gpt pro 5.2 attempt on Math-5965

FORMAL RESTATEMENT. Fix $n\ge 1$. A set $S\subseteq\{1,2,\dots,n\}$ is admissible if \[ \forall a\ne b\in S,\qquad \gcd(a,b)=1. \] Define \[ G(n)=\max_{\substack{S\subseteq\{1,\dots,n\}\\ S\text{ admissible}}}\ \sum_{a\in S} a. \] Define \[ H(n)=\sum_{\substack{p\text{ prime}\\ p<n}} p\ +\ n\,\pi(\sqrt{n}), \] where $\pi(x)=\#\{p\le x: p\text{ prime}\}$.

Questions in the problem text:

• Is it true that $G(n)>H(n)-n^{1+o(1)}$?
• Is it true that for every fixed $k\ge 2$, for all sufficiently large $n$, an admissible set achieving $G(n)$ contains an integer with at least $k$ prime factors? (The problem text does not specify whether "prime factors" means with multiplicity $\Omega(\cdot)$ or distinct prime factors $\omega(\cdot)$.)

The problem text states that Erdős and van Lint proved bounds \[ H(n)-n^{3/2-o(1)}<G(n)<H(n) \quad\text{and}\quad \frac{H(n)-G(n)}{n}\to\infty, \] and proved the second claim for $k=2$.

QUICK LITERATURE/CONTEXT CHECK. I only use the statements explicitly given in the problem text (quoted above) as context. The proofs below are elementary and self-contained.

ATTACK PLAN.

1. Prove an elementary upper bound of the shape $G(n)\le H(n)+O(1)$ by decomposing admissible sets into primes and composites and using that every composite has a small prime factor.
2. Identify forced elements that must appear in any maximizing set (e.g. primes in $(n/2,n]$).
3. Compute exact values for small $n$ via dynamic programming on prime-factor supports.

WORK.

Lemma 1 (elementary $H$-type upper bound). For every $n\ge 4$ and every admissible $S\subseteq\{1,\dots,n\}$, \[ \sum_{a\in S} a \le \sum_{p<n} p\ +\ n\,\pi(\sqrt{n})\ +\ 1. \] In particular, $G(n)\le H(n)+1$ for all $n\ge 4$.

Proof. Write $S=\{1\}\cup S'$ if $1\in S$, otherwise $S=S'$. Let \[ S_{\le \sqrt n}:=\{a\in S':\ a\text{ is divisible by some prime }p\le \sqrt n\}, \] and let $S_{>\sqrt n}:=S'\setminus S_{\le\sqrt n}$.

Step 1: bound $S_{\le \sqrt n}$. For each $a\in S_{\le\sqrt n}$, let $p(a)$ be the smallest prime divisor of $a$. Then $p(a)\le \sqrt a\le \sqrt n$. If $a\ne b$ are distinct elements of $S_{\le\sqrt n}$, then $\gcd(a,b)=1$, so they cannot share any prime divisor. In particular $p(a)\ne p(b)$. Hence the map $a\mapsto p(a)$ is injective from $S_{\le\sqrt n}$ into the set of primes $\le \sqrt n$. Therefore \[ |S_{\le\sqrt n}|\le \pi(\sqrt n). \] Since every element of $S_{\le\sqrt n}$ is at most $n$, we get \[ \sum_{a\in S_{\le\sqrt n}} a \le n\,\pi(\sqrt n). \]

Step 2: identify $S_{>\sqrt n}$. Any integer $a\le n$ that has no prime factor $\le \sqrt n$ is either $1$ or a prime $>\sqrt n$. Indeed, if $a$ were composite with all prime factors $>\sqrt n$, it would have at least two such prime factors and hence exceed $n$. Thus $S_{>\sqrt n}$ consists only of primes in $(\sqrt n,n]$. Consequently, \[ \sum_{a\in S_{>\sqrt n}} a \le \sum_{\substack{p\text{ prime}\\ p<n}} p, \] since the right-hand side sums all primes $<n$, which certainly dominates the sum over the subset $S_{>\sqrt n}$.

Combining Step 1 and Step 2, and adding back a possible $1\in S$, we obtain \[ \sum_{a\in S} a \le \Big(\sum_{p<n} p\Big) + n\pi(\sqrt n) + 1. \] $\square$

Lemma 2 (primes in $(n/2,n]$ are forced in any maximizing set). Let $n\ge 2$ and let $S\subseteq\{1,\dots,n\}$ be admissible with $\sum_{a\in S}a=G(n)$. Then every prime $p$ with $n/2<p\le n$ must belong to $S$.

Proof. Fix a prime $p$ with $n/2<p\le n$. Then $2p>n$, so the only multiple of $p$ in $\{1,\dots,n\}$ is $p$ itself. Therefore no element of $\{1,\dots,n\}\setminus\{p\}$ is divisible by $p$. Hence $\gcd(p,a)=1$ for every $a\in\{1,\dots,n\}\setminus\{p\}$.

If $p\notin S$, then $S\cup\{p\}$ is still admissible and has strictly larger sum, contradicting maximality of $S$. Thus $p\in S$. $\square$

VERIFICATION (FAST REALITY CHECK). For small $n$ I computed $G(n)$ exactly by dynamic programming over prime-support masks (weighted set packing). Below are values of $G(n)$ and $H(n)$ (as defined in the problem), together with one maximizing admissible set $S$ found by reconstruction.

 n   G(n)   H(n)  H(n)-G(n)   example maximizing S
10    30     37       7       {5,7,8,9}
15    55     71      16       {7,8,11,13,15}
20    99    117      18       {7,11,13,15,16,17,19}
30   193    219      26       {11,13,17,19,23,25,27,28,29}
40   275    317      42       {11,13,17,19,23,27,29,31,32,35,37}
50   446    528      82       {13,17,19,23,25,27,29,31,37,41,43,44,47,49}

(These sets are admissible: all pairs have gcd $1$.) The difference $(H(n)-G(n))/n$ increases over this range (e.g. $0.7$ at $n=10$ and $1.64$ at $n=50$), consistent with the statement in the problem text that $(H(n)-G(n))/n\to\infty$.

For the "$k$ prime factors" question: in these examples the maximum number of distinct prime factors among selected elements is $2$ (e.g. $15=3\cdot 5$), while the maximum number of prime factors counted with multiplicity can be larger (e.g. $16=2^4$).

FINAL. UNRESOLVED.

(i) Strongest proved partial result here. Lemma 1: a completely elementary $H$-type upper bound $G(n)\le H(n)+1$ (for $n\ge 4$). Lemma 2: any maximizing set must contain all primes in $(n/2,n]$.

(ii) First gap (crisp). Prove (unconditionally) a lower bound of the form $G(n)\ge H(n)-n^{1+o(1)}$ or disprove it by constructing $n$ with $H(n)-G(n)$ larger than $n^{1+o(1)}$.

(iii) Top 3 next moves (concrete).

1. Improve Lemma 1 from $H(n)+1$ toward $H(n)$ by sharpening the treatment of primes and composites (the crude upper bound counts all primes $<n$ regardless of whether they can coexist with the chosen composites).
2. Explore explicit constructions for large admissible sets that use many near-$n$ composites with carefully chosen disjoint prime supports, and quantify how close their sums get to $H(n)$.
3. For the "$k$ prime factors" problem, search computationally for maximizing sets for larger $n$ using heuristic/ILP methods and track the maximum of $\omega(a)$ or $\Omega(a)$ among chosen elements.

(iv) Minimal counterexample structure. A counterexample to $G(n)>H(n)-n^{1+o(1)}$ would be an infinite sequence $n_j$ where every admissible set has sum at most $H(n_j)-n_j^{1+\delta}$ for some fixed $\delta>0$. Such an $n_j$ would force a strong incompatibility between taking many large primes and simultaneously taking $\pi(\sqrt{n_j})$ large composite "almost-$n_j$" numbers with disjoint prime supports.