Reused writeup on Math-1005 (gpt pro 5.2)

Erdős Problem 970 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/970

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_5.2/970.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/970.tex.

Let $h(k)$ be Jacobsthal's function, defined to as the minimal $m$ such that, if $n$ has at most $k$ prime factors, then in any set of $m$ consecutive integers there exists an integer coprime to $n$. Determine the order of magnitude of $h(k)$. In particular, is it true that\[h(k) \ll k^2?\] That $h(k)\ll k^2$ is a conjecture of Jacobsthal. Iwaniec [Iw78] proved\[h(k) \ll (k\log k)^2.\]The best lower bound known is\[h(k) \gg \frac{(\log k)(\log\log\log k)}{(\log\log k)^2}k,\]due to Ford, Green, Konyagin, Maynard, and Tao [FGKMT18]. This is a more general form of the function considered in [687] . References [FGKMT18] Ford, Kevin and Green, Ben and Konyagin, Sergei and Maynard, James and Tao, Terence, Long gaps between primes . J. Amer. Math. Soc. $2018$, 65-105. [Iw78] Iwaniec, Henryk, On the problem of {J}acobsthal . Demonstratio Math. $1978$, 225--231.

1) FORMAL RESTATEMENT
Fix $k\in\mathbb{N}$. For an integer $n\ge 1$, let \[ j(n) := \min\{m\in\mathbb{N}: \text{every interval of $m$ consecutive integers contains an integer coprime to $n$}\}. \] (Here "interval of $m$ consecutive integers" means a set of the form $\{t+1,t+2,\dots,t+m\}$ for some $t\in\mathbb{Z}$. "Coprime" means $\gcd(\cdot,\cdot)=1$.) Define \[ h(k) := \max\{ j(n): n\ge 1\text{ and $n$ has at most $k$ prime factors}\}. \] Ambiguity. "At most $k$ prime factors" can mean either $\Omega(n)\le k$ (counted with multiplicity) or $\omega(n)\le k$ (distinct prime divisors). In either convention, the value of $j(n)$ depends only on the distinct prime divisors of $n$ (Lemma lem:970-rad), so $h(k)$ is unchanged by replacing $\Omega$ with $\omega$.

The question asks for the order of magnitude of $h(k)$ as $k\to\infty$; in particular whether there exists an absolute constant $C$ such that $h(k)\le C k^2$ for all sufficiently large $k$.

2) QUICK LITERATURE/CONTEXT CHECK
The problem statement records: Jacobsthal conjectured $h(k)\ll k^2$; Iwaniec proved $h(k)\ll (k\log k)^2$; and the best known lower bound is linear in $k$ up to slowly varying factors. I do not use any additional literature facts beyond what is explicitly in the problem statement.

3) ATTACK PLAN
Proof track: (a) reduce to squarefree $n$ and interpret $j(n)$ as a longest "covered" interval by residue classes $0\pmod p$; (b) try to show extremizers use small primes; (c) seek a universal upper bound in terms of $k$ by bounding the longest run covered by $k$ prime moduli.

Disproof track: try to construct, for some $k$, an integer $n$ with $\le k$ prime factors and a run of length $>c k^2$ in which every integer shares a prime factor with $n$ (a covering-system style construction).

I only reach structural lemmas + small-case computation; the main asymptotic bound remains open here.

4) WORK
FAST REALITY CHECK (small cases). For squarefree $n$ with prime set $P(n)$, $j(n)$ can be computed by scanning one full period mod $n$ (Lemma lem:970-period). I computed $j(n)$ for $n$ equal to the product of the first $k$ primes $P_k:=\prod_{i=1}^k p_i$ for $1\le k\le 8$: \[ \begin{array}{c|cccccccc} k & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\\hline P_k & 2 & 6 & 30 & 210 & 2310 & 30030 & 510510 & 9699690\\ j(P_k) & 2 & 4 & 6 & 10 & 14 & 22 & 26 & 34 \end{array} \] Moreover, in a brute-force search over all squarefree $n$ whose prime divisors are contained in $\{2,3,5,7,11,13,17,19\}$, for each $k\le 6$ the maximum $j(n)$ among those with $\omega(n)=k$ was achieved by $P_k$ and matched the table above.

Lemma 970.1 (reduction to squarefree). For every integer $n\ge 1$, letting $\operatorname{rad}(n)=\prod_{p\mid n}p$ be the squarefree kernel, one has \[ j(n)=j(\operatorname{rad}(n)). \] In particular, $h(k)$ is unchanged whether "$n$ has at most $k$ prime factors" is interpreted via $\Omega(n)$ or $\omega(n)$.

Proof. For any integer $x$, $\gcd(x,n)=1$ if and only if no prime divisor of $n$ divides $x$, which is equivalent to $\gcd(x,\operatorname{rad}(n))=1$. Therefore, for any $m$ and any interval $\{t+1,\dots,t+m\}$, the statement "there exists $x$ in the interval coprime to $n$" is equivalent to "there exists $x$ in the interval coprime to $\operatorname{rad}(n)$". Taking the minimal such $m$ gives $j(n)=j(\operatorname{rad}(n))$. Finally, if $\Omega(n)\le k$ then certainly $\omega(\operatorname{rad}(n))\le k$, and if $\omega(n)\le k$ then also $\Omega(n)\le k$ holds for $n$ squarefree; since maximizing $j(n)$ may be restricted to squarefree $n$, the two conventions agree for $h(k)$. $\square$

Lemma 970.2 (periodic gap formula). Let $n\ge 2$ be squarefree. Define the length-$n$ periodic "bad" indicator \[ b(r)=\begin{cases}1,&\gcd(r,n)>1,\\0,&\gcd(r,n)=1,\end{cases}\qquad (r\in\mathbb{Z}). \] Let $L(n)$ be the maximum length of a cyclic run of $1$'s in $\{b(0),b(1),\dots,b(n-1)\}$ (allowing wrap-around from $n-1$ to $0$). Then \[ j(n)=L(n)+1. \]

Proof. Since $b(r)$ depends only on $r\bmod n$, the pattern of integers coprime to $n$ is periodic with period $n$.

If there is a run of $L$ consecutive integers all not coprime to $n$, then any $m\le L$ fails the defining property of $j(n)$, so $j(n)\ge L+1$. Conversely, if $L(n)$ is the longest cyclic run of residues mod $n$ with $\gcd(\cdot,n)>1$, then any $L(n)+1$ consecutive residues mod $n$ must include at least one residue coprime to $n$. Lifting back to integers, any interval of $L(n)+1$ consecutive integers contains an integer coprime to $n$, so $j(n)\le L(n)+1$. Combining the two inequalities yields $j(n)=L(n)+1$. $\square$

Lemma 970.3 (CRT lower bound $h(k)\ge k+1$). For every $k\ge 1$, one has $h(k)\ge k+1$.

Proof. Fix distinct primes $p_1,\dots,p_k$ and let $n=\prod_{i=1}^k p_i$, which is squarefree with $\omega(n)=k$. By the Chinese remainder theorem, there exists an integer $t$ such that \[ t+i\equiv 0\pmod{p_i}\qquad\text{for each }i=1,2,\dots,k. \] Then for each $i$ the integer $t+i$ is divisible by $p_i$, hence $\gcd(t+i,n)>1$. Therefore the block $\{t+1,\dots,t+k\}$ contains no integer coprime to $n$, showing $j(n)\ge k+1$. Since $h(k)$ is the maximum of $j(n)$ over all $n$ with $\le k$ prime factors, $h(k)\ge j(n)\ge k+1$. $\square$

5) VERIFICATION
Quantifiers/boundaries. Lemma lem:970-rad covers $n\ge 1$; for $n=1$ one has $j(1)=1$, which is compatible but irrelevant for maximizing $h(k)$. Lemma lem:970-period assumes $n$ squarefree; by Lemma lem:970-rad this is without loss for $h(k)$. Lemma lem:970-crt-lower uses distinct primes; the CRT system is consistent because the moduli are coprime.

Computation sanity. For $k\le 8$ the computed $j(P_k)$ values satisfy $j(P_k)\ge k+1$ as guaranteed by Lemma lem:970-crt-lower. The computation used the periodic formula of Lemma lem:970-period.

6) FINAL
UNRESOLVED

(i) Strongest proved partial result. One may restrict to squarefree $n$ (Lemma lem:970-rad) and compute $j(n)$ as a maximum cyclic bad run modulo $n$ (Lemma lem:970-period). Universally $h(k)\ge k+1$ by a CRT construction (Lemma lem:970-crt-lower). Small-case computation gives $j(\prod_{i=1}^k p_i)=(2,4,6,10,14,22,26,34)$ for $1\le k\le 8$.

(ii) First gap (crisp). Prove (or disprove) that there exists an absolute $C$ such that for all sufficiently large $k$, $h(k)\le Ck^2$.

(iii) Top 3 next moves.

• Prove a monotonicity/exchange lemma: replacing a prime factor of $n$ by a larger prime does not increase $j(n)$; if true, this would reduce $h(k)$ to $j(P_k)$ with $P_k$ the product of the first $k$ primes.
• Translate the problem to a covering-system formulation: a run of length $L$ corresponds to residue classes $0\pmod p$ covering an interval of length $L$; attempt to bound maximum cover length with $k$ primes.
• Compute $h(k)$ exactly for moderate $k$ by optimizing over prime sets and locating extremal runs (using Lemma lem:970-period), to guide conjectures about growth.

(iv) Minimal counterexample structure. A disproof of $h(k)\ll k^2$ would require some $k$ and a squarefree $n$ with $\omega(n)=k$ for which there is a block of length $\gg k^2$ covered by the union of residue classes $0\pmod p$ for $p\mid n$; heuristically such an $n$ would likely involve many small primes to make the bad set dense and a highly structured alignment of their multiples to produce an unusually long covered interval.

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.