GPT 6 Astra Ultra attempt on Math-784
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Erdős Problem #371: independent continuation
Date: 2026-09-23. Research attempt; no claim of a new published result.1) TARGET AND SOURCE AUDIT
Does $\{n:P^+(n)<P^+(n+1)\}$ have natural density $1/2$?All supplied versions considered: 371.tex. The source cites logarithmic averaging and almost-all-scale results. I test whether those conclusions can bridge the remaining limit gap using only properties of counting functions. They cannot, as the following explicit model demonstrates.
2) WORK
Let $M_j=2^{2^j}$ and define a set $A$ by starting with the even positive integers and adjoining every integer in $[M_j,2M_j]$. Its logarithmic density is $1/2$: the added harmonic mass up to $X$ is at most $O(\#\{j:M_j\le X\})=O(\log\log X)=o(\log X)$. However, at $X=2M_j$, the latest block adds $M_j/2+O(1)$ formerly absent odd integers. All earlier blocks have total size $o(M_j)$, since $M_{j-1}=\sqrt{M_j}$. Thus \[ {|A\cap[1,2M_j]|\over2M_j}\longrightarrow\frac34. \] At $X=M_j-1$, earlier additions have density $o(1)$, giving a subsequential limit $1/2$.This model even has density $1/2$ at almost all logarithmic scales in the quantitative sense that for every fixed $\epsilon>0$, the scales with deviation greater than $\epsilon$ have logarithmic density zero. Indeed additions up to $X$ are bounded by a constant times the largest $M_j\le X$. A deviation forces $X\le C_\epsilon M_j$, so the exceptional scales lie in the union of intervals $[M_j,C_\epsilon M_j]$, each of bounded logarithmic length. Their count is $O(\log\log X)$. Therefore the generic counting-function structure, even together with those two averaging properties, cannot yield a full natural-density limit.