Erdős Problem #421

Erdős Problem 421

Statement and scope.

A construction that retains every prime.

Proof that no equal consecutive products occur.

The sets $A_p$ are nested and future stages only append larger elements. Every pair of finite blocks of $A$ is therefore already present as the same pair of blocks in some $A_p$. They cannot collide. Also every prime is retained, so $A$ is infinite. Initial stages are \[ \begin{split} A_5&=\{2,3,4,5\},\\ A_7&=\{2,3,4,5,7\},\\ A_{13}&=\{2,3,4,5,7,8,9,10,11,13\},\\ A_{17}&=A_{13}\cup\{17\}. \end{split} \] Here $6$ is rejected because $2\cdot3=6$, $12$ because $3\cdot4=12$, and the interval $(13,17)$ because $3\cdot4\cdot5\cdot7\cdot8=14\cdot15\cdot16$. These stages and the absence of other collisions were checked exactly.

The exact density estimate still needed.

One can separate $1$ into gaps of length at most $p^\theta$ and longer gaps, with $\theta=5/72$. It suffices to prove that each of the two corresponding weighted sums is $o(X)$. Counting rejected gaps without their lengths would not suffice. Pratt's proof supplies the necessary estimates through results on primes in almost all short intervals and bounds for integer points on algebraic curves, together with a recursive analysis of rejection. Those estimates are the missing work here; the elementary construction alone does not establish density one.

Sources and attribution.

Completion estimate.

ROUND-3 OBJECTIVE

We pursue path (C) obstruction/correction. Round-2 already ruled out the original dyadic "hit the current collision family" argument. The most promising continuation is therefore to audit the revised January 2026 Moser--Tardos formulation, because that is the latest publicly described route claiming to overcome the Round-2 instability.

The new goal of this round is to prove a sharper, theorem-level obstruction: the revised collision-cylinder encoding still fails near the left boundary of each dyadic block. More precisely, the admissible pattern sets used to define the bad events are automatically empty for endpoints in a boundary strip $[X,X+2T(X)]$. Hence the bad-event family cannot possibly encode all logarithmic collisions with maximal element in $[X,2X)$. This is not the same as the Round-2 same-scale deletion obstruction; it is a separate defect in the revised local-pattern/LLL module.

I also isolate the minimal correction needed to make that module logically coherent.

Round-2 FOUNDATION USED

I use the following Round-2 results as fixed inputs.

1. Future-scale immutability. Deleting elements $\ge 2X$ cannot create or destroy collisions all of whose values are $<2X$.
2. Same-scale instability. Deleting inside $[X,2X)$ can create a new collision whose maximal value still lies in $[X,2X)$.
3. Corrected dyadic implication theorem. If one can choose dyadic deletion sets $B_X\subset[X,2X)$ so that after stage $X$ there are no collisions with values $<2X$, and if $|B_X|=o(X)$, then the final set has density $1$ and no collisions.

I also use the Round-1 collision normal form, which rewrites any collision as an equality of products of two disjoint consecutive blocks.

NEW INSIGHT / TOOL (ROUND-3)

The new ingredient is a boundary-strip emptiness lemma for current-scale-only local pattern encodings.

The revised January 2026 proof introduces, for each dyadic $X$, a variable set \[ V_X=[X,2X)\cap A_0, \] windowed local patterns near an endpoint $u$, and an admissibility condition requiring that every integer interval $J$ of length $L(X)$ inside the local window contain at least a positive proportion of currently deleted points from $V_X$. The collision bad events are indexed only by such admissible patterns.

I prove that any such scheme is automatically empty on the boundary strip $u\in[X,X+2T(X)]$: there exists a length-$L(X)$ test interval $J$ lying entirely to the left of $X$, so $J\cap V_X=\varnothing$, and the required positive lower bound is impossible. Consequently the collision-cylinder family cannot encode collisions with an endpoint in that strip.

This is a strictly new obstruction beyond Round-2, and it rules out an entire class of boundary-buffer Moser--Tardos encodings that only record the current-scale deletion bits.

ATTACK PLAN (ROUND-3)

Round-2 left the following possibility open:

Replace the invalid one-shot hitting-set step by a Moser--Tardos/LLL construction on local pattern events.

The exact gap to check now is whether the bad-event family really covers all logarithmic collisions with maximal element in $[X,2X)$.

The plan is:

1. Prove an abstract boundary-strip emptiness lemma for current-scale-only admissible pattern sets.
2. Apply it to the January 17, 2026 preprint's definitions of $\Sigma_u$ and $T_X$.
3. Deduce that the collision-cylinder family misses an entire boundary strip of possible endpoints.
4. Record the second, formal mismatch: the scale theorem quoted in the preprint states only an upper bound on $|B_X\cap I|$ for $I\subset[X,2X)$, whereas the coverage lemma assumes a two-sided capacity condition on $D_{<X}\cup B_X$ for all $I\subset[X/2,2X)$.
5. Give the minimal corrected statement consistent with the Round-2 corrected dyadic implication theorem.

This overcomes the previous obstacle in the sense that it identifies a concrete, local, rigorously provable failure mode in the revised proof, rather than merely saying that the combinatorics is delicate.

WORK (ROUND-3)

Abstract boundary-strip emptiness for current-scale pattern families

Fix positive integers $X,L,T$ with $T\ge 1$. Let \[ V_X\subseteq [X,2X) \] be any set of "current-scale eligible variables". For $u\in[X,2X)$ suppose one defines a family $\Sigma_u$ of local patterns on a subset of the current-scale window near $u$, with the following property:

($*$) For every $\sigma\in\Sigma_u$ and every integer interval $J\subset [u-(L+2T),u)$ of length $L$, one has \[ \sum_{n\in J\cap V_X}\sigma(n)\ge cT \] for some fixed constant $c>0$.

Here the sum is taken only over the current-scale positions $J\cap V_X$.

Lemma 5.1 (boundary-strip emptiness). Assume ($*$). Then for every \[ u\in [X,X+2T] \] one has $\Sigma_u=\varnothing$.

Proof. Fix $u\in[X,X+2T]$. Consider the interval \[ J:=[X-L,X). \] It has length $L$. Also, \[ u-(L+2T)\le X-L \] because $u\le X+2T$, and clearly $X\le u$. Hence \[ J\subset [u-(L+2T),u). \] But by definition $V_X\subseteq [X,2X)$, so \[ J\cap V_X=\varnothing. \] Therefore for every pattern $\sigma$ one has \[ \sum_{n\in J\cap V_X}\sigma(n)=0. \] Since $T\ge1$ and $c>0$, this contradicts the lower bound $0\ge cT>0$ required by $(*)$. Hence no such $\sigma$ exists and $\Sigma_u=\varnothing$. $\square$

The point is that the contradiction is structural: it does not depend on number theory, on the geometry input, or on whether a collision actually exists. It follows solely from using a positive lower bound on current-scale deletions inside a buffered window that reaches left of $X$.

Application to the January 17, 2026 preprint

In the January 17, 2026 preprint, the scale-$X$ Moser--Tardos module introduces \[ V_X=[X,2X)\cap A_0, \] and for $u\in[X/2,2X)$ it defines admissible local patterns $\sigma_u\in\Sigma_u$ by the condition that for every integer interval \[ J\subset [u-(L+2T),u) \qquad\text{with}\qquad |J|=L, \] one has \[ \frac{T}{200} \le \sum_{n\in J\cap V_X}\sigma_u(n) \le \frac{T}{2}. \] The collision bad events $A_\tau$ are indexed by tuples \[ \tau=(x,y,\ell,k,\sigma_x,\sigma_y) \] with $\sigma_x\in\Sigma_x$ and $\sigma_y\in\Sigma_y$.

Corollary 5.2 (preprint boundary emptiness). For the preprint's pattern definition, one has \[ \Sigma_u=\varnothing \qquad\text{for every }u\in[X,X+2T]. \]

Proof. Apply Lemma 5.1 with $c=1/200$. $\square$

Corollary 5.3 (the cylinder family misses a boundary strip). In the preprint's collision index set $T_X$, no tuple $\tau=(x,y,\ell,k,\sigma_x,\sigma_y)$ can have \[ x\in[X,X+2T] \qquad\text{or}\qquad y\in[X,X+2T]. \] Hence the bad-event family $\{A_\tau:\tau\in T_X\}$ does not encode any putative logarithmic collision whose endpoint $x$ or $y$ lies in that boundary strip.

Proof. If $x\in[X,X+2T]$, then Corollary 5.2 gives $\Sigma_x=\varnothing$, so no legal tuple $\tau\in T_X$ can involve that $x$. The same argument applies to $y$. $\square$

Consequence for the coverage lemma

The preprint's coverage lemma asserts:

If global capacity holds and there is a logarithmic collision in the final kept set with maximal element in $[X,2X)$, then there exists a tuple $\tau\in T_X$ such that the corresponding cylinder event $A_\tau$ occurs.

The proof says: take the actual endpoint patterns $\sigma_x,\sigma_y$, define $\tau=(x,y,\ell,k,\sigma_x,\sigma_y)$, and conclude $\tau\in T_X$.

Proposition 5.4 (the coverage step is not justified as written). The above proof step fails whenever $x\in[X,X+2T]$ or $y\in[X,X+2T]$.

Proof. By Corollary 5.2, in that situation one has $\Sigma_x=\varnothing$ or $\Sigma_y=\varnothing$. Therefore the actual pattern pair cannot satisfy the legality requirement \[ \sigma_x\in\Sigma_x, \qquad \sigma_y\in\Sigma_y \] needed to define a tuple in $T_X$. So the step "define $\tau$ by these endpoints, lengths, and patterns" is invalid in that boundary regime. $\square$

This proposition does not assert that a boundary-strip collision actually occurs in the final set. What it proves is the sharper statement needed for proof-checking: the current bad-event family does not cover that case. Thus the revised Moser--Tardos argument is incomplete unless the boundary case is handled separately or the pattern encoding is changed.

A second formal mismatch: theorem statement vs. coverage hypothesis

The same preprint defines, for every length-$L$ interval $I\subset[X/2,2X)$, the two-sided capacity bad event \[ B_I=B_I^+\cup B_I^- \] where \[ B_I^+ := \{ |(D_{<X}\cup B_X)\cap I|>T\}, \qquad B_I^- := \{ |(D_{<X}\cup B_X)\cap I|<T/100\}. \] The coverage lemma assumes \[ \bigcap_I \neg B_I, \] that is, two-sided capacity for all such intervals $I\subset[X/2,2X)$.

However, the scale theorem is stated only as:

for every interval $I\subset[X,2X)$ of length $L$, one has $|B_X\cap I|\le T/2$.

Proposition 5.5 (statement-level implication failure). The scale theorem's stated conclusion does not imply the hypothesis of the coverage lemma.

Proof. The coverage lemma requires two independent features:

1. an upper and lower bound on $|(D_{<X}\cup B_X)\cap I|$;
2. these bounds for all length-$L$ intervals $I\subset[X/2,2X)$.

1. it gives only an upper bound, and only for $|B_X\cap I|$, not for $|(D_{<X}\cup B_X)\cap I|$;
2. it restricts $I$ to lie inside $[X,2X)$, omitting intervals that meet the left buffer $[X/2,X)$.

This is weaker than Proposition 5.4: Proposition 5.5 is only a formal implication mismatch, while Proposition 5.4 gives a concrete structural reason the current event family cannot cover the boundary strip.

Minimal corrected statement

Round-2 already provided the right global target: a dyadic stage theorem must directly imply that after stage $X$, no collision with maximal value in $[X,2X)$ remains.

The minimal local correction consistent with the present Moser--Tardos strategy is:

At scale $X$, define local patterns on the full buffered window \[ \widetilde W_u:=[u-(L+2T),u)\cap [X/2,2X)\cap A_0, \] recording the total deletion pattern of $D_{<X}\cup B_X$ on $\widetilde W_u$ (earlier-scale bits are fixed constants, current-scale bits are random variables). Define admissibility using the total deletion pattern on every length-$L$ interval in the window. Then prove:

1. a corrected coverage lemma for these full-window patterns;
2. a two-sided capacity statement on all length-$L$ intervals in $[X/2,2X)$;
3. $|B_X|=o(X)$.

By the Round-2 corrected dyadic implication theorem, such a corrected scale theorem would be sufficient for the whole dyadic construction.

I do not prove that this correction works. The point of this round is narrower and rigorous: the current module does not yet provide it.

ADVERSARIAL VERIFICATION

I now try to break the new claims.

1. Did Lemma 5.1 accidentally use information about $A_0$? No. The proof uses only the set inclusion $V_X\subseteq [X,2X)$, so it is independent of any arithmetic property of $A_0$.

2. Is the interval $J=[X-L,X)$ always legal? Yes. It has length $L$. The only inclusion needed is $J\subset [u-(L+2T),u)$, and this follows exactly from $u\in[X,X+2T]$.

3. Could the preprint avoid the problem because no collision endpoint can lie in $[X,X+2T]$? No such restriction is stated anywhere in the scale theorem or coverage lemma. The theorem aims to remove all logarithmic collisions with maximal element in $[X,2X)$.

4. Is Proposition 5.5 merely cosmetic? No. The proof of the coverage lemma explicitly assumes two-sided capacity on the larger buffered interval range $[X/2,2X)$, while the scale theorem as stated guarantees only an upper bound for $B_X$ on the smaller range $[X,2X)$. This is a genuine logical gap in the printed implication.

5. Does this round disprove the original Erdős problem? No. It disproves neither the problem nor the possibility that a corrected Moser--Tardos implementation could work. It proves only that the current revised proof is not yet complete.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

This round makes a mathematically substantive advance beyond Round-2:

1. it proves a general boundary-strip emptiness theorem for current-scale-only local pattern encodings with a positive lower deletion requirement;
2. it applies that theorem to the January 17, 2026 Moser--Tardos preprint and deduces that the collision-cylinder family omits every endpoint in $[X,X+2T(X)]$;
3. it pinpoints the exact failure in the coverage lemma: the actual endpoint patterns need not lie in the admissible sets $\Sigma_x,\Sigma_y$;
4. it isolates a second, purely formal mismatch between the theorem statement and the hypothesis needed by the coverage lemma;
5. it gives the minimal corrected local-pattern statement that would need to be proved to revive the dyadic Moser--Tardos strategy.

So the investigation is still unresolved, but the current best publicly described proof route has now been narrowed to a sharper and more explicit obstruction than in Round-2.

COMPLETION ESTIMATE

COMPLETION: 70%

REFERENCES

1. T. F. Bloom, Erdős Problem #421, problem page and discussion thread.
2. P. Chojecki, A Density-One Sequence with Distinct Consecutive-Block Products, January 17, 2026 preprint.

ROUND-2 OBJECTIVE

We pursue path (C) obstruction/correction. The strongest Round-1 progress isolated the main unresolved issue as a stability problem for deletion-based constructions. In this round I sharpen that point and prove that the currently proposed dyadic deletion argument is not valid as written: there are two independent fatal obstructions.

1. The "capacity-constrained hitting set" lemma used in the January 16, 2026 draft is false.
2. Even if one deletes a hitting set for all currently visible collisions at a dyadic stage $[X,2X)$, that deletion can create new collisions whose maximal value still lies in $[X,2X)$.

Thus the current continuation strategy does not yet yield a proof of Erdős problem #421. The natural corrected statement is a stronger dyadic implication theorem, proved below, which isolates exactly what a future construction must establish.

Round-1 FOUNDATION USED

I use the following Round-1 results without reproving them.

1. Collision normal form. Any collision can be rewritten as an equality of products of two disjoint consecutive blocks \[ \prod_{i=M-\ell+1}^{M} d_i = \prod_{i=N}^{N+k} d_i, \qquad M<N, \] after cancelling overlap.

2. Deletion is non-monotone. Round-1 already proved that deleting terms can create new collisions.

3. Analytic input is not the current bottleneck. Round-1 verified a uniform polylogarithmic point-count for the non-line separated-variable curves in the logarithmic-degree regime, and also proved that affine line components are harmless in the collision region $x<y$, $x,y>0$.

The new work below strengthens item 2 in a way that directly breaks the dyadic stage argument, and separates the true combinatorial obstruction from the already-treated Diophantine point-counting input.

NEW INSIGHT / TOOL (ROUND-2)

The new ingredient is a precise decomposition of the stability issue.

1. Future-scale deletions are harmless. Once a dyadic stage has produced a set with no collisions whose values are $<2X$, later deletions above $2X$ cannot change that fact.

2. Same-scale deletions are dangerous. A deletion inside $[X,2X)$ can create a brand-new collision whose maximal value is still in $[X,2X)$. Hence a one-shot "hit all current collisions in $[X,2X)$" rule is insufficient.

3. The hitting-set lemma used in the draft is false. The lemma is false as a pure combinatorial statement, and the probabilistic proof also contains two separate errors.

This strictly advances the Round-1 analysis: the gap is no longer a vague "prefix-stability problem", but a concrete obstruction with explicit counterexamples and a corrected conditional theorem.

ATTACK PLAN (ROUND-2)

The Round-1 gap was:

Give a rigorous stagewise construction showing that deleting a sparse set in each dyadic interval both preserves local deficiency and eliminates all collisions permanently.

The exact subclaims that must be checked are now:

1. $G1$ Does there exist a low-capacity hitting set $B_X\subset [X,2X)$ for the relevant collision family?
2. $G2$ If such a set is chosen from the current collision family, does deleting it actually guarantee that no collisions with maximal value in $[X,2X)$ remain afterwards?

I prove:

1. a positive lemma showing that later scales are harmless;
2. an explicit same-scale counterexample disproving the naive stagewise stability needed for $G2$;
3. an explicit combinatorial counterexample disproving the hitting-set lemma used for $G1$;
4. a corrected dyadic implication theorem that states the minimal property a future construction would need.

WORK (ROUND-2)

Future-scale deletions do not affect earlier collisions

Lemma 1 (future-scale immutability). Let $A\subset \mathbb N$ be infinite, let $X\ge 1$, and let $B\subset [2X,\infty)$. Let \[ a_1<a_2<\cdots \qquad\text{and}\qquad a'_1<a'_2<\cdots \] be the increasing enumerations of $A$ and $A\setminus B$, respectively. Then the elements of $A$ below $2X$ and the elements of $A\setminus B$ below $2X$ coincide in the same order. Consequently every consecutive block product involving only values $<2X$ is identical in the two sequences. In particular, the existence or nonexistence of collisions all of whose values are $<2X$ is unchanged by deleting $B$.

Proof. Since every element of $B$ is at least $2X$, deleting $B$ removes no element of $A\cap [1,2X)$. Therefore \[ A\cap [1,2X)=(A\setminus B)\cap [1,2X) \] as sets. Because both sequences are increasing enumerations, their values below $2X$ appear in exactly the same order. Hence if \[ a_i,a_{i+1},\dots,a_j < 2X, \] then these same values form a consecutive block in the enumeration of $A\setminus B$ as well, with the same product, and conversely. Thus every consecutive block product formed entirely from values $<2X$ is the same before and after deleting $B$.

Therefore a collision whose participating values are all $<2X$ exists in the sequence enumerating $A$ if and only if it exists in the sequence enumerating $A\setminus B$. $\square$

This lemma shows that the only genuine stability problem is within the current dyadic scale. Later dyadic stages cannot revive a collision already eliminated below $2X$.

A same-scale instability counterexample

The next proposition strengthens Round-1's non-monotonicity example. It shows that deleting inside one dyadic block can create a new collision whose maximal value remains in that very same dyadic block.

Proposition 2 (same-scale instability). There exists a finite strictly increasing sequence with pairwise distinct consecutive-block products such that deleting a single term $b\in [X,2X)$ creates a new collision whose maximal value also lies in $[X,2X)$.

Proof. Take \[ A=(2,8,9,10,12,15). \] Its $\frac{6\cdot 7}{2}=21$ consecutive-block products are: \[ \begin{array}{llllll} 2, & 8, & 9, & 10, & 12, & 15,\\ 16, & 72, & 90, & 120, & 180,\\ 144, & 720, & 1080, & 1800,\\ 1440, & 8640, & 16200,\\ 17280, & 129600,\\ 259200. \end{array} \] When listed in increasing order these are \[ 2<8<9<10<12<15<16<72<90<120<144<180<720<1080<1440<1800<8640<16200<17280<129600<259200, \] so they are pairwise distinct. Hence $A$ has the distinct consecutive products property.

Now delete the element $8$. The resulting sequence is \[ A'=(2,9,10,12,15). \] In $A'$ we have the collision \[ 2\cdot 9\cdot 10 = 180 = 12\cdot 15. \] So the intervals $[1,3]$ and $[4,5]$ in $A'$ have equal products.

The deleted value is $8\in [8,16)$, and the maximal value occurring in the new collision is $15\in [8,16)$. Thus the new collision lies in the same dyadic scale as the deletion. $\square$

Consequence. A dyadic algorithm cannot justify stage $X$ by saying "choose $B_X$ to hit all collisions currently visible with maximal value in $[X,2X)$" and then conclude that no such collisions remain after deleting $B_X$. Proposition 2 gives a direct counterexample to that inference.

The capacity-constrained hitting-set lemma in the 2026 draft is false

The January 16, 2026 draft introduces the following statement at dyadic scale $X$:

If $\mathcal E_X$ is a family of subsets $V(E)\subset [X,2X)$ with $|V(E)|\le L(X)$ and $|\mathcal E_X|=o(X)$, then there exists a hitting set $B_X\subset [X,2X)$ such that every interval $I\subset [X,2X)$ of length $L(X)$ satisfies $|B_X\cap I|\le T(X)/2$.

Here $L(X)=\lceil C_0\log X\rceil$ and $T(X)=\lceil (\log X)^{2/3}\rceil$.

Proposition 3 (the hitting-set lemma is false). For all sufficiently large $X$, the above statement is false.

Proof. Fix a large $X$ and an interval $I\subset [X,2X)$ of length exactly $L(X)$. Since \[ L(X)=\lceil C_0\log X\rceil \quad\text{and}\quad T(X)=\lceil (\log X)^{2/3}\rceil, \] we have $L(X)\ge T(X)$ for all large $X$. Choose $T(X)$ distinct points \[ v_1,\dots,v_{T(X)}\in I. \] Define a family $\mathcal E_X$ by taking $T(X)$ singleton vertex sets: \[ V(E_r):=\{v_r\},\qquad r=1,\dots,T(X). \] Then every $V(E_r)$ has size $1\le L(X)$, and \[ |\mathcal E_X|=T(X)=O((\log X)^{2/3})=o(X). \] So the hypotheses of the stated lemma are satisfied.

But any hitting set $B_X$ for $\mathcal E_X$ must meet every singleton $\{v_r\}$, hence must contain all the points $v_1,\dots,v_{T(X)}$. Therefore \[ |B_X\cap I|\ge T(X)>T(X)/2, \] which violates the required capacity bound.

Thus the lemma is false. $\square$

Two independent errors in the probabilistic proof of that lemma

The same draft gives a probabilistic proof by choosing one random point $v(E)\in V(E)$ for each $E\in \mathcal E_X$. There are two separate problems.

Proposition 4 (the probability estimate used in the draft is false). In the notation above, the draft's estimate \[ \mathbb P\left(v(E)\in I\right)\le \frac{|I|}{X}=\frac{L(X)}{X} \] is false in general.

Proof. Take the family from Proposition 3. For each $r$ one has $V(E_r)=\{v_r\}\subset I$. Since $v(E_r)$ is chosen uniformly from the singleton $V(E_r)$, it is forced that \[ \mathbb P\left(v(E_r)\in I\right)=1. \] But $L(X)/X<1$ for every sufficiently large $X$. Hence the claimed upper bound fails. $\square$

Proposition 5 (the union-bound asymptotic in the draft is false). Even if one had a tail estimate of the form \[ \mathbb P\left(|B_X\cap I|>T(X)/2\right)\le e^{-cT(X)} \] for each interval $I$ of length $L(X)$, the subsequent claim \[ 2X e^{-cT(X)}\to 0 \qquad (X\to\infty) \] with $T(X)=(\log X)^{2/3}$ is false.

Proof. We have \[ 2X e^{-cT(X)}= \exp\!\left(\log(2X)-c(\log X)^{2/3}\right). \] As $X\to\infty$, the exponent tends to $+\infty$ because $\log X$ dominates $(\log X)^{2/3}$. Therefore \[ 2X e^{-cT(X)}\to \infty, \] not to $0$. $\square$

So the hitting-set step fails at three levels:

1. the combinatorial statement is false,
2. the probability estimate used to prove it is false,
3. the final union-bound asymptotic is false.

Minimal corrected dyadic implication

The previous propositions do not disprove the original Erdős problem. They show that the current stagewise deletion argument is incomplete. The right correction is to formulate exactly what a dyadic construction would actually need to prove.

Theorem 6 (corrected dyadic implication). Let $L(X)$ and $T(X)$ be positive functions on dyadic integers $X=2^t$. Suppose one can choose sets \[ B_X\subset [X,2X)\cap \mathbb Z \qquad (X=2,4,8,\dots) \] such that:

1. (i) for every interval $I\subset [X,2X)$ of length $L(X)$, \[ |B_X\cap I|\le T(X)/2; \]

2. (ii) if \[ A_{\le X}:=\mathbb N\setminus \bigcup_{Y\le X} B_Y, \] then in the increasing enumeration of $A_{\le X}$ there is no collision all of whose values are $<2X$;

3. (iii) $|B_X|=o(X)$ as $X\to\infty$ through dyadic values.

Then the final set \[ A:=\mathbb N\setminus \bigcup_X B_X \] has asymptotic density $1$, and the increasing enumeration of $A$ has pairwise distinct consecutive-block products.

Proof. First we prove density $1$. Write $X_j=2^j$ and $\varepsilon_j:=|B_{X_j}|/X_j$. By hypothesis, $\varepsilon_j\to 0$. Let $N\ge 1$ and choose $m$ with $2^m\le N<2^{m+1}$. Then \[ \#\left(([1,N]\cap \mathbb N)\setminus A\right) \le \sum_{j\le m} |B_{2^j}| =\sum_{j\le m} \varepsilon_j 2^j. \] Fix $\eta>0$. Choose $j_0$ so that $\varepsilon_j\le \eta$ for all $j\ge j_0$. Then \[ \sum_{j\le m} \varepsilon_j 2^j \le \sum_{j<j_0} \varepsilon_j 2^j + \eta \sum_{j=j_0}^{m}2^j \le C(\eta)+2\eta N. \] Dividing by $N$ and letting $N\to\infty$ gives \[ \frac{\#(([1,N]\cap \mathbb N)\setminus A)}{N}\le 2\eta+o(1). \] Since $\eta>0$ is arbitrary, the complement of $A$ has density $0$, so $A$ has density $1$.

Now suppose, for contradiction, that the increasing enumeration of $A$ has a collision. Let $M$ be the maximal value among all entries occurring in the two equal consecutive blocks. Choose dyadic $X$ with \[ X\le M<2X. \] All values appearing in the collision are then $<2X$. By Lemma 1, deleting elements above $2X$ cannot affect any collision all of whose values are $<2X$. Therefore the same collision already exists in the increasing enumeration of $A_{\le X}$. This contradicts property (ii).

Hence no collision exists in $A$. $\square$

Interpretation. Theorem 6 shows exactly what a dyadic construction must establish. The difficult part is not merely counting current collisions. One must choose $B_X$ so that after the deletion is performed, no collision with maximal value in $[X,2X)$ remains. Proposition 2 shows that hitting only the currently visible collisions is not enough.

ADVERSARIAL VERIFICATION

I now try to break the new work.

1. Does Lemma 1 overlook index changes? No. The statement is about collisions whose values are below $2X$. Deleting only values $\ge 2X$ leaves the list of surviving values below $2X$ unchanged and in the same order, so every consecutive block among those values is unchanged.

2. Could Proposition 2 fail because I missed a duplicate in the original sequence? No. All $21$ consecutive-block products were written explicitly, and when sorted they form a strictly increasing list. Therefore the original sequence has no collision.

3. Could the new collision in Proposition 2 already have existed before deletion in some disguised form? No. The original sequence had no collision at all. After deletion, the equality $2\cdot 9\cdot 10=12\cdot 15$ is a genuinely new equality of consecutive-block products.

4. Does Proposition 3 use a family of vertex sets too special to matter? It does not need to model actual collisions. The draft's lemma was stated for all families of subsets $V(E)\subset [X,2X)$ with $|V(E)|\le L(X)$ and $|\mathcal E_X|=o(X)$. A single counterexample therefore disproves the lemma exactly as stated.

5. Could Proposition 5 be repaired by changing constants? No. For any fixed $c>0$, \[ \log(2X)-c(\log X)^{2/3}\to +\infty, \] so the claimed limit cannot hold with $T(X)=(\log X)^{2/3}$.

6. Does Theorem 6 smuggle in the conclusion? No. It is a genuine implication theorem. Condition (ii) is strictly stronger than the invalid "hit all currently visible collisions" rule, and Proposition 2 proves that such strengthening is genuinely necessary.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

Round-2 does not solve Erdős problem #421, but it makes a mathematically substantive advance:

1. it proves that the currently proposed dyadic deletion proof is not valid as written;
2. it exhibits an explicit same-scale instability counterexample, showing that deleting a hitting set for the current collision family can create new collisions in the same dyadic block;
3. it disproves the draft's capacity-constrained hitting-set lemma by an explicit combinatorial counterexample;
4. it supplies the minimal corrected dyadic implication theorem (Theorem 6), which isolates the exact combinatorial task still required.

So the analytic point-counting side may be in good shape, but the combinatorial deletion mechanism remains unresolved and is now sharply formulated.

COMPLETION ESTIMATE

COMPLETION: 62%

REFERENCES

1. T. F. Bloom, Erdős Problem #421, problem page and discussion thread, accessed March 7, 2026.
2. P. Chojecki, Density-One Sequences with Distinct Consecutive Products, draft dated January 16, 2026.
3. W. Castryck, R. Cluckers, Ph. Dittmann, and K. H. Nguyen, The dimension growth conjecture, polynomial in the degree and without logarithmic factors, Algebra & Number Theory 14 $2020$, no. 8, 2261--2294.