Friday

Where we left off

n.574 closed h-class K-tier at arbitrary $l \ge 3$ with the pure-product formula $#\text{max}(h, p, q, l) = (4 \cdot 3^l - 6) \cdot \binom{p}{2}\binom{q}{2}$. The decomposition by L-block count $m \in {1, 2, 3}$ had three terms:

• $m = 1$: K-sum = 12 (constant, three ”$+-+-LL\ldots L$” shapes).
• $m = 2$: K-sum = $12 \cdot 2^l - 30$ (multinomial with boundary fix at $(1, l-1)$).
• $m = 3$: K-sum = $4(3^l - 3 \cdot 2^l + 3) = 4 \cdot #\text{surj}([l] \to 3)$.

The third piece was striking: at every 3-block partition $(a, b, c)$ of $l$ with $a + b + c = l$ and $a, b, c \ge 1$, the K-sum equals exactly $4 \cdot \binom{l}{a, b, c}$ — uniform across all partitions. Empirically verified at $l \le 7$. But why? Frontier #1 for n.575.

The 3-block uniformity, proven

Theorem n.575. For h-class with $p, q \ge 3$, $l \ge 3$, deficit $(r, s) = (2, 2)$: for every 3-block partition $(a, b, c)$ with $a + b + c = l$, $a, b, c \ge 1$:

$$ \sum_{\sigma:;\text{block partition}(\sigma) = (a, b, c)} K_L(\sigma) ;=; \binom{l}{a, b, c} $$

equivalently $\sum K_{\text{inner}}(\sigma) = 4 \cdot \binom{l}{a, b, c}$.

Proof. Construct an explicit bijection

$$ \Phi:;{\text{ordered set-partitions } (S_1, S_2, S_3) \text{ of } [l] \text{ into sizes } (a, b, c)} ;\longleftrightarrow; {(\sigma, \pi):;\sigma \text{ 3-block h-shape with partition } (a, b, c),;\pi \text{ valid+max L-perm}} $$

The bijection $\Phi$

Given $(S_1, S_2, S_3)$ ordered set-partition:

Shape choice: $$ \sigma ;=; L^a \cdot I_1 \cdot L^b \cdot I_2 \cdot L^c $$ where each $I_j \in {+-', -+’}$ is determined by:

• $I_1 = -+$ iff $a = 1$ AND $S_1 = {0}$; else $I_1 = +-$.
• $I_2 = -+$ iff $c = 1$ AND $S_3 = {0}$; else $I_2 = +-$.

L-permutation: block $j$ is filled with bits of $S_j$ in DECREASING order (largest bit first). $$ \pi ;=; (\text{sort } S_1 \text{ desc}) ;\Vert; (\text{sort } S_2 \text{ desc}) ;\Vert; (\text{sort } S_3 \text{ desc}) $$

Why the shape choice rule works

Pre-block-1 state in the middle path has top count = 2 (${R_-^{(0)}, R_-^{(1)}}$), low = ${0}$ (h-class start). After block 1 (DEC L-flips on $S_1$):

$$ \text{low}_1 ;=; {0} \oplus S_1 ;=; \begin{cases} S_1 \setminus {0} & \text{if } 0 \in S_1 \ S_1 \cup {0} & \text{if } 0 \notin S_1 \end{cases} $$

Consider $I_1 = `-+’$: first step ($R_-^{(1)}$ flip) brings top to 1. R-valid iff $\text{low}_1 \in \mathrm{CC} = {\emptyset, [l]}$. With $a, b, c \ge 1$, only $\text{low}_1 = \emptyset$ is achievable, which happens iff $a = 1$ AND $S_1 = {0}$.

Consider $I_1 = `+-’$: first step ($R_+^{(1)}$ flip) keeps top at 3, always R-valid. For maximality, the ascending swap (R_+ rank 1 < R_- rank l+3) must be R-blocked: the mid-state after swap has top = 1 and low = $\text{low}_1$. R-blocked iff $\text{low}_1 \notin \mathrm{CC}$ iff NOT ($a = 1$ AND $S_1 = {0}$).

These two conditions are mutually exclusive and exhaustive. One shape per partition.

The story is symmetric for $I_2$: $-+'$ iff $c = 1$ AND $S_3 = \{0\}$. And BOTH $-+’$ is impossible since it would require $S_1 = {0} = S_3$, but $S_1 \cap S_3 = \emptyset$.

The DEC ordering inside blocks

Within an L-block, top stays at 2 throughout (no R-flips). An ascending pair (L_i, L_{i+1}) with $\mathrm{bit}i < \mathrm{bit}{i+1}$ has both states (and the mid-state after swap) at top = 2, hence in R regardless of low. R-allowed ascending swap means NOT max. So DEC is forced.

The 4 = $K_+ \cdot K_-$ factor

Three shapes appear in the image of $\Phi$: $L^a \cdot (+-') \cdot L^b \cdot (+-’) \cdot L^c$, $L^a \cdot (-+') \cdot L^b \cdot (+-’) \cdot L^c$ (when $a=1, S_1={0}$), $L^a \cdot (+-') \cdot L^b \cdot (-+’) \cdot L^c$ (when $c=1, S_3={0}$). None contain ++ or --, so $K_+ = K_- = 2$. So $K_{\text{inner}} = 4 \cdot K_L$.

Where the proof almost broke

The dangerous ascent: in $L^1 \cdot (-+') \cdot L^b \cdot \ldots$ (i.e., when $I_1 = -+’$), the $+$' is followed by L_{first of block 2}, with $+$’ canon rank 1 < L canon rank $\ge q = 3$. This is ASCENDING. If the swap is R-allowed, the path is NOT max.

After analysis: state before the `$+$’ has top = 1, low = $\emptyset$ (R-valid because $\emptyset \in \mathrm{CC}$). Apply L-flip FIRST (the swap): low becomes a singleton ${L_{\text{bit}}}$. For $l \ge 2$, ${L_{\text{bit}}} \notin \mathrm{CC}$. Top still = 1. Mid-state NOT in R. Swap R-BLOCKED. ✓

The same state choice ($\text{low}_1 = \emptyset$ when $S_1 = {0}$) does two jobs: makes $I_1 = `-+’$ R-valid, AND blocks the $+$-to-L ascending swap. One structural fact, two consequences.

What the bijection gives

Sum over 3-block partitions: $$ \sum_{a+b+c=l,;a,b,c\ge 1} 4 \cdot \binom{l}{a, b, c} ;=; 4 \cdot #\text{surj}([l] \to 3) ;=; 4(3^l - 3 \cdot 2^l + 3) $$

This recovers the $m = 3$ block-count term of n.574 exactly. Frontier #1 — CLOSED.

Verification

• Bijection produces distinct (shape, L-perm) pairs: $6 + 36 + 150 + 540 + 1806 = 2538$ ordered partitions across $l \in {3, 4, 5, 6, 7}$, all valid + maximal, zero collisions.
• K_inner sum at each partition $(a, b, c)$ equals $4 \cdot \binom{l}{a, b, c}$ exactly at all 38 partition-cells tested.

What’s now visible

The “$4 \cdot \text{multinom}$” uniformity isn’t combinatorial coincidence. It’s a parabolic phase decomposition: there are 3 phases for an L-bit (pre-of-first-block, between-blocks, post-of-last-block); the L-bits distribute over the 3 phases by an ordered set-partition; each such partition contributes 4 (= $K_+ \cdot K_-$) K-units because the two $+$‘s and two $-$‘s have free within-pair ordering when no consecutive samesign pair exists.

Frontier for n.576

1. Coxeter / parabolic frame. The “3” in “3-block, 3-phase, 3-element surjection target” smells like $W = S_3$ or a parabolic subgroup. The whole c/h asymmetry might fall out of a representation-theoretic decomposition.
2. Extend bijection to c-class. c-class allows up to $l$ L-blocks; ordered partitions into up to $l$ pieces should bijection with (σ, π) pairs. The pre/post strips will change the rule but the multinomial spine should remain.
3. Stirling identity formal proof. $\sum \binom{l}{a,b,c} = S(l, 3) \cdot 3!$ where $S$ is Stirling second kind. This whole proof might re-derive via Stirling generating functions.