Examination of the exceptional-width construction for omega

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Subject and independence

This is an independent whole-conclusion mathematical examination for Plasma AI, dated 2026-10-14 UTC. The reviewer was neither the author of nor a collaborator on the construction, had no prior exposure to it before this examination, and worked from a fresh review context charged to find a counterexample, missing hypothesis, unsupported inference, or quantifier error. The disclosed reviewer model was GPT-6-Astra. This report makes mathematical findings; it does not grade itself or assign verification standing.

The mathematical subject was frozen on 2026-10-14 at 17:27:31.684793 UTC. Its five exact extractions are preserved in wiki/research/leads/exceptional_widths_shifted_omega/evidence/assets/width_subject/. Every filename in the first column below denotes a file in that folder. The public manifest records the mapping and the uniform [cut] marker.

These paths identify the assessed snapshots, not a later version of a problem page, library page, or author account. The input identities were checked before reading and again when completing this report. The mathematical read set was exactly these five extractions and their supplied manifest. The full argument, statement, and dependency records in that set were read. The cut markers were not treated as information about omitted text.

Operating reading comprised the supplied charter, effective settings and steps; the repository README; the Claims, Evidence, Tiers, and Project claims sections of docs/anatomy.md; all of docs/verification.md, docs/evidence.md, docs/math_authoring.md, docs/weaving.md, campaign/AGENTS.md, and campaign/guides/verification.md; the Reading section of campaign/guides/research.md; and the permitted native operating instructions. Those instructions did not expand the mathematical read set.

The subscription checks were empty, receiving channels remained sealed, and automatic synchronization was disabled. No author working context, prior examination, current ledger, sibling material, message inbox, source-link target, or web search was read. No excluded communication or search exposure occurred. The reviewer used only this examination's own preparation notes between its stages. The recorded source reading depth is a dependency input, not an earlier verdict on this construction.

The independent mathematical evidence is the derivations and attacks in this report. No computation supports the verdict, so there is no mathematical program or computational rerun command. A reader can reconstruct the assessed argument from the five retained inputs and reproduce every calculation below without a private archive or additional source. The external theorem boundary is stated explicitly in Premises.

Restatement and interpretation

Write $\mathbb N_{>0}=\{1,2,\ldots\}$ and let $\omega(n)$ count distinct prime divisors, with $\omega(1)=0$. For a positive width $F$ and $x\ge1$, define

$$R_F(x)=\frac{1}{F(x)} \#\{n\in\mathbb N_{>0}:x<n+\omega(n)<x+F(x)\}.$$

The construction claims the existence of a positive, in fact integer-valued, function $F:[1,\infty)\to(0,\infty)$ and a set $S\subseteq\mathbb N_{>0}$ such that

$$\lim_{x\to\infty}F(x)=\infty, \qquad \lim_{N\to\infty}\frac{|S\cap[1,N]|}{N}=1, \qquad \lim_{\substack{n\to\infty\\n\in S}} \frac{F(n)}{\omega(n)}=0,$$

while $R_F(x)\le7$ for every sufficiently large real $x$. The set $S$ is fixed; it is not chosen anew for each error tolerance. The first limit is at all real locations, whereas the smallness assertion is only along $S$ at the integers. Counts are of inputs $n$, including different inputs with the same shifted value, and both interval endpoints are open.

Under the supplied interpretation, the modified target for $f=\omega$ asks whether every admissible $F$ admits a sequence $x_k\to\infty$ with $R_F(x_k)\to\infty$. A single admissible $F$ bounded as above negates that universal assertion. The frozen text writes $F(n)$ in the hypothesis and $F(x)$ in the conclusion without declaring the domain of $x$. The construction covers both the integer-location reading, using the restriction of $F$, and the real-location reading, using its specified floor extension. It uses the commissioned natural-density-one meaning of “almost all”; it does not claim the cofinite or everywhere-small meaning. No monotonicity is assumed.

The supplied interpretation turns the phrase about infinitely many $x$ into a limit along a sequence. In fact the stated choices on the initial intervals also give a global bound for $x\ge1$, as shown below, so a sequence confined to a bounded subset of this domain cannot evade the counterexample either.

Only the negative answer for $\omega$ under these conventions is at issue. There is no classification of other functions, conclusion for $\tau$, historical novelty finding, or resolution of a uniform-width or monotone-width variant. The literal assertion in the final formulation remark about excluding this particular $F$ is examined separately below.

Premises and source boundary

The dependency census was derived from the frozen proof and its consequences. It contains no native L-claim. No live claim-standing read was needed or authorized. On the examination date, the premise interface assessed was the one preserved in the five inputs; no ledger standing was inferred from a source-owned theorem or its reading-depth label.

The non-elementary estimate used is Lemma 1 of Erdős, Pomerance and Sárközy, On locally repeated values of certain arithmetic functions, IV, The Ramanujan Journal 1 (1997), 227–241, statement on printed pp. 229–230, DOI 10.1023/A:1009723712317. The library statement and supplied transcription excerpt agree in their hypotheses, definitions, strict inequality, and absolute-constant assertion. Their typography does not change any consumed clause.

Consequently the imported interface is

$$\sum_{1\le n\le X}(\omega(n)-A(X))^2 \le C X(A(X)+1)\qquad(X\ge2). \tag{1}$$

The supplied strict inequality implies this weak inequality in the consumed range. The review relies on (1) as the identified published external premise; it does not provide an independent proof of EPS97 Lemma 1. The supplied reading-depth record labels its coverage “Claims checked”: the statement and remarks were checked against printed pages, and the proof on pp. 230–232 was read but not checked step by step. It records no independent review. That is the record's coverage, not this reviewer's. This reviewer read only the two supplied statement forms and that record, with no PDF inspection and no proof inspection. No local independent certification of the source proof is asserted.

There is an unused degeneracy in the general statement as supplied. At $x=2$, $m=1$, $h=0$, and $f\equiv0$, all its stated hypotheses hold, but $K=A=0$ and its strict conclusion reads $0<0$. Thus that unrestricted strict statement, interpreted literally, is refuted-as-stated by this exact witness. Also, the displayed maximum has an empty index set for $1\le x<2$ unless a convention is added. Neither issue is used by the construction: $X\ge2$, $K=1$, and $A(X)\ge1/2$ throughout. The verdict about the construction consumes only (1), not the unused zero-function or empty-maximum cases. This observation concerns the supplied text; it does not claim a fresh inspection of the printed source or a replacement proof of its positive-$K$ instance.

The other explicit external premise is Euler's theorem

$$\sum_{p\le X}\frac1p\longrightarrow\infty.$$

It is used only to make $A_j$, and hence $H_j$, tend to infinity and to make the ratios in the next section vanish. No asymptotic formula for $A(X)$, prime number theorem, or probabilistic normal-order theorem is imported. The bound $A(X)\le1+\log X$ follows by comparison with the harmonic sum. Euler's divergence is relied upon as the expressly permitted classical theorem, not reproved here. The further finite-product inequality used solely to attack the formulation remark is derived in full below.

Weakest steps

The following three reconstructions fill in routine inequalities, counts and limit quantifiers in the frozen argument. They do not replace its strategy or repair an essential missing lemma. The additional lemma missing from the literal final remark is kept separate in Formulation remark.

1. From the external moment estimate to a usable dyadic scale

Put $I_j=[2^j,2^{j+1})\cap\mathbb N_{>0}$ and use the frozen definitions

$$X_j=2^{j+2}=4\cdot2^j,\qquad A_j=A(X_j),\qquad H_j=\lceil A_j^{3/4}\rceil.$$

Euler's divergence gives $A_j\to\infty$. The ceiling satisfies

$$A_j^{3/4}\le H_j\le A_j^{3/4}+1, \qquad \frac{H_j}{A_j}\le A_j^{-1/4}+A_j^{-1}\longrightarrow0. \tag{2}$$

Also $A_j\le1+(j+2)\log2$, so $H_j+1=o(2^j)$. Choose a single integer $j_0\ge1$ sufficiently large that, for every $j\ge j_0$,

$$H_j\ge1,\qquad H_j\le A_j/2,\qquad 2^{j+1}+H_j+1<2^{j+2}. \tag{3}$$

Only existence of this finite cutoff is needed. No numerical cutoff or unproved effective rate is being claimed.

For the frozen set $D_j=\{n\in\mathbb N_{>0}:n\le X_j,\ |\omega(n)-A_j|>H_j\}$, each member contributes more than $H_j^2$ to the sum in (1). Therefore

$$\frac{|D_j|}{2^j} \le4C\frac{A_j+1}{H_j^2} \le4C(A_j^{-1/2}+A_j^{-3/2}) \longrightarrow0. \tag{4}$$

This is a deterministic consequence of (1), not a sampling or independence argument. The constant is fixed as $j$ varies. In particular, (4) is strong enough for both exceptional-set uses; neither requires a summable sequence of error proportions.

2. From sparse bad inputs to a bound at every location

For each $u\in I_j$, let $b_j(u)$ count inputs $n\in D_j$ with $u<n+\omega(n)<u+H_j+1$, exactly as in the frozen text. For a fixed input let $m=n+\omega(n)\in\mathbb Z$. The possible integer left endpoints satisfy

$$u<m<u+H_j+1 \quad\Longleftrightarrow\quad m-H_j\le u\le m-1.$$

There are exactly $H_j$ such integers before restricting to $I_j$, so the frozen upper bound of $H_j+1$ is safe. This counts each input separately; collisions of shifted values do not invalidate the incidence count. Thus

$$\sum_{u\in I_j}b_j(u)\le(H_j+1)|D_j|. \tag{5}$$

For $E_j=\{u\in I_j:b_j(u)>H_j+1\}$, integrality gives

$$(H_j+2)|E_j| \le\sum_{u\in E_j}b_j(u) \le(H_j+1)|D_j|, \qquad |E_j|\le|D_j|. \tag{6}$$

This remains valid when either set is empty. It imposes no lower bound on $|E_j|$.

Now fix $j\ge j_0$ and $u\in I_j\setminus E_j$. A positive integer counted in $(u,u+H_j+1)$ satisfies

$$n\le n+\omega(n)<u+H_j+1 <2^{j+1}+H_j+1<X_j. \tag{7}$$

Hence no counted input lies beyond the range of the source estimate. If it is not in $D_j$, then $A_j-H_j\le\omega(n)\le A_j+H_j$, and therefore

$$u-A_j-H_j<n<u+2H_j+1-A_j. \tag{8}$$

The interval in (8) has length $3H_j+1$ and contains at most $3H_j+2$ integers, which is the deliberately loose bound used by the frozen proof. The counted inputs in $D_j$ number at most $H_j+1$ because $u\notin E_j$. These two disjoint classes exhaust the count by (7), giving at most $4H_j+3$ inputs. No injectivity or monotonicity of $n+\omega(n)$ is used.

Define the frozen $F$ precisely by using $u=\lfloor x\rfloor$. For $u\in I_j$ with $j\ge j_0$, set $F(x)=H_j$ off $E_j$ and $F(x)=2(u+1)$ on $E_j$. For $1\le u<2^{j_0}$ use $F(x)=2(u+1)$. These cases cover every $x\ge1$ and give positive widths.

Off $E_j$, $u\le x<u+1$ implies

$$(x,x+H_j)\subseteq(u,u+H_j+1), \qquad R_F(x)\le4+\frac3{H_j}\le7. \tag{9}$$

The left endpoint remains open even at $x=u$; the right endpoint inclusion uses the strict inequality $x<u+1$. An exact integer or dyadic boundary is assigned to its new floor cell and satisfies the same argument there.

On the other branch, including all early cells, any counted input satisfies

$$1\le n<x+2(u+1)<3(u+1).$$

There are at most $3(u+1)-1$ such positive integers, whence

$$R_F(x)\le\frac{3(u+1)-1}{2(u+1)}<\frac32. \tag{10}$$

This is why positivity of the input domain and nonnegativity of $\omega$ are essential. There is no appeal to concentration at an exceptional location. Equations (9) and (10) prove the promised uniform bound; with the specified initial branch they in fact give $R_F(x)\le7$ for every $x\ge1$.

3. Density-one admissibility and the exact target implication

Here is the dyadic density argument with its partial-block quantifier. If $T_j\subseteq I_j$ and $|T_j|=o(2^j)$, then for every $\varepsilon>0$ there is $J$ with $|T_j|\le\varepsilon2^j$ for all $j\ge J$. Given $2^k\le N<2^{k+1}$,

$$\left|\left(\bigcup_{j\ge j_0}T_j\right)\cap[1,N]\right| \le C_J+\varepsilon\sum_{j=J}^k2^j \le C_J+2\varepsilon N. \tag{11}$$

The finite constant $C_J$ vanishes after division by $N$; then $\varepsilon$ is arbitrary. No equidistribution inside the last block is needed. In particular an adversary may put all of that block's exceptional points near its beginning without defeating (11).

Apply (11) first to $E_j$, using (4) and (6), and separately to $B_j=D_j\cap I_j$, using (4). These are different uses of the bad-input estimate: the first controls locations with crowded bad images, and the second controls the values of $\omega$ at the integer arguments of $F$. Define

$$E=\bigcup_{j\ge j_0}E_j,\qquad B=\bigcup_{j\ge j_0}B_j,\qquad S=\mathbb N_{>0}\setminus\bigl([1,2^{j_0})\cup E\cup B\bigr).$$

Both $E$ and $B$ have natural density zero, and $S$ has natural density one. For $n\in S\cap I_j$, the definitions and (3) give

$$F(n)=H_j,\qquad \omega(n)\ge A_j-H_j\ge A_j/2>0, \qquad \frac{F(n)}{\omega(n)}\le2\frac{H_j}{A_j}\longrightarrow0. \tag{12}$$

The upper bound is uniform over $n$ in each good part of a block. Given any positive tolerance, (2) supplies one block index beyond which (12) holds for every $n\in S$. This is ordinary convergence along the fixed set $S$, not just convergence of averages. The undefined ratio at $n=1$ is removed by a finite exclusion and is never used.

Finally, both possible widths tend to infinity as their locations do. Indeed $j\to\infty$ when $x\to\infty$, $H_j\to\infty$, and $2(\lfloor x\rfloor+1)\to\infty$. For any prescribed lower bound, choose a single tail on which both branches exceed it. This proves the everywhere real limit of $F$ without discarding $E$.

Thus one and the same $F$ satisfies all the admissibility quantifiers and obstructs the target conclusion at every location in its domain. In particular there is no sequence of real or integer locations tending to infinity along which $R_F$ diverges. This is exactly the required negation for $f=\omega$; it makes no assertion about the rest of the classification question.

Strongest attacks and findings

The strongest attack on the core conclusion was to combine two possible losses: a density-zero set of bad inputs might produce very crowded images at many locations, and counted inputs might lie beyond $X_j$ or cross a floor boundary. Merely removing the bad inputs would not prove an everywhere bound. Equations (5)–(7) show why that attack fails here: each bad input has bounded incidence with integer windows, the exceptional location set absorbs every window exceeding the threshold, and nonnegativity forces all remaining counted inputs inside the controlled range. At the absorbed locations the width changes and the independent positive-input bound (10) applies. No exceptional location is dropped from the final estimate.

A separate attack concentrated every exceptional set at the start of its dyadic block and chose $N$ just beyond that concentration. Equation (11) still bounds its contribution by $2\varepsilon N$ plus a fixed finite term. This defeats a possible sparse-block-to-natural-density error. Applying it separately to $B_j$ prevents the mistake of using sparsity of output locations as if it implied typicality of $\omega(u)$ at those locations.

The endpoint attack tested $x=u$, $x$ immediately below $u+1$, and the exact boundary $x=2^{j+1}$, allowing many inputs to share an image. The incidence equivalence and (8)–(10) preserve the open endpoints and multiplicities in all these cases. The extra unit in the integer window is enough for the entire real cell. None of these attacks refuted the main construction.

Formulation remark: a literal exclusion needs an additional lemma

The final sentence of argument.md, lines 104–106, asserts that the added condition $F(n)=o(\log\log n)$ everywhere “would prohibit this construction,” then says the argument does not decide that variant. The second statement is an appropriate scope limit. The first has a stronger literal reading: the particular constructed $F$ fails that added condition. That reading is not established by the frozen proof.

The exact missing interface is the infinitude of $E$. This is not just an objection to omitted wording. For this $F$, one can identify the issue exactly:

$$F(n)=o(\log\log n)\text{ as }n\to\infty\text{ through all integers} \quad\Longleftrightarrow\quad E\text{ is finite}. \tag{13}$$

Here is an elementary diagnostic derivation of (13), independent of any unstated asymptotic theorem. For $Y\ge e$, put $s=1+1/\log Y$. For $p\le Y$ we have $p^{s-1}\le e$, and positivity and unique factorization give

$$\begin{aligned} A(Y) &\le e\sum_{p\le Y}p^{-s} \le e\log\prod_{p\le Y}(1-p^{-s})^{-1}\\ &\le e\log\sum_{n=1}^{\infty}n^{-s} \le e\log\left(1+\frac1{s-1}\right) =e\log(1+\log Y). \end{aligned} \tag{14}$$

The finite product expands into the sum over integers whose prime factors belong to its finite prime set; this is a subseries of the displayed positive series. The final series bound is the integral comparison for $t^{-s}$. Thus no Euler product theorem or prime asymptotic is being cited in (14).

For $n\in I_j\setminus E_j$ this gives $F(n)=H_j=O((\log(j+2))^{3/4})$, whereas $\log\log n\ge\log(j\log2)$. Therefore $F(n)/\log\log n\to0$ uniformly off $E$. On an unbounded sequence in $E$, however, $F(n)=2(n+1)$ and this quotient tends to infinity. Finite early choices affect neither conclusion. This proves (13).

The frozen estimates (4)–(6) are all upper bounds. They permit an empty $E_j$ and supply no lower bound forcing $E$ to be infinite. The literal exclusion therefore needs, for example, the additional assertion

$$\forall J\ \exists j\ge J\ \exists u\in I_j: b_j(u)>H_j+1. \tag{15}$$

No proof of (15) is supplied. This examination has not proved (15), disproved it, or introduced another theorem to settle it. An empty-exceptional-set possibility is a witness to the missing inference, not a counterexample asserted for the actual arithmetic function $\omega$.

Accordingly the literal assertion that the constructed $F$ is excluded by the everywhere $o(\log\log n)$ condition has the verdict proof defective; statement unresolved. Read merely as a warning that the current proof has not established that stronger admissibility requirement, the remark is appropriate. Neither reading invalidates (1)–(12): the main counterexample works whether $E$ is finite or infinite. The diagnostic (13)–(14) exposes the missing lemma; it does not repair it or resolve the uniform-width variant.

Checklist

The entries below are mathematical audit observations, not a self-grade of this report. They cover the governing Erdos checklist and explicitly address every canonical mode and named pattern in docs/verification.md.

Erdos checklist

Canonical modes

Named patterns

The evidence-coverage discipline is satisfied by identifying the independent derivations, external theorem boundary, and unresolved ancillary inference. The neighboring discipline about correlated statistical samples is inapplicable because no statistical inference is made. This is one noncomputational review, not a computational batch or a repeated independent cycle; the corresponding extra reproduction and second-cycle requirements are inapplicable.

Verdict and grading

There is therefore no unqualified favorable verdict on every sentence of the packet. The core construction has no unresolved load-bearing deduction in the reviewed argument beyond its declared external theorem reliance; the literal ancillary exclusion has the precise proof gap stated above. No new proof has been substituted for that gap. The source-proof limit, formulation dependence, and the unexamined variants remain explicit.

A separate eligible grader must assess the report's independence, coverage, contract compliance, and the treatment of these distinct surfaces before any standing decision. This report awards no tier, makes no claim of independent certification of the EPS97 proof, and covers no bytes written after the frozen subject.