Reused writeup on Math-919 (gpt pro 5.2)

Erdős Problem 726 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/726

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_5.2/726.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/726.tex.

1) FORMAL RESTATEMENT

For an integer $n\ge 2$ define \[ S(n) := \sum_{p\le n} \frac{1}{p}\,\mathbf 1\big( n\bmod p \in (p/2,\,p)\big), \] where the sum runs over primes $p$, and $\mathbf 1(\cdot)$ is the indicator of the stated condition. Explicitly, $\mathbf 1(n\bmod p\in(p/2,p))=1$ iff there exists an integer residue $r$ with $p/2<r<p$ and $n\equiv r\pmod p$.

Conjecture (Erdős--Graham--Ruzsa--Straus): as $n\to\infty$ over integers, \[ S(n) \sim \frac{1}{2}\log\log n. \]

2) QUICK LITERATURE/CONTEXT CHECK

The problem statement records this as a conjecture, and compares it to Mertens' estimate \[ \sum_{p\le n}\frac{1}{p}\sim \log\log n. \] I do not use additional literature.

3) ATTACK PLAN

Proof track ideas: - Rewrite $S(n)=\sum_{p\le n}(1/p)\,\mathbf 1(\{n/p\}>1/2)$ and try to show the indicator behaves like a fair coin for most primes. - Prove first the mean value over $n\le N$ is $(1/2)\log\log N+O(1)$, then attempt to bound fluctuations.

Disproof track ideas: - Construct a sequence $n$ for which $n\bmod p$ lies mostly in $[0,p/2]$ for a large range of primes, forcing $S(n)$ to be $o(\log\log n)$.

I can rigorously prove the mean order over $n\le N$; the pointwise asymptotic remains open.

4) WORK

FAST REALITY CHECK.

A direct computation of $S(n)$ gives (values rounded):

- $S(50)\approx 0.679$, while $(1/2)\log\log 50\approx 0.682$. - $S(200)\approx 0.850$, while $(1/2)\log\log 200\approx 0.834$. - Near $n=10^6$, $S(n)$ varies noticeably (e.g. $S(1{,}000{,}004)\approx 1.416$ and $S(1{,}000{,}015)\approx 0.688$) while $(1/2)\log\log(10^6)\approx 1.313$.

These are consistent with a slowly growing main term and $O(1)$-scale fluctuations at this range.

Lemma 1 (local density of the condition mod $p$). Let $p$ be prime.

- If $p=2$, then for all integers $n$, $\mathbf 1(n\bmod 2\in(1,2))=0$.

- If $p$ is odd, then among the residues $r\in\{0,1,\dots,p-1\}$ there are exactly $(p-1)/2$ residues satisfying $p/2<r<p$. Equivalently, for odd $p$, \[ \frac{1}{p}\sum_{r=0}^{p-1} \mathbf 1(r\in(p/2,p)) = \frac{p-1}{2p} = \frac12-\frac{1}{2p}. \]

Proof. For $p=2$ the open interval $(p/2,p)=(1,2)$ contains no integer residue class mod $2$, so the indicator is always $0$.

Now assume $p$ is odd. Then $p/2$ is not an integer, and the integers $r$ with $p/2<r<p$ are exactly \[ \frac{p+1}{2},\ \frac{p+3}{2},\ \dots,\ p-1. \] This is an arithmetic progression with first term $(p+1)/2$ and last term $p-1$, hence it contains \[ (p-1) - \frac{p+1}{2} + 1 = \frac{p-1}{2} \] integers. Dividing by $p$ gives the stated average. $\square$

Lemma 2 (mean order over $n\le N$). Let $N\ge 3$. Then \[ \frac{1}{N}\sum_{n=1}^N S(n) = \frac12\sum_{p\le N}\frac{1}{p} + O(1), \] and consequently (using Mertens' estimate from the statement) \[ \frac{1}{N}\sum_{n=1}^N S(n) = \frac12\log\log N + O(1). \]

Proof. Start from the definition and swap summations: \[ \sum_{n=1}^N S(n) = \sum_{n=1}^N\sum_{p\le n} \frac{1}{p}\,\mathbf 1(n\bmod p\in(p/2,p)) = \sum_{p\le N}\frac{1}{p}\sum_{n=p}^N \mathbf 1(n\bmod p\in(p/2,p)). \] (The inner sum starts at $n=p$ since for $n<p$ the prime $p$ does not appear in $S(n)$.)

Fix a prime $p\le N$. Consider the sequence of residues $n\bmod p$ as $n$ runs over an interval of length $p$. In each full block of $p$ consecutive integers, the residues cover $\{0,1,\dots,p-1\}$ exactly once. By Lemma 1, in a full block of length $p$ there are exactly $(p-1)/2$ values with residue in $(p/2,p)$ for odd $p$, and $0$ such values for $p=2$.

Write $N = ap + b$ with integers $a=\lfloor N/p\rfloor$ and $0\le b<p$. Then among $\{1,2,\dots,N\}$, the number of $n$ with $n\bmod p\in(p/2,p)$ equals \[ a\cdot \frac{p-1}{2} + O(p), \] where the $O(p)$ term accounts for the incomplete final block of length $b$ and the initial shift from starting at $n=1$ instead of $n=0$. The same estimate holds if we restrict to $n\in\{p,p+1,\dots,N\}$ since removing the first $p-1$ values changes the count by at most $p$. Thus \[ \sum_{n=p}^N \mathbf 1(n\bmod p\in(p/2,p)) = a\cdot \frac{p-1}{2} + O(p). \] Multiply by $1/p$ and sum over primes $p\le N$: \[ \sum_{n=1}^N S(n) = \sum_{p\le N} \frac{1}{p}\left(\frac{p-1}{2}\left\lfloor\frac{N}{p}\right\rfloor + O(p)\right). \] Use $\lfloor N/p\rfloor = N/p + O(1)$ to get \[ \sum_{n=1}^N S(n) = \sum_{p\le N}\left(\frac{N}{2p} - \frac{N}{2p^2}\right) + O\left(\sum_{p\le N}1\right). \] Divide by $N$: \[ \frac{1}{N}\sum_{n=1}^N S(n) = \frac12\sum_{p\le N}\frac{1}{p} - \frac12\sum_{p\le N}\frac{1}{p^2} + O\left(\frac{\pi(N)}{N}\right). \] The sum $\sum_{p\le N}1/p^2$ is bounded uniformly in $N$ (since $\sum_p1/p^2<\infty$), and $\pi(N)/N\to 0$. Hence \[ \frac{1}{N}\sum_{n=1}^N S(n) = \frac12\sum_{p\le N}\frac{1}{p} + O(1). \] Finally, by Mertens' estimate $\sum_{p\le N}1/p\sim\log\log N$, so the mean is $(1/2)\log\log N+O(1)$. $\square$

5) VERIFICATION

- Lemma 1 carefully treats $p=2$ separately; for odd primes the residue count is exact. - In Lemma 2, all error terms are controlled explicitly by $O(p)$ per prime and the fact that $\sum_{p\le N}1/p^2$ converges. - The computation examples in the FAST REALITY CHECK are consistent with a $(1/2)\log\log n$ main term but are far too small to be diagnostic.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result: The mean value over $n\le N$ satisfies $\frac{1}{N}\sum_{n\le N}S(n)=\frac12\log\log N+O(1)$ (Lemma 2), and the exact local density of the condition modulo each prime is given by Lemma 1.

(ii) First gap (crisp): Prove or disprove the pointwise asymptotic $S(n)\sim \tfrac12\log\log n$ as $n\to\infty$ (not just on average).

(iii) Top 3 next moves: 1. Prove a uniform bound on fluctuations, e.g. show $S(n)=\tfrac12\log\log n + o(\log\log n)$ for all $n$, or at least for a density-$1$ set of $n$. 2. Study correlations of the events $\{n\bmod p\in(p/2,p)\}$ across primes (large-sieve type estimates) to control the deviation from the mean. 3. Run computations for much larger $n$ and analyze the empirical distribution of $S(n)-\tfrac12\log\log n$ to guess the correct fluctuation scale.

(iv) Minimal counterexample structure: If the conjecture fails, a minimal counterexample would be a sequence $n\to\infty$ for which the residues $n\bmod p$ land in $[0,p/2]$ for a positive proportion of primes up to $n^\alpha$ (for some fixed $\alpha>0$), creating a deficit of order $\log\log n$ in $S(n)$.

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.