Reused writeup on Math-5683 (gpt pro 5.2)

Erdős Problem 486 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/486

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_5.2/486.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/486.tex.

FORMAL RESTATEMENT

Let $A\subseteq\mathbb{N}$ (with $\mathbb{N}=\{1,2,3,\dots\}$). For each $n\in A$, fix a subset $X_n\subseteq \mathbb{Z}/n\mathbb{Z}$. Define \[ B := \left\{ m\in\mathbb{N}:\ \forall n\in A\ (n<m\implies m\bmod n\not\in X_n)\right\}. \] Equivalently, $m\in B$ iff for every modulus $n\in A$ that is smaller than $m$, the residue class of $m$ modulo $n$ is not one of the forbidden classes $X_n$.

Define the (upper/lower) logarithmic density of $B$ (if it exists) by \[ \delta_{\log}(B) := \lim_{x\to\infty}\frac{1}{\log x}\sum_{\substack{m<x\\m\in B}}\frac{1}{m}, \] provided this limit exists.

Question: For all choices of $A$ and $\{X_n\}_{n\in A}$, must this limit exist?

Edge cases:

• If $A=\varnothing$, then $B=\mathbb{N}$ and $\delta_{\log}(B)=1$.
• If $1\in A$ and $X_1=\mathbb{Z}/1\mathbb{Z}$, then $B\subseteq\{1\}$ (indeed $m\in B$ only possible for $m\le 1$), so $\delta_{\log}(B)=0$.

QUICK LITERATURE/CONTEXT CHECK

The problem statement records that Davenport--Erdős proved the answer is yes when $X_n=\{0\}$ for all $n\in A$ (i.e. $B$ is the set of integers not divisible by any $n\in A$ with $n<m$), and that Besicovitch exhibited examples without a natural density even in that special case, motivating the logarithmic density question. Per the integrity rules for this project, I do not use any external literature beyond what is explicitly stated above.

ATTACK PLAN

Proof track:

• Prove the statement for restricted families (e.g. $A$ finite; $X_n$ empty/full; periodic setups).
• Try to approximate the harmonic sum for $B$ via finite truncations $A\cap[1,T]$ and control the tail $n>T$.

Disproof track:

• Attempt to encode oscillations on logarithmic scales by turning on new congruence restrictions at sparse moduli $n\in A$.

I only reach restricted positive results (finite $A$, and trivial extremal $X_n$); the general case remains open here.

WORK

Lemma 1 (finite $A$ $\Rightarrow$ eventual periodicity $\Rightarrow$ logarithmic density exists).

If $A$ is finite, then $B$ has a natural density (hence also a logarithmic density), and this density can be computed from one period modulo $L:=\mathrm{lcm}(A)$.

Proof.

Assume $A$ is finite and let $M:=\max(A)$ (if $A=\varnothing$ the conclusion is trivial). Let $L:=\mathrm{lcm}(A)$. For any integer $m>M$, the defining condition for membership $m\in B$ reads: \[m\in B \iff (\forall n\in A)\ \left(m\bmod n\not\in X_n\right),\] because $m>M\ge n$ ensures the implication "$n<m$" holds for all $n\in A$. Now, for each fixed $n\in A$, the residue $m\bmod n$ is determined by the residue $m\bmod L$ (since $n\mid L$). Therefore, for $m>M$, membership $m\in B$ depends only on $m\bmod L$.

Formally, define the subset $S\subseteq \mathbb{Z}/L\mathbb{Z}$ by \[S := \{ r\in \mathbb{Z}/L\mathbb{Z} : (\forall n\in A)\ (r\bmod n \not\in X_n)\}. \] Then for all integers $m>M$, \[m\in B \iff (m\bmod L)\in S. \] Hence $B\cap\{M+1,M+2,\dots\}$ is a union of residue classes modulo $L$. Such a set has a natural density equal to $|S|/L$ (because each residue class has density $1/L$), and adding/removing finitely many elements (those $\le M$) does not change the density. Therefore $B$ has a natural density $|S|/L$, and in particular the logarithmic density exists and equals the same value. $\square$

Lemma 2 (two extremal families of forbidden sets).

1. If there exists $n_0\in A$ with $X_{n_0}=\mathbb{Z}/n_0\mathbb{Z}$, then $B\subseteq\{1,2,\dots,n_0\}$ and $\delta_{\log}(B)=0$.
2. If $X_n=\varnothing$ for every $n\in A$, then $B=\mathbb{N}$ and $\delta_{\log}(B)=1$.

Proof.

$1$ If $X_{n_0}$ is all residue classes, then for any $m>n_0$ we have $m\bmod n_0\in X_{n_0}$, so the defining condition for $B$ fails at $n=n_0$. Thus no $m>n_0$ lies in $B$, i.e. $B\subseteq[1,n_0]$ is finite. For any finite set, the weighted sum $\sum_{m<x,m\in B}1/m$ is bounded as $x\to\infty$, so dividing by $\log x\to\infty$ gives limit $0$.

$2$ If every $X_n$ is empty, then the condition "$m\bmod n\not\in X_n$" holds for all $m,n$, so $B=\mathbb{N}$. Then \[\frac{1}{\log x}\sum_{m<x,\,m\in B}\frac{1}{m}=\frac{1}{\log x}\sum_{m<x}\frac{1}{m}\to 1\] because $\sum_{m<x}1/m = \log x + O(1)$. $\square$

FAST REALITY CHECK (one concrete example).

Take $A=\{2\}$ and $X_2=\{0\}$. Then $m\in B$ iff either $m\le 2$ or $m$ is odd, so up to finitely many exceptions $B$ is the odd integers. Numerically, using the exact identity $\sum_{\substack{m\le X\\ m\ \mathrm{odd}}} \frac1m = H_X - \tfrac12 H_{\lfloor X/2\rfloor}$ (with $H_n$ the $n$th harmonic number), I computed \[ \frac{1}{\log X}\sum_{\substack{m<X\\ m\in B}}\frac1m\approx \begin{cases} 0.5459759644 & X=10^6,\\ 0.5344819734 & X=10^8,\\ 0.5229879822 & X=10^{12}, \end{cases} \] which drifts slowly downward toward the expected limit $1/2$.

VERIFICATION

• Lemma 1: checked that the $m>n$ condition becomes "all $n\in A$" once $m>\max(A)$, so only finitely many initial terms are exceptional.
• Lemma 2: for the "full forbidden set" case, verified directly that all $m>n_0$ are excluded.
• The example computation matches the well-known fact that the odd integers have logarithmic density $1/2$ (here demonstrated numerically, not assumed).

FINAL

UNRESOLVED

(i) Strongest proved partial result. If $A$ is finite then $B$ is eventually periodic modulo $L=\mathrm{lcm}(A)$, hence $B$ has a natural (therefore logarithmic) density given by $|S|/L$ where $S\subseteq\mathbb{Z}/L\mathbb{Z}$ is the set of residues avoiding all $X_n$. Also, in trivial extremal cases (some $X_n$ is all residues, or all $X_n$ are empty) the logarithmic density exists and equals $0$ or $1$.

(ii) First gap (crisp). For infinite $A$ and arbitrary subsets $X_n\subseteq\mathbb{Z}/n\mathbb{Z}$, prove or disprove that the limit \[\lim_{x\to\infty}\frac{1}{\log x}\sum_{\substack{m<x\\m\in B}}\frac1m\] exists. In particular, I do not have a method to control the effect of infinitely many congruence restrictions turning on at different scales.

(iii) Top 3 next moves.

1. Try to approximate $B$ by truncated conditions $A\cap[1,T]$ and prove a quantitative estimate that the harmonic weight contributed by constraints with $n>T$ is $o(\log x)$ uniformly in $x$.
2. Attempt an explicit oscillatory construction: choose a sparse increasing sequence $n_j\in A$ and forbidden sets $X_{n_j}$ designed to alternately remove/almost preserve integers on logarithmic intervals $[e^{t_j},e^{t_{j+1}}]$.
3. Special cases: resolve the question when each $X_n$ is a single residue class (the "$|X_n|=1$" regime referenced in the statement), or when $|X_n|/n$ has a summable tail.

(iv) Minimal counterexample structure. A counterexample would require two sequences $x_j\to\infty$ along which the normalized harmonic sums for $B$ approach two different limits. Structurally, this would likely come from designing $A$ and $X_n$ so that new moduli $n$ in certain ranges erase a proportion of $B$ that is significant on the logarithmic scale.

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.