Erdős Problem #41

Erdős Problem 41 (Round 4)

Problem

An infinite set $A\subset \mathbb{N}$ is a $B_3$-set if all sums \[ a_1+a_2+a_3\qquad (a_1\le a_2\le a_3,\ a_i\in A) \] are distinct. The target statement is \[ \liminf_{N\to\infty}\frac{|A\cap[1,N]|}{N^{1/3}}=0. \] Write \[ A(N):=|A\cap[1,N]|. \]

1) ROUND-4 OBJECTIVE

Path (C): obstruction/correction.

Round 3 showed that the Round 2 "full proof" was not gap-free, and reduced the original Erdős problem to the oscillatory regime \[ 0<\liminf_{N\to\infty}\frac{A(N)}{N^{1/3}} < \limsup_{N\to\infty}\frac{A(N)}{N^{1/3}} <\infty. \] The most promising Round 4 advance is therefore not to restart the whole proof, but to sharpen that obstruction: we prove that no such oscillation can stabilize on an asymptotically dense multiplicative scale. This rules out an entire class of possible counterexamples.

2) Round-3 FOUNDATION USED

We use the following vetted results from Round 3.

• Round 1 Lemma 41.1 (retained in Round 3). For every infinite $B_3$-set, \[ A(N)\le (18N)^{1/3}+2, \] hence \[ \limsup_{N\to\infty}\frac{A(N)}{N^{1/3}}<\infty. \]
• Round 3 Proposition 5.1. If the limit \[ \lim_{N\to\infty}\frac{A(N)}{N^{1/3}} \] exists for an infinite $B_3$-set, then it must equal $0$. Equivalently, no infinite $B_3$-set satisfies \[ A(N)\sim \alpha N^{1/3} \qquad (\alpha>0). \]
• Round 3 Proposition 5.2. Given only the verified inputs, the original Erdős problem is equivalent to ruling out the oscillatory regime \[ 0<\liminf_{N\to\infty}\frac{A(N)}{N^{1/3}} < \limsup_{N\to\infty}\frac{A(N)}{N^{1/3}} <\infty. \]

3) NEW INSIGHT / TOOL (ROUND-4)

Dense-subsequence rigidity by monotonicity.

Let $B(N)$ be any nondecreasing counting function. If $B(N)N^{-\gamma}$ converges along a sequence $N_k$ with $N_{k+1}/N_k\to 1$, then in fact $B(N)N^{-\gamma}$ converges along all integers $N$. Applied with $\gamma=1/3$, this means that a $B_3$-set cannot even look pseudo-cubic along an asymptotically dense multiplicative subsequence.

This is genuinely new in Round 4: Round 3 ruled out only global pseudo-cubic asymptotics; here we rule out subsequential pseudo-cubic asymptotics on scales $N_{k+1}=(1+o(1))N_k$.

4) ATTACK PLAN (ROUND-4)

The exact gap left after Round 3 is the possibility of a $B_3$-set with \[ 0<\liminf f(N)<\limsup f(N)<\infty, \qquad f(N):=\frac{A(N)}{N^{1/3}}, \] but with no full limit.

Round 4 proves the following claims.

1. A general rigidity lemma: if $f(N_k)$ converges along a sequence with $N_{k+1}/N_k\to 1$, then $f(N)$ converges globally.
2. Using Round 3 Proposition 5.1, this implies that no infinite $B_3$-set admits a subsequence $N_k$ with $N_{k+1}/N_k\to 1$ and $f(N_k)\to \alpha>0$.
3. Therefore any hypothetical counterexample must oscillate on every asymptotically dense multiplicative scale; equivalently, every positive cluster value of $f$ can only be attained along multiplicatively sparse subsequences.

This does not settle the full problem, but it excludes a broad family of oscillatory profiles that remained possible after Round 3.

5) WORK (ROUND-4)

Lemma 5.1 (dense-subsequence rigidity).

Let $B:\mathbb{N}\to[0,\infty)$ be nondecreasing, let $\gamma>0$, and define \[ g(N):=\frac{B(N)}{N^{\gamma}}. \] Suppose there is an increasing sequence $N_1<N_2<\cdots$ such that \[ \frac{N_{k+1}}{N_k}\to 1 \qquad\text{and}\qquad g(N_k)\to \alpha \] for some $\alpha\in[0,\infty)$. Then \[ g(N)\to \alpha\qquad (N\to\infty). \]

Proof.

Fix large $N$, and choose the unique index $k=k(N)$ such that \[ N_k\le N < N_{k+1}. \] Since $B$ is nondecreasing, \[ B(N_k)\le B(N)\le B(N_{k+1}). \] Dividing by $N^{\gamma}$ gives \[ g(N_k)\left(\frac{N_k}{N}\right)^{\gamma} \le g(N) \le g(N_{k+1})\left(\frac{N_{k+1}}{N}\right)^{\gamma}. \] Because $N_k\le N < N_{k+1}$, we also have \[ \frac{N_k}{N}\ge \frac{N_k}{N_{k+1}} \qquad\text{and}\qquad \frac{N_{k+1}}{N}\le \frac{N_{k+1}}{N_k}. \] Hence \[ g(N_k)\left(\frac{N_k}{N_{k+1}}\right)^{\gamma} \le g(N) \le g(N_{k+1})\left(\frac{N_{k+1}}{N_k}\right)^{\gamma}. \] Now $N\to\infty$ implies $k(N)\to\infty$, and by hypothesis \[ g(N_k)\to \alpha, \qquad g(N_{k+1})\to \alpha, \qquad \frac{N_{k+1}}{N_k}\to 1. \] Therefore both the lower and upper bounds tend to $\alpha$. By the squeeze theorem, \[ g(N)\to \alpha. \] This proves the lemma.

Corollary 5.2 (no positive asymptotic on asymptotically dense scales).

Let $A$ be an infinite $B_3$-set and define \[ f(N):=\frac{A(N)}{N^{1/3}}. \] Then there is no increasing sequence $N_1<N_2<\cdots$ such that \[ \frac{N_{k+1}}{N_k}\to 1 \qquad\text{and}\qquad f(N_k)\to \alpha>0. \]

Proof.

Assume such a sequence exists. Since $A(N)$ is nondecreasing, Lemma 5.1 applies with $B=A$ and $\gamma=1/3$, and yields \[ \frac{A(N)}{N^{1/3}}\to \alpha. \] But Round 3 Proposition 5.1 says that if the full limit $\lim_{N\to\infty}A(N)N^{-1/3}$ exists for an infinite $B_3$-set, then it must equal $0$. This contradicts $\alpha>0$. Therefore no such sequence can exist.

Corollary 5.3 (positive cluster values are multiplicatively sparse).

Let $A$ be an infinite $B_3$-set and let $\alpha>0$. If $N_k$ is any increasing sequence satisfying \[ \frac{A(N_k)}{N_k^{1/3}}\to \alpha, \] then \[ \frac{N_{k+1}}{N_k}\not\to 1. \] Equivalently, \[ \limsup_{k\to\infty}\frac{N_{k+1}}{N_k}>1. \]

Proof.

If $N_{k+1}/N_k\to 1$, then Corollary 5.2 gives a contradiction. So $N_{k+1}/N_k\not\to 1$. Since every ratio is at least $1$, this is equivalent to \[ \limsup_{k\to\infty}\frac{N_{k+1}}{N_k}>1. \]

Proposition 5.4 (sharpened obstruction to a counterexample).

Assume, for contradiction, that the original Erdős statement fails for some infinite $B_3$-set $A$. Then, by Round 3 Proposition 5.2, \[ 0<\ell:=\liminf_{N\to\infty}f(N) < L:=\limsup_{N\to\infty}f(N) <\infty. \] Let \[ K:=\left\{\beta\in\mathbb{R}:\ \exists N_k\to\infty\text{ with }f(N_k)\to\beta\right\} \] be the cluster set of $f$. Then $K\subset[\ell,L]\subset(0,\infty)$, and for every $\alpha\in K$, every increasing sequence $N_k$ with $f(N_k)\to\alpha$ must be multiplicatively sparse in the sense of Corollary 5.3.

In particular, no hypothetical counterexample can stabilize near any positive level $\alpha$ on a scale $N_{k+1}=(1+o(1))N_k$.

Proof.

The inequality \[ 0<\ell<L<\infty \] is exactly the unresolved regime identified in Round 3 Proposition 5.2. By definition of cluster set, every $\alpha\in K$ is obtained as the limit of $f(N_k)$ along some sequence $N_k\to\infty$. Since $K\subset[\ell,L]\subset(0,\infty)$, Corollary 5.3 applies to every such sequence and shows that $N_{k+1}/N_k\not\to1$. This is precisely the asserted multiplicative sparseness, and the final sentence is just a restatement.

6) ADVERSARIAL VERIFICATION

We now test the new argument for weaknesses.

• Use of Round 3 results. The only inherited nontrivial input beyond monotonicity is Round 3 Proposition 5.1, i.e. the verified "no pseudo-cubes" consequence of Helm. We do not reuse the unsupported stronger oscillation theorem from Round 2.
• Hidden regularity assumptions. Lemma 5.1 uses only that $B(N)$ is nondecreasing. No convexity, differentiability, or regular variation is assumed.
• Boundary case $\alpha=0$. Lemma 5.1 remains true for $\alpha=0$. Corollary 5.2 only excludes $\alpha>0$, exactly because Round 3 Proposition 5.1 leaves $\alpha=0$ possible.
• Could the ratios oscillate? Yes, and our statement allows that. Corollary 5.3 says only that the ratios cannot converge to $1$ if a positive cluster value is present. This is the precise conclusion proved.
• Does this settle the original Erdős problem? No. A counterexample could still oscillate between positive levels along sufficiently sparse subsequences. The remaining open regime is now narrower: any such oscillation must occur on genuinely macroscopic multiplicative jumps.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 3 reduced the problem to the regime \[ 0<\liminf_{N\to\infty}\frac{A(N)}{N^{1/3}} < \limsup_{N\to\infty}\frac{A(N)}{N^{1/3}} <\infty. \] Round 4 proves a new rigidity theorem showing that such a regime cannot stabilize along any asymptotically dense multiplicative subsequence $N_{k+1}=(1+o(1))N_k$. Equivalently, every positive cluster value of $A(N)/N^{1/3}$ must be realized only along multiplicatively sparse subsequences. This rules out an entire class of possible counterexamples and is strictly stronger than any result established in earlier rounds.

8) COMPLETION ESTIMATE

COMPLETION: 78%

9) REFERENCES

• P. Erdős, Problems and results on combinatorial number theory III, in Number Theory Day (Rockefeller Univ., 1976), Lecture Notes in Math. 626, Springer, 1977, pp. 43--72.
• M. Helm, On the Distribution of $B_3$-Sequences, J. Number Theory 58 $1996$, 124--129.
• K. O'Bryant, A Complete Annotated Bibliography of Work Related to Sidon Sequences, Electron. J. Combin. DS11 $2004$. Used for the recorded abstract of Helm's paper.
• J. Cilleruelo, Conjuntos de Sidon, in Aritmética, Grupos y Análisis, AGRA II lecture notes (2015; later online circulation checked in 2022). Used for the statement that the odd-$h$ liminf problem remains open there, and that Helm proves only the pseudo-cube obstruction.

Erdős Problem #41: independent continuation

Date: 2026-09-23. Research attempt; no claim of a new published result.

1) TARGET AND SOURCE AUDIT

If $A$ is an infinite $B_3$-set (unordered triple sums, with repetitions, are unique), must $\liminf A(N)/N^{1/3}=0$?

All supplied versions considered: 41.tex. The source proves a valid monotone interpolation lemma and invokes an earlier no-positive-limit result. Here I prove an additional unconditional restriction on the cluster set; I do not use the inherited theorem without its proof.

2) WORK

Put $f(N)=A(N)N^{-1/3}$. If $t=A(N)$, the $\binom{t+2}{3}$ triple multisets have distinct integer sums between $3$ and $3N$, hence $t^3/6\le3N$ and $f(N)\le18^{1/3}$. Also $A(N+1)-A(N)\in\{0,1\}$, giving \[ |f(N+1)-f(N)|\le (N+1)^{-1/3} +A(N)\left(N^{-1/3}-(N+1)^{-1/3}\right)=o(1). \] The second term is $O(N^{-1})$ by the mean value theorem and the preceding bound.

The entire interval is realized.

Write $\ell=\liminf f(N)$ and $L=\limsup f(N)$. The cluster set is exactly $[\ell,L]$. For $\ell<c<L$, there are arbitrarily late indices below $c$ and above $c$. Choose a crossing after index $M$. At its first crossing index $n$, one has $|f(n)-c|\le |f(n)-f(n-1)|$. Letting $M\to\infty$ proves that $c$ is a cluster value. The endpoints are cluster values by boundedness, and no other values can occur by the definitions of liminf and limsup.

Thus the oscillatory alternative in the source cannot have just two separated cluster values; it must realize every intervening value. This is a general small-increment observation, not a new $B_3$ density bound.

3) ADVERSARIAL VERIFICATION

An interval of cluster values does not supply a single convergent subsequence with successive scale ratios tending to one. Crossings of the same level may be far apart. Consequently it gives no contradiction to the supplied dense-subsequence lemma. The bound above allows repeated summands exactly as the definition requires.

4) FINAL

UNRESOLVED. Nothing forces the left endpoint $\ell$ to vanish. The missing ingredient must use uniqueness of triple sums more strongly than the elementary counting and small-increment bounds.

5) SOURCES CHECKED

Read all of 41.tex; its earlier-round Helm theorem is not needed above. Direct statement retrieval from https://www.erdosproblems.com/41 returned 403 on 2026-09-23.

6) COMPLETION ESTIMATE

This is a conservative subjective measure of progress toward the original question, not a probability of correctness or a claim that the remaining work is routine. COMPLETION: 8%

Erdős Problem #41 (Round 3)

Problem

An infinite set $A\subset\mathbb N$ is a $B_3$-set if all sums \[ a_1+a_2+a_3\qquad (a_1\le a_2\le a_3,\ a_i\in A) \] are distinct. The Erdős question is whether every infinite $B_3$-set satisfies \[ \liminf_{N\to\infty}\frac{|A\cap[1,N]|}{N^{1/3}}=0. \]

1) ROUND-3 OBJECTIVE

Path (C): obstruction/correction + strict progress. Round 2 claimed a full proof by invoking "Helm obstructions" $H1$--$H2$ as if they were unconditional. In this round we (i) identify the precise missing hypothesis in Helm's Theorem 1 (the proof only works for monotonically distributed sequences), (ii) record the strongest correct consequence obtainable from Helm (pseudo-cubes and monotone distribution are ruled out), and (iii) add a new structural lemma (a local version of Lemma 41.1) and a quantitative "early-gap  $\Rightarrow$ small ratio" criterion. This isolates a concrete remaining obstruction: to prove Erdős' liminf statement it would suffice to force Helm's long gap blocks to begin sufficiently early.

2) Round-2 foundation used

We rely on the following vetted Round 1 facts.

• Lemma 41.1 (Round 1). If $A\subseteq[1,N]$ is $B_3$ and $m:=|A|$, then $\binom{m+2}{3}\le 3N-2$, hence $m\le (18N)^{1/3}+2$.
• Lemma 41.2 (Round 1). Every infinite $B_3$-set is Sidon ($B_2$).

3) New insight / tool (Round 3)

1. Correct reading of Helm $1996$. Helm's key non-$B_3$ criterion (his Theorem 1) assumes the sequence is monotonically distributed. This hypothesis was omitted in Round 2, so the Round 2 "full proof" does not go through.
2. Helm's Theorem 3 (gap blocks). Under the (still hypothetical) assumption $\liminf A(n)n^{-1/3}>0$, Helm proves the existence of long consecutive blocks of unusually sparse length-$N$ intervals ("$\omega$-gaps").
3. Strengthening Lemma 41.1 to a local interval bound. Any translate of a finite $B_3$ set is $B_3$, hence the $O(N^{1/3})$ bound holds in every interval of length $N$.
4. New quantitative criterion. If Helm-type gap blocks occur early enough relative to their length, then the Erdős liminf conclusion follows.

4) Attack plan (Round 3)

• Gap after Round 2. Round 2 treated Helm's Theorem 1 as an unconditional obstruction $H2$. In fact Helm assumes monotone distribution. There is no proof (and it is not true in general) that every $B_3$ set is monotonically distributed.
• What we can still do. We state the correct Helm theorems and derive rigorous partial consequences. We then combine them with the local interval bound to obtain a clean "if the gap block starts early, then the normalized counting function dips to 0" lemma.
• Why this is progress. It isolates one sharply formulated missing step: show that Helm's gap blocks (which exist at all large scales under $\liminf>0$) can be taken to start within $o(\log^{1/6}N)$ blocks for infinitely many scales. That condition would settle Erdős Problem 41.

5) Work (Round 3)

5.1. Local $O(N^{1/3})$ bound on every interval.

5.2. What Helm actually proves (and what he does not).

We record the parts of Helm's paper that are directly relevant.

5.3. Correct partial consequence of Helm + Lemma 41.1.

5.4. Helm's gap-block theorem (what any counterexample must look like).

5.5. New lemma: early gap blocks force small normalized density.

Theorem Theorem (Helm, Theorem 3) does not control where the gap block begins. The next lemma shows that if the gap block begins early enough (relative to its length), then Erdős' liminf conclusion follows.

6) Adversarial verification

• Translation invariance (Lemma Lemma (Local version of Lemma 41.1)). Triple-sum distinctness is preserved under $a\mapsto a-M$ since $a+b+c=a'+b'+c'$ iff $(a-M)+(b-M)+(c-M)=(a'-M)+(b'-M)+(c'-M)$. Hence the local bound is rigorous.
• The Round 2 flaw. Helm Theorem 1 explicitly assumes "monotonically distributed". Dropping this hypothesis changes the statement and is not justified by any argument in Round 2.
• Lemma lem:early-gap constants. The proof only uses (i) Lemma 41.1 (Round 1) at $l_0N$, and (ii) the integral comparison $\sum_{l=l_0+1}^{l_0+K}l^{-2/3}\ll K^{1/3}$. No hidden hypotheses.
• Edge cases. If $l_0=0$, then the first term in (eq:ratio-bound) vanishes and we get $A(KN)/(KN)^{1/3}\ll 1/\omega(N)$, consistent with the formula.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

• The Round 2 "full proof" is not correct: it relies on an unconditional form of Helm's Theorem 1 that omits the needed monotonic distribution hypothesis.
• What is proved: Erdős' liminf statement holds for the important regularity classes covered by Helm (monotonically distributed sequences and pseudo-cubes), and any putative counterexample with $\liminf>0$ must exhibit Helm's long consecutive gap blocks at every scale.
• New progress: Lemma Lemma (Local version of Lemma 41.1) (local interval bound) and Lemma lem:early-gap (early gap blocks force the normalized counting function to dip to $0$) isolate a concrete missing step that would settle the conjecture.

8) Completion estimate

COMPLETION: 65%

9) References

• M. Helm, On the distribution of $B_3$-sequences, J. Number Theory 58 $1996$, 124--129.
• (For current-status bookkeeping) Erdős Problems website entry for Problem #41.