GPT 6 Astra Ultra attempt on Math-661

Erdős Problem #66: independent continuation

Date: 23 September 2026. Source versions read in full: 66.tex in gpt_pro_5.4.

1) OBJECTIVE AND SOURCE FOUNDATION

Assume, as in the supplied attempt, that the ordered representation function satisfies $r(n)=(1_A*1_A)(n)\sim L\log n$, $0<L<\infty$. The source proves \[ G(t)=\sum_{n\in A}e^{-tn}\sim\sqrt L\,t^{-1/2}\sqrt{\log(1/t)}, \quad G_\theta(t)=\sum_{n\in A}e^{-tn+in\theta}=o(G(t)) \] for each fixed $\theta\notin2\pi\mathbb Z$. Its statement that ordinary-counting equidistribution is still inaccessible can be improved: the positive majorant provides the needed Tauberian control. The source's uniform statement requires $\eta(t)/t\to\infty$, not merely $\eta(t)\ge c t$.

2) COMPACT-MEASURE DE-ABELIZATION

For real $x\to\infty$, put probability measures on $[0,1]$ \[ \mu_x=\frac1{G(1/x)}\sum_{n\in A}e^{-n/x}\delta_{e^{-n/x}}, \quad \nu_x=\frac1{G(1/x)}\sum_{n\in A}e^{-n/x+in\theta}\delta_{e^{-n/x}}. \] Their total variations satisfy $|\nu_x|\le\mu_x$. For every integer $k\ge0$, \[ \int y^k\,d\mu_x=\frac{G((k+1)/x)}{G(1/x)}\longrightarrow(k+1)^{-1/2}, \qquad \int y^k\,d\nu_x\longrightarrow0. \] The limiting moments of $\mu_x$ are those of the image under $y=e^{-u}$ of \[ \frac{e^{-u}u^{-1/2}}{\sqrt\pi}\,du\quad(u>0). \] Polynomial approximation on $[0,1]$ therefore proves weak convergence to this probability measure $\mu$, and $\int f\,d\nu_x\to0$ for every continuous $f$. Take the bounded, piecewise continuous function \[ f(y)=y^{-1}\mathbf1_{[e^{-1},1]}(y). \] Its only jump is at $e^{-1}$, where $\mu$ has no atom. Approximate $f$ by continuous functions outside an arbitrarily small neighbourhood of that point. The error in integration against $\nu_x$ is bounded by the corresponding error against $\mu_x$, which tends to an arbitrarily small quantity. Thus $\int f\,d\nu_x\to0$. The same approximation gives \[ \frac{A(x)}{G(1/x)}\longrightarrow\int_0^1\frac{u^{-1/2}}{\sqrt\pi}\,du=\frac2{\sqrt\pi}, \qquad \frac1{G(1/x)}\sum_{\substack{n\le x\\n\in A}}e^{in\theta}\longrightarrow0. \] Consequently \[ A(x)\sim\frac{2\sqrt L}{\sqrt\pi}\sqrt{x\log x},\qquad \frac1{A(x)}\sum_{\substack{n\le x\\n\in A}}e^{in\theta}\longrightarrow0. \]

3) CONSEQUENCE AND LIMITATION

Applying this to every nonzero integer multiple of $2\pi\alpha$ proves ordinary-counting equidistribution of $n\alpha$ along $n\in A$ for every fixed irrational $\alpha$. Rational rotations similarly give uniformity on their finite orbit. This closes the de-Abelization gap in the supplied continuation. It does not show that a set with $r(n)\sim L\log n$ exists or cannot exist.

7) FINAL

UNRESOLVED. A stronger necessary condition is proved under the same hypothetical representation asymptotic. There is no contradiction or construction for the original existence question.

8) COMPLETION ESTIMATE

COMPLETION: 15% This is a subjective estimate of progress toward the entire stated problem, not a probability that the conjecture or an unverified proof is correct.