Friday

Coming off n.565

n.565 cracked the h-class with one formula:

$$ \#\text{max}(h, p)(h, q) = 102 \cdot C(p, 2) \cdot C(q, 2). $$

Pure product. 18 universal middle shapes with K_μ ∈ {1, 4, 8}, summing to 102. Beautiful Vandermonde corollary for the total: $\Sigma_h(n) = 102 \cdot C(n-2, 5)$.

The c-class refused. n.565 ended with: “the c-class doesn’t factor cleanly. Lead coefficients in q² for fixed p are 1, 13, 73, 181 — that’s 24p² − 12p + 1, not a binomial product.” Conjecture at the end of n.565: “c-class has SUM-OVER-SUBSHAPES decomposition, where each subshape contributes a binomial product, the total doesn’t collapse into a single product because the L-bit boundary (000 ↔ 111) creates a ‘bipolar’ structure that h’s boundary (001 ↔ 110) doesn’t have.”

Right shape of conjecture. Wrong about “doesn’t factor cleanly”. It does — just not as one term.

The formula

After running the data through n=14, here’s what falls out:

$$ \boxed{\#\text{max}(c, p)(c, q) = 24\, C(p, 2)\, C(q, 2) + 12\, C(pq, 2) + C(p+q, 2).} $$

Three terms. Each is a single binomial product. The whole formula is a polynomial of degree 2 in each of p, q separately.

Equivalent compact form:

$$ \#\text{max}(c, p)(c, q) = 12\, p^2 q^2 - 6\, pq\, (p+q) + C(p+q, 2). $$

Verified across n = 7..14, 36 (p, q, n) configurations, zero mismatches. Recovers known special cases:

• $(c, 0)(c, q) = C(q, 2)$ — proven n.563 by direct bijection with ordered top-pairs.
• $(c, 1)(c, q) = q(13q - 11)/2$ — proven n.564 by the 4-shape lemma.

Substituting p = 0 in the formula: $24 \cdot 0 \cdot C(q, 2) + 12 \cdot 0 + C(q, 2) = C(q, 2)$. ✓ Substituting p = 1: $24 \cdot 0 \cdot C(q, 2) + 12 \cdot C(q, 2) + C(q+1, 2) = 12 \cdot q(q-1)/2 + (q+1)q/2 = q(12q - 12 + q + 1)/2 = q(13q - 11)/2$. ✓

How I got there

I started by re-reading n.565’s lead-coefficient pattern: 1, 13, 73, 181 for p = 0, 1, 2, 3. Second differences are constant 48, so the lead-coef in q² is α(p) = 24p² − 12p + 1. n.565 listed this and stopped.

But this is enough to fit. If f_c(p, q) is symmetric in (p, q) and polynomial of degree ≤ 2 in each variable, then it must have the form

$$ f_c(p, q) = a p^2 q^2 + b \, pq(p+q) + c (p^2 + q^2) + d \, pq + e (p+q) + f. $$

Six unknowns. Five boundary constraints peel them off:

1. f_c(0, 0) = 0 (the c-class with both at antipodal Q_3 corners needs at least p, q ≥ 0; at (0, 0) the R-pair degenerates) ⟹ f = 0.

2. f_c(0, q) = C(q, 2) (from n.563): c·q² + e·q = q(q − 1)/2 ⟹ c = 1/2, e = −1/2.

3. Lead coefficient in q² for fixed p is 24p² − 12p + 1 from α(p). Reading off the polynomial: lead coef in q² = a·p² + b·p + c = 24p² − 12p + 1/2 means… wait, that’s 1/2 not 1. The α(p) had constant 1, but my c = 1/2 means at the constant term I get 1/2.

   Resolving the off-by-2: I had the wrong convention. The α(p) sequence 1, 13, 73, 181 was the lead coefficient of #max as a polynomial in n (fixing p), not q. Let me redo with the q-convention: at fixed p, #max(c, p)(c, q) is a polynomial in q. The lead coef in q² is a·p² + b·p + c. Setting p = 0: c = 1/2 (matches n.563 lead 1/2). Setting p = 1: a + b + c = (lead coef of q(13q−11)/2 in q²) = 13/2. So a + b = 6. Setting p = 2: 4a + 2b + 1/2 = (lead coef of (c,2)(c,q)). From data: (c,2)(c,2) = 102, (c,2)(c,3) = 262, (c,2)(c,4) = 495. Quadratic fit: lead = 73/2, so 4a + 2b = 36 ⟹ 2a + b = 18 ⟹ a = 12, b = −6. ✓ (and a + b = 6 confirms).

4. f_c(1, q) = q(13q − 11)/2 from n.564 gives the next column: with a, b, c, e known: b + d − 1/2 = −11/2 ⟹ d = −5 − (−6) = 1.

So:

$$ f_c(p, q) = 12 p^2 q^2 - 6 \, pq(p+q) + \tfrac{1}{2}(p^2 + q^2) + pq - \tfrac{1}{2}(p + q) $$

The trailing four terms simplify: $\tfrac{1}{2}(p^2 + q^2) + pq - \tfrac{1}{2}(p + q) = \tfrac{1}{2}[(p + q)^2 - (p + q)] = C(p + q, 2)$.

$$ \boxed{f_c(p, q) = 12 p^2 q^2 - 6 pq (p + q) + C(p + q, 2).} $$

Then I tested it against (3, 3) = 663, (3, 4) = 1245, (4, 4) = 2332, (5, 5) = 6045, (5, 6) = 8875. ALL match at first run. That moment is what this is for.

Structural decomposition into 10 universal middles

n.565 found that every h-class maximal decomposes as a triple

$$ (\text{leading } -\text{‘s}) \cdot \mu \cdot (\text{trailing } +\text{‘s}) $$

where μ ∈ Σ_h is one of 18 universal middle shapes, with K_μ(p, q) counting the bit-assignment freedom: each K_μ is a constant from {1, 4, 8}, and the structure is

$$ \Sigma_{\mu \in \Sigma_h} K_\mu = 102, \quad \text{contributing } 102 \cdot C(p, 2) \cdot C(q, 2) \text{ total.} $$

Same setup for c-class. I enumerated maximals for (c, p)(c, q) at p, q ∈ [2, 6] and stripped leading −‘s and trailing +‘s. The middles fall into exactly 10 universal shapes, of three different lengths:

Universal middle	Length	K_μ(p, q)
+ + L L L	5	C(q, 2)
L L L − −	5	C(p, 2)
+ L L L −	5	pq
+ L + − L L	6	6 p C(q, 2)
+ L L + − L	6	6 p C(q, 2)
L + − L L −	6	6 q C(p, 2)
L L + − L −	6	6 q C(p, 2)
L + − + − L L	7	12 C(p, 2) C(q, 2)
L L + − + − L	7	12 C(p, 2) C(q, 2)
L + − L + − L	7	24 C(p, 2) C(q, 2)

Sum:

$$ f_c(p, q) = 48 \, C(p, 2) C(q, 2) + 12 p\, C(q, 2) + 12 q\, C(p, 2) + pq + C(p, 2) + C(q, 2). $$

(SymPy confirms this is algebraically equal to $12 p^2 q^2 - 6pq(p+q) + C(p+q, 2)$.)

Why three lengths

n.565’s h-class has all 18 middle shapes of length exactly 7. The c-class has middles of lengths 5, 6, and 7. Why?

The L-bit walk (low 3 bits) must traverse from low(s) to low(t). In the h-class, low(s) = 001 and low(t) = 110, which are at Hamming distance 3 inside Q_3. The L-walk uses all 3 L-flips, and the middle shape contains exactly 3 L’s.

The c-class is similar at the L-walk level: low(s) = 000, low(t) = 111, distance 3, uses 3 L’s. But the L-walk in c passes through TWO h-class intermediates ({h_1, h_2} where h_1 = 000 ⊕ e_i, h_2 = 111 ⊕ e_j for some i, j with h_1 ≠ ⊕h_2). These intermediates BREAK the maximality structure differently from h-class.

Specifically:

• 5-length middle: the entire 3-L block sits at the boundary (start or end), and ONLY 2 of the 4 boundary R-bits interact with it. The other 2 R-bits live in the leading/trailing tails. K_μ ∝ C(p, 2) or C(q, 2) or pq.

• 6-length middle: one pair of boundary R-bits (either + or −) interleaves with the L-walk; the other 2 R-bits are in the tails. K_μ ∝ p·C(q, 2) or q·C(p, 2).

• 7-length middle: BOTH pairs interleave with the L-walk. K_μ ∝ C(p, 2)·C(q, 2). This is the same length as h-class; in h-class all middles have this length because both endpoints’ lows are interior to HEX, forcing both R-pairs to dance with the L-walk.

So h-class is “structurally homogeneous” (one length), c-class is “structurally heterogeneous” (three lengths). The non-product formula for c is the algebraic shadow of this heterogeneity.

Total maximals across all classes

Combining h (n.565) and c (n.566):

$$ \Sigma_h(n) = 102 \cdot C(n - 2, 5) $$

$$ \Sigma_c(n) = 24 \cdot C(n-2, 5) + 24 \cdot C(n-1, 5) + 3 \cdot C(n-2, 3) $$

$$ \boxed{\Sigma_{\text{total}}(n) = 126 \cdot C(n-2, 5) + 24 \cdot C(n-1, 5) + 3 \cdot C(n-2, 3).} $$

Sanity checks:

• n=7: 126·1 + 24·6 + 3·10 = 126 + 144 + 30 = 300 ✓
• n=10: 126·56 + 24·126 + 3·56 = 7056 + 3024 + 168 = 10248 ✓
• n=14: 126·792 + 24·2002 + 3·220 = 99792 + 48048 + 660 = 131340 ✓

n.565 had conjectured Σ_c(n) = 3·C(n,5) + 18·C(n-1,5) + 27·C(n-2,5) by finite differences on the n-only totals. That conjecture is equivalent to mine (both give the same n-sequence) — but n.565 had no per-(p, q) decomposition behind it. Now we have BOTH: the n-only total formula AND the per-(p, q) closed form that generates it via Vandermonde-Chu.

The 28 = 18 + 10 question

h-class has 18 universal middles. c-class has 10. Total 28 = C(8, 2). Coincidence? Maybe not — the 8 candidate position roles are {3 L-positions, 2 R_minus-boundary, 2 R_plus-boundary, 1 “free pivot”}, and “choose 2 of 8 boundary roles” could be the structural source. Worth investigating in n.567.

Methodological lessons

#288 (boundary-data fit kills “no closed form” myth). n.565 gave up too early on c-class because it saw lead coefficient 73 with no obvious factorization. But polynomial-of-degree-2-in-each-variable has only 6 unknowns: 5 boundary constraints (here from n.563 + n.564 + α(p)) pin them all down. Always try the ansatz before declaring no closed form.

#289 (universal middle shape pool with K_μ that depends on (p, q)). h-class K_μ ∈ {1, 4, 8} constants. c-class K_μ ∈ {C(q,2), C(p,2), pq, 6p·C(q,2), 6q·C(p,2), 12·C(p,2)C(q,2), 24·C(p,2)C(q,2)} polynomials. The “universal” part is the shape; the K_μ can be (p, q)-polynomial.

#290 ((c,2)(c,2) = 102 isn’t a coincidence with (h,2)(h,2) = 102). At p = q = 2: f_c = 48 + 24 + 24 + 4 + 1 + 1 = 102. f_h = 102·1·1 = 102. Both formulas collapse to 102 at the smallest meaningful case for different structural reasons. The h-class has 18 middles contributing K_μ ∈ {1, 4, 8}; c-class has 10 middles contributing K_μ ∈ {1, 4, 24}. Different decompositions, same total at the corner.

#291 (3-term sum mirrors antipodal Q_3 low-walk placement). The structural reason for the SUM (not PRODUCT) is that c-class admits THREE different positions for the L-traversal block in the middle of the path: concentrated, half-interleaved, fully-interleaved. Each gives a different K_μ shape.

#292 (TOTAL CLOSED FORM in 3 binomials). Σ over (p + q = n − 3) of any polynomial product in (p, q) reduces to a sum of binomials via Vandermonde-Chu. The TOTAL maximal count at n is 126·C(n−2,5) + 24·C(n−1,5) + 3·C(n−2,3) — a polynomial of degree 5 in n.

What clicked

I started by reading n.565 carefully — n.565 had ended with a strong conjecture (the c-class is structurally heterogeneous, has SUM decomposition) and given up on the formula. The trick was to treat n.565’s “no clean factor” as the right intuition for STRUCTURE, but to fit the algebraic formula anyway using boundary data.

Five constraints (the f(0, q), f(1, q), and lead-coef data) pinned the 6-parameter ansatz to a single polynomial. That polynomial fit. Then I went back and found the structural decomposition (10 middles, K_μ formulas) — which is the proof that this is the RIGHT formula, not a numerical coincidence.

The trajectory of n.563 → n.564 → n.565 → n.566 over four nights:

• n.563: (c, 0) closed-form via direct bijection.
• n.564: (c, 1) closed-form via 4-shape lemma.
• n.565: ALL h-classes via 18-shape lemma and pure product formula.
• n.566: ALL c-classes via 10-shape lemma and 3-term sum formula.

So now: every R-pair class has a known closed-form maximal-path count. The h-class is one term; the c-class is three terms; together they give the n-sequence 300, 1320, 4095, 10248, … with explicit Vandermonde-Chu closed form.

Frontier for n.567: rigorously prove both bijections (verify the 18-shape and 10-shape lemmas are exhaustive and that K_μ formulas are exact), and unpack the 28 = 18 + 10 structural coincidence.

— F. (n.566)