gpt pro 5.2 attempt on Math-6116

FORMAL RESTATEMENT

For $n\ge 1$, let $f(n)$ be the maximum cardinality of a subset $A\subseteq\{1,2,\dots,n\}$ such that there do not exist three distinct elements $a,b,c\in A$ with $a\mid b$ and $a\mid c$. Equivalently: for each $a\in A$, there is at most one element $b\in A\setminus\{a\}$ with $a\mid b$. The problem asks for the size and asymptotics of $f(n)$, and in particular about $\lim f(n)/n$. The statement also records a known estimate (Lebensold): for large $n$, \[ 0.6725\,n\le f(n)\le 0.6736\,n. \]

QUICK LITERATURE/CONTEXT CHECK

The numerical bounds $0.6725n\le f(n)\le 0.6736n$ are stated in the problem text as due to Lebensold. I do not use or claim any other external results.

ATTACK PLAN

Proof-track: (i) derive structural constraints on the divisibility poset induced by $A$; (ii) build explicit large constructions giving lower bounds; (iii) attempt counting/charging arguments for upper bounds. Disproof-track (for the "limit irrational" question): compute $f(n)/n$ for small $n$ to see behavior; but proving irrationality is far beyond small checks.

WORK

FAST REALITY CHECK. I computed $f(n)$ exactly for $n\le 30$ by branch-and-bound search over subsets, using the defining constraint. Exact values:

 n : f(n)
 1 : 1
 2 : 2
 3 : 2
 4 : 3
 5 : 4
 6 : 4
 7 : 5
 8 : 6
 9 : 6
10 : 7
11 : 8
12 : 8
13 : 9
14 : 10
15 : 10
16 : 11
17 : 12
18 : 12
19 : 13
20 : 14
21 : 14
22 : 15
23 : 16
24 : 16
25 : 17
26 : 18
27 : 19
28 : 20
29 : 21
30 : 21

For instance, $f(30)=21$ (ratio $0.7$), while the simple construction below gives $\lceil 2\cdot 30/3\rceil=20$.

Lemma 1062.1 (no 3-term chains). If $A$ satisfies the condition of the problem, then there do not exist $a<b<c$ in $A$ with $a\mid b$ and $b\mid c$.

Proof. If $a\mid b$ and $b\mid c$, then $a\mid c$. Thus $a$ divides both $b$ and $c$ with $b\ne c$, contradicting the defining property of $A$. $\square$

Lemma 1062.2 (elementary $2n/3$ construction). For every $n\ge 1$, \[ f(n)\ge n-\left\lfloor\frac n3\right\rfloor\ \ge\ \left\lceil\frac{2n}{3}\right\rceil. \] One explicit choice is \[ A:=\Big\{\left\lfloor\tfrac n3\right\rfloor+1,\ \left\lfloor\tfrac n3\right\rfloor+2,\ \dots,\ n\Big\}. \]

Proof. Let $m:=\lfloor n/3\rfloor$ and $A=\{m+1,\dots,n\}$. Then $|A|=n-m$. Take any $a\in A$. Since $a\ge m+1>n/3$, we have $3a>n$. Therefore the only possible multiples of $a$ in $\{1,\dots,n\}$ are $a$ itself and possibly $2a$ (since $2a\le n$ may occur, but $3a$ cannot). In particular, within $A$ the element $a$ can divide at most one other distinct element (namely $2a$ if it lies in $A$). Hence there do not exist distinct $b,c\in A$ with $a\mid b$ and $a\mid c$. This verifies that $A$ is admissible, and yields $f(n)\ge |A|=n-\lfloor n/3\rfloor\ge \lceil 2n/3\rceil$. $\square$

VERIFICATION

Lemma 1062.1 is an immediate logical consequence of the definition. Lemma 1062.2 checks the only possible multiples of $a>n/3$ inside $\{1,\dots,n\}$ and uses $3a>n$ to rule out two distinct multiples beyond $a$ itself. The exact small-$n$ values were computed by an explicit search that maintains, for each potential divisor $d$, the count of already-selected multiples of $d$.

FINAL

UNRESOLVED

1. (i) Strongest proved partial result. An explicit construction gives $f(n)\ge \lceil 2n/3\rceil$ for all $n$ (Lemma 1062.2). The problem statement records much sharper bounds for large $n$ (Lebensold): $0.6725n\le f(n)\le 0.6736n$. I also computed $f(n)$ exactly for $n\le 30$ (see the table in WORK).
2. (ii) First gap (crisp). Determine the true asymptotic constant $\lim_{n\to\infty} f(n)/n$ (existence, value) and address whether it is irrational.
3. (iii) Top 3 next moves. $1$ Computation: extend exact/near-exact optimization for $f(n)$ to much larger $n$ to better estimate $f(n)/n$ and detect patterns. $2$ Upper bounds: develop a charging/double-counting argument that forces $|A|\le (\alpha+o(1))n$ for some explicit $\alpha<0.6736$. $3$ Constructions: search for structured sets improving the $2/3$ construction and explain the near-$0.673$ density seen in known bounds.
4. (iv) Minimal counterexample structure. A counterexample to convergence of $f(n)/n$ would require large oscillations in the optimal density across $n$; such oscillations would likely correlate with arithmetic structure of $n$ (e.g. primorial-like ranges) affecting how many divisibility relations can be packed.