> Full-text translation from Chinese. The original remains authoritative for its research claims; this translation is not independent mathematical verification.

# Legendre Base-Space Program Phase 6
## Dynamic Obstruction Horizon, Factor Shells, and the Sieve-Critical Gap
### Dynamic Obstruction Horizon, Factor Shells, and the Sieve-Critical Gap

**Version:** v0.1  
**Date:** 2026-09-30  
**Author:** Neo.K × AI collaboration  
**Institution:** EveMissLab / 一言諾科技有限公司  
**Research line:** Legendre Base-Space Prime-Fitting Program  
**Prerequisites:** Phase 1–5  
**Status:** Phase 6 formalization + finite audit + literature-boundary audit; not a completed proof of Legendre's conjecture

---

# 0. Conclusions of This Phase

Phase 5 compressed the problem to:

$$
\text{Position Fit}
+
\text{Low-Obstruction Avoidance}
+
\text{High-Obstruction Closure}.
$$

Phase 6 directly connects ‘Position Fit’ and ‘High Obstruction’ for the first time.

If:

$$
q=n^2+k
$$

lies on the left side of the square cell, composite obstruction cannot come from arbitrary:

$$
a\le n.
$$

The offset $k$ itself forces a lower upper bound on the largest necessary factor:

$$
\boxed{
B(n,k)
<
n.
}
$$

For:

$$
k=o(n),
$$

we further have:

$$
\boxed{
B(n,k)
=
n-\sqrt n
+
O\left(
1+\frac{k}{\sqrt n}
\right).
}
$$

Thus, a fit candidate near the left square boundary is automatically immune to an entire obstruction strip closest to $n$.

This phase also establishes:

1. An exact shell coordinate for factor pairs crossing the square root;
2. A discriminant-square characterization of composite obstruction;
3. A roughness ladder under $z=n^\alpha$ presieving;
4. The critical gap between the standard linear sieve and $P_3/P_2$ structuring at $\alpha=\frac12$.

---

# 1. Factor-Pair Coordinates in a Square Cell

Let:

$$
q=n^2+k,
\qquad
1\le k\le2n.
$$

Suppose $q$ is composite.

Take a nontrivial factor:

$$
a\le\sqrt q.
$$

Since:

$$
q<(n+1)^2,
$$

we have:

$$
a<n+1.
$$

Therefore:

$$
a\le n.
$$

Write:

$$
\boxed{
a=n-u,
\qquad
u\ge0.
}
$$

Let the other factor be:

$$
b=\frac qa.
$$

Since:

$$
q>n^2
$$

and:

$$
a\le n,
$$

necessarily:

$$
b>n.
$$

Write:

$$
b=n+v,
\qquad
v\ge1.
$$

---

# 2. Directionality of Factors Crossing the Square Root

From:

$$
(n-u)(n+v)>n^2
$$

we obtain:

$$
n(v-u)>uv.
$$

Thus:

$$
v>u.
$$

Let:

$$
\boxed{
\delta=v-u\ge1.
}
$$

Any factor pair crossing $n$ can therefore be written uniquely as:

$$
\boxed{
(a,b)
=
(n-u,\,
n+u+\delta).
}
$$

---

# 3. Factor-Shell Identity

Substitute the expression above into:

$$
q=ab
$$

to obtain:

$$
n^2+k
=
(n-u)(n+u+\delta).
$$

Expanding:

$$
\boxed{
k
=
(n-u)\delta-u^2.
}
$$

That is:

$$
\boxed{
k
=
n\delta-u(u+\delta).
}
$$

This is the first core identity of Phase 6.

It converts ‘divisibility’ into:

$$
(n,u,\delta)
$$

three geometric quantities.

---

# 4. The Dynamic Obstruction Horizon

From:

$$
\delta\ge1,
$$

we have:

$$
k
=
(n-u)\delta-u^2
\ge
n-u-u^2.
$$

Thus:

$$
u^2+u
\ge
n-k.
$$

When:

$$
k<n,
$$

define:

$$
\boxed{
u_*(n,k)
=
\left\lceil
\frac{
\sqrt{1+4(n-k)}-1
}{2}
\right\rceil.
}
$$

Then:

$$
u\ge u_*(n,k).
$$

Hence:

$$
a=n-u
\le
n-u_*(n,k).
$$

Define:

$$
\boxed{
B(n,k)
=
n-u_*(n,k).
}
$$

Call it:

## Candidate-Dependent Obstruction Horizon

---

# 5. Obstruction Horizon Theorem

## Theorem 5.1

Let:

$$
1\le k<n
$$

and:

$$
q=n^2+k\in M6^*.
$$

If $q$ is composite, it has an atomic factor:

$$
a\le B(n,k).
$$

In other words:

$$
\boxed{
q\text{ composite}
\Longrightarrow
\exists a\in\mathfrak A:
a\mid q,\;
a\le B(n,k).
}
$$

Thus:

$$
\boxed{
\gcd
\left(
q,
\prod_{\substack{a\in\mathfrak A\\a\le B(n,k)}}a
\right)=1
\Longrightarrow
q\text{ prime}.
}
$$

This is an exact candidate-dependent horizon for a prime certificate.

---

# 6. The Asymptotic Horizon for Left-Boundary Candidates

If:

$$
k=o(n),
$$

then:

$$
u_*(n,k)
=
\sqrt n
-
\frac{k}{2\sqrt n}
+
O(1)
+
O\left(
\frac{k^2}{n^{3/2}}
\right).
$$

Therefore:

$$
\boxed{
B(n,k)
=
n-\sqrt n
+
\frac{k}{2\sqrt n}
+
O(1)
+
O\left(
\frac{k^2}{n^{3/2}}
\right).
}
$$

In particular, if:

$$
k=O(\log n),
$$

then:

$$
\boxed{
B(n,k)
=
n-\sqrt n+O(1).
}
$$

V3-type left-boundary fit candidates therefore do not need to consider:

$$
(n-\sqrt n+O(1),\,n]
$$

factor obstructions in this interval.

---

# 7. High-Top Strip Exclusion

More generally, let:

$$
0<c<1.
$$

If a factor satisfies:

$$
a>n-c\sqrt n,
$$

then:

$$
u<c\sqrt n.
$$

From:

$$
k\ge n-u-u^2,
$$

we obtain:

$$
\boxed{
k
>
(1-c^2)n-c\sqrt n.
}
$$

Thus:

> The highest-order factors, closest to $n$, can hit only the right-hand region of the square cell.

If:

$$
k=o(n),
$$

then for any fixed:

$$
c<1,
$$

these factors cannot hit the candidate at all for sufficiently large $n$.

---

# 8. Discriminant Shell

From:

$$
k
=
n\delta-u(u+\delta),
$$

we obtain the quadratic equation:

$$
u^2+\delta u+(k-n\delta)=0.
$$

Its discriminant is:

$$
\boxed{
D_\delta(n,k)
=
\delta^2+4n\delta-4k.
}
$$

By the factor-shell identity:

$$
\boxed{
D_\delta(n,k)
=
(2u+\delta)^2.
}
$$

Composite obstruction therefore leaves, in:

$$
(n,k,\delta)
$$

this space, a perfect-square discriminant certificate.

---

# 9. Discriminant-Square Characterization

If there exists:

$$
\delta\ge1
$$

such that:

$$
D_\delta(n,k)
$$

is a perfect square:

$$
r^2,
$$

and:

$$
r\equiv\delta\pmod2,
$$

then:

$$
u=\frac{r-\delta}{2}
$$

is an integer.

If, in addition:

$$
1<n-u<n^2+k,
$$

then:

$$
\boxed{
n^2+k
=
(n-u)(n+u+\delta),
}
$$

so the candidate is composite.

Divisibility obstruction can therefore be reinterpreted as:

$$
\boxed{
\text{square-discriminant shell hit}.
}
$$

---

# 10. The First Two Shells

## $\delta=1$

$$
k
=
n-u(u+1).
$$

Thus:

$$
\boxed{
4(n-k)+1
=
(2u+1)^2.
}
$$

The first shell is a triangular-number / odd-square curve.

## $\delta=2$

$$
k
=
2n-u(u+2)
=
2n+1-(u+1)^2.
$$

Thus:

$$
\boxed{
2n+1-k
=
(u+1)^2.
}
$$

The second shell is directly a square curve.

This shows that high-order factor obstructions closest to $n$ are not arbitrary residue noise, but lie on explicit quadratic shells.

---

# 11. Equivalence to the Phase 4 Residue Field

The Phase 4 obstruction condition is:

$$
a\mid n^2+k
$$

equivalent to:

$$
k\equiv-n^2\pmod a.
$$

When Phase 6 writes:

$$
a=n-u
$$

we have:

$$
n\equiv u\pmod a,
$$

so:

$$
k
\equiv
-u^2
\pmod{n-u}.
$$

Thus:

$$
\boxed{
\text{residue obstruction}
\equiv
\text{factor-shell obstruction}.
}
$$

Phase 4 examines the modular space;

Phase 6 examines the geometry of factor pairs crossing the square root.

---

# 12. The Actual Connection Between V3 Fit and the Obstruction Horizon

Phase 5 obtains:

$$
x_3(n)-n^2
=
O(\log n).
$$

Nearest to $x_3(n)$, the $M6^*$ candidate differs from $x_3(n)$ by at most a constant amount.

The rank-1 V3 candidate offset is therefore:

$$
k_3(n)
=
O(\log n).
$$

Thus:

$$
\boxed{
B(n,k_3(n))
=
n-\sqrt n+O(1).
}
$$

This is the first time in this series that we obtain:

$$
\boxed{
\text{Position Fit}
\Longrightarrow
\text{Obstruction Domain Shrinkage}.
}
$$

---

# 13. Roughness Ladder

Now consider the low-order presieving scale:

$$
z(n)=n^\alpha,
\qquad
0<\alpha<1.
$$

Suppose the candidate:

$$
q<(n+1)^2
$$

has no prime factor:

$$
p\le n^\alpha
$$

in this range.

If:

$$
\Omega(q)\ge r
$$

with prime factors counted with multiplicity, then:

$$
q
>
n^{\alpha r}.
$$

If:

$$
\alpha r>2,
$$

then for sufficiently large $n$:

$$
n^{\alpha r}
>
(n+1)^2,
$$

a contradiction.

Therefore:

## Roughness Ladder Theorem

For any integer $r\ge2$, if:

$$
\alpha r>2,
$$

then for sufficiently large $n$:

$$
\boxed{
\Omega(q)\le r-1.
}
$$

---

# 14. Two Important Thresholds

If:

$$
\alpha>\frac12,
$$

take:

$$
r=4,
$$

Then:

$$
\boxed{
\Omega(q)\le3.
}
$$

Thus, a survivor is at most a $P_3$.

If:

$$
\alpha>\frac23,
$$

take:

$$
r=3,
$$

Then:

$$
\boxed{
\Omega(q)\le2.
}
$$

If the candidate is not prime, it can only be a semiprime.

---

# 15. Relationship to the 2026 $P_3$ Frontier

Peter J. Campbell proved in 2026:

$$
\boxed{
\text{Every }(n^2,(n+1)^2)
\text{ contains an integer with at most three prime factors.}
}
$$

This improves the $P_4$ result of Dudek–Johnston.

The roughness ladder in this section is not an alternative proof of Campbell's theorem.

It only indicates:

> If a square-cell survivor can be sieved to roughness beyond $n^{1/2}$, its remaining multiplicative complexity naturally falls into the $P_3$ layer.

Campbell's actual work unconditionally guarantees that such an almost-prime object exists in **every cell**, rather than merely giving a conditional factor-count implication.

---

# 16. The Critical Gap of the Linear Sieve

For the square-interval sequence:

$$
\mathcal A_n
=
\{n^2+1,\dots,n^2+2n\},
$$

for squarefree $d$ we have:

$$
|\mathcal A_{n,d}|
=
\frac{2n}{d}
+
O(1).
$$

If a dimension-one linear sieve is used, let the sieve level be:

$$
D
=
\frac{n}{(\log n)^B},
$$

Then:

$$
\log D
=
(1+o(1))\log n.
$$

If:

$$
z=n^\alpha,
$$

the linear sieve parameter is:

$$
s
=
\frac{\log D}{\log z}
=
\frac1\alpha+o(1).
$$

The standard lower linear sieve function satisfies:

$$
f(s)=0
\quad
(0<s\le2),
$$

and:

$$
f(s)>0
\quad
(s>2).
$$

A positive standard lower-sieve main term therefore requires:

$$
\boxed{
\alpha<\frac12.
}
$$

---

# 17. Sieve-Critical Gap

However, for the roughness ladder to yield a $P_3$ automatically:

$$
\boxed{
\alpha>\frac12.
}
$$

Thus:

$$
\boxed{
\alpha=\frac12
}
$$

separates the two requirements exactly:

$$
\text{standard lower-sieve positivity}
\quad
\alpha<\frac12,
$$

$$
\text{automatic }P_3\text{ residual structure}
\quad
\alpha>\frac12.
$$

This is the **sieve-critical gap** of this problem in these scale coordinates.

For semiprimes:

$$
\alpha>\frac23,
$$

the gap is even larger.

---

# 18. The Meaning of This Gap

This is not merely an unfortunate choice of parameters.

It is a direct projection of the classical parity / sifting barrier into this square base space.

Thus, any strategy relying only on:

$$
\boxed{
\text{‘Sieving deeply enough’}
}
$$

will encounter a critical gap between ‘guaranteeing a survivor’ and ‘guaranteeing extremely low multiplicative order of that survivor’.

To cross it, one must add:

- Weighted sieves;
- Bilinear structure;
- Parity-breaking information;
- Fitting non-alignment;
- Spectral / residue structure;
- Or other information beyond a simple density sieve.

---

# 19. Finite Machine Audit: Horizon

This round checks:

$$
2\le n\le600
$$

all $M6^*$ composite candidates in this range exhaustively.

The number of Obstruction Horizon Theorem failures is:

$$
\boxed{
0.
}
$$

Additionally, for:

$$
20\le n\le5000
$$

the V3 rank-1 candidates in this range:

- Number of composites: 3929;
- Horizon violations: 0;
- Median width of the highest-order top strip automatically excluded: about 50;
- Median:

$$
\frac{B(n,k)}n
\approx
0.980070.
$$

The finite data are therefore consistent with:

$$
n-B(n,k)
\asymp
\sqrt n
$$

this relationship.

---

# 20. Finite Machine Audit: Roughness Ladder

This round uses the V3 center as the position score, first excluding candidate factors:

$$
p\le n^\alpha
$$

and then selecting the nearest candidate.

| $\alpha$ | rank-1 prime rate | residual composite max $\Omega$ | median $\Omega$ |
|---:|---:|---:|---:|
| 0.40 | 40.594% | 4 | 2 |
| 0.49 | 47.440% | 3 | 2.0 |
| 0.51 | 48.805% | 3 | 2.0 |
| 0.60 | 56.394% | 3 | 2.0 |
| 0.67 | 61.694% | 2 | 2.0 |
| 0.70 | 63.883% | 2 | 2 |
| 0.75 | 68.239% | 2 | 2.0 |

The most important finite observations are:

- After $\alpha$ crosses $1/2$, the maximum residual-composite $\Omega$ falls to $3$;
- After $\alpha$ crosses approximately $2/3$, all residual composites in the finite sample have only $\Omega=2$.

This is consistent with the theoretical roughness-ladder thresholds.

However, the rank-1 prime rate itself is only finite empirical evidence, not an existence theorem.

---

# 21. The High-Obstruction Domain Actually Reduced after Phase 6

The Phase 5 high-order domain was originally:

$$
z(n)<a\le n.
$$

Now, if the candidate offset satisfies:

$$
k<n,
$$

it can be reduced directly to:

$$
\boxed{
z(n)<a\le B(n,k).
}
$$

For V3 left-boundary candidates:

$$
\boxed{
z(n)
<
a
\le
n-\sqrt n+O(1).
}
$$

The topmost atomic region, of width approximately:

$$
\sqrt n
$$

is automatically eliminated by position geometry.

---

# 22. The New High-Obstruction Residue Field

Define:

$$
\boxed{
\mathcal O_{\mathrm{hor}}
=
\left\{
(n,a):
z(n)<a\le B(n,\kappa_{\mathcal F}(n))
\right\}.
}
$$

The residue non-hit condition actually remaining to be proved is only:

$$
\boxed{
n^2+\kappa_{\mathcal F}(n)
\not\equiv0\pmod a
}
$$

for:

$$
(n,a)\in\mathcal O_{\mathrm{hor}}.
$$

This is strictly smaller than Phase 5's:

$$
z(n)<a\le n
$$

domain.

---

# 23. The Remaining Obligation in Shell Form

Equivalently, instead of directly quantifying over $a$, quantify over:

$$
\delta\ge1.
$$

We need to exclude:

$$
\boxed{
D_\delta
\left(
n,\kappa_{\mathcal F}(n)
\right)
}
$$

every case in which this is an admissible perfect square.

The remaining composite obstruction can therefore be written as:

$$
\boxed{
\exists\delta:
D_\delta(n,\kappa_{\mathcal F}(n))
=
r^2.
}
$$

Prime-fit closure is:

$$
\boxed{
D_\delta(n,\kappa_{\mathcal F}(n))
\neq
r^2
}
$$

for all admissible shells.

---

# 24. What Should Be Tackled in the Next Phase?

Phase 7 should not further increase presieving depth.

What actually needs testing is:

> Does a V3 / fit-derived offset $\kappa_{\mathcal F}(n)$ have a provable non-alignment with the discriminant-square shells?

That is, study:

$$
\boxed{
\delta^2
+
4n\delta
-
4\kappa_{\mathcal F}(n)
}
$$

the possibility that it is a perfect square.

Only if this can be converted into:

- Finitely many shells;
- Controllable shell density;
- Incompatible congruence families;
- Or a global discriminant gap;

is there a chance of genuinely crossing the classical sieve parity barrier.

---

# 25. Research Boundaries

As of 2026, Legendre's conjecture remains unsolved.

The closest proved unconditional result preserving the ‘every consecutive-square interval’ quantifier currently gives a $P_3$ almost-prime in every cell; Campbell 2026 improves the $P_4$ result of Dudek–Johnston.

Thus, this paper's:

$$
\text{obstruction horizon},
$$

$$
\text{factor shells},
$$

$$
\text{sieve-critical gap}
$$

should be understood as research reformulation and compression of proof obligations, not as having crossed the prime-existence barrier.

---

# 26. References

1. Peter J. Campbell, *On the Existence of Integers with at Most 3 Prime Factors Between Every Pair of Consecutive Squares*, arXiv:2603.10356, 2026; Journal of Number Theory, DOI: 10.1016/j.jnt.2026.07.014.
2. Adrian W. Dudek and Daniel R. Johnston, *Almost Primes Between All Squares*, Journal of Number Theory 278 (2026), 726–745; DOI: 10.1016/j.jnt.2025.05.009.
3. J. Friedlander and H. Iwaniec, *Opera de Cribro*, AMS Colloquium Publications 57, 2010.
4. Standard Rosser–Iwaniec linear sieve: lower sieve function $f(s)=0$ for $s\le2$ and $f(s)>0$ for $s>2$.

---

# 27. Phase 6 Status

```text
Legendre Base-Space Program
Phase 6 — Dynamic Obstruction Horizon and Factor Shells

Factor-pair square-crossing coordinates:   DEFINED
Factor-shell identity:                     PROVED
Candidate-dependent obstruction horizon:  PROVED
Left-edge horizon asymptotic:              DERIVED
High-top obstruction strip exclusion:     PROVED
Discriminant shell:                        DEFINED
Discriminant-square obstruction:           PROVED
Route-4 residue/shell equivalence:         DERIVED
V3 position -> obstruction shrinkage:      DERIVED
Roughness ladder:                          PROVED ASYMPTOTICALLY
P3 threshold alpha > 1/2:                  DERIVED
P2 threshold alpha > 2/3:                  DERIVED
Linear-sieve positivity alpha < 1/2:       IDENTIFIED
Sieve-critical gap at alpha = 1/2:         IDENTIFIED
Finite horizon audit:                      PASS
Finite roughness audit:                    CONSISTENT
Legendre proof status:                     OPEN
Next phase:                                DISCRIMINANT-SHELL NON-ALIGNMENT
```