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OP-8878OpenMathematics · general

Erdős #177 · discrepancy / arithmetic progressions

Find the smallest h(d)h(d) such that the following holds. There exists a function f:N→{−1,1}f:\mathbb{N}\to\{-1,1\} such that, for every d≥1d\geq 1,

max⁡Pd∣∑n∈Pdf(n)∣≤h(d),\max_{P_d}\left\lvert \sum_{n\in P_d}f(n)\right\rvert\leq h(d),

where $P_d…

Source reports OPENAs of 2026-09-08Source & evidence ↓
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Problem statement

Find the smallest h(d)h(d) such that the following holds. There exists a function f:N→{−1,1}f:\mathbb{N}\to\{-1,1\} such that, for every d≥1d\geq 1,

max⁡Pd∣∑n∈Pdf(n)∣≤h(d),\max_{P_d}\left\lvert \sum_{n\in P_d}f(n)\right\rvert\leq h(d),

where PdP_d ranges over all finite arithmetic progressions with common difference dd.

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Background

Cantor, Erdős, Schreiber, and Straus [Er66] proved that h(d)≪d!h(d)\ll d! is possible. Van der Waerden's theorem implies that h(d)→∞h(d)\to \infty. Beck [Be17] has shown that h(d)≤d8+ϵh(d) \leq d^{8+\epsilon} is possible for every ϵ>0\epsilon>0. Roth's famous discrepancy lower bound [Ro64] implies that h(d)≫d1/2h(d)\gg d^{1/2}.

Progress & source assessment

The source labels this entry OPEN in the 2026-09-08 snapshot. Editorial progress notes are reproduced as source commentary, not independently verified results. Prize amounts and Lean badges at the source are not local awards or verification.

References
  1. [1]Thomas Bloom (ed.), Erdős Problems #177 link

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