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Math-5960-A2 · revision r1 · on Math-5960

gpt pro 5.2 attempt on Math-5960

No claim recorded · LLM-generated · gpt pro 5.2 ·

Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Problem #873 (Erdős--Szemerédi: small LCM of consecutive terms)
  2. 1. Formal restatement
  3. 2. Quick literature/context check (browsing available)
  4. 3. Attack plan
  5. 4. Detailed work
  6. 4.1 Elementary monotonicity and trivial bounds
  7. 4.2 Consequence of the Erdős--Szemerédi $k=3$ bound
  8. 4.3 "Easy" cases where the conjectured conclusion holds
  9. 5. Verification (gap check, edge cases)
  10. 6. Final

Problem #873 (Erdős--Szemerédi: small LCM of consecutive terms)

1. Formal restatement

Let $A=\{a_1<a_2<\cdots\}\subseteq\mathbb{N}$ be infinite. For real $X\ge1$ and integer $k\ge1$, define \[ F(A,X,k) := \#\Big\{ i\ge1 : \operatorname{lcm}(a_i,a_{i+1},\dots,a_{i+k-1}) < X\Big\}. \]

Problem. Is it true that for every $\varepsilon>0$ there exists an integer $k\ge1$ (possibly depending on $A$ and $\varepsilon$) for which \[ F(A,X,k) < X^{\varepsilon} \] holds for all sufficiently large $X$? (The original prompt does not specify whether "for all $X$" or "for all large $X$"; most analytic-number-theory inequalities are meant for large $X$, and this is the interpretation I adopt.)

2. Quick literature/context check (browsing available)

The Erdős Problems website (problem #873) records an Erdős--Szemerédi theorem: for every infinite $A$, \[ F(A,X,3) \ll X^{1/3}\log X, \] and that there exist sets $A$ with $F(A,X,3)\gg X^{1/3}\log X$ for infinitely many $X$. The website lists the problem status as open.

3. Attack plan

  1. Record basic monotonicity properties of $F(A,X,k)$ in $X$ and $k$.
  2. Use the known $k=3$ result to deduce that the conjectured statement is true for every $\varepsilon>1/3$.
  3. Identify "easy" classes of sequences $A$ for which the statement is true (very sparse sequences, or sequences with large growth).
  4. Explain what additional input seems needed to reach arbitrarily small exponents $\varepsilon\to0$ for general $A$.

4. Detailed work

4.1 Elementary monotonicity and trivial bounds

Lemma (Monotonicity): For fixed $A$ and $X$, the function $k\mapsto F(A,X,k)$ is nonincreasing. For fixed $A$ and $k$, the function $X\mapsto F(A,X,k)$ is nondecreasing.
Proof: If $k'\ge k$, then $\operatorname{lcm}(a_i,\dots,a_{i+k'-1})\ge \operatorname{lcm}(a_i,\dots,a_{i+k-1})$ for every $i$, so the condition "$<X$" becomes harder as $k$ grows; hence $F(A,X,k')\le F(A,X,k)$. Monotonicity in $X$ is immediate from the definition. ∎
Lemma (Trivial counting bound): Let $A(X):=|A\cap[1,X)|$. Then for all $k\ge1$, \[ F(A,X,k)\le \max\{A(X)-k+1,\,0\}\le A(X)\le X. \]
Proof: If $\operatorname{lcm}(a_i,\dots,a_{i+k-1})<X$, then in particular $a_{i+k-1}<X$, so $i+k-1\le A(X)$, which implies $i\le A(X)-k+1$. This gives the first inequality. The bounds $A(X)\le X$ and $\max\{A(X)-k+1,0\}\le A(X)$ are immediate. ∎

4.2 Consequence of the Erdős--Szemerédi $k=3$ bound

Assume the known theorem: there exists an absolute constant $C>0$ such that for every infinite $A\subset\mathbb{N}$, \begin{equation} F(A,X,3) \le C\,X^{1/3}\log X\qquad (X\ge 2). \label{wsref-0-65-71-3a-45-53-33} \end{equation}

Then by monotonicity in $k$, for every $k\ge3$, $F(A,X,k)\le F(A,X,3)$, hence \begin{equation} F(A,X,k) \ll X^{1/3}\log X\qquad (k\ge3). \label{wsref-0-65-71-3a-6b-67-65-33} \end{equation}

In particular, for any fixed $\varepsilon>1/3$, taking $k=3$ yields $F(A,X,3)=o(X^{\varepsilon})$ as $X\to\infty$. Thus the problem is only genuinely difficult for $\varepsilon\le 1/3$.

4.3 "Easy" cases where the conjectured conclusion holds

  • If $A(X)\ll X^{\varepsilon}$ for some $\varepsilon>0$, then taking $k=1$ already gives $F(A,X,1)=A(X)\ll X^{\varepsilon}$. In particular, for extremely sparse sequences (e.g. exponential growth), the statement is trivial.
  • For the full set $A=\mathbb{N}$, one expects (and can prove with more work) that $F(\mathbb{N},X,k)$ is about $X^{1/k}$ up to polylog factors, so choosing $k>1/\varepsilon$ would suffice. This suggests the conjecture is plausible, but the difficulty is achieving a bound of the form $X^{c_k+o(1)}$ uniformly for all sets $A$, with $c_k\to 0$ as $k\to\infty$.

5. Verification (gap check, edge cases)

  • The interpretation "for all sufficiently large $X$" is not explicitly stated in the prompt; if the intent is uniform in all $X$, then even the meaning of "$<X^{\varepsilon}$" at small $X$ would need conventions. All standard asymptotic statements are intended for large $X$.
  • The deduction "$k\ge3\Rightarrow F(A,X,k)\le F(A,X,3)$" is correct by lcm monotonicity.
  • No attempt was made to reprove (eq:ES3) here; it is taken as known context.

6. Final

UNRESOLVED.

  1. Furthest point reached: Explained that the problem is open as stated; proved monotonicity and trivial bounds; deduced that the desired conclusion holds for every $\varepsilon>1/3$ using the known Erdős--Szemerédi estimate for $k=3$.
  2. Blocking issue: The known uniform bound is only at exponent $1/3$ (up to logs). To settle the question one needs bounds for larger windows $k$ that beat $X^{1/3}$ and in fact tend to $X^{o(1)}$ as $k\to\infty$, uniformly in $A$.
  3. Most plausible next steps: Develop a theory controlling how many long blocks can have small lcm, perhaps via: (i) estimates for the distribution of $y$-smooth numbers inside an arbitrary set $A$, (ii) structure theory for sets whose consecutive elements have large gcd overlap, (iii) adapting sum-product or multiplicative energy methods to the lcm constraint.
  4. Small experiments/checks: For model sets ($A=\mathbb{N}$, primes, smooth numbers, products of two primes, etc.) one can numerically estimate $F(A,X,k)$ to guess the correct exponent as a function of $k$, but this does not resolve the general worst-case problem.