../

---

This page transports the mechanisms reconstructed in this folder to the problems they touch. Every statement on it is author-recorded: written and checked by the compilation; it changes no problem status and assigns no tier. The statements were independently reviewed as they stood on 2026-09-28T13:50:48Z by nine focused reviews graded in the connections grade, which passes all nine and accepts seventeen corrections, since applied; it assigns no tier. After the review, the Target paragraphs were updated to the standing their problem pages record, and the next investigations of the Problem 304, 293 and 1099 sections to match; no statement or proof changed. Each connection section names its target with the exact question and the status the target's own page records, states what the mechanism gives when transported, with a proof or with the missing bridge named, what it does not give and why, and the next investigation. Inputs are consumed at their recorded standing: the hughes_ pages are proved modulo the second-hand import of the Berend–Harmse estimate, the doorn_ pages are claimed, and the factorial bound of question (b) is the Conjectures.io-accepted Lean proof described on the Problem 18 page (kernel checked by the bounty site, not refereed, not built here).

Conventions. For practical $N$, $h(N)$ is the least $k$ such that every integer $1\le m\le N$ is a sum of at most $k$ distinct divisors of $N$, a fresh set for each $m$; erdosproblems.com's $h$ ranges over $m<N$ and is at most this one, so every upper bound below holds for erdosproblems.com's $h$ as well. $\tau(N)$ counts the divisors of $N$, $v_p$ is the $p$-adic valuation, $L_x=\mathrm{lcm}(1,\dots,x)$, and $\varepsilon_j$ and the windows $[\sqrt{(j-1)!},\sqrt{j!}]$ are those of the Corollary 3 page. For $1\le a<b$, $N(a,b)$ is the least number of distinct unit fractions with denominators above $1$ summing to $a/b$ and $N(b)=\max_{1\le a<b}N(a,b)$; $D_k$ is the set of denominators occurring in some decomposition $1=1/n_1+\cdots+1/n_k$ with $1\le n_1<\cdots<n_k$, and $v(k)=\min\{m>1:m\notin D_k\}$. For $\alpha>1$, $h_\alpha(n)=\sum_i(d_{i+1}/d_i-1)^\alpha$ over the consecutive divisors $1=d_1<\cdots<d_{\tau(n)}=n$.

Shared lemmas

The four lemmas below are used by several sections. Each is author-recorded with its proof; none is in the frozen reconstruction pages.

Lemma A: a valuation criterion for the ratio-two property

Statement. Let $N\ge1$ satisfy $v_2(N)\ge\lfloor\log_2p\rfloor$ for every odd prime $p$ dividing $N$. Then consecutive divisors of $N$ have ratio at most $2$. Consequently, by the consequence clause of Lemma 4, $N$ is practical, the greedy expansion of every $1\le m\le N$ uses distinct divisors, and $h(N)\le\log_2N+1$. The hypothesis holds for $N=n!$, for $N=L_x$, and for $N=p^jM$ whenever it holds for $M$ and $p$ is an odd prime with $\lfloor\log_2p\rfloor\le v_2(M)$.

Proof. Let $d<N$ divide $N$ and put $q=N/d>1$. If $q$ is even, $2d$ divides $N$. If $q$ is odd, choose an odd prime $p\mid q$ and let $2^c<p<2^{c+1}$ (strict, since $p$ is odd). Then $v_2(d)=v_2(N)\ge c$ because $q$ is odd, and $v_p(d)<v_p(N)$ because $p\mid q$. So $d'=dp/2^c$ is an integer dividing $N$, and $d<d'<2d$ since $1<p/2^c<2$. Hence some divisor lies in $(d,2d]$, which is the ratio property. For the consequence: a nonterminal greedy step at $R$ with bracketing divisors $d<R<b$ has $d\ge b/2>R/2$, so the remainder $R-d$ is below $R/2$; after $s$ nonterminal steps the remainder is below $m/2^s$, so there are at most $\lfloor\log_2m\rfloor$ nonterminal steps and one terminal divisor. For $n!$, $v_2(n!)\ge\lfloor n/2\rfloor\ge\lfloor\log_2p\rfloor$ for every prime $p\le n$ when $n\ge4$, and $n\le3$ is checked by inspection; for $L_x$, $v_2(L_x)=\lfloor\log_2x\rfloor\ge\lfloor\log_2p\rfloor$ for $p\le x$; and multiplying $M$ by an odd prime $p$ leaves $v_2$ unchanged, so the hypothesis for $p^jM$ is the hypothesis for $M$ together with $\lfloor\log_2p\rfloor\le v_2(M)$. $\square$

Lemma B: the greedy count from a gap profile

This is the step count of the Theorem 1 page with the factorial windows replaced by an arbitrary profile; only the two symmetry facts (i) and (ii) of that page and Lemma 4 are used, and both hold for every $N$.

Statement. Let $N$ have the ratio-two property, let $T\ge2$ with $2T\le\sqrt N$, and let $\delta\colon[T,\sqrt N]\to(0,1/8]$ be nonincreasing. Suppose that every pair of consecutive divisors $d<b$ of $N$ with $T\le\sqrt{db}\le\sqrt N$ satisfies

$$\log\frac bd\le\tfrac12\,\delta\bigl(\sqrt{db}\bigr).$$

Put $\sigma(\ell)=\log\bigl(1/\delta(e^\ell/2)\bigr)$ for $\log(2T)\le\ell\le\tfrac12\log N$, a nondecreasing function with $\sigma\ge\log8$. Then

$$h(N)\le3\int_{\log(2T)}^{\frac12\log N}\frac{d\ell}{\sigma(\ell)} +\log_2\sigma\bigl(\tfrac12\log N\bigr)+2\log_2(2T)+5 .$$

Proof. Fix $1\le m\le N$ and run the greedy expansion $R_0=m>R_1>\cdots$; by the ratio property the divisors used are distinct (Lemma 4). Count the nonterminal steps by the remainder $R$ at which they start.

Lower range $2T\le R\le\sqrt N$. Let $d<R<b$ bracket $R$. As on the Theorem 1 page, $d>R/2$ and $b<2R$ give $R/\sqrt2<\sqrt{db}<\sqrt2R$, and fact (i) ($R\le\sqrt N$ forces $db\le N$) gives $\sqrt{db}\le\sqrt N$; also $\sqrt{db}>R/\sqrt2\ge T$. The hypothesis and the monotonicity of $\delta$ give $\log(b/d)\le\tfrac12\delta(\sqrt{db})\le\tfrac12\delta(R/2)$, and Lemma 4 gives $R'=R-d\le2R\log(b/d)\le R\,\delta(R/2)$, that is, $\log R'\le\log R-\sigma(\log R)$. Since $\sigma$ is nondecreasing, the charging integral of the Theorem 1 page applies verbatim: each step's interval $[\log R',\log R]$ carries $\int d\ell/\sigma(\ell)\ge1$, the intervals have disjoint interiors and lie in $[\log(2T),\tfrac12\log N]$ except possibly the last, so this range has at most $1+\int_{\log(2T)}^{\frac12\log N}d\ell/\sigma$ steps.

Upper range $\sqrt N<R\le N/(2T)$. Put $U=N/R\in[2T,\sqrt N)$. The mirror pair $N/b<U<N/d$ consists of consecutive divisors with the same ratio $b/d$, and its geometric mean $N/\sqrt{db}$ lies in $(U/\sqrt2,\sqrt2U)$ and is at most $\sqrt N$ by fact (ii) ($R>\sqrt N$ forces $db\ge N$) and at least $T$. Hence $\log(b/d)\le\tfrac12\delta(U/2)$ and, by Lemma 4 applied to $R$, $R'\le R\,\delta(U/2)$, i.e. $\log U'-\log U\ge\sigma(\log U)$, where $U$ increases along the expansion. Partition $[\log(2T),\tfrac12\log N)$ into the sets $J_k=\{\ell:2^k\log8\le\sigma(\ell)<2^{k+1}\log8\}$, $0\le k<K$, each an interval or empty, with $K\le\log_2\bigl(\sigma(\tfrac12\log N)/\log8\bigr)+1$. Consecutive starts in $J_k$ are separated in $\log U$ by at least $2^k\log8$, while $1/\sigma>1/(2^{k+1}\log8)$ on $J_k$; so $J_k$ holds at most $1+|J_k|/(2^k\log8)\le1+2\int_{J_k}d\ell/\sigma$ starts, and the range has at most $K+2\int_{\log(2T)}^{\frac12\log N}d\ell/\sigma$ steps.

Endgames. For $R>N/(2T)$ and for $R<2T$ each step halves the remainder, so each endgame has at most $\log_2(2T)+1$ steps. Adding the terminal divisor gives the bound. $\square$

Lemma C: the unit-fraction transfer (Lemma 5.1 of the van Doorn note)

Lemma 5.1 is the one step of the note's applications that the folder did not reconstruct; it is elementary and is reconstructed here from physical p. 6 of the note held by van Doorn (2026), read in the extracted text. The note calls it well known; it is the transfer behind Erdős's 1950 bound and behind Vose's, as the Problem 304 page records.

Statement. Let $n\ge b\ge2$ be integers with $n$ practical, and let $1\le a<b$. Then $a/b$ is a sum of at most $2h(n)$ distinct unit fractions, all with denominators above $1$; the denominators divide $n$ or have the form $bn/e$ with $e\mid n$ and exceed $n$. If $b\nmid n$, none of them equals $b$.

Proof. Write $an=bq+r$ with $0\le r<b$. Then $q\le an/b<n$ and $r<b\le n$, so $q$ (if positive) is a sum of $s\le h(n)$ distinct divisors $d_1,\dots,d_s$ of $n$, and $r$ (if positive) is a sum of $t\le h(n)$ distinct divisors $e_1,\dots,e_t$ of $n$. Dividing by $bn$,

$$\frac ab=\sum_{i=1}^{s}\frac1{n/d_i}+\sum_{j=1}^{t}\frac1{bn/e_j}.$$

The $n/d_i$ are distinct integers above $1$ (as $d_i\le q<n$) and at most $n$; the $bn/e_j$ are distinct integers above $n$ (as $e_j\le r<b$); so all $s+t\le2h(n)$ denominators are distinct and exceed $1$. If $b\nmid n$, no $n/d_i$ equals $b$, and every $bn/e_j$ exceeds $n\ge b$. $\square$

Lemma C′ (avoiding a prescribed denominator with factorials). Let $b\ge2$ be large, let $n$ be the least integer with $n!\ge b$, and let $p$ be the least prime not dividing $n$ (so $p\le n-1$: the primes dividing $n$ have product at most $n$, whence $p\ll\log n$). Put $N=n!$ if $b\nmid n!$, and $N=p^{j}(n-1)!$ with $j=\lceil\log n/\log p\rceil$ otherwise. Then $N$ is practical, $N\ge b$, $b\nmid N$, and

$$h(N)\le\max\bigl(h(n!),\ h((n-1)!)+\lceil\log_2n\rceil\bigr).$$

Proof. In the first case there is nothing to show beyond Lemma A. In the second, $p^j\ge n$ gives $N\ge n!\ge b$. Every residue $c$ modulo $p$ with $1\le c\le p-1\le n-2$ is itself a divisor of $(n-1)!$, hence of $p^{i}(n-1)!$, not divisible by $p$, with total $c\le p^{i}(n-1)!$; the residue $0$ is the empty sum. So Lemma 3.1 with $A=p$ and $L=1$, applied $j$ times from the practical number $(n-1)!$, shows that $N$ is practical with $h(N)\le h((n-1)!)+j$, and $j\le\lceil\log_2n\rceil$. Finally, if $b\mid N$ then $b$ divides $\gcd(n!,p^j(n-1)!)=(n-1)!\gcd(n,p^j)=(n-1)!$, contradicting the minimality of $n$, which gives $(n-1)!<b$. $\square$

Lemma D: a counting floor for iterated extensions

Statement. Let $n_0$ be practical and let $n_{j+1}=A_jn_j$ ($0\le j<J$) be a chain in which each step is certified by Lemma 3.1 with cost $L_j$: every residue modulo $A_j$ is a sum of at most $L_j$ distinct divisors of $n_j$. Then

$$\sum_{j<J}L_j\ \ge\ \sum_{j<J}\frac{\log A_j}{\log\bigl(\tau(n_j)+1\bigr)} \ \ge\ \frac{(\log\log n_J)^2-(\log\log n_0)^2}{2(\log2+\eta)}-C(\eta)$$

for every $\eta>0$, with $C(\eta)$ depending on $\eta$ alone. Hence, for chains from a bounded seed, $n_0\le C$ with $C$ fixed, the bound $h(n_0)+\sum_jL_j$ that such a chain certifies for $h(n_J)$ is at least $(1/(2\log2)-o(1))(\log\log n_J)^2$, where the $o(1)$ tends to $0$ as $n_J\to\infty$ uniformly over the chains with $n_0\le C$. Without a bound on the seed the sum $\sum_jL_j$ has no floor in $n_J$ alone: the one-step chain $n_0=2^k$, $A_0=3$, $L_0=1$ satisfies the hypothesis of Lemma 3.1 and has $\sum_jL_j=1$.

Proof. Distinct residues modulo $A_j$ need distinct representing subsets, so $A_j\le\sum_{i\le L_j}\binom{\tau(n_j)}i\le(\tau(n_j)+1)^{L_j}$, which is the first inequality. For the second, Wigert's maximal order of the divisor function, $\log\tau(n)\le(\log2+\eta(n))\log n/\log\log n$ with $\eta(n)\to0$, gives $\log(\tau(n_j)+1)\le(\log2+\eta'(n_j))\log n_j/\log\log n_j$ with $\eta'\to0$. With $\ell_j=\log n_j$ and $f(\ell)=\log\ell/\ell$, which decreases for $\ell>e$, let $n(\eta)$ be a threshold beyond which $\eta'(n)\le\eta$ and $\log n>e$, and let $j(\eta)$ be the least index with $n_{j(\eta)}\ge n(\eta)$; if there is none, then $n_J<n(\eta)$ and the bound holds with any $C(\eta)\ge(\log\log n(\eta))^2/(2\log2)$. The steps $j\ge j(\eta)$ contribute at least $(\log2+\eta)^{-1}\sum(\ell_{j+1}-\ell_j)f(\ell_j)$, and a left Riemann sum of a decreasing function dominates its integral, so this is at least

$$\frac1{\log2+\eta}\int_{\ell_{j(\eta)}}^{\ell_J}f(\ell)\,d\ell =\frac{(\log\ell_J)^2-(\log\ell_{j(\eta)})^2}{2(\log2+\eta)}.$$

The step $j(\eta)-1$, if it exists, costs at least $\log(n_{j(\eta)}/n(\eta))/\log(2n(\eta))$ by the first inequality and $\tau(n)\le n$, which exceeds $(\log\ell_{j(\eta)})^2$ once $n_{j(\eta)}$ is large in terms of $\eta$, and the other steps contribute nonnegatively. If $j(\eta)=0$ the subtracted term is $(\log\ell_0)^2=(\log\log n_0)^2$ and nothing compensates it. In either case $\sum_jL_j\ge\bigl((\log\ell_J)^2-(\log\ell_0)^2\bigr)/(2(\log2+\eta))-O_\eta(1)$, which is the second inequality. For $n_0\le C$ the subtracted term is $O_C(1)$, and letting $\eta\to0$ gives the floor $(1/(2\log2)-o(1))(\log\log n_J)^2$. $\square$

This floor is an obstruction to the method, not a lower bound for $h$: it says that no certificate built from Lemma 3.1 extensions alone, from a fixed seed, can prove $h(N)\le(\log\log N)^{2-\eta}$ or even a constant below $1/(2\log2)\approx0.72$ in front of $(\log\log N)^2$. The van Doorn chain pays $4$ per step where the first inequality demands $\log A_j/\log\tau(n_j)\approx6/7$, and its primes of size $k^6\log k$ keep $\tau(n_j)$ a factor about $6$ below the maximal order in the logarithm; both are constant losses. The reconstructed $c_0=14/\log2\approx20.2$ sits a factor $28$ above the floor.

Problem 304: the number of unit fractions for a/b

Target. Problem 304 asks to estimate $N(b)=\max_{1\le a<b}N(a,b)$ and whether $N(b)\ll\log\log b$. Its page records the problem as solved, with the question proved: Theorem 1.1 of the OpenAI mathematics release's manuscript of 25 September 2026 gives $N(b)\ll\log\log b$, accepted on its Lean evidence as its claim page records, so $N(b)\asymp\log\log b$ with Erdős's 1950 lower bound. The note's Theorem 1.2 is a pending partial claim that the accepted bound implies. That bound supersedes (304.1)--(304.3) below, which stay recorded as consequences of the transfer, as the closed lead factorial_unit_fraction_transfer records. The maximum runs over all $1\le a<b$ with no coprimality condition, and Lemma C covers every $a$.

What transports. Lemma C turns any practical $N\ge b$ into $N(b)\le2h(N)$, so every bound on $h$ along a sequence that reaches every scale becomes a bound on $N(b)$. Three inputs are available here.

(304.1) From the accepted factorial bound (author-recorded; premise: the Conjectures.io-accepted proof of question (b), at its recorded standing). For every $\varepsilon>0$ there is $b_0(\varepsilon)$ with

$$N(b)\le2(\log b)^{\varepsilon}\qquad(b\ge b_0(\varepsilon)).$$

Proof. Let $n$ be the least integer with $n!\ge b$. Then $n\to\infty$ with $b$, and $(n-1)!<b$ gives $n<\log b$ for large $b$, since $\log(n-1)!\ge(n-1)(\log(n-1)-1)>n$ once $n\ge14$. Lemma C with the practical number $n!$ (Lemma A) gives $N(a,b)\le2h(n!)$ for every $a$, and the accepted statement, $h(n!)<n^{\varepsilon}$ for all $n\ge n_0(\varepsilon)$, gives $N(b)\le2n^{\varepsilon}\le2(\log b)^{\varepsilon}$ once $b>(n_0-1)!$. The accepted statement is exactly this $h$ (the fresh-set reading over $1\le m\le n!$), as the accepted target page records. $\square$

This is below Vose's $\sqrt{\log b}$ for every $\varepsilon<\tfrac12$ and is the first consequence of the accepted part (b) drawn in this corpus; no search for an earlier one was made. Its standing is that of part (b): documented independent acceptance of the formal statement by the bounty site, a single kernel, no refereeing, nothing built here; the transfer itself is Lemma C.

(304.2) From Hughes's Theorem 1 (author-recorded; imports the Berend–Harmse estimate second-hand through Corollary 3). With the same $n$, $n\log n=(1+o(1))\log b$ gives $n/\log n=(1+o(1))\log b/(\log\log b)^2$, so Theorem 1 and Lemma C give

$$N(b)\le(4\log2+o(1))\,\frac{\log b}{(\log\log b)^2}.$$

This improves Erdős's 1950 bound $c_1\log b/\log\log b$ by a factor $\log\log b$ and is weaker than Vose's; it is recorded for completeness and as a check of the transfer against a known route.

(304.3) From the claimed van Doorn chain (claimed; every input labeled claimed on its page). Proposition 4.1 at $x=b$ with $p_*=3$ gives a practical $n\in[b,b^2)$ with $h(n)\le c_0(\log\log b)^2-1$, and Lemma C gives $N(b)\le2c_0(\log\log b)^2-2$ for $b\ge x_0$, which is the note's Theorem 1.2 as recorded on its result page. The reconstruction of Lemma 5.1 above closes the one unreconstructed step of that application; the chain stays claimed.

What it does not give, and why. The conjecture $N(b)\ll\log\log b$ is out of reach of both reconstructed mechanisms, for reasons that can be made exact. Lemma C gives $N(b)\le2\min\{h(N):N\ \text{practical},\ N\ge b\}$, so the conjecture would follow from practical numbers $N$ with $b\le N\le\exp((\log b)^{O(1)})$ and $h(N)\ll\log\log N$, that is, from question (a) of Problem 18 with exponent $1$ at every scale. Along factorials this is impossible: Remark 6 gives $h(n!)\gg(\log n)^2$, and with $n!\approx b$ this is $\gg(\log\log b)^2$, so no bound on $h(n!)$, however sharp, gives less than $(\log\log b)^2$ through Lemma C. Along the extension chains of the van Doorn note it is impossible too: Lemma D shows that every certificate built from Lemma 3.1 extensions from a seed bounded independently of the scale, as the note's chain is, is at least $(1/(2\log2)-o(1))(\log\log N)^2$. So both mechanisms stop at exponent $2$ for this target, and the exponent, not the factor $2$ of Lemma C, is the obstruction. A bound $h(N)\ll(\log\log N)^2$ along some sequence reaching every scale, which is what (304.3) claims, would leave the conjecture open by a factor $\log\log b$.

Falsification attempts. The transfer is uniform in $a$ because Lemma C is; the quantifier "for all large $b$" follows from "for all large $n$" because $n$ is nondecreasing in $b$; the denominators are distinct and exceed $1$ as Problem 304 requires; erdosproblems.com's $m<N$ convention only lowers $h$; and Lemma C uses representations of $q\le n-1$ and $r\le b-1$, inside the range of the accepted statement. No defect was found.

Next investigation. None toward the target, which the accepted bound settles; the lead factorial_unit_fraction_transfer, which held the route, is closed as superseded. The mathematical question left by this page is the exponent-one question: is there, for every large $b$, a practical $N\in[b,\exp((\log b)^{O(1)})]$ with $h(N)\ll\log\log N$? A positive answer would give a second proof of the bound through Lemma C. Remark 6 shows that factorials cannot supply it and Lemma D that Lemma 3.1 chains from a bounded seed cannot; the greedy count of Lemma B on another ratio-two family, such as $L_x$ under Lemma A, is excluded by neither, since its counting floor $h(N)\ge\log N/\log(\tau(N)+1)$ is only of order $\log\log N$ for $N=L_x$, but no gap input of the required strength is known.

Problem 293: the least denominator missing from k-term decompositions of 1

Target. Problem 293 asks for the growth of $v(k)$. Its page records the status open, since every claim is partial, with $e^{e^{k/600}}\le v(k)$ for all large $k$, Corollary 1.3 of the OpenAI mathematics release's manuscript of 25 September 2026, accepted on its Lean evidence as its claim page records, and $v(k)\le c_0^{(2/5+o(1))2^k}$ (van Doorn–Tang; the Elsholtz–Planitzer count), so $\log\log v(k)=\Theta(k)$, the doubly exponential growth the monograph guessed. The earlier $e^{ck^2}\le v(k)$ (van Doorn–Tang, Theorem 1.1) is an accepted partial claim, and the note's Theorem 1.3 is a pending one. The accepted bound supersedes (293.1)--(293.3) below, which stay recorded as consequences of the transfer, as the closed lead factorial_unit_fraction_transfer records.

What transports. Two facts drive every deduction: the nesting $D_k\subseteq D_{k+1}$ for $k\ge2$ (van Doorn–Tang, Lemma 2.1, refereed, recorded on the Theorem 1.1 page), and the observation that a practical $N\ge b$ with $b\nmid N$ turns Lemma C, applied to $(b-1)/b$ and followed by adjoining $1/b$, into a decomposition of $1$ with at most $2h(N)+1$ terms containing $b$. Hence $b\in D_k$ for every $k\ge2h(N)+1$.

(293.1) From the accepted factorial bound (author-recorded; premise: the Conjectures.io-accepted proof of question (b)). For every $\varepsilon>0$ there is $k_0(\varepsilon)$ with

$$v(k)\ \ge\ \exp\bigl((k/5)^{1/\varepsilon}\bigr)\qquad(k\ge k_0(\varepsilon)),$$

so $v(k)\ge\exp(k^{A})$ for every fixed $A>0$ and all large $k$.

Proof. Let $b\ge2$ be large and let $N$ be the practical number of Lemma C′, so $N\ge b$, $b\nmid N$ and $h(N)\le\max(h(n!),h((n-1)!)+\lceil\log_2n\rceil)$, with $n<\log b$ the least integer with $n!\ge b$. The accepted statement gives $h(N)\le n^{\varepsilon}+\log_2n+2\le2n^{\varepsilon}$ for $n\ge n_1(\varepsilon)$, so $b\in D_{k}$ whenever $k\ge4n^{\varepsilon}+1$, in particular whenever $k\ge5(\log b)^{\varepsilon}$. Thus every $b$ with $b_0(\varepsilon)\le b\le\exp((k/5)^{1/\varepsilon})$ lies in $D_k$, and every $2\le b<b_0(\varepsilon)$ lies in $D_k$ for $k\ge k_1(\varepsilon)$ because $v(k)\ge e^{ck^2}\to\infty$. Hence $v(k)>\exp((k/5)^{1/\varepsilon})$ for $k\ge k_0(\varepsilon)$. $\square$

This lies strictly between the published $e^{ck^2}$ (the case $A=2$) and the claimed doubly exponential bound. Its standing is that of part (b) plus the elementary Lemmas C and C′.

(293.2) From Hughes's Theorem 1. The same route with $h(N)\le(2\log2+o(1))n/\log n+O(\log n)$ places $b$ in $D_k$ once $(4\log2+o(1))\log b/(\log\log b)^2\le k$, giving only $\log v(k)\ge(1/(4\log2)-o(1))\,k(\log k)^2$, weaker than $e^{ck^2}$; it is recorded so that the transfer is calibrated against the known bound.

(293.3) From the claimed chain. Proposition 4.1 at $x=b$, with $p_*$ an odd prime factor of $b$ (or $b$ a power of two above $2^E$, when $2^E\parallel n$ gives $b\nmid n$), Lemma C and the nesting give the note's Theorem 1.3, $v(k)\ge\exp\exp\sqrt{k/(2c_0)}$, exactly as its result page sketches; Lemma C′ is not needed there because the proposition arranges $b\nmid n$ itself. The chain stays claimed.

What it does not give, and why. Doubly exponential growth $v(k)\ge e^{e^{ck}}$ needs $b\in D_k$ for $\log\log b\ll k$, that is, decompositions of $1$ containing $1/b$ with $O(\log\log b)$ terms, which through this route means $h(N)\ll\log\log b$: the exponent-one question of the Problem 304 section again. The factorial route is capped at $\exp\exp(C\sqrt k)$: the counting argument of Remark 6, applied to $N$ (for $N=p^{j}(n-1)!$ one has $\tau(N)\le(j+1)\tau((n-1)!)$ and $\log N\ge\log(n-1)!$), gives $2h(N)+1\ge2c(\log n)^2=(2c+o(1))(\log\log b)^2$ for the $N$ of Lemma C′, so it can only place $b$ in $D_k$ when $\log\log b\le(1+o(1))\sqrt{k/(2c)}$. Certificates built from Lemma 3.1 extensions of a bounded seed are capped at the same shape by Lemma D, which is why the claimed Theorem 1.3 has the form it has. The converse implication of van Doorn–Tang's Section 3 ($b<v(k)$ gives $N(b-1,b)\le k-1$) turns (293.1) into $N(b-1,b)\le5(\log b)^{\varepsilon}$, which (304.1) already gives; nothing new flows back.

Falsification attempts. The case $b\mid n!$ is where a naive transfer fails (the divisor $n!/b$ can appear in a greedy representation of $(b-1)n!/b$, which would repeat $1/b$); Lemma C′ avoids it by changing the practical number rather than the representation, and its three requirements ($N\ge b$, $b\nmid N$, the cost $\lceil\log_2n\rceil$) were each checked. The nesting lemma is applied only from $k\ge2$, and the constructed decomposition has at least two terms. Small $b$ are handled by the published growth of $v(k)$, not by the accepted statement, whose threshold $n_0(\varepsilon)$ is unknown. No defect was found.

Next investigation. None through this route, which the accepted bound supersedes; the same lead is closed. A computation cannot test (293.1), whose content is asymptotic; the known finite values $v(3),\dots,v(8)$ on the problem page are consistent with every bound here.

Problem 18, question (a), along the least common multiples

Target. Question (a) of Problem 18: are there infinitely many practical $m$ with $h(m)<(\log\log m)^{O(1)}$? The page records it as claimed (the Price claim and the van Doorn note, neither accepted; erdosproblems.com showed OPEN), with Vose's $h(m)\ll(\log m)^{1/2}$ the best accepted bound. Erdős's own candidate sequences for small divisor ratios were $n!$, $L_x=\mathrm{lcm}(1,\dots,x)$ and the primorials (Tenenbaum 1987, p. 2, read as described in the Problem 1099 section), and the erdosproblems.com discussion of 14 September 2026 reduces $h(L_x)=O((\log x)^2)$ to a divisor-gap bound it does not prove. Since $\log\log L_x=\log x+o(1)$, such a bound answers (a) with exponent $2$ along an explicit classical sequence.

What transports. Lemma A shows that $L_x$ is practical with consecutive divisors in ratio at most $2$ and $h(L_x)\le\log_2L_x+1=(1/\log2+o(1))\,x$, the analog of Erdős's $h(n!)\le n$. Lemma B then converts a gap profile for the divisors of $L_x$ into a step count. Write the hypothesis in the form Lemma B consumes:

$(\mathrm H_c)$ There are $c>0$ and $x_0$ such that for every $x\ge x_0$, every pair of consecutive divisors $d<b$ of $L_x$ with $x\le\sqrt{db}\le\sqrt{L_x}$ satisfies $\log(b/d)\le(db)^{-c/(2\log x)}$.

(18a.1) Transfer (author-recorded; premise $(\mathrm H_c)$, unproved). Under $(\mathrm H_c)$, with $c\le4$ after decreasing $c$ if necessary,

$$h(L_x)\le\Bigl(\frac3c+o(1)\Bigr)(\log x)^2 =\Bigl(\frac3c+o(1)\Bigr)(\log\log L_x)^2 .$$

Proof. Take $T=x^{4/c}$ and $\delta(g)=2g^{-c/\log x}$ on $[T,\sqrt{L_x}]$; then $\delta$ is nonincreasing, $\delta\le2e^{-4}<1/8$, $2T\le\sqrt{L_x}$ for large $x$ because $\log L_x=\psi(x)\sim x$, and $(\mathrm H_c)$ is the hypothesis of Lemma B with this $\delta$. Here $\sigma(\ell)\ge(c/\log x)\ell-2$, so with $\kappa=c/\log x$ the integral is at most

$$\frac1\kappa\log\frac{\kappa\cdot\tfrac12\log L_x-2}{\kappa\log(2T)-2} \le\frac{\log x}{c}\log\frac{cx}{\log x}\le\frac{(\log x)^2}{c},$$

using $\kappa\log(2T)-2\ge2$ and $\tfrac12\log L_x\le x$, while $\log_2\sigma(\tfrac12\log L_x)=O(\log x)$ and $2\log_2(2T)=O(\log x/c)$. $\square$

The comment's form of the hypothesis, and where it fails. The erdosproblems.com comment writes $\Delta_x(Y)=\max_{d_i<Y}(\min\{d_{i+1},Y\}-d_i-1)$ for the largest gap between consecutive divisors of $L_x$ below $Y$ and proposes $\Delta_x(Y)+1\le Y^{1-c/\log x}$ for all large $x$ and all $x<Y\le L_x$. As stated this is false in the top range: at $Y=L_x$ the pair $(L_x/2,L_x)$ gives $\Delta_x(L_x)+1\ge L_x/2$, while $L_x^{1-c/\log x}=L_x\,e^{-c\psi(x)/\log x}$ is far smaller, and the same failure occurs on $(L_x/2+2L_x^{1-c/\log x},L_x]$ and near every $L_x/k$ with bounded $k$. The lower-half form $(\mathrm H_c)$ repairs it, because the mirror symmetry $d\mapsto L_x/d$ (facts (i) and (ii) inside Lemma B) makes the lower half suffice; the comment's recurrence $R'\le R^{1-c/\log x}$ is, up to a constant factor, the lower-range step of Lemma B and gives the constant $1/c$ there; the mirror recurrence on $U=L_x/R$ gives another $1/c$ in the upper range, so the direct count gives $2/c$, and only the factor $3/2$ between it and the $3/c$ above is a bookkeeping loss. The comment's appeal to strictly decreasing greedy divisors for distinctness is supplied by Lemma A.

(18a.2) What the known gap inputs give. Tenenbaum's 1987 paper (Ann. Inst. Fourier 37, no. 2, 1–18, open access, not held here; read in the Numdam text layer, where the displayed formulas are garbled and the page images were not read; its corrigendum, Ann. Inst. Fourier 50 (2000), no. 1, 317–319, cited by Hughes for the deduction on pp. 6–7 and Theorem 4, was not read) deduces its Théorème 1 from the admissibility, for every $\beta<3/2$, of $\varepsilon_k=\exp(-(\log k)^{\beta})$ for each of the three sequences, which it then establishes: for large $k$, every $z\in(\sqrt{L_{k-1}},\sqrt{L_k}]$ has a divisor of $L_k$ in $(z,(1+\varepsilon_k)z]$. Fed into Lemma B as the profile $\sigma(\ell)\approx(\log\ell)^{\beta}$ at scale $\ell\approx\tfrac12\psi(k)$, this gives $h(L_x)\ll_\beta x/(\log x)^{\beta}$ for every $\beta<3/2$, the analog of the Tenenbaum–Yokota factorial bound and a first improvement of Lemma A's $x/\log2$; a Berend–Harmse-strength profile, $\sigma(\ell)\asymp(\log\ell)^2$, would give order $x/(\log x)^2$ as Theorem 1 does for $n!$, but Berend and Harmse state (p. 572) that their method does not apply to $L$ at all. Neither input approaches $(\mathrm H_c)$, whose descent $\sigma(\ell)=c\ell/\log x$ is linear rather than polylogarithmic in $\ell$.

Scale of the hypothesis. $\log\tau(L_x)=(\log2+o(1))\,x/\log x$, so the window $[\sqrt{L_x}/x,\sqrt{L_x}]$, of logarithmic width $\log x$, holds a consecutive pair with $\log(b/d)\ge\log x/\tau(L_x)=\exp(-(\log2+o(1))x/\log x)$, while $(\mathrm H_c)$ allows such a pair at most $\exp(-(c/2+o(1))x/\log x)$; so necessarily $c\le2\log2+o(1)$. With $c$ near $2\log2$, $(\mathrm H_c)$ is at the mean-gap scale near $\sqrt{L_x}$, where the exponent $-(c/2)x/\log x$ it allows and the exponent $-(\log2)x/\log x$ of the mean gap agree to first order; at scale $x^m$ with $m$ fixed it allows gaps $e^{-cm}$ of constant size against a mean gap of order $x^{-m}$ up to powers of $\log x$, so below the top it asks much less than mean-gap matching. Near the top it is a statement of the kind no factorial gap estimate reaches (Berend and Harmse's Remark 2 says their bound is the best their method gives up to a bounded power of $\log_2n$).

Exploratory computation (author-recorded; a finite sample proves nothing). For $x\le64$ every divisor of $L_x$ was enumerated and, over the consecutive pairs with $x\le\sqrt{db}\le\sqrt{L_x}$, the least value of $\kappa(d,b)=(\log x)\log(1/\log(b/d))/\log\sqrt{db}$, the largest $c$ for which $(\mathrm H_c)$ holds at that pair, was computed: $\kappa_{\min}(x)=$ $1.116$, $1.257$, $1.311$, $1.298$, $1.348$, $1.358$ at $x=20,30,40,50,60,64$ ($\tau(L_{64})=4128768$). It shows no downward trend over this range (it dips once, from $1.311$ at $x=40$ to $1.298$ at $x=50$) and approaches $2\log2=1.386\ldots$ from below; the binding pairs sit at the top of the range, within a factor of about $2.6$ of $\sqrt{L_x}$ (the middle of the full divisor range of $L_x$), where the largest $\log(b/d)$ in the top tenth falls from $2.8\cdot10^{-2}$ at $x=20$ to $4.5\cdot10^{-5}$ at $x=64$. This was observed after the fact, not predicted. The script and command are retained on the lead page.

What it does not give. Nothing about $h(n!)$, hence nothing for questions (b) and (c); nothing unconditional beyond $h(L_x)\ll x/(\log x)^{3/2-\eta}$; and no bound on $h_\alpha(L_x)$ for Problem 1099, which needs gaps $\ll g^{-\eta}$ at every scale rather than $g^{-c/\log x}$ (and is settled by Tenenbaum anyway, see below).

Next investigation. The lead lcm_divisor_gap_greedy_route holds the route. A proof of $(\mathrm H_c)$ for any $c>0$ gives exponent $2$ in question (a) along $L_x$, independently of the two AI-generated chains; a disproof (a sequence of $x$ with $\kappa_{\min}(x)\to0$) shows that no hypothesis of the form $(\mathrm H_c)$ with a fixed $c$ holds, so (18a.1) as stated gives nothing along that sequence; whether a scale-dependent profile, or one that tolerates a few exceptional pairs, still gives $(\log x)^2$ through Lemma B would remain open. The computation can be pushed to $x$ near $100$ by enumerating only the divisors in a window around each scale instead of all of them.

Problem 18, question (c): what the greedy route would need

Target. Question (c): is $h(n!)<(\log n)^{O(1)}$? The page records it as open, between Remark 6's $(\log n)^2$ and the accepted $n^{o(1)}$ of question (b).

What transports. Lemma B is the only mechanism of this folder aimed at $n!$. To obtain $h(n!)\le(\log n)^{C}$ from it one needs $\int_{\ell_0}^{\frac12\log n!}d\ell/\sigma(\ell)\ll(\log n)^{C}$ over a range of length $\tfrac12\log n!\sim\tfrac12n\log n$; the natural profile is $\sigma(\ell)=\ell/(\log n)^{C-1}$, that is, consecutive divisors with geometric mean $g$ in ratio $1+O(g^{-1/(\log n)^{C-1}})$. At the middle, $g=\sqrt{n!}$, this is a gap $\exp(-n/(2(\log n)^{C-2}))$. But the window of index $n$ has logarithmic width $\tfrac12\log n$ and holds at most $\tau(n!)$ divisors, so some consecutive pair there has $\log(b/d)\ge\tfrac12\log n/\tau(n!)=\exp(-O(n/\log n))$, because $\log\tau(n!)\asymp n/\log n$ (the upper bound is on the Remark 6 page; the primes in $(n/2,n]$ give the lower bound). Hence the natural profile with $C=2$ is inadmissible, and no admissible profile does better than exponent $2$: the pair above forces $\sigma\le\log\tau(n!)$ on all but the last $\tfrac12\log n+O(1)$ of the range, so the integral is at least $(\tfrac12\log n!-O(\log n)-\log(2T))/\log\tau(n!)\gg(\log n)^2$ whenever $\log T\le(\tfrac12-\eta)\log n!$, and otherwise the term $2\log_2(2T)$ is $\gg n\log n$; this is consistent with Remark 6. Whether a profile capped at $\sigma=cn/\log n$ for a small constant $c$, whose integral is $O((\log n)^2)$, satisfies the hypothesis of Lemma B is a maximal-gap question that this counting leaves open. The natural profile with $C=3$ asks the largest gap to match this mean gap up to a constant in the exponent throughout the windows of index $j\asymp n$, while asking much less below them: the factorial analog of $(\mathrm H_c)$, with the same mean-gap character near the top. The Berend–Harmse window, $\exp(-(\log n)^2/(2\log2))$, is exponentially wider, which is why Theorem 1 stops at $n/\log n$.

What it does not give. The accepted proof of (b) does not run through a gap profile: its structure, read only at the level of theorem statements on the accepted target page, is a Fourier-decay property of product measures on $\mathbb Z/2^J$, closer in spirit to the character-sum criterion of Lemma 3.2 than to the greedy count, and nothing about divisor gaps can be extracted from this folder's reading of it. Outcome: open; the greedy route to (c) is a mean-gap problem for the divisors of $n!$ at least as hard as $(\mathrm H_c)$.

Problem 1099: sums of powers of consecutive divisor ratios

Target. Problem 1099 asks whether $\liminf_nh_\alpha(n)\ll_\alpha1$ for $\alpha>1$. Its page records the problem as solved, with the question proved, on Vose's accepted claim page, and records Tenenbaum's 1987 theorem, which settles the question along the factorials, the least common multiples and the primorials, on its own claim page. The erdosproblems.com page attributes the solution to Vose's construction, quotes Erdős's remark that $n!$ or $\mathrm{lcm}(1,\dots,n)$ "would be good candidates", and states that it remains unknown whether those two sequences have the property; its discussion carries a comment of 2 July 2026 that the problem was also solved by Tenenbaum in 1987. Tenenbaum's paper (cited above; read in the Numdam text layer, page images not read) states this as its Théorème 1 with $h(u)=u^\alpha$: $F_\alpha(N)=O_\alpha(1)$ uniformly for $N$ in the factorials, the least common multiples and the primorials, "la seconde partie de la conjecture d'Erdős", and Berend and Harmse's introduction (p. 570) records the same. So the factorial and lcm cases erdosproblems.com calls unknown are theorems of 1987, and the connection below reproves the factorial case with explicit constants from the reconstructed Corollary 3; it changes no status.

What transports (author-recorded; imports the Berend–Harmse estimate second-hand through Corollary 3). Let $\alpha>1$, $j_0=2^{16}$, $T=\sqrt{(j_0-1)!}$, and

$$C(\alpha)=4\sum_{j\ge j_0}(4\varepsilon_j)^{\alpha-1} \Bigl(\tfrac12\log j+\log2\Bigr),$$

a convergent series since $\log(1/\varepsilon_j)=(1-o(1))(\log j)^2/(2\log2)$. Then for every integer $n\ge T$,

$$2\sum_{d=1}^{n-1}d^{-\alpha}\ \le\ h_\alpha(n!)\ \le\ 2\zeta(\alpha)+C(\alpha),$$

so $2\zeta(\alpha)\le\liminf_nh_\alpha(n!)\le\limsup_nh_\alpha(n!)\le2\zeta(\alpha)+C(\alpha)$. The correction is small only when $\alpha$ is bounded away from $1$: for every $j$, $(4\varepsilon_j)^{\alpha-1}\le2^{-62(\alpha-1)}$, so $C(\alpha)\le2^{62\eta}C(1+\eta)\,2^{-62(\alpha-1)}$ for $\alpha\ge1+\eta$, with $C(2)<10^{-10}$ and $C(3/2)<10^{-3}$; but $C(5/4)\approx20$ exceeds $2\zeta(5/4)$, and $C(\alpha)\to\infty$ as $\alpha\to1^+$.

Proof. Write $N=n!$ and $r=b/d$ for a pair of consecutive divisors $d<b$; $r\le2$ by Lemma A. Sort the pairs into three zones. Bottom: $d<T$. Every integer up to $\lceil T\rceil\le n$ divides $N$, so $b=d+1$ and the pair contributes $d^{-\alpha}$; the zone contributes less than $\zeta(\alpha)$. Top: $b>N/T$. The mirror pair $(N/b,N/d)$ is consecutive with $N/b<T$, so $N/d=N/b+1$ and $r-1=1/(N/b)$; the zone contributes less than $\zeta(\alpha)$. The two zones are disjoint, since a pair in both would have $r>N/T^2>2$. Middle: $d\ge T$ and $b\le N/T$. If $db\le N$ then $g=\sqrt{db}\in[T,\sqrt N]$; otherwise the mirror pair has geometric mean $N/g\in[T,\sqrt N]$ and the same ratio. The windows of index $j_0\le j\le n$ cover $[T,\sqrt N]$, so Corollary 3 gives $\log r\le3\varepsilon_j$ at the window index $j$ of $\min(g,N/g)$. Then $r-1\le e^{3\varepsilon_j}-1\le4\varepsilon_j$ and $r-1\le2\log r$, so $(r-1)^\alpha\le(4\varepsilon_j)^{\alpha-1}\cdot2\log r$. The pairs assigned to window $j$ through their own geometric mean have spans $[\log d,\log b]$ with disjoint interiors inside an interval of length $\tfrac12\log j+\log2$ (as $d>g/\sqrt2$ and $b<\sqrt2g$), and the pairs assigned through their mirror likewise; summing $\log r$ over each family gives at most that length, so window $j$ contributes at most $4(4\varepsilon_j)^{\alpha-1}(\tfrac12\log j+\log2)$ and the zone at most $C(\alpha)$. The lower bound counts the pairs $(d,d+1)$, $1\le d\le n-1$, and their mirrors, which are distinct pairs for $n\ge4$. $\square$

Corollary (Erdős's companion question (1.2) of 1981, along $n!$). Since $\sum_i(r_i-1)-\log N=\sum_i(r_i-1-\log r_i)$ and $0\le r-1-\log r\le\tfrac12(r-1)^2$,

$$-1\ \le\ \sum_i\frac{d_{i+1}}{d_i}-\tau(n!)-\log n! \ \le\ \zeta(2)+\tfrac12C(2)-1\qquad(n\ge T),$$

so the difference Erdős asked about is bounded along the factorials.

What it does not give. Nothing for the least common multiples (Corollary 3 is a factorial statement; Tenenbaum's theorem covers them) and nothing for $\liminf$ over all $n$ beyond what Vose and Tenenbaum proved. The argument is Tenenbaum's Théorème 2, the principle Tenenbaum attributes to Vose, that $F(N_k;h)\le\sum_{j\le k}(1+\varepsilon_j)h'(\varepsilon_j)\log(N_j/N_{j-1})$ for an admissible sequence along a divisor sequence, run with the Berend–Harmse $\varepsilon_j$ in place of $\exp(-(\log j)^\beta)$ and with the small divisors counted exactly instead of through $\varepsilon_j=1$; Berend and Harmse (p. 572) note that their estimate could extend Tenenbaum's class of $h$ along $n!$ without pursuing it. Only the explicit bound $2\zeta(\alpha)+C(\alpha)$, with $C(\alpha)=O_\eta(2^{-62(\alpha-1)})$ for $\alpha\ge1+\eta$, appears to be new, and no search for it in the later literature was made.

Falsification attempts. The boundary $n\ge T$ is where "every integer up to $T$ divides $n!$" is used; below it the bottom zone is not a run of consecutive integers and the bound is not claimed. The window bound is applied only to pairs whose relevant geometric mean lies in $[T,\sqrt N]$, and the mirror map is a bijection of consecutive pairs, so no pair is dropped or counted twice. The convergence of $C(\alpha)$ for $\alpha$ near $1$ uses the superpolynomial decay of $\varepsilon_j$ recorded on the Corollary 3 page, not the weaker $\varepsilon_j\le j^{-4}$. No defect was found; the imported Berend–Harmse statement is the one Hughes prints and was not checked against the 1993 paper's page images.

Next investigation. None toward the target, which is settled. Tenenbaum's 1987 theorem, which settles the two candidate sequences erdosproblems.com calls unknown, has its own claim page on the problem page, linked above; this page's reading of the exact statement should be checked on the page images of the open-access paper (doi:10.5802/aif.1083) rather than in its text layer.

Problem 673: the sum of consecutive divisor ratios

Target. Problem 673 asks whether $G(n)=\sum_{i<\tau(n)}d_i/d_{i+1}\to\infty$ for almost all $n$ and for an asymptotic formula for $\sum_{n\le X}G(n)$. Its page records the problem as solved, with both questions proved, on Erdős and Tenenbaum's accepted 1983 claim page, whose Théorème 2 gives $\sum_{n\le x}G(n)=x\log x\,(1+o(1))$; the erdosproblems.com page records Erdős's 1982 remark that the almost-all statement is trivial, an erdosproblems.com comment's bounds $\tau(n/m)/m\le G(n)\le\tau(n)$ for $1<m\mid n$ (the lower bound fails at $m=1$, since $G(n)<\tau(n)$ for every $n$), and the Erdős–Tenenbaum result that $G(n)/\tau(n)$ has a continuous distribution function.

Outcome: fails. Both questions quantify over almost all $n$ or average over $n\le X$, while every mechanism of this folder concerns a single thin sequence: $n!$, $L_x$, or the constructed $2^EV_j$, all inside the practical numbers, which have density zero (their counting function is $\sim cx/\log x$ by Weingartner's theorem). Along $n!$ the statement is trivial without any gap estimate: Lemma A gives $d_i/d_{i+1}\ge\tfrac12$ for every pair, so $G(n!)\ge(\tau(n!)-1)/2\to\infty$, and the erdosproblems.com comment's bound already gives $\tau(n)/4\le G(n)$ for even $n$. Nothing here bears on the average. No transfer exists, and no next investigation is proposed.

Problem 859: the density of n representing a fixed t

Target. Problem 859 asks whether the density $d_t$ of the integers $n$ for which $t$ is a sum of distinct divisors of $n$ satisfies $d_t\sim c_1/(\log t)^{c_2}$. Its page records the problem as solved, with the answer disproved, on Kruer and Kohlmeyer's accepted Lean disproof, which the bounty site approved on 21 September 2026 and certified on 23 September 2026: $-\log d_t/\log\log t$ has lower limit $\delta=0.0860\ldots$ while $(\log t)^{\delta}d_t\to0$.

Outcome: fails. The target is resolved, and the mechanisms do not bear on it in either direction. A practical $N\ge t$ makes every multiple of $N$ a member of the set counted by $d_t$, so the constructions of this folder give $d_t\ge1/N$ for the least practical $N\ge t$ of the construction, which is at least $1/t^{2}$ (Proposition 4.1 places $N$ in $[t,t^2)$ once $t\ge x_0$) and at most $1/t$ for any seed $N\ge t$, so polynomial in $1/t$ whatever the construction, and says nothing at the scale $(\log t)^{-\delta}$. The accepted disproof's lower estimate does use "practical seeds" inside a tilted prime-pattern law (read only at the level of structure, as the problem page says), but as seeds of a density argument, not through $h$; the reconstructed bounds on $h$ neither supply nor replace that step. No transfer exists.

Other divisor problems examined

The search of the problems, research and theory trees for the objects of the proofs (practical numbers, sums of distinct divisors, consecutive divisor ratios, divisors of $n!$ and of $L_x$, residue covering by short divisor sums, greedy expansions) surfaced the following further pages; each is a resemblance without a transfer, for the reason given.

• Problem 144 (proved): almost every $n$ has divisors $d_1<d_2<2d_1$. For $n,x\ge3$ the divisors $2<3<4$ of $n!$ and $L_x$ give the property outright (Lemma A's non-strict ratio bound alone does not, as $N=2^k$ shows); the target is a density-one statement over all $n$.
• Problem 886 and Problem 887 (open): the number of divisors of $n$ in $(\sqrt n,\sqrt n+n^{1/2-\eta})$ or $(\sqrt n,\sqrt n+Cn^{1/4})$. For $n!$ these are multiplicative windows of relative width $(n!)^{-\eta}$ or $(n!)^{-1/4}$, exponentially narrower than the Berend–Harmse window $\varepsilon_n=\exp(-(\log n)^2/(2\log2)(1-o(1)))$; the reconstructed estimate says nothing at those scales. (Berend and Harmse's Theorems 3 and 4 concern the most balanced factorization of $n!$ and are not reconstructed here.)
• Problem 1100 (open): the number of coprime consecutive divisor pairs. The pairs $(d,d+1)$, $d<n$, give $\tau_\perp(n!)\ge n-1$, far below the question's $\exp((\log n!)^{o(1)})$; the mechanisms bound gaps, not coprimality.
• Problem 468, Problem 469, Problem 470 and Problem 1054: representation by partial sums of the divisor chain, pseudoperfect and weird numbers, and sums of the smallest divisors. Each concerns whether $n$ itself, or a given $n$, has one representation of a prescribed shape, not whether every $m\le N$ has a short one; the practical-number mechanism does not address them.
• Problem 350 (proved) and Problem 882 (solved): distinct subset sums and divisibility among subset sums, the opposite of the many-representations property that practicality asserts.
• Problem 446, Problem 448, Problem 449 and Problem 450: density statements about divisors of typical $n$ in short multiplicative intervals; the sequences here have density zero.

Leads created

Two connections amount to routes someone could take up now, and each has a lead folder: factorial_unit_fraction_transfer records (304.1) and (293.1), the conditional deductions from the accepted factorial bound with Lemma C and Lemma C′, and the fresh-context review they need before the problem pages can carry them; lcm_divisor_gap_greedy_route records the greedy route to exponent $2$ in question (a) along $L_x$ with its explicit premise $(\mathrm H_c)$, the retained exploratory script, and the obstruction at the mean-gap scale. The Problem 1099 reproof is not a lead: its target is settled, and its literature question is a reading task noted above.

Sources read for this page

Beyond the folder's reconstruction pages and the six problem pages, this page read: physical p. 6 of the van Doorn note in the text extracted from the held PDF (Lemma 5.1 and the proofs of Theorems 1.2 and 1.3); the erdosproblems.com pages of Problems 1099 and 673 and the discussion threads of Problems 18 and 1099 on 2026-09-28; and the text layers of Tenenbaum's 1987 paper (doi:10.5802/aif.1083) and of Berend and Harmse's 1993 paper (doi:10.5802/aif.1348) on Numdam, both open access, neither held in the library, with displayed formulas garbled and page images not read. Statements taken from those text layers are marked as such above and should be verified on the page images before any problem page relies on them. The 2000 corrigendum to Tenenbaum's paper (Ann. Inst. Fourier 50, no. 1, 317–319), which Hughes cites for the deduction on pp. 6–7 and Theorem 4 of the 1987 paper, was not read.