Erdős Problem #66

Problem 66 --- Round 4 (continuation & spectral-rigidity mode)

1) ROUND-4 OBJECTIVE

Path (C): obstruction/correction direction. Round 3 proved that if \[ \frac{r(n)}{\log n}\to L\in(0,\infty),\qquad r(n):=(1_A\ast 1_A)(n), \] then $A$ is asymptotically equidistributed in every fixed residue class. The exact gap left after Round 3 is that this only controls rational Fourier modes (roots of unity). The most promising next step is therefore to pass from rational frequencies to all fixed frequencies on the unit circle.

In this round I prove a strictly stronger rigidity statement: the twisted Laplace transform \[ G_\theta(t):=\sum_{n\ge 1}1_A(n)e^{-tn}e^{in\theta} \] is lower-order for every fixed $\theta\notin 2\pi\mathbb Z$, and in fact this decay is uniform away from an $o(1)$-arc around $\theta=0$. As a consequence, the exponentially weighted measures supported on $\{n\alpha\}$ with $n\in A$ converge to Haar measure on the orbit closure of $\alpha$; in particular, for irrational $\alpha$, the set $A$ is Abel-weightedly equidistributed modulo $1$ along the rotation $n\mapsto n\alpha$.

This is a genuine extension of Round 3 from finite cyclic Fourier analysis to the full torus.

2) ROUND-3 FOUNDATION USED

I use the following previously established results as black boxes.

• Round 2 Lemma 66.3: \[ \sum_{n\ge2}(\log n)e^{-tn}\sim \frac1t\log\frac1t. \]
• Round 2 Lemma 66.4: with \[ F(t):=\sum_{n\ge 2} r(n)e^{-tn}, \] we have \[ F(t)\sim \frac{L}{t}\log\frac1t\qquad(t\to 0^+). \]
• Round 2 Lemma 66.5: with \[ G(t):=\sum_{n\ge 1}1_A(n)e^{-tn}, \] we have \[ G(t)\sim \sqrt{L}\,t^{-1/2}\sqrt{\log\tfrac1t}. \]
• Round 3 Theorem 66.10 and Corollary 66.11: every fixed modulus $q$ is equidistributed, equivalently every nontrivial rational Fourier mode is asymptotically negligible in natural counting. I use this as the structural benchmark that the present round must strictly strengthen.

3) NEW INSIGHT / TOOL (ROUND-4)

New tool: uniform summation-by-parts for twisted Laplace sums.

For \[ S_\theta(t):=\sum_{n\ge 2}(\log n)e^{-tn}e^{in\theta}, \] I prove the quantitative bound \[ S_\theta(t)\ll \frac{\log(1/t)}{\operatorname{dist}(\theta,2\pi\mathbb Z)} \] uniformly for $\theta\notin 2\pi\mathbb Z$. Since the main untwisted size is \[ \sum_{n\ge2}(\log n)e^{-tn}\asymp \frac1t\log\frac1t, \] this shows that twisting by any angle separated from $0$ by more than $\gg t$ forces a power-sized drop in the Laplace transform.

Inserted into the identity \[ \sum_{n\ge2} r(n)e^{-tn}e^{in\theta}=G_\theta(t)^2, \] this gives a uniform spectral exclusion principle: the only place where the generating function can have the principal singular size is inside a vanishing arc around $z=1$.

4) ATTACK PLAN (ROUND-4)

The remaining gap after Round 3 is:

Round 3 controls only rational phases $e^{2\pi i m/q}$; it does not rule out a persistent irrational or quasiperiodic bias of $1_A$.

I address exactly that gap by proving the following claims.

1. For the model coefficients $(\log n)$, twisted Laplace sums at angle $\theta$ are only of size $O\big(\log(1/t)/\delta\big)$, where $\delta=\operatorname{dist}(\theta,2\pi\mathbb Z)$.
2. Therefore the twisted Laplace transform of the representation function, \[ R_\theta(t):=\sum_{n\ge2}r(n)e^{-tn}e^{in\theta}, \] is $o\big((1/t)\log(1/t)\big)$ uniformly whenever $\delta\gg t$.
3. Since $R_\theta(t)=G_\theta(t)^2$, the twisted transform $G_\theta(t)$ is $o\big(t^{-1/2}\sqrt{\log(1/t)}\big)$ uniformly away from a $t$-scale neighborhood of $0$.
4. Normalizing by the untwisted mass $G(t)$ then yields convergence of the exponentially weighted orbital measures \[ \mu_t^{(\alpha)}:=\frac1{G(t)}\sum_{n\ge1}1_A(n)e^{-tn}\,\delta_{\{n\alpha\}} \] to Haar measure on the compact subgroup generated by $\alpha$.

This strictly strengthens Round 3: rational frequencies become a special case, while irrational frequencies are genuinely new.

5) WORK (ROUND-4)

Throughout, write \[ a_n:=1_A(n),\qquad r(n):=\sum_{k=1}^{n-1}a_ka_{n-k}, \] and assume \begin{equation} \label{wsref-0-65-71-3a-52-34-2d-61-73-73-75-6d-70-74-69-6f-6e} \frac{r(n)}{\log n}\to L\in(0,\infty). \end{equation} Also put \[ M(t):=\frac1t\log\frac1t\qquad(t\in(0,e^{-1}]). \]

Lemma 66.12 (uniform twisted model bound). Let \[ S_\theta(t):=\sum_{n\ge2}(\log n)e^{-tn}e^{in\theta} \qquad(t>0,\ \theta\in\mathbb R), \] and let \[ \delta(\theta):=\operatorname{dist}(\theta,2\pi\mathbb Z)\in[0,\pi]. \] Then for every $t\in(0,e^{-1}]$ and every $\theta\notin 2\pi\mathbb Z$, \[ S_\theta(t)\ll \frac{\log(1/t)}{\delta(\theta)}, \] with an absolute implied constant.

Proposition 66.13 (uniform twisted cancellation for the representation transform). Define \[ R_\theta(t):=\sum_{n\ge2} r(n)e^{-tn}e^{in\theta} \qquad(t>0,\ \theta\in\mathbb R). \] Let $\eta:(0,e^{-1}]\to(0,\pi]$ satisfy \[ \frac{\eta(t)}{t}\to\infty\qquad(t\to0^+). \] Then \[ \sup_{\delta(\theta)\ge \eta(t)} \frac{|R_\theta(t)|}{M(t)}\to 0 \qquad(t\to0^+). \] In particular, for every fixed $\theta\notin2\pi\mathbb Z$, \[ R_\theta(t)=o\!\left(\frac1t\log\frac1t\right). \]

Theorem 66.14 (uniform spectral exclusion away from $1$). Let \[ G_\theta(t):=\sum_{n\ge1} a_n e^{-tn}e^{in\theta} \qquad(t>0,\ \theta\in\mathbb R). \] Then \[ R_\theta(t)=G_\theta(t)^2 \qquad\text{for all }t>0,\ \theta\in\mathbb R. \] Moreover, for every function $\eta(t)$ with $\eta(t)/t\to\infty$, \[ \sup_{\delta(\theta)\ge \eta(t)} \frac{|G_\theta(t)|}{t^{-1/2}\sqrt{\log(1/t)}}\to0 \qquad(t\to0^+). \] In particular, for every fixed $\theta\notin 2\pi\mathbb Z$, \[ G_\theta(t)=o\!\left(t^{-1/2}\sqrt{\log\tfrac1t}\right). \]

Corollary 66.15 (Abel-weighted orbit equidistribution on the circle). For $\alpha\in\mathbb R$, define a probability measure on $\mathbb T:=\mathbb R/\mathbb Z$ by \[ \mu_t^{(\alpha)}:=\frac{1}{G(t)}\sum_{n\ge1} a_n e^{-tn}\,\delta_{\{n\alpha\}} \qquad(t>0), \] where $\{x\}$ denotes the class of $x$ mod $1$. Let \[ H_\alpha:=\overline{\{n\alpha\pmod 1:n\in\mathbb Z\}}\le \mathbb T \] be the closed subgroup generated by $\alpha$, and let $m_{H_\alpha}$ denote Haar probability measure on $H_\alpha$.

Then \[ \mu_t^{(\alpha)}\Longrightarrow m_{H_\alpha} \qquad(t\to0^+) \] in the weak-* topology. Equivalently, for every continuous $1$-periodic function $f$, \[ \frac{\sum_{n\ge1} a_n e^{-tn}f(n\alpha)}{\sum_{n\ge1} a_n e^{-tn}} \to \int_{H_\alpha} f\,dm_{H_\alpha}. \] In particular:

• if $\alpha\notin\mathbb Q$, then $H_\alpha=\mathbb T$ and \[ \frac{\sum_{n\ge1} a_n e^{-tn}f(n\alpha)}{\sum_{n\ge1} a_n e^{-tn}} \to \int_0^1 f(x)\,dx; \]
• if $\alpha=p/q\in\mathbb Q$ in lowest terms, then $H_\alpha=\{0,1/q,\dots,(q-1)/q\}$ and the limit measure is uniform on this $q$-point orbit.

6) ADVERSARIAL VERIFICATION

• Interaction with Round 3. Round 3 handled rational frequencies by discrete Fourier analysis modulo $q$. The present round does not re-prove that theorem; it extends the cancellation mechanism to all real frequencies via a different summation-by-parts argument. Rational phases become a special case of Theorem 66.14.
• Uniformity near $\theta=0$. The bound in Lemma 66.12 necessarily degenerates like $1/\delta(\theta)$. Accordingly, Theorem 66.14 only claims uniform decay when $\delta(\theta)\gg t$; no false uniformity across a $t$-scale arc around $0$ is asserted.
• Ordered convolution. The identity $R_\theta(t)=G_\theta(t)^2$ depends on the ordered representation function. For unordered representations the transform would differ by a diagonal correction, so the present constants and formulas are specific to the ordered problem, consistently with earlier rounds.
• Complex square-root issue. No branch choice is used. We only infer \[ |G_\theta(t)|^2=|R_\theta(t)|, \] so the uniform $o$-estimate follows from absolute values alone.
• Haar-limit corollary. The proof uses only Fourier coefficients of probability measures on the compact group $\mathbb T$ and density of trigonometric polynomials in $C(\mathbb T)$. No unjustified de-Abelization is claimed.
• What is and is not proved. Corollary 66.15 is an Abel-weighted equidistribution theorem. It does not imply that \[ \frac{1}{A(x)}\sum_{\substack{n\le x\\ n\in A}} e^{2\pi i mn\alpha}\to0 \] for irrational $\alpha$; such a Tauberian de-Abelization step remains open here.
• Consistency with earlier rounds. If $\alpha=p/q$, Corollary 66.15 gives only exponential-weight equidistribution on the $q$-point orbit. Round 3 Theorem 66.10 is stronger in that special case, because it upgrades the conclusion to ordinary counting in each residue class mod $q$.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 4 proves a new rigidity theorem beyond Round 3: \[ \sum_{n\ge1}1_A(n)e^{-tn}e^{in\theta} =o\!\left(t^{-1/2}\sqrt{\log\tfrac1t}\right) \qquad(\theta\notin2\pi\mathbb Z), \] and in fact the decay is uniform outside every arc of width $\eta(t)$ with $\eta(t)\gg t$. Equivalently, the principal singular behavior of the generating function is spectrally localized at $z=1$; every other fixed boundary phase is negligible.

As a consequence, for every fixed $\alpha$, the exponentially weighted orbital measures supported on $\{n\alpha:n\in A\}$ converge to Haar measure on the subgroup generated by $\alpha$. In particular, every irrational rotation is uniformly distributed along $A$ in Abel means. This rules out an entire class of constructions with persistent fixed-frequency or quasiperiodic bias.

The original existence/nonexistence problem remains open: I have not produced either a contradiction or a construction of such a set $A$.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 63%

9) REFERENCES

• N. H. Bingham, C. M. Goldie, J. L. Teugels, Regular Variation, Cambridge Univ. Press.
• W. Rudin, Fourier Analysis on Groups, Wiley.

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.

Problem 66 --- Round 2 (continuation & gap-closure mode)

1) ROUND-2 OBJECTIVE

Path (C): obstruction/correction direction. Starting from the Round 1 status UNRESOLVED, I pursue the "disproof/rigidity" track by extracting a strictly stronger necessary consequence of the existence of a nonzero limit \[ \frac{(1_A\ast 1_A)(n)}{\log n}\to L\in(0,\infty). \] Round 1 only obtained the order of growth $A(x)=\Theta(\sqrt{x\log x})$. In this round I close that gap: assuming the limit exists, I prove that the counting function in fact has a full asymptotic with an explicit constant depending on $L$: \[ A(x)\sim 2\sqrt{\frac{L}{\pi}}\,\sqrt{x\log x}. \] This is a rigid constraint on any potential example and is a substantial strengthening of Round 1.

2) ROUND-1 FOUNDATION USED

I rely on the following Round 1 items (as black boxes):

• The formal definitions: $a_n:=1_A(n)$ and $r(n):=(1_A\ast 1_A)(n)=\sum_{k=1}^{n-1}a_ka_{n-k}$ counts ordered representations.
• Corollary 66.1 only as context (I strengthen it).

No Round 1 probabilistic computations are used in the new proof.

3) NEW INSIGHT / TOOL (ROUND-2)

New tool: Abelian--Tauberian passage through generating functions/Laplace transforms.

Let \[ F(t):=\sum_{n\ge 2} r(n)e^{-tn}=\Big(\sum_{m\ge 1} a_m e^{-tm}\Big)^2\qquad(t>0), \] which is the Laplace transform of $r(n)$ and equals the square of the Laplace transform of $a_m$.

From the assumed pointwise asymptotic $r(n)\sim L\log n$ I first derive an Abelian asymptotic for $F(t)$ as $t\to 0^+$, then take a square root, and finally apply Karamata's Tauberian theorem (Laplace transform form) to recover an explicit asymptotic for the counting function $A(x)=\sum_{m\le x}a_m$.

4) ATTACK PLAN (ROUND-2)

Round 1 gap addressed: Corollary 66.1 only yields two-sided bounds $c\sqrt{x\log x}\le A(x)\le C\sqrt{x\log x}$.

Goal in this round: upgrade this to an explicit asymptotic with constant.

Plan:

1. Prove an Abelian lemma: from $r(n)\sim L\log n$, show \[ F(t)=\sum_{n\ge 2} r(n)e^{-tn}\sim \frac{L}{t}\log\frac1t\qquad(t\to 0^+). \]
2. Take square roots to obtain the Laplace transform asymptotic for $a_n$: \[ G(t):=\sum_{m\ge 1} a_m e^{-tm}=\sqrt{F(t)}\sim \sqrt{L}\,t^{-1/2}\,\sqrt{\log\tfrac1t}. \]
3. Apply Karamata's Tauberian theorem to the nondecreasing function $A(x)$ to conclude \[ A(x)\sim \frac{\sqrt{L}}{\Gamma(3/2)}\,x^{1/2}\,\sqrt{\log x}=2\sqrt{\frac{L}{\pi}}\,\sqrt{x\log x}. \]

5) WORK (ROUND-2)

Throughout, assume \begin{equation} \label{wsref-0-65-71-3a-6c-69-6d-69-74-2d-61-73-73-75-6d-70-74-69-6f-6e} \frac{r(n)}{\log n}\to L\in(0,\infty),\qquad r(n)=(1_A\ast 1_A)(n). \end{equation}

Lemma 66.3 (model Laplace asymptotic for $\sum \log n\,e^{-tn}$). Define \[ S(t):=\sum_{n\ge 2} (\log n)e^{-tn}\qquad(t>0). \] Then as $t\to 0^+$, \[ S(t)=\frac{1}{t}\log\frac{1}{t}+O\Big(\frac{1}{t}\Big). \] In particular $S(t)\sim \frac{1}{t}\log\frac{1}{t}$.

Lemma 66.4 (Abelian step: Laplace asymptotic for $r(n)$). Let \[ F(t):=\sum_{n\ge 2} r(n)e^{-tn}\qquad(t>0). \] Under (eq:limit-assumption), as $t\to 0^+$, \[ F(t)\sim \frac{L}{t}\log\frac1t. \]

Lemma 66.5 (square-root transfer to the Laplace transform of $A$). Define \[ G(t):=\sum_{m\ge 1} a_m e^{-tm}=\sum_{m\in A} e^{-tm} \qquad(t>0). \] Then $F(t)=G(t)^2$ and, under (eq:limit-assumption), \[ G(t)\sim \sqrt{L}\,t^{-1/2}\,\sqrt{\log\tfrac1t}\qquad(t\to 0^+). \]

Proposition 66.6 (Tauberian sharpening of Round 1 Corollary 66.1). Assume (eq:limit-assumption). Then the counting function satisfies the full asymptotic \[ A(x):=|A\cap[1,x]|\sim 2\sqrt{\frac{L}{\pi}}\,\sqrt{x\log x}\qquad(x\to\infty). \] Equivalently, \[ \lim_{x\to\infty}\frac{A(x)^2}{x\log x}=\frac{4L}{\pi}. \]

6) ADVERSARIAL VERIFICATION

• Ordered vs unordered representations. The identity $F(t)=G(t)^2$ uses ordered convolution. If one used unordered representations, $F(t)$ would instead correspond to $(G(t)^2+G(2t))/2$; constants would change. Round 1 and this round consistently use ordered pairs.
• Asymptotic of $\sum (\log n)e^{-tn}$. Lemma 66.3 only needs a leading term. The discretization error is at most $O(\log(1/t))$, far smaller than the main term $(1/t)\log(1/t)$, so the $\sim$ claim is stable.
• Taking square roots. For $t>0$, $G(t)\ge 0$ and $F(t)=G(t)^2$, so $G(t)=\sqrt{F(t)}$ with no sign ambiguity.
• Karamata hypotheses. The function $A(x)$ is nondecreasing and right-continuous; $\ell(u)=\sqrt{\log u}$ is slowly varying. Also $\rho=1/2>0$, so the theorem applies.
• Boundary cases. If $L=0$ the same method gives $G(t)=o(t^{-1/2}\sqrt{\log(1/t)})$, hence $A(x)=o(\sqrt{x\log x})$, consistent with Round 1. If $L=\infty$ the method is inapplicable; the statement of the problem asks $L$ finite.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 2 closes a concrete Round 1 gap: under the existence of a finite nonzero limit $r(n)/\log n\to L$, the counting function must satisfy the precise asymptotic \[ A(x)\sim 2\sqrt{\frac{L}{\pi}}\,\sqrt{x\log x}, \] which is substantially stronger than $A(x)=\Theta(\sqrt{x\log x})$.

The core existence/nonexistence question remains open: I did not derive a contradiction nor construct such an $A$.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 40%

9) REFERENCES

• N. H. Bingham, C. M. Goldie, J. L. Teugels, Regular Variation, Cambridge Univ. Press (for Karamata Tauberian theorem, Laplace-transform form).
• W. Feller, An Introduction to Probability Theory and Its Applications, Vol. II (Tauberian theorems chapter).