Independent review of the analytic aliquot-collision results

Subject and independence

Reviewer: independent mathematical reviewer, commissioned model GPT-6-astra, with no authorship of or collaboration on the assessed argument. Examination date: 2026-10-16 UTC. Freeze date: 2026-10-16 UTC. The local mathematics under examination is attributed to Plasma AI.

The native subject is wiki/research/leads/aliquot_collision_density/_index.md, from Target and conventions, through The density-zero form is false and An equivalent form of the problem, ending with the semiprime separation inequality before Exact census. The independently reconstructed supporting countermodel is from wiki/problems/arithmetic_functions/E0955/_index.md, Exact scope of the arithmetic reduction. Only the supplied frozen extractions were examined; the full pages were not opened.

The retained subject is wiki/research/leads/aliquot_collision_density/evidence/assets/analytic_conclusions/subject.md. Its line numbers below refer to that extraction, not to the current native page. The accompanying countermodel.md preserves the supplied countermodel extraction, and source_interfaces.md records the exact mathematical interfaces used from the supplied literature statements, with their provenance and reading limits. The first two are unchanged copies of the supplied extractions; source_interfaces.md is the reviewer's restatement, not an original-source facsimile. All three belong in that same assets directory.

The charge was to find a counterexample, error, unsupported essential inference, or canonical failure mode in every commissioned mathematical conclusion. The finite census, later sieve route, and any assertion of a full solution of E0955 are outside this examination. Condition (D) is reviewed as an antecedent of implications, not as an established bound.

The review began in a fresh context separated from the subject's construction. The allowed mathematical reading consisted of the subject extraction, its source manifest, the PNT, Mertens and Pollack--Pomerance--Thompson excerpts, and the countermodel extraction. All six were read completely. The governing corpus and commissioned operating instructions were also read at their allowed scope. No original mathematical page, source card, native claim ledger, prior review, author working notes, inbox, live website or original PDF was opened. No broad search or upstream update was performed. The supplied subject was kept fixed. An empty subscription table was checked at the initial boundary and again on resuming the examination; this supports channel isolation but is not itself proof of what was read.

Exposure requiring the grader's ruling. General motivational advocacy accompanied the assignment before mathematical reading: it listed unrelated reported AI mathematical achievements and urged ambitious full solutions. That is advocacy of the kind excluded by the verification standard. It was disclosed when the exclusion was identified. It supplied no step of these aliquot arguments, and none of its announcements is used as a premise or as support for a favorable verdict. This disclosure does not adjudicate its materiality. A separate grader must rule on that issue before treating this report as an independent warrant.

This is a noncomputational examination. The independent derivations, exact inequalities, countermodels used as attacks, and findings are contained below. No executable evidence, finite census, statistical sample or cached output supports a verdict, so there is no computational rerun command. The report neither grades itself nor asserts a tier.

Mathematical result: refutation-failed for Theorems A and B and the scoped analytic consequences. The literature-status characterization and the independence exposure retain the qualifications stated below.

Conventions, claim surfaces and external premises

Inputs are positive integers; $s(n)=\sigma(n)-n$ and $s(1)=0$. For real $x\ge1$, $n\le x$ means positive integer inputs. The index m in a fiber sum is a nonnegative integer, and $\sum_m r_x(m)=\lfloor x\rfloor$. Here $r_x(m)=\#\{n\le x:s(n)=m\}$; $r_{S,x}(m)$ restricts that count to $n\in S$, and $R_S(x)=\#(S\cap[1,x])$. The admitted set S may depend on x. The only input of value zero is 1, since every $n>1$ has proper divisor 1. Thus the two conventions about whether zero belongs to the target universe change a preimage count by at most one.

For an admitted set S, the equality $C_S(x)=\sum_m r_{S,x}(m)^2$ counts ordered pairs, including exactly $R_S(x)$ diagonal pairs. Interchanging coordinates pairs up the remaining pairs without fixed points. Hence $C_S-R_S$ and $(C_S-R_S)/2$ have precisely the ordered and unordered meanings stated in the extraction. A finite target B captures at most $\sqrt{|B|C_S(x)}$ admitted inputs, by Cauchy--Schwarz. An infinite sparse target needs the additional output-tail argument below.

The report separates five groups, with every subsidiary consequence recorded:

No native L-claim is consumed as an established premise. This was checked from the frozen statements, proofs and consequence sentences on the examination date. The local arithmetic assertions and countermodel are rederived here; no author's tier is assumed and no live ledger reading was required or made.

The exact external interfaces are as follows. The fuller source identities and formulas are retained in source_interfaces.md in the assets directory named above.

These are identified external theorem interfaces, not local reconstructions of the literature. Ordinary reliance on them is the commissioned boundary. The PNT excerpt uses the conventional logarithmic integral; the singular integral notation is understood as principal value. Changing its finite additive normalization would not change the asymptotic used here. Nothing depends on the other explicit estimates printed in that excerpt.

There is one unsupported characterization: the closed packet supplies no literature-wide evidence for the sentence that (D) is not known to be equivalent. The argument establishes sufficiency and does not establish necessity. That mathematical limitation is checked; the separate assertion about the state of knowledge is not certified. No disproof of equivalence is inferred from absence of its proof here.

A: removal of a density-zero family

Restatement

For every family $E_x\subseteq\{1,\ldots,\lfloor x\rfloor\}$, with no consistency or monotonicity assumed between different x, if $|E_x|/x\to0$, then

$$\frac{C_{[1,x]\setminus E_x}(x)}x\longrightarrow+\infty.$$

The conclusion is eventual divergence for each such family, not a bound uniform in all families' starting points or a prescribed rate of divergence.

Weakest steps, independently rederived

A1. The multiplicative class really has small outputs. Fix $0<\delta\le1/10$, put $z=1/\delta$, $Z=100z$, and let $P_Z=\prod_{q\le Z}q$. For a positive integer $k$ coprime to $P_Z$, with $f(k)=\sum_{q\mid k}1/(q-1)\le\delta/10$, unique factorization gives

$$\frac{\sigma(k)}k \le\prod_{q\mid k}(1-1/q)^{-1} \le\exp\!\left(\sum_{q\mid k}\frac1{q-1}\right).$$

The logarithmic bound follows from $-\log(1-u)\le u/(1-u)$ for $0\le u<1$; prime powers cause no extra factor. For $0\le t\le1/100$, the exponential series gives $e^t\le(1-t)^{-1}$. Consequently $e^{\delta/10}\le1+(10/99)\delta\le1+0.11\delta$. If $z/1.8<p\le z/1.1$, then $1.1\delta\le1/p<1.8\delta$. Since p is smaller than Z and k is Z-rough, p does not divide k. Multiplicativity now yields

$$\delta\le\frac{s(pk)}{pk} \le1.91\delta+0.198\delta^2 \le1.9298\delta<2\delta.$$

The prime p is the only prime divisor of pk at most Z, so two different choices of p cannot produce the same input. The case $k=1$ gives a prime input, with positive value 1, and satisfies the same ratio bounds. In particular no value zero enters the pigeonhole count.

A2. Weighted rough-number counting retains positive mass. Every full interval of length $P_Z$ contains $\varphi(P_Z)$ integers coprime to $P_Z$. Finite inclusion--exclusion gives $\varphi(P_Z)/P_Z=m_Z$. Thus, uniformly for real $t\ge0$,

$$R_Z(t):=\#\{k\le t:(k,P_Z)=1\}=m_Zt+O_Z(1).$$

For a prime $q>Z$, multiplication by q preserves coprimality with $P_Z$. Counting each prime divisor once, including numbers divisible by its higher powers, gives the exact finite identity

$$\sum_{\substack{k\le y\\(k,P_Z)=1}}f(k) =\sum_{Z<q\le y}\frac{R_Z(y/q)}{q-1}.$$

The accumulated residue error is bounded by a constant depending only on Z multiplied by $\sum_{q\le y}1/(q-1)$. The latter is $O(\log\log y)$ from the supplied Mertens statement, since

$$\sum_{q\le y}\frac1{q-1} \le\sum_{q\le y}-\log(1-1/q) +\sum_{m=2}^{\infty}\frac1{m(m-1)} =\log(1/m_y)+1.$$

The main term is at most $m_Zy\sum_{m>Z}1/[m(m-1)]=m_Zy/\lfloor Z\rfloor$. For real $Z\ge1000$, $Z/\lfloor Z\rfloor\le1000/999<1.01$. Markov's inequality at threshold $\delta/10$ therefore excludes at most

$$0.101m_Zy+O_\delta(\log\log y) \le0.11m_Zy+o_\delta(y)$$

rough integers. The retained count is at least $0.89m_Zy-o_\delta(y)$. There are finitely many allowed p for fixed delta. Summing this count at $y=x/p$, using A1's uniqueness, proves exactly the stated lower bound $(d_\delta-o_\delta(1))x$, with

$$d_\delta=0.89m_Z\sum_{5z/9<p\le10z/11}\frac1p.$$

The constants implicit in the error can be enormous as delta decreases. No uniformity in delta was used in this calculation.

A3. The prime-window constant and two successive limits. For fixed $0<a<b$, partial summation gives

$$\sum_{az<p\le bz}\frac1p =\frac{\pi(bz)}{bz}-\frac{\pi(az)}{az} +\int_{az}^{bz}\frac{\pi(t)}{t^2}\,dt.$$

The PNT approximation is uniform on this fixed-ratio interval as $z\to\infty$, simply because its relative error is eventually small for every $t\ge az$. The endpoint difference is $o(1/\log z)$, whereas the integral is $(\log(b/a)+o(1))/\log z$. Taking $(a,b)=(5/9,10/11)$ gives the coefficient $\log(18/11)$. Also $m_{100z}\sim e^{-\gamma}/\log z$ by Mertens, since $\log(100z)/\log z\to1$. Hence

$$d_\delta\log^2(1/\delta) \longrightarrow0.89e^{-\gamma}\log(18/11)>0.24.$$

For completeness, the numerical strict inequality needs no floating-point calculation. Let $H_n=\sum_{j\le n}1/j$. From the definition of gamma,

$$H_n-\log n-\gamma =\sum_{k=n}^{\infty}\int_k^{k+1} \left(\frac1t-\frac1{k+1}\right)dt >\frac1{2(n+1)}.$$

For the last inequality, each integral is at least $1/[2(k+1)^2]$, and the sum of these lower bounds exceeds $\tfrac12\int_{n+1}^{\infty}t^{-2}dt$.

The four-term positive series for $2\operatorname{artanh}(2/3)=\log5$ is $122492/76545>8/5$. Taking $n=5$ in the preceding inequality gives $\gamma<11/5-\log5<3/5$. The alternating exponential series through degree 5 gives $e^{-3/5}>0.548752>5487/10000$. Also

$$\log(18/11)>2\left(\frac7{29}+\frac{(7/29)^3}{3}\right) =\frac{36008}{73167}>\frac{123}{250}.$$

Thus the coefficient exceeds $ (89/100)(5487/10000)(123/250)=60066189/250000000>0.24$. In particular $d_\delta\ge0.2/\log^2(1/\delta)$ for all sufficiently small delta; no effective cutoff or all-delta version follows from this step.

For any one such fixed delta, removal of $o(x)$ inputs leaves at least $(d_\delta-o(1))x$ members of $V_\delta$. Their values are positive integers at most $2\delta x$. There are at most $\lfloor2\delta x\rfloor$ such values, so Cauchy--Schwarz gives

$$\liminf_{x\to\infty}\frac{C_{[1,x]\setminus E_x}(x)}x \ge\frac{d_\delta^2}{2\delta} \ge\frac{0.02}{\delta\log^4(1/\delta)}.$$

This holds for every sufficiently small fixed delta. Given any finite M, choose delta making the last expression larger than $2M$, then take x large for that delta and the given exclusion family. Eventually the collision ratio exceeds M. This proves actual divergence, without interchanging two limits.

Strongest attacks and findings

The main attempted failure was to let the rough-number residue error grow with delta while an adaptive exclusion follows $V_{\delta(x)}$. The proof defeats that attack by retaining every fixed positive-density class eventually; the residue error is sublinear at each fixed delta. It never takes a simultaneous limit with delta depending on x.

An explicit boundary test shows why the asymptotic qualification matters. At $\delta=1/10$, the prime window contains only 7, and $d_{1/10}=0.89m_{1000}/7\le89/2625$. But $\log10<12/5$ (the first six positive exponential-series terms at $12/5$ exceed 10), so $0.2/\log^2 10>5/144>89/2625$. Thus the tempting bound for every delta in the displayed interval is false. The frozen proof expressly excludes that reading, and the counterexample does not refute it. Prime powers, the prime case $k=1$, real cutoffs, and the zero output convention were also checked above.

Premises: the exact PNT and Mertens interfaces and the elementary inequalities rederived in A1--A3. Mathematical verdict: refutation-failed. The checklist below applies separately to A and its consequences.

A-epsilon: density order and quantitative cost of exclusions

Restatement

For sufficiently small fixed delta, the class $V_\delta$ has input proportions bounded above and below by positive constant multiples of $1/\log^2(1/\delta)$, after taking lower and upper limits in x. This order-of-magnitude claim does not require an assertion that its exact natural density exists. The distinct-semiprime example is a density-zero set.

For each fixed $\varepsilon>0$ and any family with $|E_x|\le\varepsilon x$ eventually, the claimed collision lower bound is

$$\liminf_x\frac{C_{[1,x]\setminus E_x}(x)}x \ge\sup_{\substack{0<\delta\le1/10\\d_\delta>\varepsilon}} \frac{(d_\delta-\varepsilon)^2}{2\delta}.$$

The range of delta is inherited from its definition in A. For all sufficiently small positive epsilon this is at least $F(\varepsilon)=(\varepsilon^2/2)\exp(\sqrt{0.1/\varepsilon})$. If the indexing set is empty, the display carries no nontrivial lower bound; the subsequent quantitative claims concern small epsilon, where it is nonempty. Therefore any choice of admissible finite constants $K(\varepsilon)$ in B(3), if such constants exist, must satisfy that lower bound and grow faster than every fixed power of $1/\varepsilon$ as epsilon decreases to zero. This is not a claim that the optimal constant is attained, equals F, or exists unconditionally.

Weakest steps and strongest attacks

AE1. Upper and lower proportions. A2 supplies the lower bound $0.89m_Z\sum_p1/p$. Dropping the f restriction gives the upper bound $m_Z\sum_p1/p$ for the upper limiting proportion, by the same finite residue count. A3 makes both of order $1/\log^2(1/\delta)$ with absolute comparison constants for sufficiently small delta. No exact density is silently assumed. For semiprimes, fix w. Those with a prime factor $p\le w$ number at most $\sum_{p\le w}\pi(x/p)=o_w(x)$; those with both factors above w are w-rough and number at most $m_wx+O_w(1)$. Taking x first, then $w\to\infty$, proves their density zero, including squares if desired.

AE2. Supremum and constant dependence. For each delta with $d_\delta>\varepsilon$, at least $(d_\delta-\varepsilon-o_\delta(1))x$ constructed inputs remain. Squaring this eventually positive lower bound gives the stated term. A lower limit bounded below by every term is bounded below by their supremum; no maximizing delta, compactness or uniform x-threshold is assumed. Take $\delta_\varepsilon=\exp(-\sqrt{0.1/\varepsilon})$. As epsilon tends to zero this enters A3's valid range, and $0.2/\log^2(1/\delta_\varepsilon)=2\varepsilon$. Thus $(d_{\delta_\varepsilon}-\varepsilon)^2/(2\delta_\varepsilon) \ge F(\varepsilon)$. Any eventual upper bound K must dominate this lower limit. For every fixed $a\ge0$,

$$\log\bigl(F(\varepsilon)\varepsilon^a\bigr) =\sqrt{0.1/\varepsilon}+(a+2)\log\varepsilon-\log2 \longrightarrow+\infty.$$

The strongest attack was to let the family $E_x$, its starting point, or $K(\varepsilon)$ depend adversely on epsilon, or to demand a uniform bound at delta equal to $1/10$. Each epsilon is fixed before x tends to infinity; the construction then fixes a sufficiently small delta depending on it. These dependences are allowed. The false all-delta extension exhibited above is not used. The derived bound is in the claimed ratio $C/x$, not in an unrelated absolute count, and supplies only a necessary lower bound on K.

Premises: A1--A3, elementary counting, and the PNT/Mertens specializations explicitly used in AE1. Mathematical verdict: refutation-failed for the density-zero example, the order statement, the supremum inequality, and the superpolynomial consequence separately.

B: the four equivalent formulations

Restatement

Write $F_x(T)=x^{-1}\#\{n\le x:r_x(s(n))>T\}$. For this particular proper-divisor-sum map, the following four assertions are equivalent:

1. Every fixed set A of natural density zero has a preimage of natural density zero.
2. $\lim_{T\to\infty}\limsup_{x\to\infty}F_x(T)=0$.
3. For each $\varepsilon>0$, some finite $K\ge0$ and $x_0$ work for every real $x\ge x_0$: one may remove at most $\varepsilon x$ inputs, with the chosen set depending on x, leaving ordered collision count at most Kx.
4. For each $\varepsilon>0$, some $\eta>0$ and $x_0$ work simultaneously for all $x\ge x_0$ and all nonnegative integer target sets B of cardinality at most $\eta x$: their input capture is at most $\varepsilon x$.

The constants may depend on epsilon. There is no bound on the locations of B's values and no requirement that B stay fixed with x. These are equivalences; none of the four assertions is declared true by the theorem.

Weakest steps, independently rederived

B1. Necessity of fiber tightness: lower-output control and one sparse union. For every prime $p\mid n$, the divisor $n/p$ is proper, so $s(n)/n\ge1/p$. Let $L\ge4$. If $n>x/\sqrt L$ and $s(n)<x/L$, then $s(n)/n<1/\sqrt L$, so n has no prime divisor at most $\sqrt L$. The residue count in A2 therefore gives

$$\#\{n\le x:s(n)<x/L\} \le (L^{-1/2}+m_{\sqrt L})x+O_L(1).$$

This is an arithmetic statement about s, not an assumption that outputs are uniformly comparable to inputs. Mertens lets its coefficient be made arbitrarily small by fixing L sufficiently large.

If (2) fails, the decreasing function $H(T)=\limsup_x F_x(T)$ has a positive limit. Choose eta strictly below that limit. For every T there are arbitrarily large x with $F_x(T)>\eta$. Choose L with $L^{-1/2}+m_{\sqrt L}\le\eta/4$ and a nonnegative constant $c_L$ bounding the residue error. Successively choose integers $x_k$ with

$$F_{x_k}(k)>\eta,\qquad x_k\ge k^2Lx_{k-1},\qquad c_L\le\eta x_k/4, \qquad x_0=1.$$

Arbitrarily large witnesses exist at each stage; this construction does not assume its conclusion or a termination property. Real witness points could also be replaced by their integer parts, where the numerator stays the same and the ratio does not decrease.

Set $A_k=\{m\ge x_k/L:r_{x_k}(m)>k\}$ and $A=\bigcup_{k\ge1}A_k$. At most $\eta x_k/2$ inputs lie below the output threshold. At least $\eta x_k/2$ of the heavy-fiber inputs consequently have their value in $A_k\subseteq A$. Thus $s^{-1}(A)$ has positive upper density.

Each $A_k$ is finite and $|A_k|<x_k/k$, because its fibers are disjoint and each has more than k members. For any sufficiently large real y, choose the largest k for which $x_k/L\le y$. No later $A_j$ has a value at most y. Although values in earlier sets may be arbitrarily large, counting their entire finite cardinalities is an upper bound. Hence

$$\frac{|A\cap[1,y]|}{y} \le\frac{L}{x_k}\left(\frac{x_k}{k} +\sum_{j<k}x_j\right) \le\frac Lk+\frac{Lk x_{k-1}}{x_k} \le\frac Lk+\frac1k\longrightarrow0.$$

Here $k\to\infty$ with y. This is a bound for all sufficiently large y, not just the sequence $x_k$. Overlap between the $A_j$ only reduces their union's cardinality. Since $x_1\ge L$, the union contains positive values, so it is a valid target under either convention for zero. This contradicts (1), proving (1) implies (2).

B2. Sufficiency of trimmed second moments: upper-output control. By pairing complementary divisors and interchanging finite sums,

$$\sum_{n\le x}\frac{\sigma(n)}n =\sum_{n\le x}\sum_{d\mid n}\frac1d =\sum_{d\le x}\frac1d\left\lfloor\frac xd\right\rfloor \le x\sum_{d\ge1}\frac1{d^2}<2x.$$

The last strict inequality follows by comparison of the tail with $\int_1^\infty t^{-2}dt$; no evaluation of $\zeta(2)$ is needed. If $s(n)>Lx$ and $n\le x$, then $\sigma(n)/n>1+L$, so these inputs number at most $2x/(L+1)\le2x/L$.

For a fixed density-zero A, fix epsilon and its constant K from (3), then fix $L\ge1$. Outside the removed inputs and the preceding upper tail, Cauchy--Schwarz bounds the capture by $\sqrt{Kx\,|A\cap[0,Lx]|}=o(x)$, because L is fixed. Therefore

$$\limsup_x x^{-1}\#\{n\le x:s(n)\in A\} \le\varepsilon+2/L.$$

Letting L increase and epsilon decrease proves (1). The chosen exclusions may vary arbitrarily with x; the estimate needs no nesting and no global excluded set. Crucially K is fixed while x tends to infinity.

B3. The finite-fiber implications and an independent cross-check. From (2), choose T with $\limsup_x F_x(T)<\varepsilon$, using strict slack, so $F_x(T)\le\varepsilon$ eventually. Removing all heavy fibers leaves counts at most T, and $\sum_m r_m^2\le T\sum_m r_m\le Tx$. This is (3). To get (4), instead use eventual heavy input mass at most $\varepsilon x/2$ and put $\eta=\varepsilon/(2T)$, taking $T\ge1$. Every target B captures at most that heavy mass plus $T|B|$ other inputs.

Conversely, for (4), use its eta and set $T=1/\eta$. The heavy-value set $B_x=\{m:r_x(m)>T\}$ has fewer than $x/T=\eta x$ elements and hence captures at most $\varepsilon x$ inputs. Monotonicity in T gives (2). This uses exactly the uniformity over target sets promised by (4).

As an independent check on the composition, (3) also directly implies (2) without either arithmetic output-tail estimate. For its excluded set, let $t_m$ denote surviving counts. Since $|B_x|<x/T$ and $\sum_m t_m^2\le Kx$, Cauchy--Schwarz gives

$$F_x(T)\le\varepsilon+\sqrt{K/T}\qquad(x\ge x_0).$$

Take the upper limit in x, then let T grow with epsilon fixed, and finally let epsilon decrease. This independently checks the connection between arbitrary partial trimming and whole-fiber tightness.

Strongest attacks and findings

The strongest attack on B1 was that the $A_k$ have no upper bound on their values, so a sparse union estimate valid only at selected endpoints might conceal dense intermediate intervals. The all-y calculation in B1 answers that attack: later sets have a lower support cutoff, and the total cardinality of each earlier set is included regardless of where its values lie.

Two explicit abstract maps tested whether the argument had silently transported an equivalence that fails outside the arithmetic setting:

• For $g(n)=n^2$, every fiber has size at most 1 and the collision count is $\lfloor x\rfloor$, yet the density-zero set of squares captures every input. Thus tightness or (3) alone cannot imply (1) for an arbitrary map. This g fails B2's large-output estimate: for every fixed L, almost all $n\le x$ have $g(n)>Lx$.
• For $f(n)=\lfloor\sqrt n\rfloor$, every density-zero A still has a density-zero preimage: with $y=\lfloor\sqrt x\rfloor$, its capture is at most $(2y+1)|A\cap[1,y]|=o(x)$. But completed fibers have sizes $2m+1$, so for each fixed T all but $O_T(1)+O(\sqrt x)$ of the inputs lie in fibers larger than T. Tightness fails. This f fails B1's small-output estimate: for every fixed L, all outputs at scale x are smaller than $x/L$ once x is sufficiently large.

These are attack models, not counterexamples to any assertion about s. They show that both arithmetic interfaces are consequential and cannot be omitted. The supplied proof establishes them independently, so neither attack breaks Theorem B. The zero fiber, limsup slack, unrestricted target locations and the order of all limits were checked explicitly.

Premises: Mertens' limit $m_w\to0$, finite residue counting, the local prime-divisor bound and mean identity, and elementary finite inequalities. Neither Theorem A's divergence conclusion nor PPT is needed for B; A2's residue-count derivation is merely reused as an elementary calculation. Mathematical verdict: refutation-failed for the four-way equivalence and the statement that its trimmed second-moment route loses no logical strength.

C: comparison with the finite-target theorem and its countermodel

Restatement

The supplied PPT theorem gives, for each fixed $h(x)\to0$, a uniform $o_h(x)$ capture bound on positive targets of size at most $x^{1/2+h(x)}$. Theorem B(4) instead asks that, for every capture tolerance epsilon, a fixed positive fraction eta of x be an admissible target size eventually. The small-target theorem, even combined with asymptotically linear output size, does not imply density-zero preservation for arbitrary maps. This last nonimplication concerns a class of abstract maps; it is not a disproof for s or a prohibition on arguments using additional arithmetic information.

Weakest steps and strongest attack

C1. Exact interface and limit order. For s, define the maximal capture proportion $M_x(\eta)$ over sets of at most $\eta x$ target values. It is nondecreasing in eta. Only finitely many values occur among inputs at scale x, so this maximum exists. Condition B(4) is equivalent to

$$\lim_{\eta\downarrow0}\limsup_{x\to\infty}M_x(\eta)=0,$$

using strict slack when passing from a limsup to an eventual bound. PPT instead controls $M_x(x^{-1/2+h(x)})$ along a scale depending on x. It gives no uniform error rate over choices of h. Adding target value zero changes capture by at most one. Thus the phrase about the same kind of uniform control is faithful, but it does not exchange the two limiting operations.

C2. Reconstruction with a uniform fiber constant. Put $a_j=\lfloor j^{3/2}\rfloor$ for $j\ge1$. These integers increase strictly and $a_1=1$, so $T(n)=\max\{a_j:a_j\le n\}$ is defined for every positive integer n. Its fiber over $a_j$ is the integer interval $[a_j,a_{j+1}-1]$. The mean value theorem gives

$$a_{j+1}-a_j\le\tfrac32\sqrt{j+1}+1.$$

If $a_j\le x$, then $j<(x+1)^{2/3}$, and for $x\ge1$ the displayed gap is at most $4x^{1/3}$. This also bounds the last, truncated fiber at x. Consequently, for every target B with $|B|\le x^{1/2+h(x)}$,

$$\#\{n\le x:T(n)\in B\} \le4x^{5/6+h(x)}=o_h(x).$$

For example, eventually $h(x)\le1/12$, so the right side is at most $4x^{11/12}$. The bound is genuinely uniform over all B and needs no independence assumption.

For $A=\{a_j:j\ge1\}$, its counting function is at most $(x+1)^{2/3}$, hence A has density zero. The same gap estimate at x=n gives $0\le n-T(n)\le4n^{1/3}$, so $T(n)/n\to1$. Nevertheless $T^{-1}(A)$ is every positive integer. Indeed, for every fixed eta, the whole image up to x has at most $(x+1)^{2/3}\le\eta x$ values eventually, and it captures all inputs; thus its analogue of $M_x(\eta)$ tends to 1.

The strongest attempted challenge was that the example might fail uniformity in B, the prescribed varying exponent, an endpoint fiber, or the promised linear output behavior. The explicit uniform constant and the last-fiber bound settle all four. The example meets the abstract hypotheses and violates the conclusion, so it supplies a concrete obstruction to the proposed general inference. Splitting a sparse target into a growing number of small sets cannot remove this obstruction merely by multiplying unspecified little-o errors.

Premises: the exact PPT statement only for the comparison; elementary calculus and counting for the countermodel; B for the linear-target equivalence. Mathematical verdict: refutation-failed for the comparison and for the countermodel's stated nonimplication separately. The implication that this countermodel defeats is not a claim of the frozen subject.

D: the sufficient condition and its consequences

Restatement

For every fixed $\delta>0$, set $D(\delta)=\{n\text{ composite}:s(n)\ge\delta n\}$ and $m_{1/\delta}=\prod_{p\le1/\delta}(1-1/p)$, with an empty product equal to 1. The distinct claims are:

1. The complement has upper natural density at most $m_{1/\delta}$, which tends to zero as delta decreases to zero.
2. If for each fixed delta there is a finite $K_\delta$ with $C_{D(\delta)}(x)\le K_\delta x$ for all sufficiently large x, then B(3), and therefore density-zero preservation, follows.
3. Any constants satisfying that bound for sufficiently small delta must obey $K_\delta\ge0.02/[\delta\log^4(1/\delta)]$; thus any such family of constants tends to infinity as delta decreases to zero.
4. For fixed delta, the semiprimes in $D(\delta)$ have at most $M_\delta=\pi(3/\delta)+1$ representatives per value, and their own collision count is at most $M_\delta x$. Their interactions with the remaining inputs are separated using the stated factor-two inequality.

No assertion that condition (D) holds, or that it is necessary, is made or verified. The literature-status sentence is subject to the explicit source limitation above.

Weakest steps, independently rederived

D1. Complement and sufficiency. If a composite n has a prime factor $p\le1/\delta$, then $s(n)\ge n/p\ge\delta n$, so n belongs to $D(\delta)$. In fact, every integer outside D is either $(1/\delta)$-rough or one of the finitely many primes at most $1/\delta$; 1 is already rough. This slightly sharper inclusion proves the asserted upper density bound by finite residue counting alone. Alternatively the printed union with all primes works because the PNT gives prime density zero.

For any epsilon choose delta so that $m_{1/\delta}<\varepsilon$. The strict inequality supplies the slack needed to turn an upper-density bound into $|[1,x]\setminus D(\delta)|\le\varepsilon x$ for every sufficiently large x. Combine that threshold with the hypothesized collision threshold for this fixed delta. The resulting exclusions and $K=K_\delta$ satisfy B(3). There is no uniform bound in delta and no claim that the complement has density zero for one fixed delta.

D2. Necessary size of an admissible constant. Every composite member of $V_\delta$ satisfies $s(n)/n\ge\delta$ by A1. Its prime members arise only from $k=1$ and belong to a finite prime window for fixed delta. Removing that finite set preserves A2's lower input proportion and the positive-value range $[1,2\delta x]$. A3's pigeonhole calculation therefore gives

$$\liminf_x\frac{C_{D(\delta)}(x)}x \ge\frac{d_\delta^2}{2\delta} \ge\frac{0.02}{\delta\log^4(1/\delta)}$$

for sufficiently small delta. Any eventual upper constant must dominate this number. Its divergence proves exactly the asserted necessary growth; it says nothing about the existence of those upper constants.

D3. Semiprimes and mixed collisions. For distinct primes $p<q$,

$$\frac{s(pq)}{pq}=\frac1p+\frac1q+\frac1{pq}\le\frac3p.$$

Membership in D therefore implies $p\le3/\delta$. Given a value m and one such p, the other prime would have to be $q=m-p-1$, so there is at most one distinct-prime semiprime for that p. A square contributes value $p+1$, which determines p, and adds at most one more representative. Thus the semiprime fiber count u satisfies $u(m)\le M_\delta$. It follows that

$$\sum_m u(m)^2\le M_\delta\sum_m u(m)\le M_\delta x.$$

Writing v for the remaining admitted fiber counts, the full count obeys

$$\sum_m(u(m)+v(m))^2 \le2M_\delta x+2\sum_m v(m)^2.$$

Thus the semiprimes can be separated when deciding whether the whole count is $O_\delta(x)$: conversely, $\sum_m v(m)^2\le C_{D(\delta)}(x)$. The bound $M_\delta x$ is for their own sum of squares; it is not a separate bound on every mixed pair involving a semiprime. The frozen sentence supplies precisely the inequality that accounts for those mixed terms.

Strongest attacks and findings

I tested whether an upper-density estimate was used without eventual slack, whether the positive-density construction contained excluded primes, whether prime squares shared the distinct-prime parametrization, and whether mixed collisions were silently discarded. D1 uses the strict inequality $m_{1/\delta}<\varepsilon$; D2 removes only finitely many primes; D3 counts squares separately and retains the factor two. These attacks fail against the stated implications. Treating (D) as proved or as necessary would strengthen the subject and is not supported by this report.

Premises: elementary divisor inequalities and residue counting, Mertens' limit, B for D1's final implication, and A1--A3 for D2; no external theorem is needed for D3. Mathematical verdict: refutation-failed for each of the four numbered claims in this section's restatement: the complement estimate, conditional sufficiency, necessary growth, and semiprime separation. The claim about what is known in the literature remains an uncertified characterization, not a refuted mathematical statement.

Checklist verdicts

The groups A, A-epsilon, B, C and D above include their individually stated subsidiary conclusions. Each row below applies to every group: a universal inapplicability is stated explicitly; otherwise the group-specific examination or finding is given. These are mathematical checklist findings, not a grade of this report. The source qualifications and independence exposure are retained rather than converted into favorable checklist findings.

Canonical failure modes

Named canonical patterns

Erdos-specific checklist

The neighboring statistical discipline is inapplicable: there are no correlated samples or statistical conclusions. Evidence coverage is matched to the actual analytic derivations, statement-only external reading and excluded census as required by the evidence standard.

Premises and verification order

The following order records substantive mathematical consumption, not merely shared notation:

1. Fix the conventions and the exact external PNT and Mertens statements. Establish finite rough-number residue counting, divisor-product bounds, Cauchy--Schwarz and Markov specializations, and the elementary inequalities used above.
2. Establish A1--A3 and Theorem A. Their conclusion does not consume B, PPT, D, or the countermodel.
3. Derive the A-epsilon consequences from the same construction. The definition of a constant admissible in B(3) is used, but B(3)'s truth is not assumed.
4. Establish B from its elementary finite-fiber arguments and its two arithmetic tail estimates. It consumes the residue-count calculation and the Mertens limit, not Theorem A's divergence conclusion.
5. Check C's comparison using the independently established B equivalence and PPT's exact supplied statement. Reconstruct its countermodel directly; its proof does not consume PPT.
6. Establish D's complement estimate, invoke B for the sufficient implication, use A's construction for necessary growth, and prove the semiprime estimate directly.

The source-owned statements are ordinary identified literature premises, with reading depth and exact specializations recorded above. Their uninspected proofs are not silently relabeled as local proof coverage. All remaining support consists of elementary integer factorization, finite counting, inequalities and calculus as displayed. No native L-claim, pending local grade, unavailable local lemma, announcement or prior author's computation is used. Accordingly there is no current or prospective native dependency-tier reading to substitute for a live ledger check; the consumed-native-claim list is empty on the examination date. The conditional theorem keeps (D) as an assumption, not as an independently warranted result.

Verdict and limits

The strongest attempted failures were the nonuniform rough-number errors, adaptively varying exclusions, loss of density control between diagonal thresholds, transport to maps lacking the two arithmetic output bounds, an invalid extension of small-target uniformity, and missing mixed collisions. The retained arguments explain why each fails against the frozen mathematical claims. No counterexample to those claims or unproved essential inference in their supplied analytic proofs was found within the exact literature interfaces authorized by the commission.

This is not an unqualified certification of every characterization in the packet. The assertion about what is known concerning equivalence of (D) remains unsupported by the closed source set. The distinction between an expository Mertens statement and primary-proof inspection also remains explicit. These limitations do not constitute a disproof of an analytic theorem or a proof of an absent equivalence. Neither (D) nor a full resolution of E0955 is established by this review. No numerical census, later sieve route, external theorem proof, original-source transcription fidelity, or subsequently changed subject text is covered.

The complete mathematical examination is retained at the stated scope. A separate grader must judge the report contract and independence, including the disclosed motivational exposure, before any acceptance or tier assertion.