Student or educator? Sign up with your school email and get 6 months of Pro free.
Cited by 3semigroups and automata theoryAdvanced Combinatorial MathematicsAlgorithms and Data Compression

The DOI matches its Crossref record; 13 of the 13 references it cites were found in a registry or an index; none of the 7 sources its claims cite is checked for support yet.

Avoiding 5/4-Powers on the Alphabet of Nonnegative Integers

Open access
Rowland, Eric, Stipulanti, Manon·2020·The Electronic Journal of Combinatorics

Abstract

We identify the structure of the lexicographically least word avoiding $5/4$-powers on the alphabet of nonnegative integers. Specifically, we show that this word has the form $\mathbf{p} \, \tau ( \varphi(\mathbf{z}) \varphi^2(\mathbf{z}) \cdots)$ where $\mathbf{p},\mathbf{z}$ are finite words, $\varphi$ is a $6$-uniform morphism, and $\tau$ is a coding. This description yields a recurrence for the $i$th letter, which we use to prove that the sequence of letters is $6$-regular with rank $188$. More generally, we prove $k$-regularity for a sequence satisfying a recurrence of the same type.

Overview

CoChat Summary

TL;DR

The paper identifies the lexicographically least infinite word \(w_{5/4}\) on nonnegative integers avoiding exact \(5/4\)-powers, showing it has the morphic form \(p\,\tau(\varphi(z)\varphi^2(z)\cdots)\) with \(p\) length 6764, \(z\) length 20226, \(\varphi\) a 6-uniform morphism, and \(\tau\) a coding. It derives a recurrence for its letters and proves the sequence of letters is 6-regular with rank 188, plus a general \(k\)-regularity theorem for similar recurrences.

Key findings

  • The lexicographically least \(5/4\)-power-free word is \(w_{5/4}=p\,\tau(\varphi(z)\varphi^2(z)\cdots)\), where \(|p|=6764\), \(|z|=20226\), \(\varphi\) is 6-uniform, and \(\tau\) is a coding.
  • The letters \(w(i)\) of \(w_{5/4}\) satisfy a recurrence (Corollary 5) involving the constants \(123061=6\cdot20510+1\) and \(5920\).
  • The sequence of letters in \(w_{5/4}\) is 6-regular with rank 188 (Theorem 33).
  • More generally, integer sequences satisfying a recurrence of the same type are \(k\)-regular (Theorem 6).

Methods

  • The proof shows \(p\,\tau(\varphi(z)\varphi^2(z)\cdots)\) is \(5/4\)-power-free and lexicographically least, the latter by proving that decreasing any nonzero letter introduces a \(5/4\)-power ending at that position.
  • Because \(\varphi\) is not \(5/4\)-power-free, the authors introduce pre-\(5/4\)-power-freeness, use locating lengths, and work with subscript-increasing words on an 8-letter-per-integer alphabet.
  • The analysis uses extensive computation, including Mathematica checks and an explicit length-331040 prefix of \(w_{5/4}\).

Limitations

  • The paper states open questions: the structures of \(w_{7/5}\), \(w_{7/6}\), and \(w_{11/6}\) are not known; it gives conjectures for five additional words and no conjecture for \(w_{11/6}\) and \(w_{11/8}\).
  • The paper notes there is no systematic way to guess the structure of \(w_{a/b}\) even with many terms, and whether each \(w_{a/b}\) is \(k\)-regular remains open.
  • Evident from the text: the proof relies on lengthy computer computations and a very long prefix, and the morphism \(\varphi\) does not preserve \(5/4\)-power-freeness, making the argument significantly more intricate.

Model-written summary. Check it against the paper before relying on it.

Integrity signals

Registry

Crossref · 10.37236/9581

Cites

13 of 13 references verified0 citation errors · 0 not checkable

Statistics

Nothing to recompute.

Figures

Figure 1 Portions of w 3 / 2 (left) and w 5 / 4 (middle and right), partitioned into rows of width 6. The word w 3 / 2 is shown from the beginning. The word w 5 / 4 = w (0) w (1) · · · is shown beginning from w ( i ) i ⩾ 6756 (middle) and w ( i ) i ⩾ 6758 (right). In the middle image, we have chopped off the first 6756 / 6 = 1126 rows to show where five columns become periodic. The term w (6759) (top row, second column on the right) is the last entry in w (6 i +3) i ⩾ 0 that is not 1.
Figure 1 Portions of w 3 / 2 (left) and w 5 / 4 (middle and right), partitioned into rows of width 6. The word w 3 / 2 is shown from the beginning. The word w 5 / 4 = w (0) w (1) · · · is shown beginning from w ( i ) i ⩾ 6756 (middle) and w ( i ) i ⩾ 6758 (right). In the middle image, we have chopped off the first 6756 / 6 = 1126 rows to show where five columns become periodic. The term w (6759) (top row, second column on the right) is the last entry in w (6 i +3) i ⩾ 0 that is not 1.From the PDF
Figure 2 A prefix of w 5 / 4 , partitioned into rows of width 72.
Figure 2 A prefix of w 5 / 4 , partitioned into rows of width 72.From the PDF
Figure 3 A plot of prefixes of the sequences w (72 i + 31) i ⩾ 0 (left) and w 5 / 4 = w ( i ) i ⩾ 0 (right).
Figure 3 A plot of prefixes of the sequences w (72 i + 31) i ⩾ 0 (left) and w 5 / 4 = w ( i ) i ⩾ 0 (right).From the PDF
Image 4
From the PDF
Case 2.1. Suppose that xyx starts after p in p τ ( u ′ ), that is, xyx is a suffix of τ ( u ′ ). Write x = τ ( x ′ ) = τ ( x ′′ ) and y = τ ( y ′ ) where x ′ y ′ x ′′ is the corresponding subscriptincreasing suffix of u ′ . Since | xy | is divisible by 4, the subscripts in x ′ and x ′′ are either equal or differ by 4. Since d = 2, the subscripts of the last letters of x ′ and x ′′ are odd. If | x | = 1, then x = c -2 and ϕ ( x ′ ) = ϕ ( x ′′ ) by definition of ϕ . Now if | x | ⩾ 2, the subscripts of the penultimate letters of x ′ and x ′′ are even and either equal each other or differ by 4. Since the subscripts cannot differ by 4 by Part 3 of Lemma 9, they must be equal. Since τ ( x ′ ) = τ ( x ′′ ), we have x ′ = x ′′ , so ϕ ( x ′ ) = ϕ ( x ′′ ). Then p τ ( ϕ ( z )) w ′ = p τ ( ϕ ( z u ′ ))
Case 2.1. Suppose that xyx starts after p in p τ ( u ′ ), that is, xyx is a suffix of τ ( u ′ ). Write x = τ ( x ′ ) = τ ( x ′′ ) and y = τ ( y ′ ) where x ′ y ′ x ′′ is the corresponding subscriptincreasing suffix of u ′ . Since | xy | is divisible by 4, the subscripts in x ′ and x ′′ are either equal or differ by 4. Since d = 2, the subscripts of the last letters of x ′ and x ′′ are odd. If | x | = 1, then x = c -2 and ϕ ( x ′ ) = ϕ ( x ′′ ) by definition of ϕ . Now if | x | ⩾ 2, the subscripts of the penultimate letters of x ′ and x ′′ are even and either equal each other or differ by 4. Since the subscripts cannot differ by 4 by Part 3 of Lemma 9, they must be equal. Since τ ( x ′ ) = τ ( x ′′ ), we have x ′ = x ′′ , so ϕ ( x ′ ) = ϕ ( x ′′ ). Then p τ ( ϕ ( z )) w ′ = p τ ( ϕ ( z u ′ ))From the PDF

Verification

Claims the paper makes

0 Valid · 2 Not valid · 7 Not checked · 66 Own results and interpretations

53 more claims

Create a free account to see all 75 claims and to verify the 3 references still open.

See all claims — free account
  • It may state two separate claims as one. 10
  • The quote only partly states this claim. 4
  • Probably not a substantial claim. 4

Create a free account to see these claims.

Citations

Cites

Show all 13 references

Cited by

See all 3

Metadata

Authors
Rowland, E., & Stipulanti, M.
Affiliations
Hofstra University (US)
Venue
The Electronic Journal of Combinatorics
Published
2020-08-21
Volume
27
Issue
3
ISSN
1077-8926 (ISSN-L)
OpenAlex
W3022614150
Cited by
3
References
13
Citations by year
2023: 1 · 2021: 1 · 2020: 1
FWCI
0.60
Type
journal article
Language
en
Open access
Licence
cc-by
Locations
  • The Electronic Journal of Combinatorics (publishedVersion, cc-by)
  • arXiv (Cornell University) (submittedVersion)
  • ORBi (University of Liège) (submittedVersion, other-oa)
  • arXiv (Cornell University)
Funders
Belgian American Educational Foundation
Topics
  • Physical Sciences › Computer Science › Computational Theory and Mathematics › semigroups and automata theory
  • Physical Sciences › Mathematics › Discrete Mathematics and Combinatorics › Advanced Combinatorial Mathematics
  • Physical Sciences › Mathematics › Theoretical Computer Science › Algorithms and Data Compression
Keywords
integer alphabet, coding, recurrence relations, combinatorics on words

About these checks

Registry, retraction, Cites and statistics results are computed, not judged, with one exception: a printed reference that no index lists is read once by a model before it counts as a citation error. Claims are extracted by a model and kept only when their quote appears verbatim in the text; each cited source is then checked, with a model’s help, for whether it says what the claim relies on. Those verdicts can be wrong — read the claim’s quote and the cited source before relying on one.

A language model drew the claims from the paper on 2026-11-14, under version 3 of the claim definition (model deepseek-v4.1-flash, prompt extract-v3).

A second model labelled each claim on 2026-11-14 (model jev-1.13-20260917, definition revision gold-guideline-v0.2).

Statistics and arithmetic are recomputed, not judged.

Confidence bands are shown only for fields calibrated on our annotated test set. None is calibrated yet, so no band is shown.

What counts as a claim ›How we measure confidence ›How we check this ›

Are you an author, or do you see a problem with this page? Request a review

How to cite

Cite CoChat for the checks on this page, and the paper for its own findings.

This page
CoChat. (2026). Citation check: Avoiding 5/4-Powers on the Alphabet of Nonnegative Integers. https://app.cochat.ai/papers/10.37236/9581
The paper
Rowland, E., & Stipulanti, M. (2020). Avoiding 5/4-Powers on the Alphabet of Nonnegative Integers. The Electronic Journal of Combinatorics. https://doi.org/10.37236/9581

Data sources

Bibliographic data: Semantic Scholar, OpenAlex, Crossref.

Contains information from the Semantic Scholar Academic Graph Datasets, made available under the ODC Attribution License (ODC-BY 1.0).

Semantic Scholar’s data is described in: Kinney, R., et al. (2023). The Semantic Scholar Open Data Platform. arXiv:2301.10140.