GPT 6 Astra Ultra attempt on Math-5620

FORMAL RESTATEMENT

Erdős #391; prime-forced upper bound and reconstruction of the limit, 2026-09-24. Let $t(n)$ be the largest possible smallest factor in a factorization $n!=a_1\cdots a_n$ into $n$ positive integers. Prove $t(n)/n\to1/e$, and determine whether $t(n)/n\leq1/e-c/\log n$ for some fixed $c>0$ and infinitely many $n$.

QUICK LITERATURE AND CONTEXT CHECK

Alexeev, Conway, Rosenfeld, Sutherland, Tao, Uhr and Ventullo established a sharper asymptotic with explicit second-order constant. The proof below independently establishes a positive logarithmic deficit for all sufficiently large $n$. For the matching limit it explicitly imports only the lower-bound consequence of their construction.

OBJECTIVE AND ATTACK PLAN

A product bound gives $1/e$ as the limiting upper bound. A positive density of primes in $(3n/4,n]$ forces extra logarithmic mass in the factors containing them, creating a definite deficit below $n/e$.

WORK

Write $t=t(n)$ and choose an optimizing factorization, so $a_i\geq t$. From $t^n\leq n!$ and Stirling's formula, \[ \frac tn\leq\frac{(n!)^{1/n}}n\longrightarrow e^{-1}. \] In particular $t\leq n/2$ for all sufficiently large $n$.

Let $\mathcal P=\{p:3n/4<p\leq n,\ p\text{ prime}\}$. Each such prime has exponent exactly one in $n!$, so occurs in exactly one factor $a_i$. If that factor contains $r_i\geq1$ primes from $\mathcal P$, then \[ \log(a_i/t)\geq\sum_{\substack{p\in\mathcal P\\p\mid a_i}}\log p-\log t \geq\sum_{\substack{p\in\mathcal P\\p\mid a_i}}\log(p/t), \] since the difference is $(r_i-1)\log t\geq0$. Factors containing none contribute nonnegatively. Summing therefore gives the exact inequality \[ \log(n!)-n\log t\geq\sum_{p\in\mathcal P}\log(p/t) \geq\#\mathcal P\log(3/2). \tag{1} \] This remains valid when several large primes occur in the same factor; counting such a factor multiple times without the displayed inequality would be unjustified.

The prime number theorem gives $\#\mathcal P\sim n/(4\log n)$, and hence $\#\mathcal P\geq n/(5\log n)$ eventually. Put $a=\log(3/2)/5$. Stirling's upper estimate and $1$ imply \[ n\log(n/t)-n+O(\log n)\geq a n/\log n. \] For all sufficiently large $n$, absorbing the $O(\log n)$ term yields \[ \log(n/t)\geq1+\frac a{2\log n},\qquad \frac tn\leq\frac1e\exp\!\left(-\frac a{2\log n}\right) \leq\frac1e-\frac{\log(3/2)}{20e\log n}. \tag{2} \] The last inequality uses $e^{-u}\leq1-u/2$ for $0\leq u\leq1$. Thus the second requested assertion holds for every sufficiently large $n$, with the explicit nonoptimal constant $c=\log(3/2)/(20e)$.

Declared external input for the lower bound. The lower-bound part of Theorem 1.3(iv) of Alexeev et al. implies that for some absolute $C$ and all sufficiently large $n$, \[ t(n)\geq n/e-Cn/\log n. \tag{3} \] The prime-allocation and small-prime rebalancing construction proving $3$ is not reconstructed here. Combining $3$ with the elementary product upper bound proves $t(n)/n\to1/e$.

For comparison, the same published theorem identifies the sharper expansion $t(n)/n=1/e-c_0/\log n+O((\log n)^{-1-\gamma})$ with $c_0=0.30441901\ldots$ and some $\gamma>0$. Inequality $2$ does not claim that optimal constant.

RESULT AND LIMITATIONS

Attempt status: solved relative to the stated lower-bound construction. The affirmative answers to both questions follow. The logarithmic deficit is proved here using Stirling and the prime number theorem; the matching lower bound is the explicit published dependency. This is a reconstruction of known results, without a claim of novelty.

SOURCES

https://www.erdosproblems.com/391; Alexeev et al., Theorem 1.3(iv): https://arxiv.org/html/2503.20170.

Completion rate estimate: 100% relative to the stated external lower bound.