For each k and every sufficiently large n with k dividing ((n-1) choose (k-1)), there is a universal cycle for the k-sub

Definitions and mathematical statement

Shared notation.

$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.

For integers $n\ge k\ge1$, write $[n]=\{1,\ldots,n\}$ and $\binom{[n]}k$ for its family of $k$-element subsets. A universal cycle for this family is a cyclic word $a_0a_1\cdots a_{L-1}$ over $[n]$, of length $L=\binom nk$, such that every cyclic window \[\{a_j,a_{j+1},\ldots,a_{j+k-1}\}\qquad(0\le j<L),\] with subscripts reduced modulo $L$, has $k$ distinct symbols and these $L$ sets are exactly the members of $\binom{[n]}k$, each once.

Statement recorded in the supplied source.

For every integer $k\ge1$, does there exist an integer $n_0(k)\ge k$ such that for every $n\ge n_0(k)$ satisfying $k\mid\binom{n-1}{k-1}$, a universal cycle for $\binom{[n]}k$ exists? [S2]

Short English statement

For each fixed subset size and all sufficiently large ground sets, is the necessary divisibility condition sufficient for a cyclic word whose windows represent every subset exactly once?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, UnsolvedMath: A Curated Collection of Open Mathematics Problems, record 3100078 (AMR-030-0078). CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Yevgeniy Rudoy, An inductive approach to constructing universal cycles on the $k$-subsets of $[n]$ (2013 revision), Section1, definition, Lemma1.1 and Chung--Diaconis--Graham conjecture, pages1--2; Section2, cycles and universal cycles, pages3--4. https://arxiv.org/abs/1209.4662.