GPT 6 Astra Ultra attempt on Math-758
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Erdős Problem #319: independent continuation
Date: 23 September 2026. Source versions read in full:319.tex in gpt_pro_5.4.1) OBJECTIVE AND SOURCE AUDIT
The problem maximizes the size of a minimal signed vanishing reciprocal sum, with no proper nonempty zero-sum subcollection. The source settles the asymptotic density of the singleton-sign construction. Its proof applies a fixed-$r$ asymptotic with $r=1/p$, although $p$ can depend on $N$. The following finite inequality fixes that uniformity issue and extends the obstruction to every bounded size of the smaller sign class.2) WORK: BOUNDED SIGN CLASSES CANNOT APPROACH DENSITY ONE
Write $A=P\sqcup Q$ and suppose the two reciprocal sums agree. Put $h=\min(|P|,|Q|)$ and, by exchanging signs, assume $|P|=h$, $|Q|=m$. Since the largest possible reciprocal mass of $h$ distinct positive integers is $H_h$, while the smallest mass of $m$ members of $[1,N]$ is a terminal harmonic sum, \[ H_h\ge\sum_{p\in P}1/p=\sum_{q\in Q}1/q \ge H_N-H_{N-m} \ge\log\frac{N+1}{N-m+1}. \] The last inequality follows by integrating $1/t$ over $[N-m+1,N+1]$. Hence the uniform finite bound is \[ |A|=h+m\le h+(N+1)(1-e^{-H_h}). \tag{1} \] For $h=1$ this recovers the source's upper density $1-e^{-1}$ without any parameter-dependent error term. For each fixed $h$, it gives the upper density $1-e^{-H_h}<1$. Thus a sequence of admissible sets with $|A|/N\to1$ must have both sign-class sizes tending to infinity.More quantitatively, using $H_h\le1+\log h$ in $1$, \[ N-|A|\ge\frac{N+1}{eh}-h-1. \] This is nontrivial, for example, when $h=o(\sqrt N)$. It isolates a necessary growth requirement on the number of positive and negative terms for any construction improving strongly on the singleton mechanism.