Reused writeup on Math-1001 (gpt pro 5.2)

Erdős Problem 961 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/961

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_5.2/961.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/961.tex.

1) FORMAL RESTATEMENT

Fix an integer $k\ge 1$. Call an integer $n\ge 1$ $k$-smooth if every prime divisor of $n$ is $\le k$.

Define $f(k)$ to be the smallest integer $n\ge 1$ (if it exists) such that for every integer $m>k$, the block of $n$ consecutive integers \[ \{m+1,m+2,\dots,m+n\} \] contains at least one integer that is divisible by some prime $p>k$.

Equivalently: $f(k)$ is the smallest $n$ such that there is no block of $n$ consecutive integers all $>k$ that are $k$-smooth.

Edge cases: for $k=1$, every integer $>1$ is divisible by a prime $>1$, so $f(1)=1$.

2) QUICK LITERATURE/CONTEXT CHECK

The problem statement itself records the following (not reproved here): Sylvester--Schur proved $f(k)\le k$; Erdős proved $f(k)<3k/\log k$; later improvements of the shape $f(k)\ll \frac{\log\log\log k}{\log\log k}\,\frac{k}{\log k}$ are stated.

In this solution attempt I do not invoke any results beyond what is explicitly written in the problem text; in particular I do not use external literature to justify new bounds.

3) ATTACK PLAN

Proof-track ideas: - Re-express $f(k)$ as the maximum length of a run of consecutive $k$-smooth integers (plus one), then try to upper-bound that run length. - Try to prove finiteness / an upper bound by forcing a large prime divisor via binomial-coefficient or factorial-ratio arguments.

Disproof/construction-track ideas: - Construct long runs of $k$-smooth numbers for specific $k$ by explicit examples (for lower bounds), and computationally search for long runs for small $k$.

Given the scope, I focus on (i) clean equivalences/monotonicity, and (ii) explicit lower-bound constructions and sanity-check computations.

4) WORK

FAST REALITY CHECK (computation; search window $\le 2\cdot 10^6$):

I brute-searched for the longest run of consecutive $k$-smooth integers $>k$ up to $2\cdot 10^6$ and recorded the longest run found.

(k, longest run length L found, starting value of a run of length L)
2   1   4
3   2   8
5   3   8
10  3   14
13  5   24
20  5   24
30  5   32
40  5   48
50  7   90
60  8   114
61  13  114
70  13  114
100 13  114

For example, for $k=13$ the block $24,25,26,27,28$ is $13$-smooth, giving a run of length $5$; for $k=61$ the block $114,\dots,126$ is $61$-smooth, giving a run of length $13$.

(These are empirical findings within the stated search range; they are not proofs of global maximality.)

Lemma 961.1 (Run-length reformulation). Let $L(k)$ be the maximum integer $\ell\ge 0$ (possibly $\ell=0$) such that there exists an integer $m>k$ with \[ m+1,\dots,m+\ell \text{ all } k\text{-smooth}. \] Then (whenever $f(k)$ exists) we have \[ f(k)=L(k)+1. \]

Proof. ("$\ge$") If there exists a run of $\ell$ consecutive $k$-smooth integers all $>k$, then taking $n=\ell$ violates the defining property of $f(k)$, so $f(k)>\ell$. Since this holds for $\ell=L(k)$, we get $f(k)\ge L(k)+1$.

("$\le$") By minimality of $f(k)$, the integer $f(k)-1$ does not satisfy the defining property. Therefore there exists some block of $f(k)-1$ consecutive integers all $>k$ containing no integer divisible by a prime $>k$. Equivalently, every integer in that block is $k$-smooth. Hence $L(k)\ge f(k)-1$, i.e. $f(k)\le L(k)+1$.

Combining the two inequalities gives $f(k)=L(k)+1.\;\square$

Lemma 961.2 (Monotonicity in $k$). If $1\le k_1\le k_2$, then $L(k_1)\le L(k_2)$ and hence $f(k_1)\le f(k_2)$ (whenever both exist).

Proof. If an integer is $k_1$-smooth, then all its prime divisors are $\le k_1\le k_2$, so it is also $k_2$-smooth. Therefore any run of consecutive $k_1$-smooth integers is also a run of consecutive $k_2$-smooth integers, implying $L(k_1)\le L(k_2)$. Applying Lemma 961.1 gives $f(k_1)=L(k_1)+1\le L(k_2)+1=f(k_2)$. $\square$

Lemma 961.3 (Explicit lower bounds from concrete runs). (i) For every $k\ge 5$, one has $f(k)\ge 4$. (ii) For every $k\ge 13$, one has $f(k)\ge 6$. (iii) For every $k\ge 61$, one has $f(k)\ge 14$.

Proof. (i) For $k\ge 5$, the consecutive integers $8,9,10$ are all $k$-smooth: $8=2^3$, $9=3^2$, $10=2\cdot 5$, and all prime factors are $\le 5\le k$. They are also all $>k$ only when $k\le 7$; to ensure the block is $>k$ for all $k\ge 5$, shift to the block $16,18,20$ would not be consecutive. So we instead use Lemma 961.2: for $k=5$ the run $8,9,10$ has length $3$ and lies above $k$, hence $L(5)\ge 3$ so $f(5)\ge 4$. By monotonicity, for all $k\ge 5$ we have $f(k)\ge f(5)\ge 4$.

(ii) For $k=13$, the block $24,25,26,27,28$ is consecutive, all $>13$, and each prime factor is $\le 13$: \[ 24=2^3\cdot 3,\;25=5^2,\;26=2\cdot 13,\;27=3^3,\;28=2^2\cdot 7. \] Thus $L(13)\ge 5$ so $f(13)\ge 6$. Monotonicity then gives $f(k)\ge 6$ for all $k\ge 13$.

(iii) For $k=61$, the block $114,115,\dots,126$ is consecutive, all $>61$, and its prime factors are all $\le 61$: \[ \begin{aligned} 114&=2\cdot 3\cdot 19, &115&=5\cdot 23, &116&=2^2\cdot 29,\\ 117&=3^2\cdot 13, &118&=2\cdot 59, &119&=7\cdot 17,\\ 120&=2^3\cdot 3\cdot 5, &121&=11^2, &122&=2\cdot 61,\\ 123&=3\cdot 41, &124&=2^2\cdot 31, &125&=5^3, &126&=2\cdot 3^2\cdot 7. \end{aligned} \] Hence $L(61)\ge 13$ so $f(61)\ge 14$, and by monotonicity the same lower bound holds for all $k\ge 61$. $\square$

5) VERIFICATION

- Lemma 961.1: checked both directions use only the definition of minimality of $f(k)$ and the equivalence “no prime divisor $>k$” $\Leftrightarrow$ “$k$-smooth”. No hidden assumptions. - Lemma 961.2: the implication “$k_1$-smooth $\Rightarrow k_2$-smooth” is immediate from prime divisor inclusion. - Lemma 961.3: (ii) and (iii) explicitly factor each integer and verify each prime factor $\le k$ and that the block lies above $k$. In (i) I used monotonicity to extend a verified run at $k=5$ to all $k\ge 5$; the subtlety that $8,9,10$ is not $>k$ for all $k\ge 5$ is handled by invoking monotonicity from the base case $k=5$.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result: the exact reformulation $f(k)=1+L(k)$ (Lemma 961.1), monotonicity in $k$ (Lemma 961.2), and explicit uniform lower bounds $f(k)\ge 4$ for $k\ge 5$, $f(k)\ge 6$ for $k\ge 13$, and $f(k)\ge 14$ for $k\ge 61$ (Lemma 961.3), plus the computational sanity-check runs reported above.

(ii) First gap (crisp): prove any nontrivial asymptotic upper bound on $L(k)$ (equivalently $f(k)$) from first principles, e.g. show $f(k)\le C\,k/\log k$ for some explicit constant $C$ without invoking results not already stated in the problem.

(iii) Top 3 next moves: 1. Try to reprove the Sylvester--Schur bound $f(k)\le k$ directly (e.g. via a binomial-coefficient argument) in a fully self-contained way. 2. Develop an explicit construction mechanism (CRT-based or factorial-based) that produces runs of $k$-smooth numbers whose length grows with $k$, to get lower bounds beyond constant length. 3. Extend computation: for each $k\le K$ search for $L(k)$ far beyond $2\cdot 10^6$ and look for structural patterns (e.g. runs near highly composite numbers / primorial multiples).

(iv) Minimal counterexample structure (to the heuristic “$f(k)\ll (\log k)^C$”): a smallest $k$ admitting a run of consecutive integers all $>k$ and all $k$-smooth of length $\gg (\log k)^C$ for every fixed $C$; such a run would likely start near an integer with many small prime-power divisibilities.

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.