gpt 5.2 attempt on Math-646

OUTPUT 1) FORMAL RESTATEMENT A positive integer $m$ is practical if every integer $1\le n < m$ can be written as a sum of distinct divisors of $m$. If $m$ is practical, define $h(m)$ to be the least integer $H$ such that every $1\le n < m$ can be written as a sum of at most $H$ distinct divisors of $m$. Questions (as stated): (a) Are there infinitely many practical $m$ with $h(m) < (\log\log m)^{O(1)}$? (b) Is it true that $h(n!)<n^{o(1)}$? Or even $h(n!)<(\log n)^{O(1)}$?

2) QUICK LITERATURE/CONTEXT CHECK Browsing is not available. I only record what the problem statement itself claims. I have not verified those results here.

3) ATTACK PLAN Proof track: 1. For explicit families of practical numbers, compute/estimate $h(m)$ by a greedy algorithm. 2. For $m=n!$, exploit the density of divisors to represent all $<m$ using few divisors.

Disproof track: 1. Produce lower bounds on $h(m)$ for large classes of practical $m$ (e.g. via information-theoretic counting). 2. Attempt to show $h(n!)$ must grow at least like a power of $\log n$.

Chosen path: work out a fully explicit family ($m=2^k$) and its exact $h(m)$.

4) WORK Lemma 1 ($2^k$ is practical). For every integer $k\ge 1$, the number $m=2^k$ is practical.

Proof. The divisors of $2^k$ are $1,2,4,\dots,2^k$. Every integer $0\le n<2^k$ has a binary expansion $n=\sum_{j=0}^{k-1} \epsilon_j 2^j$ with $\epsilon_j\in\{0,1\}$, i.e. a sum of distinct divisors of $2^k$. Thus $2^k$ is practical. $\square$

Lemma 2 (Upper bound for $h(2^k)$). For $m=2^k$, one has $h(m)\le k$.

Proof. By Lemma 1, every $n<2^k$ is a sum of distinct powers of $2$ less than $2^k$, and there are at most $k$ such powers. So $k$ divisors always suffice. $\square$

Lemma 3 (Lower bound for $h(2^k)$). For $m=2^k$, one has $h(m)\ge k$.

Proof. The integer $2^k-1$ has binary expansion with all $k$ lower bits equal to $1$: $2^k-1=1+2+\cdots+2^{k-1}$. Any sum of distinct divisors of $2^k$ uses distinct powers of $2$. To obtain $2^k-1$ one must use each of $1,2,\dots,2^{k-1}$, hence at least $k$ divisors. Thus $h(2^k)\ge k$. $\square$

Corollary 4 (Exact value). For $m=2^k$, $h(m)=k$.

Proof. Combine Lemma 2 and Lemma 3. $\square$

5) VERIFICATION - Lemmas 1--3 are purely about binary expansion and use no external results. - Edge cases: $k=1$ gives $m=2$, and indeed $h(2)=1$.

6) FINAL UNRESOLVED

(i) Strongest fully proved partial result: An explicit infinite family of practical numbers has exact growth $h(2^k)=k=\log_2(2^k)$ (Corollary 4).

(ii) First gap: No argument here produces practical numbers with $h(m)$ as small as a polylogarithm of $\log m$, or controls $h(n!)$ beyond the trivial constructions.

(iii) Top 3 next moves: 1. Prove general lower bounds on $h(m)$ in terms of divisor structure (e.g. via counting possible subset sums). 2. For $m=n!$, design a constructive algorithm representing all $<m$ using very few divisors and prove its correctness. 3. Search for families of practical numbers with exceptionally dense divisors (e.g. highly composite numbers) and estimate $h(m)$.

(iv) Minimal counterexample structure (if $h(n!)<(\log n)^{O(1)}$ were false): A sequence $n_j$ for which there exists some $t<n_j!$ that requires many divisors in any subset-sum representation, forcing $h(n_j!)$ to grow faster than any power of $\log n_j$.