Reused writeup on Math-832 (gpt pro 5.2)

Erdős Problem 489 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/489

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/489.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/489.tex.

FORMAL RESTATEMENT

Let $A\subseteq\mathbb{N}$ be a set such that \[|A\cap[1,x]| = o(x^{1/2})\quad\text{as }x\to\infty. \] Define \[B := \{ n\ge 1 : \forall a\in A,\ a\nmid n\}. \] List the elements of $B$ in increasing order as $B=\{b_1<b_2<\cdots\}$ (if $B$ is finite, the sequence terminates). Consider the quantity \[M(x):=\frac{1}{x}\sum_{\substack{b_i<x}}(b_{i+1}-b_i)^2,\] where the sum is over all indices $i$ with $b_i<x$ and $b_{i+1}$ denotes the next element of $B$ (so this sum involves gaps between consecutive elements of $B$ up to scale $x$).

Question: Does $\lim_{x\to\infty} M(x)$ exist, and is it finite?

Edge cases:

• If $1\in A$ then $B=\varnothing$ and $M(x)=0$ for all $x$.
• If $A$ is finite then $B$ is periodic modulo $\mathrm{lcm}(A)$, so the limit should exist (proved below).

QUICK LITERATURE/CONTEXT CHECK

The statement notes: when $A=\{p^2: p\text{ prime}\}$, $B$ is the squarefree numbers, and Erdős proved existence of the limit in that case. Per the integrity rule, I do not use any other external results.

ATTACK PLAN

Proof track:

• Prove basic consequences of the sparsity assumption $|A\cap[1,x]|=o(\sqrt x)$.
• Prove the limit exists for finite $A$ via periodicity.
• Try (unsuccessfully here) to control the second moment of gaps for infinite $A$ from the sparsity assumption.

Disproof track:

• Try to build a sparse $A$ whose excluded multiples force very irregular gaps in $B$.

I only reach partial results.

WORK

Lemma 1 (the sparsity assumption forces a convergent reciprocal sum).

If $A\subseteq\mathbb{N}$ satisfies $|A\cap[1,x]|=o(\sqrt x)$, then \[\sum_{a\in A}\frac{1}{a}<\infty. \]

Proof.

For $j\ge 0$ let $I_j:=(2^j,2^{j+1}]$ and let $A(x):=|A\cap[1,x]|$. Then \[ \sum_{a\in A}\frac{1}{a} = \sum_{j\ge 0}\sum_{a\in A\cap I_j}\frac{1}{a} \le \sum_{j\ge 0}\frac{|A\cap I_j|}{2^j} \le \sum_{j\ge 0}\frac{A(2^{j+1})}{2^j}. \] Since $A(x)=o(\sqrt x)$, for every $\varepsilon>0$ there exists $J(\varepsilon)$ such that for all $j\ge J$ we have \[A(2^{j+1}) \le \varepsilon\,\sqrt{2^{j+1}}=\varepsilon\,2^{(j+1)/2}. \] Therefore for $j\ge J$, \[\frac{A(2^{j+1})}{2^j} \le \varepsilon\,\frac{2^{(j+1)/2}}{2^j}=\varepsilon\,2^{-(j-1)/2}. \] The tail series $\sum_{j\ge J} 2^{-(j-1)/2}$ converges (it is geometric with ratio $2^{-1/2}<1$), so the tail of $\sum_{j\ge 0} A(2^{j+1})/2^j$ is bounded by a constant multiple of $\varepsilon$. Since $\varepsilon$ was arbitrary and the initial segment $j<J$ is finite, the full series converges. Thus $\sum_{a\in A} 1/a<\infty$. $\square$

Lemma 2 (finite $A$ $\Rightarrow$ eventual periodicity of $B$ $\Rightarrow$ the limit exists and is finite).

If $A$ is finite, then the limit $\lim_{x\to\infty}M(x)$ exists and is finite.

Proof.

Assume $A$ is finite and let $L:=\mathrm{lcm}(A)$. Then divisibility by any $a\in A$ depends only on the residue modulo $L$. Hence membership in $B$ depends only on $n\bmod L$, so $B$ is a union of residue classes modulo $L$. Consequently the indicator sequence $\mathbf{1}_B(n)$ is periodic with period $L$.

From periodicity, the pattern of gaps between consecutive elements of $B$ is eventually periodic as well: once $n$ is large enough to include at least one full period, the sequence of residues in $B$ repeats every $L$. Therefore the multiset of gaps occurring in one period is finite, and for large $x$ the sum $\sum_{b_i<x}(b_{i+1}-b_i)^2$ is equal to \[Q(x)\cdot G + O(1), \] where $Q(x)=\lfloor x/L\rfloor$ is the number of full periods up to $x$, and $G$ is the sum of squared gaps contributed by one full period. (The $O(1)$ accounts for the incomplete initial/final partial period.) Dividing by $x$ gives \[M(x)=\frac{Q(x)G}{x}+O\left(\frac{1}{x}\right)=\frac{G}{L}+O\left(\frac{1}{x}\right), \] so $\lim_{x\to\infty}M(x)=G/L$ exists and is finite. $\square$

FAST REALITY CHECK (squarefree example).

For $A=\{p^2: p\text{ prime}\}$, $B$ is the squarefree numbers. Computing $M(x)$ numerically for $x=10^k$ (using exact sieving for squarefreeness up to $x$ and including the last gap crossing $x$), I obtained: \[ M(10)=1.7,\quad M(10^2)=2.04,\quad M(10^3)=2.038,\quad M(10^4)=2.0364,\quad M(10^5)=2.04128. \] These values suggest stabilization near $\approx 2.04$ at this scale, consistent with the existence of a finite limit in this special case (as stated in the problem text).

VERIFICATION

• Lemma 1: verified dyadic decomposition and used only the hypothesis $A(x)=o(\sqrt x)$.
• Lemma 2: periodicity argument is purely finite-state; the $O(1)$ remainder term comes from at most $O(L)$ boundary effects in the first/last partial periods.
• Numerical check: ensured the last gap $(b_i,b_{i+1})$ with $b_i<x<b_{i+1}$ is included by sieving slightly beyond $x$.

FINAL

UNRESOLVED

(i) Strongest proved partial result.

• Under the stated hypothesis $|A\cap[1,x]|=o(\sqrt x)$, the reciprocal sum $\sum_{a\in A}1/a$ converges.
• If $A$ is finite, then $B$ is periodic modulo $\mathrm{lcm}(A)$ and the limit $\lim_{x\to\infty}M(x)$ exists and is finite.
• For the specific infinite example $A=\{p^2\}$ (squarefree numbers), the problem statement asserts Erdős proved existence of the limit; my computation for $x\le 10^5$ gives $M(x)$ around $2.04$.

(ii) First gap (crisp). For infinite $A$ with $|A\cap[1,x]|=o(\sqrt x)$, prove that $M(x)$ converges as $x\to\infty$. Equivalently, obtain a uniform-in-$x$ control on the second moment of the gaps between consecutive integers avoiding all divisibilities by $A$.

(iii) Top 3 next moves.

1. Try to express $\sum_{b_i<x}(b_{i+1}-b_i)^2$ as a weighted count of runs of consecutive integers excluded by multiples of $A$, and bound the total contribution using the convergent reciprocal sum from Lemma 1.
2. Investigate whether a truncation argument works: define $A_T:=A\cap[1,T]$ and corresponding $B_T$ (periodic), then control the effect of adding moduli $a>T$ on the gap-square average.
3. Computationally search for sparse $A$ (e.g. sets with $a_j\approx j^2 \log j$) that might create highly irregular gaps, to test the plausibility of counterexamples.

(iv) Minimal counterexample structure. A counterexample would require an $A$ with $|A\cap[1,x]|=o(\sqrt x)$ such that the squared gaps in $B$ exhibit persistent oscillations: there would be infinitely many scales $x$ at which unusually long runs of excluded integers (divisible by some $a\in A$) contribute disproportionately to $\sum (\Delta b_i)^2$, causing $M(x)$ to have at least two distinct subsequential limits.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.