Eligibility lemma

(a) For a positive integer \(n\) and prime \(p\),

Indeed, under \(0\leq a_p \[ t=\frac{n-a_p}{p}=\left\lfloor\frac np\right\rfloor. \]

Thus \(t\geq k\) is equivalent to \(kp\leq n\). The verifier exhaustively recomputes this identity for \(1\leq k\leq6\), all primes \(p\leq97\), and \(0\leq n<500\).

Finite feasibility and the discarded-prefix function

(a) Define \(F_k(A,B)\) to mean:

> there are least residues \(a_p\) such that every \(n\in[A,B]\) matches \(a_p\bmod p\) for some prime \(p\leq n/k\).

Only primes \(p\leq B/k\) occur, so this is a finite constraint problem.

For \(B\) large enough that a one-point cover exists, define

Then \(H_k(B)\) is nondecreasing in \(B\): a cover through \(B+1\), restricted to the earlier integers and earlier eligible primes, is a cover through \(B\).

Exact compactness reduction

(a) For a fixed \(k\), the strong conclusion of #279 holds if and only if \(H_k(B)\) is bounded as \(B\to\infty\).

Proof:

• If one assignment covers every \(n\geq A\), then \(F_k(A,B)\) holds for every \(B\), hence \(H_k(B)\leq A\).
• Conversely, suppose \(H_k(B)\leq A\) for all \(B\). A witness for \(F_k(H_k(B),B)\) also witnesses \(F_k(A,B)\). Form a tree whose level \(B\) consists of assignments to primes \(p\leq B/k\) that cover \([A,B]\), with restriction as the parent map. Every level is finite and nonempty. Kőnig's infinity lemma supplies an infinite compatible path, i.e. one residue for every prime covering all \(n\geq A\).

Thus the exact missing finiteness statement in the first open case is

The original “for every \(k\)” question asks for this boundedness for every fixed \(k\geq3\).

This reduction is stronger and more precise than saying merely that long finite windows exist: the left endpoint must remain bounded.

3. CRT shadow and the prime-gap connection

Finite CRT equivalence

(a) Given residues for all primes \(p\leq B/k\), the Chinese remainder theorem gives an integer \(R\), unique modulo

such that

Consequently

Therefore \(F_k(A,B)\) is equivalent to the existence of an \(R\) such that every \(n\in[A,B]\) has

The bound \(p\leq n/k\) is triangular: the set of allowed primes grows with the index \(n\). Removing that triangular restriction recovers the standard finite prime-gap/Jacobsthal covering problem.

What the published finite-window theorem gives

(b), modulo FGKMT (2018), equation (1.2). If residues modulo \(p\leq x\) cover \([1,Y(x)]\), then those same residues are valid for #279 on

because every covering prime satisfies \(p\leq x\leq n/k\). Hence

Inverting (FGKMT) gives, for every fixed \(k\),

For completeness, choose

Then \(\log_jx\sim\log_jB\) for \(j=1,2,3\), so (FGKMT) gives \(Y(x)\geq B\) when \(C\) is sufficiently large. Equation (3.2) gives (3.3).

This is genuine uniform finite progress: \(H_k(B)=o(B)\). It is still far from the required \(H_k(B)=O(1)\).

Why longer prime gaps alone do not close the problem

(a), as a logical diagnosis. Any theorem only of the form “primes \(p\leq x\) cover a window of length \(Y(x)\)” validates the threshold \(t\geq k\) only after index \(kx\). Its certified left endpoint therefore moves to infinity with \(x\). Making \(Y(x)\) much larger improves the ratio \(H_k(B)/B\), but it does not keep \(H_k(B)\) bounded.

What is needed is a triangular/nested covering theorem that controls every index from one fixed \(A\), or equivalently a direct proof of (2.3). This is the precise uniformity step absent from the published finite-window machinery.

4. Random baseline: why independent residues do not solve the strong problem

(b), modulo Mertens' product theorem. If each \(a_p\) is independently uniform, then for fixed \(n\)

Thus the expected number of misses in a dyadic interval of length comparable to \(B\) is of order \(B/\log B\), not \(o(1)\). Independent random choices therefore do not directly produce a finite-exception cover.

This does not prove that a highly correlated or adaptive assignment cannot work. It only identifies why a naive Borel–Cantelli argument has the wrong summability and why the high-probability/nested strengthening mentioned in the public comment would be a genuinely new ingredient.

5. Repair of the listed density-one result

The public note's Theorem 7 has the endpoint gap identified in comment 10. The following deterministic argument avoids random blocks entirely.

Greedy periodic-density theorem

(a). Let \(A=\{m_1,m_2,\ldots\}\subseteq\mathbb N_{\geq2}\) consist of distinct moduli and satisfy

For every fixed integer \(k\geq1\), there are least residues \(a_m\bmod m\) such that the set of positive integers not representable as

has natural density zero.

Proof. Let \(S_0=\mathbb Z\). Having selected \(a_{m_1},\ldots,a_{m_{j-1}}\), let \(S_{j-1}\) be the integers avoiding all selected classes. It is periodic, so its density \(\delta_{j-1}\) exists. For \(0\leq r \[ d_r=d\bigl(S_{j-1}\cap\{n:n\equiv r\pmod{m_j}\}\bigr). \]

Since the \(m_j\) residue classes partition the integers,

Choose \(a_{m_j}=r\) with \(d_r\geq\delta_{j-1}/m_j\). The new survivor set has density

Inductively,

Let \(E_k\) be the final threshold-\(k\) exceptional set. For fixed \(j\), once

every match with one of the first \(j\) residues has quotient at least \(k\). Hence, apart from a finite initial segment,

Therefore \(\overline d(E_k)\leq\delta_j\) for every \(j\). Letting \(j\to\infty\) in (5.1) gives \(\overline d(E_k)=0\), hence \(E_k\) has natural density zero. ∎

This proves the live page's almost-all statement and repairs the linked note's proof. It does not upgrade density zero to finitely many exceptions.

The standalone verifier exhausts complete common periods for both coprime moduli \((2,3,5,7,11)\) and non-coprime moduli \((4,6,9,10,14)\), checking every greedy density inequality as an exact rational inequality.

6. Exact finite computation for \(k=3\)

CNF encoding

(a). For a finite interval \([A,B]\), create a Boolean variable \(x_{p,r}\) for every relevant prime/residue pair. Its meaning is “choose residue \(r\bmod p\).”

• For every prime \(p\), add

\[ \neg x_{p,r}\vee\neg x_{p,s}\qquad(r\neq s) \]

so at most one residue is selected.

• For every \(n\in[A,B]\), add

\[ \bigvee_{\substack{p\ \mathrm{prime}\\3p\leq n}}x_{p,n\bmod p}. \tag{6.1} \]

It is harmless not to require a residue for a prime unused by the model; an arbitrary residue can be filled in afterward. Thus this CNF is satisfiable exactly when \(F_3(A,B)\) holds.

An independent CP-SAT encoding instead gives every prime an integer variable \(a_p\in[0,p-1]\) and reifies the equalities \(a_p=n\bmod p\). Both encodings are rebuilt from first principles by the verifier.

Verified table

(d). The following are exact finite values. For each row the program found and directly checked a model for \([H_3(B),B]\), and proved the immediately preceding start \([H_3(B)-1,B]\) infeasible.

| \(B\) | \(H_3(B)\) | covered length \(B-H_3(B)+1\) |

|---:|---:|---:|

| 12 | 10 | 3 |

| 24 | 16 | 9 |

| 36 | 26 | 11 |

| 48 | 28 | 21 |

| 60 | 34 | 27 |

| 72 | 40 | 33 |

| 84 | 46 | 39 |

| 96 | 52 | 45 |

| 108 | 58 | 51 |

| 120 | 62 | 59 |

| 132 | 64 | 69 |

| 144 | 68 | 77 |

| 156 | 68 | 89 |

| 168 | 76 | 93 |

| 172 | 76 | 97 |

The last row follows from the explicit \([76,172]\) certificate below and the independently checked infeasibility of \([75,168]\).

Explicit 97-integer certificate

(d), with direct arithmetic checking. The following least residues cover every integer \(76\leq n\leq172\) with \(t\geq3\):

These are all primes \(p\leq172/3\). For primes not displayed, choose any residue; they are irrelevant to this finite certificate.

Examples at the tight threshold include

The verifier directly checks a valid \((p,t)\) for each of the 97 integers.

Its CRT shadow is

where the modulus is \(53\#\). The script independently checks, for every displayed prime and for \(0\leq n\leq500\), that \(n\equiv a_p\pmod p\) if and only if \(p\mid R+n\).

Exact local maximality

(d).

• \([75,168]\) is infeasible: 316 Boolean residue variables and 4382 direct pairwise CNF clauses.
• \([76,173]\) is infeasible: 329 variables and 4719 clauses.
• The displayed assignment covers \([76,172]\).

Therefore

The two infeasibility results were independently reproduced by:

1. CaDiCaL 1.9.5 on the direct Boolean encoding;

2. MiniSat 2.2 on the same freshly generated clauses;

3. OR-Tools CP-SAT on the separate integer-residue encoding, single-threaded.

The full clean run took 52.79 seconds. CP-SAT reported, respectively, 418,521 branches/202,732 conflicts and 573,511 branches/269,906 conflicts. Because no independently checked DRAT/LRAT certificate is shipped, (6.2) remains correctly labelled computational-only rather than elementary-rigorous.

7. Standalone reproduction

Complete code is in:

SHA-256:

Run the full independent audit with:

cd /home/exedev/MathDyad
python runs/erdos279_wave5p_reverify.py

The program:

1. generates primes by its own sieve;

2. exhaustively checks the least-residue eligibility identity;

3. checks all 97 integers in the displayed certificate;

4. recomputes its CRT shadow;

5. rebuilds every SAT instance in the table;

6. checks both sides of every claimed \(H_3(B)\) value;

7. reruns the two frontier infeasibility instances through two Boolean solvers and one independently encoded integer solver;

8. exhausts finite periodic examples of the density-descent lemma.

Observed final line:

ALL CHECKS PASSED in 52.79 seconds

8. Precise wall

The infinite problem is not solved here.

• (a) The exact remaining statement for the first open case is \(H_3(B)=O(1)\).
• (b) Published prime-gap machinery gives the quantitative but weaker

\[ H_3(B)\ll B\frac{\log_2 B}{\log B\,\log_3 B}. \]

• (d) The exact finite frontier reaches \(H_3(172)=76\), with a locally maximal 97-integer certificate.
• (a) No computation at any single finite \(B\), however large, can supply the Kőnig/compactness uniformity required for all \(B\).
• (c) A robust nested-covering or high-probability extension theorem might bridge this gap, as suggested in the public comment, but no such theorem was located or proved.

Heavier SAT computation is therefore not the missing mathematical step. It could extend the table, but it cannot certify boundedness of \(H_3\). The pilot already became irregular near \(B\approx190\) (one 25-second CP-SAT probe returned UNKNOWN); extrapolating this encoding to \(B=1000\) would likely require at least many core-hours and would still be only finite evidence, so that computation was not run.

PARTIAL: Rigorous reduction to boundedness of H_k, FGKMT-based sublinear bound, repaired density-one proof, and independently verified exact frontier H_3(172)=76; the required uniform O(1) bound remains open.