gpt pro 5.2 attempt on Math-913

1) FORMAL RESTATEMENT

For each $n\ge 1$, define $f(n)$ to be the minimal real number with the following property:

For every set $A=\{a_1,\dots,a_n\}\subset\mathbb{N}\cap[2,\infty)$ with $|A|=n$ and $m=\max(A)$, and for every interval $I$ consisting of $\lfloor f(n)m\rfloor$ consecutive integers, there exist distinct integers $x_1,\dots,x_n\in I$ such that $a_i\mid x_i$ for all $i$.

(There is a mild ambiguity in the problem statement about whether the interval length is exactly $f(n)m$ or an integer rounding; for the purposes of rigorous inequalities below I will prove statements for integer multiples, which implies corresponding bounds for any reasonable rounding convention.)

2) QUICK LITERATURE/CONTEXT CHECK

Using only what is stated in the problem text: Erdős--Surányi proved \[(\log n)^c \ll f(n) \ll n^{1/2}\] for some constant $c>0$.

3) ATTACK PLAN

- Prove easy universal lower/upper bounds by extremal examples and matching arguments. - For upper bounds, model the problem as a bipartite matching between divisors $a_i$ and integers in the interval.

4) WORK

Lemma 709.1 (Trivial lower bound $f(n)\ge 1$). For every $n\ge 1$, $f(n)\ge 1$.

Proof. Fix $n\ge 1$. Consider any $A$ with $m=\max(A)$. If an interval $I$ contains fewer than $m$ consecutive integers, then it is possible to choose $I$ to avoid all multiples of $m$ (e.g. $I=\{1,2,\dots,m-1\}$). In such an interval there is no integer divisible by $m$, so we cannot assign a distinct $x_i$ to the element of $A$ equal to $m$. Hence, to guarantee a solution for all intervals, we need interval length at least $m$, i.e. $f(n)\ge 1$. $\square$

Lemma 709.2 (General upper bound $f(n)\le n$). For every $n\ge 1$ and every finite $A\subset\mathbb{N}\cap[2,\infty)$ with $|A|=n$ and $m=\max(A)$, every interval $I$ of $nm$ consecutive integers contains distinct $x_1,\dots,x_n\in I$ with $a_i\mid x_i$ for all $i$. In particular, $f(n)\le n$.

Proof. Let $A=\{a_1,\dots,a_n\}$ and let $I$ be any interval of $nm$ consecutive integers. Construct a bipartite graph $H$ with left vertex set $A$ and right vertex set $I$, where we connect $a\in A$ to $x\in I$ iff $a\mid x$. We seek a matching that covers all left vertices.

By Hall's marriage theorem, it suffices to show that for every subset $S\subseteq A$, its neighborhood \[ N(S)=\{x\in I:\exists a\in S\text{ with }a\mid x\} \] has size at least $|S|$.

Fix a nonempty $S\subseteq A$ and let $d=\min S$. Every multiple of $d$ that lies in $I$ belongs to $N(S)$ (because it is divisible by $d\in S$), so $|N(S)|$ is at least the number of multiples of $d$ in $I$. Write $|I|=nm$. In any block of $nm$ consecutive integers, the number of multiples of $d$ is at least $\lfloor nm/d\rfloor$. Since $d\le m$, we have $nm/d\ge n$, hence $\lfloor nm/d\rfloor\ge n$. Therefore $|N(S)|\ge n$. Finally, because $|S|\le n$, we have $|N(S)|\ge n\ge |S|$, establishing Hall's condition. Thus $H$ contains a matching of size $n$, i.e. distinct $x_1,\dots,x_n\in I$ with $a_i\mid x_i$. $\square$

Lemma 709.3 (Exact value for $n=2$: $f(2)=2$ in the integer-multiple model). If we restrict attention to intervals of integer length $t\,\max(A)$ with $t\in\mathbb{N}$, then $t=1$ does not suffice for $n=2$, while $t=2$ always suffices. Hence in this integer-multiple model, $f(2)=2$.

Proof. Lower bound: $t=1$ fails. Take $A=\{2,3\}$, so $m=3$. Consider the interval $I=\{5,6,7\}$ of length $m$. The only multiple of $3$ in $I$ is $6$, and the only multiple of $2$ in $I$ is also $6$. So there do not exist distinct $x_1,x_2\in I$ with $2\mid x_1$ and $3\mid x_2$. Thus $t=1$ fails.

Upper bound: $t=2$ suffices. Lemma 709.2 with $n=2$ shows that any interval of $2m$ consecutive integers admits such a matching. Therefore $t=2$ suffices for all $A$ with $|A|=2$. $\square$

FAST REALITY CHECK (limited brute force for $n=2$). I brute-forced all $A$ with $|A|=2$, $\max(A)\le 20$, and all interval starts $t\le 500$. Result: $t=1$ fails (e.g. $A=\{2,3\}$ with interval $\{5,6,7\}$), while $t=2$ worked for all tested instances.

5) VERIFICATION

- Lemma 709.2: The key inequality is that any interval of length $nm$ contains at least $\lfloor nm/d\rfloor\ge n$ multiples of $d\le m$. This is correct because multiples of $d$ occur every $d$ integers, so among $nm$ consecutive integers there are at least $\lfloor nm/d\rfloor$ multiples. - Hall is applied correctly: $|N(S)|\ge n\ge |S|$ for all $S$.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result. I proved the unconditional bounds $1\le f(n)\le n$ (Lemmas 709.1 and 709.2) and, in the integer-multiple model, $f(2)=2$ (Lemma 709.3). I also ran a finite brute-force sanity check for $n=2$ (reported above).

(ii) First gap (crisp). Close the gap between $f(n)\le n$ and the much smaller upper bound $f(n)\ll n^{1/2}$ stated in the problem text, or determine the true asymptotic behavior of $f(n)$.

(iii) Top 3 next moves. 1. Strengthen the Hall lower bound $|N(S)|\ge n$ by using the full divisor structure of $S$ rather than just $\min S$, aiming for shorter intervals. 2. Search for extremal $A$ that make matching difficult (many $a_i$ sharing large common multiples) and try to prove lower bounds from explicit constructions. 3. Implement a computational optimizer for small $n$ to guess the exact $f(n)$ and the structure of extremal sets $A$.

(iv) Minimal counterexample structure. A minimal obstruction to a candidate bound $f(n)\le c$ would be a set $A$ with $m=\max(A)$ and an interval $I$ of $\approx cm$ consecutive integers such that the divisibility bipartite graph fails Hall's condition for some subset $S\subseteq A$; i.e. the union of multiples in $I$ of elements of $S$ has size $<|S|$.