gpt pro 5.2 attempt on Math-5948

FORMAL RESTATEMENT.

Fix an integer $N\ge 2$. Let $A\subseteq \{1,2,\dots,N\}$. For an integer $t\ge 2$, write $\operatorname{spf}(t)$ for the smallest prime divisor of $t$.

Assume $A$ satisfies:

There do not exist $a,b\in A$ and an integer $t\ge 2$ such that $b=a t$ and $\operatorname{spf}(t)>a$.

Define \[ F(N):= \max_{A\subseteq [N]\ \text{satisfying the condition}} \frac{1}{\log N}\sum_{n\in A}\frac{1}{n}. \] The problem asks to estimate $F(N)$ as $N\to\infty$.

Ambiguity/misstated-context note. The statement in the problem file asserts that Alexander and Erdős--Sárközy--Szemerédi proved that this maximum is $o(1)$. Under the literal finite-$N$ definition above, we give an explicit family with $F(N)\ge \tfrac12+o(1)$, so $F(N)$ cannot be $o(1)$. This suggests the cited $o(1)$ result concerns a different "maximum" (e.g. for a fixed infinite set $A$ and $\limsup_{N\to\infty}$), or the condition is different in the original source.

QUICK LITERATURE/CONTEXT CHECK.

The problem file itself cites Alexander and Erdős--Sárközy--Szemerédi as proving an $o(1)$ bound for the stated maximum, and provides an $N$-dependent example set $A$. We do not assume any additional literature beyond what is in the problem file.

ATTACK PLAN.

First, sanity-check the condition: when $a$ is large (e.g. $a\ge \sqrt N$), the constraint "$\operatorname{spf}(t)>a$" becomes impossible for any multiple $b\le N$, making the condition vacuous. This immediately yields a lower bound $F(N)\ge 1/2+o(1)$.

We then brute-force optimise $F(N)$ for small $N$ to see what the true finite-$N$ behaviour looks like.

WORK.

Lemma 858.1 (The constraint is vacuous for $a\ge \sqrt N$).

Lemma. Let $N\ge 2$. Suppose $a,b\in\{1,\dots,N\}$ and $t\ge 2$ satisfy $b=a t$. If $a\ge \sqrt N$, then $\operatorname{spf}(t)\le a$. In particular, no triple $(a,b,t)$ with $a\ge \sqrt N$ can violate the condition $\operatorname{spf}(t)>a$.

Proof. From $b=a t\le N$ we get $t\le N/a$. If $a\ge \sqrt N$ then $N/a\le \sqrt N\le a$, hence $t\le a$. For $t\ge 2$ we have $\operatorname{spf}(t)\le t$ (because the smallest prime factor cannot exceed the number itself). Therefore $\operatorname{spf}(t)\le t\le a$. $\square$

Lemma 858.2 (A concrete family gives $F(N)\ge 1/2+o(1)$).

Lemma. Let $A_N:=\{\lceil \sqrt N\rceil,\lceil \sqrt N\rceil+1,\dots,N\}$. Then $A_N$ satisfies the condition, and \[ \frac{1}{\log N}\sum_{n\in A_N}\frac{1}{n} = \frac{H_N-H_{\lceil\sqrt N\rceil-1}}{\log N} = \frac12 + o(1) \quad\text{as }N\to\infty, \] where $H_m=\sum_{j=1}^m \frac{1}{j}$.

Proof. By Lemma 858.1, any $a\in A_N$ satisfies $a\ge \sqrt N$, so no equation $b=a t$ with $b\le N$ can have $\operatorname{spf}(t)>a$. Hence $A_N$ satisfies the defining condition.

For the asymptotic, we use the standard integral bounds for harmonic numbers: for every integer $M\ge 1$, \[ \int_1^{M+1} \frac{dx}{x} \le H_M \le 1 + \int_1^M \frac{dx}{x}, \] which imply \[ \log(M+1) \le H_M \le 1+\log M. \] Apply this with $M=N$ and $M=\lceil\sqrt N\rceil-1$. Then \[ H_N - H_{\lceil\sqrt N\rceil-1} \ge \log(N+1) - (1+\log(\lceil\sqrt N\rceil-1)) = \tfrac12\log N + O(1), \] and similarly \[ H_N - H_{\lceil\sqrt N\rceil-1} \le (1+\log N) - \log(\lceil\sqrt N\rceil) = \tfrac12\log N + O(1). \] Dividing by $\log N$ gives the claimed limit $\tfrac12+o(1)$. $\square$

Lemma 858.3 (The element $1$ is incompatible with any other element).

Lemma. If $1\in A$ and $b\in A$ with $b\ge 2$, then $A$ violates the condition.

Proof. Take $a=1$ and $t=b$. Then $b=a t$ holds. Since $b\ge 2$, $\operatorname{spf}(t)=\operatorname{spf}(b)\ge 2>1=a$. Thus $(a,b,t)$ is a forbidden solution, contradicting the defining property. $\square$

Fast reality check (exact optimisation for small $N$).

Interpreting the problem literally as a finite-$N$ maximisation, we can brute-force compute $F(N)$ for $N\le 20$ (by enumerating all $2^N$ subsets and checking the forbidden condition). The computed maximisers for $N=20$ give \[ F(20)\approx 0.6968, \] attained for example by \[ A=\{2,3,4,5,7,8,9,11,12,13,16,17,19\}. \] In particular, for these small values the optimum is noticeably larger than the simple lower bound coming from $A_N=[\lceil\sqrt N\rceil,N]$.

VERIFICATION.

• Lemma 858.1 hinges only on the inequalities $t\le N/a$ and $\operatorname{spf}(t)\le t$; there are no hidden assumptions.
• Lemma 858.2 checks the forbidden condition by reducing to Lemma 858.1; this is valid because every $a\in A_N$ satisfies $a\ge\sqrt N$.
• The brute-force calculation for $N\le 20$ is consistent with Lemma 858.3 (all computed maximisers omit $1$).

FINAL. UNRESOLVED.

(i) Strongest proved partial results. For the literal finite-$N$ formulation, we proved the unconditional lower bound \[ F(N)\ge \frac{H_N-H_{\lceil\sqrt N\rceil-1}}{\log N} = \tfrac12+o(1), \] by exhibiting an explicit admissible set $A_N=[\lceil\sqrt N\rceil,N]$. We also proved the structural fact that any forbidden solution must have $a<\sqrt N$.

(ii) First gap. Determine the true asymptotic behaviour of the finite-$N$ maximum $F(N)$: does $F(N)$ converge, and if so to what constant? (The computation up to $N=20$ suggests values $>1/2$.)

(iii) Top 3 next moves. $1$ Reformalise the statement in the original Erdős source to resolve the apparent conflict between the literal lower bound $\ge 1/2$ and the cited "$o(1)$" claim. $2$ For the finite-$N$ version, model the condition as a maximum-weight independent set problem on the graph with vertices $\{1,\dots,N\}$ and forbidden edges $(a,at)$; use branch-and-bound/ILP to compute $F(N)$ for larger $N$ to guess a limit. $3$ Attempt a "swap"/charging argument comparing inclusion of small $a$ (large weight $1/a$) versus the forced exclusion of many multiples $a p$ with primes $p>a$.

(iv) What a minimal counterexample would look like. If the true statement is that $F(N)=o(1)$, then a counterexample to that would require an admissible family $A_N$ whose harmonic weight is a positive fraction of $\log N$ while still avoiding all forbidden multiplicative relations. The explicit family $[\lceil\sqrt N\rceil,N]$ already achieves fraction $\approx 1/2$ under the literal reading.