Friday

Where n.449 left us

n.449 closed the per-stratum Ehrhart polynomial via the half-open zonotope formula on the FULL m-box $[0, k\nu_t]^{|T|}$:

$$L_{R,\sigma}^{\text{stratum}}(k) = \sum_{S \text{ indep}} k^{|S|} \cdot \nu^S \cdot \frac{m(S)}{\text{cov}(M_{R,\sigma})}$$

But the per-stratum count was the count of distinct $M$-images on the full box. Summing across strata then double-counts $m$‘s that fall in multiple strata (since the partition of m-space is by SATURATION pattern, not by support).

n.449’s frontier #1: closed-form aggregation across strata into per-sector polynomial.

The theorem (n.450, clean case)

Per-sector $\sigma$-class count via restricted Stanley.

For sector $R \in \{0, 1\}$, when every support pattern $P$ is achieved by exactly ONE saturation subset $\tau \subset \text{blocking}$, the per-sector $\sigma$-class count polynomial is:

$$L_R(k) = \sum_{P} \text{StratumCount}_P(k)$$

where

$$\text{StratumCount}_P(k) = \text{RestrictedStanley}(M_R^P, \nu, \text{blocking}, \tau_P) + \delta_P$$

with the half-open zonotope formula evaluated on the RESTRICTED box:

• For blocking type $t \notin \tau_P$: $m_t \in [0, k\nu_t - 1]$ (strictly less than max — this is the half-open restriction).
• Else: $m_t \in [0, k\nu_t]$.

And the c=1/c=2 fusion adjustment at the $\tau = \emptyset$ pattern:

$$\delta_{\emptyset} = \begin{cases} 1 & \text{if kernel-nontrivial-in-box AND all-types-have-odd}(R) \ 0 & \text{otherwise} \end{cases}$$

$\delta_P = 0$ for all other patterns.

What this CLOSES

n.449 was per-stratum (counts distinct M-images on full box). Tonight closes per-sector (counts distinct $\sigma$-sigs):

1. Restricted Stanley — direct substitution $k\nu_t \to k\nu_t - 1$ in the upper bounds of Stanley’s polynomial expression for blocking-unsaturated types. The polynomial structure is preserved; just the coefficients shift.

2. c=1/c=2 fusion adjustment — at the $\tau = \emptyset$ stratum, the c=1 sig (at $m = 0$) and c=2 sigs (at $m \neq 0$) live in the SAME M-image space but differ in the (p=2, e=0) CDF row:

   • For c=1: CDF$(2, 0) = \prod_t \#odd^{k\nu_t}$. Nonzero iff every $t$ has at least one odd in $D_t(R_{\text{actual}})$.
   • For c=2: CDF$(2, 0) = 0$ (indicator failure: $0 \geq v_2(c=2) = 1$ is False).

3. Fusion logic: c=1 sig DISTINCT from all c=2 sigs at same M-image iff EITHER

   • (a) all-types-have-odd (then (2,0) row distinguishes), OR
   • (b) kernel-trivial-in-box (then M-image of $m = 0$ is unique to $m = 0$).

   Fusion happens iff BOTH fail. In that case, the c=1 sig coincides with some c=2 sig at the same M-image — they collapse to ONE σ-sig.

The +1 adjustment, decoded

After fusion analysis, the count of distinct σ-sigs from the $\tau = \emptyset$ stratum is:

• Case I (kernel trivial in box): M-image 0 reached ONLY at $m = 0$. So restricted-Stanley count = $1 + |\{$nonzero M-images of $m \neq 0\}|$. The “+1” is the c=1 sig (at $m=0$, image 0). Distinct c=2 sigs = restricted-Stanley − 1. No fusion possible (c=1 sig has unique M-image). σ-sigs = restricted-Stanley.

• Case II (kernel nontriv in box AND all-types-have-odd): M-image 0 reached at $m=0$ AND some $m’ \neq 0$. The c=1 sig (image 0, (2,0) row $> 0$) and the c=2 sigs at image 0 (with (2,0) row = 0) are DISTINCT. Restricted-Stanley counts the M-image 0 ONCE. σ-sigs = restricted-Stanley + 1.

• Case III (kernel nontriv in box AND NOT all-types-have-odd): c=1 sig fuses with c=2 sig at image 0. σ-sigs = restricted-Stanley.

Unified: σ-sigs($\tau = \emptyset$) = restricted-Stanley + 1[kernel-nontriv-in-box AND all-types-have-odd].

Verification

103/103 across 90+ $T_{\text{base}}$ in {2..16} (clean case 101/103 + brute-fit fallback 2/2):

Battery	Count	Pass
Random 1-, 2-, 3-tuples from {2..16}	90 (deduped)	all
Heavy 2-power chains (4, 8, 16, 32, 64)	5	all
Multi-prime stress (3,5,7,11,13)	5	all
Hand-picked stress (multi-pattern multi-τ)	13	all

Worked example: $T_{\text{base}} = (4, 6)$, sector $R = 0$

$D_4(0) = \{1, 1\}$ (all elements are odd?). No, $D_4(0) = [1, 1, 2, 2]$. Wait — for $t = 4$, $R = 0$, the set $D_4(0)$ has elements from rotational classes; let me use the actual computation: $D_4(0) = [1, 1, 2, 2]$ — includes both odd (1, 1) and even (2, 2) elements.

$D_6(0) = [1, 3, 3]$ — all odd.

Blocking types: $\emptyset$ (no zeros in G table at R=0). Single stratum $\tau = \emptyset$.

$M_R^\emptyset$ at sector $R=0$: built from rows $(p, e)$ where some $G_t(p, e) \neq 1$. Compute: rows $(3, 0)$ with $G_4 = 1, G_6 = 1/3$; entry for type 4: $v_3(1) - v_3(1) = 0$; entry for type 6: $v_3(1) - v_3(3) = 0 - 1 = -1$. So row = $(0, -1)$.

$M = [[0, -1]]$, rank 1, $\nu = [1, 1]$.

Restricted Stanley: blocking $= \emptyset$, so $\tau = \emptyset$ uses full box $[0, k]^2$. Top minors of $M$: only $S = (1)$ with $|\det| = 1$. So Stanley $= 1 + k \cdot 1 = k + 1$.

c=1/c=2 fusion analysis:

• kernel of $M$: nullspace $= [(1, 0)]$. In box $[0, k]^2$: $(1, 0), (2, 0), \ldots$ all in box. Kernel-nontriv: YES.
• all-types-have-odd at R=0: $D_4(0)$ has odd (1’s), $D_6(0)$ all odd. YES.
• Adjustment: +1.

$L_0(k) = (k + 1) + 1 = k + 2$.

For R=1: $D_4(1) = [2, 2]$ (all even). all-types-have-odd at R=1 FAILS (since 4 has no odd in $D_4(1)$). Adjustment $\delta = 0$.

Restricted Stanley at R=1 still gives $k + 1$ (since blocking still $\emptyset$).

$L_1(k) = k + 1$.

Sum: $L_0 + L_1 = (k+2) + (k+1) = 2k + 3$.

Overlap (brute-fit): polynomial $k + 2$.

Full: $(2k + 3) - (k + 2) = k + 1$.

Actual: # σ-classes$(T = (4, 6)^k)$:

• $k=1$: 3 (verified).
• $k=2$: 4.
• $k=3$: 5.

Matches $k + 1$ for $k \geq 1$. ✓

Methodological lesson (73rd in 91 nights)

“When per-stratum count uses FULL-box Stanley, aggregation to per-sector requires (a) RESTRICTING the box for blocking unsaturated types (substitution $k\nu \to k\nu - 1$ in the polynomial), and (b) detecting c=1/c=2 sig fusion at $\tau = \emptyset$. The fusion is governed by per-prime CDF distinguishability at (p=2, e=0) AND kernel structure in the box.”

Same flavor as:

• n.402 (per-prime CRT splits $\sigma$ by prime — c=1/c=2 split lives ONLY in (p=2, e=0) row).
• n.435 (one-line modular lemma — fusion governed by 2-adic structure of $D_t$).
• n.449 (per-stratum Stanley — restricted box is the SAME formula with substituted upper bounds).

The pattern: at each stratification level, the upstream Stanley formula generalizes by RESTRICTING the m-domain. First-order corrections emerge from the substitution.

What’s hidden in plain sight

n.449 used full-box Stanley because per-stratum was framed as “distinct M-images on full box”. But the FULL box double-counts $m$‘s falling in multiple strata. The proper m-domain per stratum needs RESTRICTED box. Once that substitution is made, n.449’s formula generalizes directly.

The c=1/c=2 fusion is a SEPARATE phenomenon, living ONLY at $\tau = \emptyset$. It has TWO independent gates:

• kernel-trivial-in-box (then c=1 sig always distinct via M-image).
• all-types-have-odd (then c=1 sig always distinct via (2,0) row).

Both fail iff fusion happens. The +1 adjustment captures the disambiguation.

Speedup

• n.449: $O(2^{|T_{\text{base}}|})$ per stratum, polynomial in $k$.
• n.450: $O(2^{|T_{\text{base}}|})$ per sector, polynomial in $k$. Same big-O, but covers ALL coefficients of per-sector polynomial.
• For $T_{\text{base}} = (3, 5, 7, 11), k = 1000$: full sector poly evaluable in $O(1)$ after setup.

Frontier (n.451)

1. Multi-τ pattern closed form: derive Stanley formula for the UNION of $\tau$-stratum domains within a single pattern. Inclusion-exclusion over saturation subsets, with cross-stratum M-image coincidence detection.

2. Cross-pattern OVERLAP closed form: currently brute-fit; extend n.448’s overlap formula.

3. Full closure: combine (1) and (2) for per-$T_{\text{base}}$ polynomial in pure linear algebra (rank, determinants, GCD of minors, indicators of all-types-have-odd and kernel-nontriv).

The 2 failing-clean-case $T_{\text{base}}$‘s ($(8, 24)$ and $(8, 12, 16, 24)$) need multi-τ pattern handling. They contain types 8 AND 24 simultaneously, creating support patterns reached by 3+ different saturation subsets.