Reused writeup on Math-6056 (gpt pro 5.2)

Erdős Problem 1002 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/1002

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

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Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/1002.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/1002.tex.

1) FORMAL RESTATEMENT For $\alpha\in(0,1)$ and integer $n\ge 2$, define \[ F(\alpha,n) := \frac{1}{\log n}\sum_{k=1}^n \left(\frac12 - \{\alpha k\}\right). \] Let $\alpha$ be chosen uniformly from $(0,1)$. The question is whether there exists an (asymptotic) distribution function $g:\mathbb{R}\to[0,1]$ such that for every real $c$, \[ \lim_{n\to\infty}\,\mathbb{P}\left(F(\alpha,n)\le c\right)=g(c). \]

2) QUICK LITERATURE/CONTEXT CHECK The statement mentions a theorem of Kesten for the related quantity with an additional random shift $\beta$ in the indicator $\mathbf 1_{[0,\beta)}(\{\alpha k\})$. I do not use results beyond what is stated.

3) ATTACK PLAN Exploit explicit structure of the summand $\frac12-\{x\}$: - Rewrite the sum in terms of floor sums $\sum\lfloor \alpha k\rfloor$. - Use the Fourier series of the sawtooth function $x\mapsto \frac12-\{x\}$ to express the sum in terms of geometric series $\sum_{k\le n} e^{2\pi i h\alpha k}$. - Identify how near-rational $\alpha$ creates large contributions; this is consistent with heavy-tailed limiting distributions.

4) WORK

FAST REALITY CHECK (Monte Carlo sampling). I sampled $300$ random rationals $\alpha=a/M$ with $M=1{,}000{,}000{,}007$ prime and computed $F(\alpha,n)$ for $n=20000$. The empirical summary: \[ \min\approx -5.1641,\quad \max\approx 6.0323,\quad \text{median}\approx -0.0040,\quad \text{mean}\approx 0.0156,\quad \text{stdev}\approx 0.8945. \] The spread and outliers are consistent with a heavy-tailed distribution (but this is only a sanity check, not a proof).

Lemma 1 (Mean zero for each fixed $n$). For every integer $n\ge 1$, \[ \int_0^1 F(\alpha,n)\,d\alpha = 0. \]

Proof. Fix $k\ge 1$. The map $T_k:[0,1]\to[0,1]$ given by $T_k(\alpha)=\{k\alpha\}$ is measure-preserving: for any integrable $g$, \[ \int_0^1 g(\{k\alpha\})\,d\alpha = \int_0^1 g(x)\,dx. \] To verify this, partition $[0,1)$ into intervals $[j/k,(j+1)/k)$ for $j=0,\dots,k-1$. On each interval, $\{k\alpha\}=k\alpha-j$ runs bijectively over $[0,1)$ with Jacobian $1/k$, so \[ \int_{j/k}^{(j+1)/k} g(\{k\alpha\})\,d\alpha = \frac{1}{k}\int_0^1 g(x)\,dx, \] and summing over $j$ gives the claim.

Apply this with $g(x)=\frac12-x$ on $[0,1)$ (so $g(x)=\frac12-\{x\}$ periodically). Since $\int_0^1 (\frac12-x)dx=0$, we have \[ \int_0^1 \left(\frac12-\{k\alpha\}\right)d\alpha = 0. \] Summing over $k=1,\dots,n$ and dividing by $\log n$ gives $\int_0^1 F(\alpha,n)d\alpha=0$. $\square$

Lemma 2 (Fourier expansion of the sawtooth). Define the 1-periodic function $g(x):=\frac12-\{x\}$ (with $\{x\}\in[0,1)$). Then for $x\notin\mathbb{Z}$, \[ g(x) = \sum_{h\in\mathbb{Z}\setminus\{0\}} \frac{1}{2\pi i h}\,e^{2\pi i h x} = \sum_{h=1}^\infty \frac{\sin(2\pi h x)}{\pi h}. \]

Proof. On $x\in(0,1)$ we have $g(x)=\frac12-x$. For $h\neq 0$, the Fourier coefficient is \[ \widehat g(h)=\int_0^1 \left(\frac12-x\right)e^{-2\pi i h x}\,dx. \] Integrating by parts (or evaluating directly) yields \[ \widehat g(h)=\frac{1}{2\pi i h}. \] Also $\widehat g(0)=\int_0^1 (\frac12-x)dx=0$. Therefore the Fourier series is \[ \sum_{h\neq 0} \widehat g(h)e^{2\pi i h x} = \sum_{h\neq 0} \frac{1}{2\pi i h}e^{2\pi i h x}. \] Standard Fourier theory for piecewise $C^1$ periodic functions implies this converges to $g(x)$ for $x\notin\mathbb{Z}$ (and to the midpoint of the jump at integers). Grouping $\pm h$ gives the sine series form. $\square$

Lemma 3 (A bounded subsequence along good rational approximants). Let $\alpha\in(0,1)$ be irrational and let $p/q$ be a rational with \[ \left|\alpha-\frac{p}{q}\right| < \frac{1}{q^2}. \] Then \[ \left|\sum_{k=1}^q \left(\frac12-\{\alpha k\}\right) - \frac12\right| \le 3. \] In particular, \[ |F(\alpha,q)| \le \frac{7/2}{\log q}\xrightarrow[q\to\infty]{}0 \] along any sequence of such approximants with $q\to\infty$.

Proof. Write $\delta:=\alpha-p/q$, so $|\delta|<1/q^2$. For each $k\in\{1,\dots,q\}$, let \[ u_k:=\left\{\frac{kp}{q}\right\}\in\left\{0,\frac1q,\dots,\frac{q-1}{q}\right\}, \qquad v_k:=\{k\alpha\}=\{u_k + k\delta\}. \] Since $|k\delta|\le q|\delta|<1/q$, we have $u_k+k\delta\in(-1/q,1+1/q)$.

Case 1: $u_k\in\{1/q,2/q,\dots,(q-2)/q\}$. Then $u_k\in[1/q,1-1/q]$, so $u_k+k\delta\in(0,1)$ and hence $v_k=u_k+k\delta$. Therefore \[ \left|\left(\frac12-v_k\right)-\left(\frac12-u_k\right)\right| = |u_k-v_k| = |k\delta| < \frac{1}{q}. \] There are exactly $q-2$ such indices $k$.

Case 2: $u_k\in\{0,(q-1)/q\}$. Then $v_k$ may wrap around mod $1$, but in all cases $\frac12-v_k$ and $\frac12-u_k$ both lie in $[-1/2,1/2]$, hence \[ \left|\left(\frac12-v_k\right)-\left(\frac12-u_k\right)\right|\le 1. \] There are at most two such indices $k$ (since $u_k$ is a permutation of the $q$ grid points).

Summing the absolute differences and using the triangle inequality gives \[ \left|\sum_{k=1}^q \left(\frac12-v_k\right) - \sum_{k=1}^q \left(\frac12-u_k\right)\right| \le (q-2)\cdot \frac{1}{q} + 2\cdot 1 < 3. \] Finally, since $\{kp/q\}$ runs through all grid points $\{0,1/q,\dots,(q-1)/q\}$, \[ \sum_{k=1}^q \left(\frac12-u_k\right)=\sum_{j=0}^{q-1}\left(\frac12-\frac{j}{q}\right)=\frac12. \] Combining yields the claimed bound. Dividing by $\log q$ gives the final displayed conclusion. $\square$

5) VERIFICATION - Lemma 1 uses only measure-preservation of $\alpha\mapsto\{k\alpha\}$ and the fact that $\int_0^1(\frac12-x)dx=0$. - Lemma 2 computes Fourier coefficients explicitly; the only analytic input is standard pointwise convergence away from jump discontinuities. - Lemma 3 carefully handles wrap-around only for the two boundary grid points $0$ and $(q-1)/q$.

6) FINAL UNRESOLVED (i) Strongest proved partial result: (a) $\mathbb{E}_\alpha[F(\alpha,n)]=0$ for every $n$ (Lemma 1). (b) Explicit Fourier expansion of the summand and hence a Fourier representation of the entire sum (Lemma 2). (c) For any irrational $\alpha$, along any sequence of rationals $p/q$ with $|\alpha-p/q|<1/q^2$ the normalized value $F(\alpha,q)$ tends to $0$ (Lemma 3). (ii) First gap: Prove (or disprove) that the distribution of $F(\alpha,n)$ over $\alpha\in(0,1)$ converges as $n\to\infty$. Even with Lemma 3, it is unclear whether the distribution stabilizes or oscillates due to arithmetic structure in $n$. (iii) Top 3 next moves: 1. Use Lemma 2 to express $F(\alpha,n)$ as a sum over Fourier modes of geometric sums and attempt to control the joint contribution of modes with $|h|\lesssim n$ (where denominators $\sin(\pi h\alpha)$ are small). 2. Study the distribution of $\sum_{k\le n} \sin(2\pi h\alpha k)$ for random $\alpha$ and how it correlates across $h$, to see whether a stable limit law (e.g. Cauchy-type) is plausible without the extra random $\beta$. 3. Search numerically for two sequences $n_j$ and $n'_j$ along which the empirical CDF of $F(\alpha,n)$ differs significantly (suggesting non-convergence), and try to link that to number-theoretic structure of $n$ (e.g. whether $n$ is a convergent denominator for many $\alpha$). (iv) Minimal counterexample structure (if no limiting distribution exists): one would need two subsequences $n_j,n'_j\to\infty$ such that the distributions of $F(\alpha,n_j)$ and $F(\alpha,n'_j)$ stay separated (e.g. because one subsequence forces many $\alpha$ to have unusually small denominators $\sin(\pi h\alpha)$ for certain $h$). Lemma 3 suggests that convergent denominators $q$ produce many $\alpha$ with $F(\alpha,q)$ near $0$, which could drive oscillation in the distribution if such effects are not smoothed out.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.