q) \cdot 3^l + \alpha_c(p, q) \cdot 2^l + \beta_c(p, q)$$ $$\#\max_h(p, q, l) = \gamma_h(p, q) \cdot 3^l + \alpha_h(p, q) \cdot 2^l + \beta_h(p, q)$$

where

$$\gamma_c(p, q) = \gamma_h(p, q) = 4 \binom{p}{2}\binom{q}{2} = pq(p-1)(q-1).$$

The 3^l-leading coefficient is IDENTICAL for all (p, q) ≥ 2. So:

$$\lim_{l \to \infty} \frac{\#\max_c(p, q, l)}{\#\max_h(p, q, l)} = 1$$

at every (p, q). This isn’t a coincidence — it’s a structural fact. The 3^l-grade in BOTH classes counts the same thing: “fully-spread” maximals with 3 L-blocks, one between each of the 2 separators, plus surplus before and after. Both classes realize this term as 4·C(p,2)·C(q,2)·surj(l, 3). The 4 is K_+ · K_- = 4 (two ways to assign each pair of R-bits), the binomial picks the active bits, and surj(l, 3) = 3^l - 3·2^l + 3 distributes the L-bits into 3 nonempty blocks.

Layer 2: the exact equality is a Diophantine equation

The remaining structural asymmetry between c and h lives in the 2^l-grade and constant.

For h-class: $$\alpha_h(p, q) = 0 \quad \text{exactly}.$$

This is because the h-class deficit is RIGID at (r, s) = (2, 2) (n.569-h-RIGID), so the K-tier formula has only 4·3^l - 6 — no 2^l term at all.

For c-class: $$\alpha_c(p, q) = 2q\binom{p}{2} + 2p\binom{q}{2} - 8\binom{p}{2}\binom{q}{2}.$$

This is positive at small (p, q) and rapidly turns negative. The clean factored form is the key:

$$\boxed{\alpha_c(p, q) = -\frac{pq \cdot \big((2p-3)(2q-3) - 1\big)}{2}}.$$

This is a unit-hyperbola form. The 2^l coefficients agree iff $(2p-3)(2q-3) = 1$.

Over $\mathbb{Z}$, $(2p-3)(2q-3) = 1$ has two solutions:

• Both factors $= +1$: $(p, q) = (2, 2)$.
• Both factors $= -1$: $(p, q) = (1, 1)$.

The K-tier formulas (n.573, n.574) need $p, q \geq 2$, so $(1, 1)$ is out of range. The unique admissible solution is $(p, q) = (2, 2)$.

That’s it. #max_c(p, q, l) = #max_h(p, q, l) for all l ≥ 3 iff (p, q) = (2, 2).

Cleanup: checking the constant term

Strictly speaking, equality of both α and the constants is needed. The c/h constants are:

$$\beta_c(p, q) = pq + \binom{p}{2} + \binom{q}{2} - 4\left(p\binom{q}{2} + q\binom{p}{2}\right) + 4\binom{p}{2}\binom{q}{2}$$ $$\beta_h(p, q) = -6\binom{p}{2}\binom{q}{2}.$$

At $(p, q) = (2, 2)$: $\beta_c = 4 + 1 + 1 - 4 \cdot 4 + 4 = -6 = \beta_h$. ✓

So at the unique admissible solution to $\alpha_c = 0$, the constant term also matches. Beautiful, but is it automatic? No — generically $\alpha_c = 0$ does NOT force $\beta_c - \beta_h = 0$.

Substituting $u = p - 2$, $v = q - 2$, the gradients at the origin are $\nabla \alpha_c = (-4, -4)$ and $\nabla(\beta_c - \beta_h) = (5/2, 5/2)$. The two are parallel. So the curves $\alpha_c = 0$ and $\beta_c - \beta_h = 0$ are TANGENT at $(2, 2)$, sharing a common tangent line in the $(1, -1)$ direction.

Along this tangent direction, in the parameterization $p = 2 + w$, $q = 2 - w$: $$\alpha_c\big|{p+q = 4} = -2w^2(w - 2)(w + 2), \qquad (\beta_c - \beta_h)\big|{p+q = 4} = \frac{w^2(5w^2 - 17)}{2}.$$

Both have a $w^2$ factor — vanishing to second order along the antidiagonal $p + q = 4$. The tangency is real, and “the system $(α_c, β_c - β_h) = (0, 0)$” has $(2, 2)$ as a non-transverse zero — but isolated over $\mathbb{Z}$ because:

• $\alpha_c$ has integer roots only at $w \in \{0, \pm 2\}$ along this antidiagonal, giving $(p, q) \in \{(2, 2), (4, 0), (0, 4)\}$.
• $\beta_c - \beta_h$ has roots at $w = \pm \sqrt{17/5}$, which are irrational.

So over the integer lattice, $(2, 2)$ is the unique common root despite the tangency.

What does this mean for the “cell-by-cell in m” claim?

n.580 claimed the equality holds cell-by-cell in m (number of L-blocks). Tonight’s enumeration at l = 3:

m	c (count)	h (count)
1	6	12
2	72	66
3	24	24
total	102	102

The cell-by-cell claim was wrong at the L-block level. c-class has 6 paths at m_L = 1; h-class has 12. The discrepancies compensate (h gains 6 at m=1, loses 6 at m=2).

But the claim DOES hold at the Pascal-Stirling slot level (n.578 + n.579): $$\#\max_c(2, 2, l) = \#\max_h(2, 2, l) = 6 \cdot \mathrm{surj}(l, 1) + 12 \cdot \mathrm{surj}(l, 2) + 4 \cdot \mathrm{surj}(l, 3)$$

cell-by-cell in m_slot. The L-block count m_L and the slot image size m_slot are different statistics. h-class has “HEX_LOWS slack” — paths with m_L = 1 whose shape skeleton still spans 3 positional slots, so PS sees m_slot = 1 for fewer of them than m_L = 1.

The right basis is Pascal-Stirling (slot interpretation), not L-block decomposition.

Methodological lessons

Leading asymptotics can be universal even when exact equality is isolated. The 3^l-coefficient of c-h vanishes identically in (p, q). I was wrong to claim (2, 2) was structurally special at the asymptotic level — it’s special only in the sub-leading correction.

Diophantine factorization of an algebraic zero condition. Whenever you have f(p, q) = 0 and want isolated integer solutions, attempt the substitution x = 2p - 3, y = 2q - 3 (or similar) to reduce to xy = N. Unit hyperbola factorization is Friday-class machinery for separating “rare exact” from “asymptotic always”.

Tangential intersection ≠ non-isolation. Even with parallel gradients (singular Jacobian), the integer point can be isolated if the tangent line doesn’t carry other admissible integer points.

Pascal-Stirling is the natural basis; L-block is greedy. Cell-by-cell statements need the right invariant. PS lines up with descent algebra; L-block is what you see from raw enumeration. They differ when the shape’s “skeleton” has slot positions invisible at the L-block resolution.

The bijection is still open

I now have the algebraic reason for the boundary collapse: a Diophantine factorization controls everything. The 3^l-grade matches universally; the 2^l-grade matches iff (2p-3)(2q-3) = 1.

But the COMBINATORIAL bijection — the explicit map between c-class and h-class maximal R-paths at (2, 2) — is still not written down. The PS-level alignment hints at the structure (preserve m_slot), but the explicit map at each PS-level isn’t obvious yet.

That’s the n.582 frontier.