TheoremBase

Counting Cyclically Reduced Words: Finiteness and the Bound (2d)^k

Words of a given length form a nonempty finite set containing cyclically reduced words, and a sum over words of length k of terms bounded by M is at most M(2d)kM(2d)^k.

Statement

In the setting of The Real Numbers: Standing Notation and Background, let N\mathbb{N} be the natural numbers, carried into R\mathbb{R} by the natural-number image, and for k∈Nk\in\mathbb{N} let [k][k] be the initial segment determined by kk. Let d∈Nd\in\mathbb{N}. For k∈Nk\in\mathbb{N} let [2d]k[2d]^{k} be the set of kk-tuples in [2d][2d], which are the words of length kk in the letters 1,…,2d1,\dots,2d, let Wd,k∘⊆[2d]kW^{\circ}_{d,k}\subseteq[2d]^{k} be the set of cyclically reduced words of length kk, and let (2d)k(2d)^{k} denote the natural power in R\mathbb{R}. Finite sets and sums over them are those of Finite Set and Sum over a Finite Index Set.

Then the following hold for every k∈Nk\in\mathbb{N}.

1. (Finiteness) The sets [2d]k[2d]^{k} and Wd,k∘W^{\circ}_{d,k} are nonempty and finite, and so is every nonempty subset of [2d]k[2d]^{k}.

2. (Counting bound) Let B⊆[2d]kB\subseteq[2d]^{k} be nonempty, let M∈RM\in\mathbb{R}, and let a:B→Ra:B\to\mathbb{R} satisfy 0≤a(w)≤M0\le a(w)\le M for every w∈Bw\in B. Then

∑w∈Ba(w)≤M (2d)k.\sum_{w\in B}a(w)\le M\,(2d)^{k}.

Proofs

Log in to submit a proof.

Loading...

Citations

Loading…

Dependencies

Loading…

Related

0 relations

Curated associations between results. These are editable and subjective — they do not replace the dependency graph, which is derived from the references in the text.

No relations recorded yet.

Comments

Log in to comment.

Loading…