gpt pro 5.2 attempt on Math-5959

FORMAL RESTATEMENT. Fix an integer $n\ge 2$ and let $U=\{2,3,\dots,n\}$. A set $A\subseteq U$ is legal if it is an antichain in the divisibility poset, i.e. \[ \forall a\neq b\in A,\qquad a\nmid b\text{ and }b\nmid a. \] A move consists of choosing $x\in U\setminus A$ such that $A\cup\{x\}$ is legal. The game ends when there is no legal move; equivalently, when $A$ is a maximal (by inclusion) divisibility antichain in $U$.

There are two players:

• Long aims to maximize the total number of moves, i.e. maximize the final $|A|$.
• Short aims to minimize the final $|A|$.

The problem text notes an ambiguity: Erdős did not specify which player goes first. Accordingly define two game values:

• $g_L(n)$: optimal final $|A|$ if Long moves first;
• $g_S(n)$: optimal final $|A|$ if Short moves first.

Question: determine (or bound) $g_L(n),g_S(n)$ asymptotically. In particular:

• Is there an $\varepsilon>0$ such that $g_L(n),g_S(n)\ge \varepsilon n$ for all large $n$?
• Is it true that for every fixed $\varepsilon>0$ and all large $n$, one has $g_L(n),g_S(n)\ge (1-\varepsilon)\frac n2$?

QUICK LITERATURE/CONTEXT CHECK. The problem text positions this as a "number theoretic variant" of Hajnal's triangle-free saturation game (and cites results for that graph game). No theorems are stated for the divisibility game itself, beyond the observation about the first-player ambiguity. I therefore treat the divisibility-game asymptotics as open from the provided information.

ATTACK PLAN.

1. Analyze extremal sizes of antichains and maximal antichains in the divisibility poset on $\{2,\dots,n\}$.
2. Identify forced elements in any maximal antichain (e.g. primes $>n/2$).
3. Compute small $n$ by minimax to detect patterns and formulate plausible bounds/strategies.

WORK.

Lemma 1 (maximum possible game length is $\lceil n/2\rceil$). Let $U=\{2,\dots,n\}$. The maximum size of a legal set (i.e. the width of this poset restricted to $U$) equals $\lceil n/2\rceil$. Moreover, the set \[ A_\mathrm{top}:=\{\lfloor n/2\rfloor+1,\ \lfloor n/2\rfloor+2,\ \dots,\ n\} \] is a maximal legal set of size $\lceil n/2\rceil$.

Proof. For each odd integer $m\le n$, consider the chain \[ C_m:=\{m,2m,4m,8m,\dots\}\cap\{1,2,\dots,n\}. \] Every positive integer $x\le n$ belongs to exactly one such chain: write $x=2^j m$ with $m$ odd. Inside a fixed chain $C_m$, divisibility totally orders the elements since $2^j m\mid 2^{j'}m$ iff $j\le j'$. Therefore any divisibility antichain intersects each chain in at most one element. The number of odd integers $m\le n$ is $\lceil n/2\rceil$, so any legal set in $U$ has size at most $\lceil n/2\rceil$.

Now consider $A_\mathrm{top}$. If $a,b\in A_\mathrm{top}$ with $a<b$, then $2a>n\ge b$, hence $b$ cannot be a multiple of $a$; also $a$ cannot be a multiple of $b$. Thus $A_\mathrm{top}$ is legal. To see maximality, take any $x\in U\setminus A_\mathrm{top}$, so $x\le \lfloor n/2\rfloor$. Choose $t=\left\lceil\frac{\lfloor n/2\rfloor+1}{x}\right\rceil$. Then $tx\ge \lfloor n/2\rfloor+1$ and also $tx\le n$ because \[ t\le \frac{\lfloor n/2\rfloor+1}{x}+1\le \frac{n}{x}, \] so $tx\le n$. Hence $tx\in A_\mathrm{top}$ and $x\mid tx$, which means $x$ is comparable to an element of $A_\mathrm{top}$ and cannot be added. Thus $A_\mathrm{top}$ is maximal. $\square$

Lemma 2 (large primes are forced in every terminal position). Let $A\subseteq U$ be a maximal legal set. Then every prime $p$ with $\frac n2<p\le n$ must belong to $A$. Consequently, every game (regardless of play) lasts at least \[ \pi(n)-\pi(\lfloor n/2\rfloor) \] moves, where $\pi(x)$ denotes the number of primes $\le x$.

Proof. Let $p$ be prime with $n/2<p\le n$. Then $2p>n$, so there is no multiple of $p$ in $U$ other than $p$ itself. Also, since $p$ is prime and $1\notin U$, there is no divisor of $p$ in $U$ other than $p$. Therefore $p$ is incomparable (under divisibility) with every element of $U\setminus\{p\}$. If $p\notin A$, then $A\cup\{p\}$ is still legal, contradicting maximality. Thus $p\in A$. Counting such primes gives the stated lower bound on $|A|$. $\square$

VERIFICATION (FAST REALITY CHECK). I computed exact minimax values for $2\le n\le 24$ by brute-force recursion with memoization. The table lists $(g_L(n),g_S(n))$ as well as:

• $m(n)$ = minimum size of a maximal legal set (computed by minimizing over all move sequences),
• $M(n)=\lceil n/2\rceil$ (Lemma 1).

 n :  g_L(n)  g_S(n)   m(n)   M(n)
 2 :     1       1      1      1
 3 :     2       2      2      2
 4 :     2       2      2      2
 5 :     3       3      3      3
 6 :     3       3      3      3
 7 :     4       4      4      4
 8 :     4       4      4      4
 9 :     5       4      4      5
10 :     5       4      4      5
11 :     6       5      5      6
12 :     6       5      5      6
13 :     7       6      6      7
14 :     7       6      6      7
15 :     7       6      6      8
16 :     7       6      6      8
17 :     8       7      7      9
18 :     8       7      7      9
19 :     9       8      8     10
20 :     9       8      8     10
21 :    10       8      8     11
22 :    10       8      8     11
23 :    11       9      9     12
24 :    11       9      9     12

These computations verify Lemma 1's upper bound and show that $g_L(n)$ and $g_S(n)$ can differ.

FINAL. UNRESOLVED.

(i) Strongest proved partial result here. Lemma 1: the game always ends in at most $\lceil n/2\rceil$ moves and this is achievable by a maximal antichain. Lemma 2: every terminal position contains all primes in $(n/2,n]$, giving a universal lower bound $\pi(n)-\pi(\lfloor n/2\rfloor)$.

(ii) First gap (crisp). Prove (or disprove) the existence of a constant $\varepsilon>0$ such that under optimal play (for either starting player) the game value is always $\ge \varepsilon n$ for all sufficiently large $n$.

(iii) Top 3 next moves (concrete).

1. Determine (or bound) $m(n)$, the minimum size of a maximal divisibility antichain in $\{2,\dots,n\}$, since game values satisfy $m(n)\le g_S(n),g_L(n)\le \lceil n/2\rceil$.
2. Develop explicit strategies: e.g. give Short a strategy producing a maximal antichain of size $\le cn$ (upper bound on game value) or Long a strategy guaranteeing size $\ge cn$ (lower bound).
3. Compute $g_L(n),g_S(n)$ for larger $n$ (beyond $24$) to guess the limiting ratio and to search for periodic/structural behavior.

(iv) Minimal counterexample structure. A counterexample to "$g(n)\ge \varepsilon n$" would be a sequence $n_j\to\infty$ together with Short strategies forcing terminal antichains of size $o(n_j)$. Structurally, this would require constructing very small maximal antichains that nevertheless dominate (by divisibility comparability) almost all numbers in $\{2,\dots,n_j\}$.