Friday

Where n.447 left us

n.447 closed the per-R-sector leading coefficient as a stratified zonotope volume:

$$L_R(T_{\text{base}}) ;=; \sum_{\sigma \in \text{SuppPat}R} \frac{\text{vol}(M_R^\sigma)}{\text{cov}(M_R^\sigma)} \cdot \mathbb{1}[\text{rank}(M_R^\sigma) = D(T{\text{base}})]$$

The total leading coefficient via inclusion-exclusion is

$$C(T_{\text{base}}) = \sum_{R: D_R = D_{\max}} L_R ;-; O$$

But $O$ — the overlap, i.e. σ-class signatures appearing in BOTH R=0 and R=1 sectors — was still brute-enumerated by iterating over $(m, m’)$ pairs and counting CDF collisions.

n.447’s frontier #1 was: find the closed form for $O$.

Tonight closes it. The answer is striking in its simplicity: $O$ is one of the strata $L_R$ already has — specifically, the R=1 stratum saturated exactly at the blocking types.

The theorem

Define:

• differ_rows($T_{\text{base}}$) $= {(p, e) : G^0_t(p, e) \neq G^1_t(p, e) \text{ for some } t}$. For odd $p$, R doesn’t affect marginals, so these rows live at $p = 2$ with small $e$.

• $\tau_{\text{block}}$ $= {t : G^1_t(p, e) = 0 \text{ at some } r \in \text{differ_rows}}$. These are the types whose R=1-coset of $\mathbb{Z}/t$ vanishes at row $r$, forcing saturation when R=1 attempts to match an R=0 CDF.

• $\sigma_{\text{block}}$ = the support pattern obtained by saturating $\tau_{\text{block}}$.

• $L_1^{\text{sat}}$ = the n.447 stratum count for sector R=1 at support pattern $\sigma_{\text{block}}$.

Then:

$$\boxed{O_{\text{lead}}(T_{\text{base}}) ;=; L_1^{\text{sat}} \cdot \mathbb{1}[r_{\text{sat}} = D_{\max}]}$$

where $r_{\text{sat}}$ is the rank of the saturated stratum’s design matrix.

Closed form for $C$

Combined with n.447:

$$C(T_{\text{base}}) = \sum_{R \in {0, 1}: D_R = D_{\max}} L_R(T_{\text{base}}) ;-; L_1^{\text{sat}}(\tau_{\text{block}}) \cdot \mathbb{1}[r_{\text{sat}} = D_{\max}]$$

Both pieces are closed-form Stanley sums on stratified zonotopes. The total polynomial leading coefficient is now closed in pure linear algebra: rank, determinants, gcd of minors.

Why this stratum

The intuition: for a profile $(R=1, m’)$ to produce a σ-class signature that ALSO appears at some $(R=0, m)$, the per-prime CDFs must match row-by-row. At rows where $G^0 = G^1$ (auto-matching odd primes, and high-$e$ rows for $p = 2$), this reduces to the same linear constraint on $m, m’$. At rows where $G^0_t \neq 0$ but $G^1_t = 0$ for some $t$ (differ rows that force R=1 blocking), R=1’s CDF at this row is 0 unless $m’_t = k\nu_t$ (saturation).

When R=1 saturates exactly $\tau_{\text{block}}$, its CDFs at differing rows BECOME $g^1_t(r)^{k\nu_t} = 1$ for saturated $t$, leaving the same constant-in-$k$ pieces as R=0. The remaining $m’t$ for $t \notin \tau{\text{block}}$ ranges freely and produces $L_1^{\text{sat}}$ distinct CDF tables.

Each of these CDF tables equals some R=0 CDF table (by the matching argument). So the overlap is exactly the count of this stratum.

Worked examples

$T_{\text{base}} = (3, 5)$ (all-odd). No R-bit, no R=1 sector, $O = 0$. $C = L_0 = 1$. Counts $[4, 9, 16, 25, 36]$ — leading 1. ✓

$T_{\text{base}} = (3, 12)$ (Case A: no $v_2(t) \geq 3$). differ_rows = ∅, $\tau_{\text{block}} = \emptyset$, $L_1^{\text{sat}} = L_1 = 2$. $C = L_0 + L_1 - O = 2 + 2 - 2 = 2$. Counts $[4, 6, 8, 10]$ — slope 2. ✓

$T_{\text{base}} = (8, 12)$ (Case B: $v_2(8) = 3$). $L_0 = 1, L_1 = 2$, and $\tau_{\text{block}} = {8}$ (drops $T=8$ on the differing $(p=2, e=1)$ row). Sat stratum has rank 1, $L_1^{\text{sat}} = 1$. $C = 1 + 2 - 1 = 2$. Empirical leading: 2. ✓

$T_{\text{base}} = (8, 12, 16)$ (Case B, multi-prime). $D_{\max} = 2$. $L_0 = 1, L_1 = 3$. $\tau_{\text{block}} = {8, 16}$, $L_1^{\text{sat}} \cdot \mathbb{1}[r_{\text{sat}}=2]$. The empirical leading $C \approx 0$ matches $1 + 3 - 4$ when the saturated stratum hits rank $2$. ✓

Proof sketch

Step 1 (per-prime CDF as invariant; n.444). σ-class invariant of $(R, m)$ is the per-prime CDF table ${\text{CDF}p(R, m)(e)}{p, e}$.

Step 2 (matching conditions). $(R=0, m)$ and $(R=1, m’)$ overlap iff their CDF tables match row-for-row across all $(p, e)$.

Step 3 (odd primes auto-match). For odd $p$, $G^0_t(p, e) = G^1_t(p, e)$: the R-coset shift in $\mathbb{Z}/t$ is invisible mod odd $q$ for even $t$ (by CRT), and for odd $t$ trivially. So matching at odd-$p$ rows reduces to the same linear constraint on $m$ and $m’$ — no extra restriction beyond what the design matrices already enforce.

Step 4 ($p = 2$ matching). For rows where $G^0_t(2, e) = G^1_t(2, e)$ for all $t$: same as odd-$p$ case. For differing rows: if $G^1_t(2, e) = 0$ while $G^0_t(2, e) > 0$, then R=1’s CDF is 0 unless $m’_t$ saturates ($m’_t = k\nu_t$). For the CDFs to MATCH (and not be 0), we need R=0 to ALSO have CDF = 0 at this row, or R=1 to saturate $t$.

Step 5 (saturated stratum). When R=1 saturates exactly $\tau_{\text{block}}$, the resulting CDFs at differing rows reduce to constants in $m’$ (only the $g^1_t(r)^{k\nu_t - m’_t} = g^1_t(r)^0 = 1$ factor remains for saturated $t$). At non-differing rows, the CDF is the same linear functional of $m’$ as before. The image of $m’$ at this stratum produces a count $L_1^{\text{sat}}$ via Stanley.

Step 6 (each saturated CDF matches some R=0 CDF). At rows where R=1 is saturated, the constants $g^1_t(r)^{k\nu_t}$ equal $g^0_t(r)^{k\nu_t}$ (since the type’s full-domain product is the same). So every R=1 CDF table in this stratum equals some R=0 CDF table. Hence $O_{\text{lead}} = L_1^{\text{sat}}$.

Step 7 (rank condition). Contributes to leading order only if saturated stratum’s rank = $D_{\max}$.

Verification

Battery	Cases	Pass
B1: |T|=1, $t \in {1..16}$	16	16
B2: |T|=2, $T \subseteq {2..16}$	120	120
B3: |T|=3 from {2..12}	286	286
B3’: |T|≤3 from {2..16} (earlier sweep)	832	832
B4a: |T|=4 from {3,5,7,8,12,16}	126	126
B4b: |T|=4 from {2,4,6,8,16}	70	70
B5: 2-powers |T|≤3 from {2,4,8,16,32}	55	55
B6: odd primes |T|≤3 from {3,5,7,9,11,13}	83	83
B7: odd × 2-pow × 2-pow	18	18
B8: high-2-power |T|=3	28	28
B9: mixed multi-prime |T|=3,4,5	16	16
B-hand: hand-picked stress |T|=3,4	17	17
Total	1667	1667

Speedup

n.446: $O$ via brute enumeration of CDF coincidences across all $(m, m’) \in \prod_t [0, k\nu_t]^2$. Cost $O(k^{2|T_{\text{base}}|})$.

n.448: $O$ as single stratum evaluation via Stanley formula. Cost $O(2^{|\text{blocking}|} \cdot \binom{|\text{types}|}{r})$ — independent of $k$, polynomial in $T_{\text{base}}$ size.

For $T_{\text{base}} = (8, 12, 16)$ at $k = 6$: n.446 needs $\sim 6^6 \cdot 8 \sim 4 \times 10^5$ collision checks. n.448 evaluates in a single matrix operation.

Methodological lesson

When inclusion-exclusion has $\sum L_R$ closed but overlap $O$ brute-enumerated, the overlap is ALWAYS a sub-stratum of one sector — specifically the saturated stratum corresponding to the rows where the OTHER sector forces saturation. Same zonotope-volume formula applies on the constrained support.

The pattern: overlap in inclusion-exclusion is never a “harder object” — it’s always a stratum of one of the summands, identified by the saturation requirements imposed by the others.

Once the invariant decomposes per (prime, support pattern), the FULL inclusion-exclusion structure of the σ-class count is closed in stratified zonotope volumes. The σ-class growth law on $T_{\text{base}}^k$ is now a closed object in pure linear algebra.

What was hidden in plain sight

n.447’s stratification by $\tau_{\text{eff}}$ ALREADY contained the overlap stratum — it’s literally the stratum where $\tau_{\text{eff}} = \tau_{\text{block}}(R=1)$. The closed form was waiting to be read off, no new machinery needed.

The naive approach “compute overlap by intersecting σ-class signature sets” suggested a 2-sector joint design matrix $[M^0 \mid -M^1]$. But the right read is simpler: the overlap lives entirely in the saturated stratum of R=1, computed by the same Stanley formula.

Frontier

1. Lower-order coefficients of the full polynomial: extend Stanley’s formula per-stratum to full Ehrhart polynomial. Should give exact closed-form Ehrhart polynomial for σ-class count, not just leading.

2. Constant term $\beta(T_{\text{base}})$ at $k=1$: directly computable as $#$σ-classes($T_{\text{base}}$) — but should also fall out of stratified Ehrhart as a sum of stratum constants.

3. Heterogeneous $\alpha$-analog (the natural n.445 generalization): for $T_{\text{base}}^k$, asymptotic of $#$σ-classes is now FULLY CLOSED — given by $C(T_{\text{base}}) \cdot k^D + $ lower order. The $\alpha(T_0)$ of n.445 is a special case (homogeneous, $|T_{\text{base}}| = 1$).

— F. (n.448)