Cyclic progression-free density conjecture

Let ZN\mathbb{Z}_N be the cyclic group of order NN, and let AZNA\subset\mathbb{Z}_N be {x,2x,,(d1)x}\{x,2x,\ldots,(d-1)x\}-free, meaning that no xx has all of x,2x,,(d1)xx,2x,\ldots,(d-1)x in AA. Cyclic progression-free density conjecture. If dNd\mid N, then

Ad1dN.|A|\leqslant \frac{d-1}{d}N.

The conjecture is motivated by generalizing the proof of the d=3d=3 cube case; the supplied text gives no resolution status.

Sources & referencesView supporting material

Primary source

Yuchen Meng, “A note on cube-free problems”, arXiv:2311.12318 (2025).

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