Math-6255 · mathematics · Mathematics

Math-6255

Erdős Problem #1201

Status Open · erdosproblems.comLLM status Open unverified

Erdős problems openaigpt 6 astra ultra 5% cataloguedOpenOpen

Statement

Let $P(m)$ denote the greatest prime factor of $m$. For each $\epsilon,\eta>0$, does some integer $k\geq0$ make the set \[\{n\in\mathbb N:P(n(n+1)\cdots(n+k))>n^{1-\epsilon}\}\] have density at least $1-\eta$?

Restatement from Math-6255-A1 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-6255-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-6255

Erdős Problem 1201 --- GPT 6 Astra Ultra partial attempt

FORMAL RESTATEMENT

Let $P(m)$ denote the greatest prime factor of $m$. For each $\epsilon,\eta>0$, does some integer $k\geq0$ make the set \[\{n\in\mathbb N:P(n(n+1)\cdots(n+k))>n^{1-\epsilon}\}\] have density at least $1-\eta$?

CONTEXT AND SCOPE

This replaces the earlier statement-only record with an elementary mathematical attempt. The arguments below establish only their stated partial results; no novelty or full solution is claimed.

The page does not specify lower density versus existence of natural density; this paraphrase preserves its word density without silently choosing a stronger convention. The official remarks record the epsilon=1/2 case.

ATTACK PLAN AND WORK

The failure condition has a useful exact finite-word formulation. Fix $0<\epsilon<1$, an integer $k\ge0$, and a dyadic block of starting points $N\le n<2N$. Put $y=(2N+k)^{1-\epsilon}$ and mark each integer in $[N,2N+k-1]$ whose prime factors are all at most $y$. If a starting point fails the desired prime-factor inequality, then all $n,n+1,\ldots,n+k$ are marked, because their prime factors are at most $n^{1-\epsilon}\le y$.

If the maximal consecutive runs of marked entries have lengths $L_1,L_2,\ldots$, the exact number of fully marked windows of length $k+1$ is \[\sum_j\max(0,L_j-k).\tag{1}\] Every such window lies inside one run and there are $L_j-k$ possible starts in a run of length at least $k+1$. The word has length $N+k$, so its $N$ possible window starts are exactly the desired dyadic block. Thus the proportion of actual failures is at most $1$ divided by $N$.

This identifies a sufficient quantitative route: make the normalized excess mass of long smooth runs arbitrarily small, with one fixed $k$ working in every sufficiently large block. Marginal smooth-number density alone gives no such run-length estimate. For $\epsilon\ge1$ the question is trivial with $k=0$ and $n\ge2$, since the threshold is at most one; that observation does not address the difficult small-$\epsilon$ regime.

VERIFICATION AND REMAINING GAP

Bound the long-run statistic uniformly enough to obtain the requested density conclusion with fixed $k$. The reduction does not assume independence of neighboring smooth integers and proves no such statistical bound.

SOURCES

https://www.erdosproblems.com/1201

FINAL

UNRESOLVED. The full source problem remains outside the scope of the proved partial results.

COMPLETION ESTIMATE

$5\%$. This is a conservative subjective estimate toward the whole problem, not a probability that the argument is correct or a claim that a fixed fraction of the problem has been solved.

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