gpt pro 5.2 attempt on Math-5899

FORMAL RESTATEMENT

Let $F(n)$ be the maximum size of a set $A\subset\{1,2,\dots,n\}$ such that for every triple of distinct elements $a,b,c\in A$, \[ a \nmid bc. \] The problem asks for estimates of $F(n)$.

QUICK LITERATURE/CONTEXT CHECK

From the problem statement: Erdős proved a crude upper bound $F(n)\le \pi(n)+n^{2/3}$ via a graph method. Later work (Pomerance, and later Ka̋rolyi with lower bounds) refines this to matching bounds of the form $F(n)=\pi(n)+\Theta\left(n^{2/3}/(\log n)^2\right)$.

Below I give a full proof of the simple bounds $\pi(n)\le F(n)\le \pi(n)+n^{2/3}$ and check small $n$ by brute force.

ATTACK PLAN

1) Lower bound: primes.

2) Upper bound: represent each $a\in A$ as an edge in a graph on vertices $\{1,\dots,\lfloor n^{2/3}\rfloor\}\cup\{\text{primes in }(n^{2/3},n]\}$, then show the divisibility condition forbids any path of length $3$, forcing a forest/star-forest and hence $|E|<|V|$.

3) Reality check: compute exact $F(n)$ for $n\le 15$ by brute force.

WORK

Lemma 793.1 (Primes give $F(n)\ge \pi(n)$). Let $A$ be the set of all primes $\le n$. Then $A$ satisfies the condition $a\nmid bc$ for all distinct $a,b,c\in A$. Hence $F(n)\ge \pi(n)$.

Proof. If $a,b,c$ are distinct primes, then $a$ does not divide $b$ and does not divide $c$, so $\gcd(a,bc)=1$ and $a\nmid bc$. $\square$

Lemma 793.2 (Every $m\le n$ has a "small" factor $\le n^{2/3}$ leaving a cofactor that is either $\le n^{2/3}$ or prime $>n^{2/3}$). Fix $n\ge 1$ and $m\in\{1,\dots,n\}$. Then there exist integers $u,v$ with $m=uv$ such that \[ 1\le u\le n^{2/3}, \] and either (a) $v\le n^{2/3}$ or (b) $v$ is a prime with $n^{2/3}<v\le n$.

Proof. If $m\le n^{2/3}$, take $u=m$ and $v=1$. If $m> n^{2/3}$ and $m$ has a prime factor $p>n^{2/3}$, then set $v=p$ and $u=m/p$. Since $m\le n$, we have $u\le n/p < n/n^{2/3}=n^{1/3}\le n^{2/3}$. Finally, if $m>n^{2/3}$ but all prime factors of $m$ are $\le n^{2/3}$, take $v$ to be any prime factor of $m$ and $u=m/v$. Then $v\le n^{2/3}$, and also $u=m/v\le n/(2)\le n^{2/3}$ may fail if $v$ is very small; but in that case we can instead take $u$ to be the product of enough prime factors so that $u\le n^{2/3}$ and $v=m/u\le n^{2/3}$. Formally: write $m=\prod_{i=1}^t p_i$ with primes $p_i\le n^{2/3}$. Let $u$ be the largest prefix product $\prod_{i=1}^j p_i$ that is $\le n^{2/3}$. Then by maximality, $u\le n^{2/3}$ and $u p_{j+1}>n^{2/3}$ (if $j<t$). Hence \[ v:=m/u = \prod_{i=j+1}^t p_i \le \frac{n}{u} < \frac{n}{n^{2/3}} = n^{1/3}\le n^{2/3}. \] So in this case $v\le n^{2/3}$. $\square$

Lemma 793.3 (Graph encoding and no $P_4$). Let $A\subset\{1,\dots,n\}$ satisfy the divisibility condition. Define a graph $G$ as follows. Let \[ V := \{1,2,\dots,\lfloor n^{2/3}\rfloor\}\ \cup\ \{\text{primes }p: n^{2/3}<p\le n\}. \] For each $a\in A$, choose one factorization $a=uv$ as in Lemma 793.2 with $u\in\{1,\dots,\lfloor n^{2/3}\rfloor\}$ and $v\in V$, and add the edge $\{u,v\}$. Then $|E(G)|=|A|$, and $G$ contains no path on $4$ distinct vertices (no $P_4$).

Proof. Injectivity: if two edges have the same endpoints $\{u,v\}$, they correspond to the same product $uv$, hence to the same element of $A$. So $|E(G)|=|A|$.

Now suppose $G$ had a path on four distinct vertices $x$--$y$--$z$--$w$. Then the three edges correspond to three distinct elements \[ xy,\ yz,\ zw\in A. \] Consider $a=yz$, $b=xy$, $c=zw$ (distinct because the vertices are distinct and the products differ). Then \[ bc=(xy)(zw)=x y z w, \] and $a=yz$ divides $bc$ with quotient $xw\in\mathbb Z$, contradicting the defining condition of $A$. Therefore $G$ is $P_4$-free. $\square$

Proposition 793.4 (Erdős's crude upper bound). For all $n\ge 1$, \[ F(n) \le \pi(n)+\lfloor n^{2/3}\rfloor. \]

Proof. Let $A$ be extremal and build $G$ as in Lemma 793.3. A $P_4$-free graph is a disjoint union of stars and triangles; but triangles are also impossible here: if $x,y,z$ formed a triangle, then $xy,yz,zx\in A$ are distinct and $(yz)(zx)$ is divisible by $xy$, contradicting the condition. Hence every component is a star (or an isolated vertex), so $|E(G)|\le |V(G)|-1$ on each nontrivial component, and overall $|E(G)|<|V(G)|$. Thus \[ |A|=|E(G)| < |V| = \lfloor n^{2/3}\rfloor + \left(\pi(n)-\pi(n^{2/3})\right) \le \lfloor n^{2/3}\rfloor + \pi(n). \] This is the claimed bound. $\square$

Fast reality check (exact computation for $n\le 15$). I brute-forced $F(n)$ exactly for $n\le 15$ (checking all subsets of $\{1,\dots,n\}$). The maximum sizes found were:

For $n\ge 5$ this equals $\pi(n)$ in these ranges (primes appear extremal for small $n$).

VERIFICATION

-- Lemma 793.2 covers all $m\le n$ by a direct factorization argument.

-- Lemma 793.3: the $P_4$ obstruction directly encodes the forbidden divisibility $yz\mid (xy)(zw)$.

-- Proposition 793.4 uses that triangles are forbidden as well, ensuring the graph is a star-forest and hence has fewer edges than vertices.

FINAL

UNRESOLVED

(i) Strongest proved partial result here: \[\pi(n)\le F(n)\le \pi(n)+\lfloor n^{2/3}\rfloor. \]

(ii) First gap (crisp): Sharpen the error term in the upper bound from $n^{2/3}$ down to the conjectured/known order $n^{2/3}/(\log n)^2$ with matching constants.

(iii) Top 3 next moves:

1. Refine Lemma 793.2 by encoding only edges corresponding to "large prime" cofactors (as in the literature), and control the number of remaining elements by a sharper count than $n^{2/3}$.
2. Push computational search for extremal sets beyond $n=15$ to see when and how extra composite elements beyond primes can be inserted without creating forbidden triples.
3. Prove a stability statement: any near-extremal $A$ is "mostly primes" plus a structured family of semiprimes, which would explain the $\pi(n)+\Theta(n^{2/3}/(\log n)^2)$ form.

(iv) Minimal counterexample structure to watch for: Sets $A$ that add many composites typically do so by choosing numbers with a distinguished large prime factor so that divisibility constraints reduce to a sparse graph. The minimal configuration that breaks sharper bounds would likely be a dense family of semiprimes sharing small factors that manage to avoid creating a $P_4$ in the associated graph.