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Transcendence for Pisot morphic words over an algebraic base

Published online by Cambridge University Press:  10 July 2026

PAVOL KEBIS
Affiliation:
Institute of Science and Technology Austria, Austria (e-mail: pavol.kebis@ist.ac.at)
FLORIAN LUCA
Affiliation:
Department of Mathematical Sciences, Stellenbosch University, Faculty of Medicine and Health Sciences, South Africa (e-mail: fluca@sun.ac.za)
JOEL OUAKNINE
Affiliation:
Max Planck Institute for Software Systems, Germany (e-mail: joel@mpi-sws.org)
ANDREW SCOONES
Affiliation:
Department of Computer Science, University of Oxford, UK (e-mail: andrew.scoones@cs.ox.ac.uk)
JAMES WORRELL*
Affiliation:
Department of Computer Science, University of Oxford, UK (e-mail: andrew.scoones@cs.ox.ac.uk)
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Abstract

It is known that for a uniform morphic sequence $\boldsymbol u = \langle u_n\rangle _{n=0}^\infty $ and an algebraic number $\beta $ such that $|\beta |>1$, the number $[\![ \boldsymbol {u} ]\!] _\beta :=\sum _{n=0}^\infty ({u_n}/{\beta ^n})$ either lies in $\mathbb Q(\beta )$ or is transcendental. In this paper, we show a similar rational–transcendental dichotomy for sequences defined by irreducible Pisot morphisms on binary alphabets. Subject to the Pisot conjecture (an irreducible Pisot morphism has pure discrete spectrum), we generalise the latter result to arbitrary finite alphabets. In certain cases, we are able to show transcendence of $[\![ \boldsymbol {u}]\!] _{\beta }$ outright. In particular, for $k\geq 2$, if $\boldsymbol u$ is the k-Bonacci word, then $[\![ \boldsymbol {u}]\!] _{\beta }$ is transcendental.

Information

Type
Original Article
Creative Commons
Creative Common License - CCCreative Common License - BY
This is an Open Access article, distributed under the terms of the Creative Commons Attribution licence (https://creativecommons.org/licenses/by/4.0), which permits unrestricted re-use, distribution and reproduction, provided the original article is properly cited.
Copyright
© The Author(s), 2026. Published by Cambridge University Press
Figure 0

Figure 1 An alternating cycle for a cycle cover C$\mathcal C$. The horizontal edges lie in C$\mathcal C$ and the non-horizontal edges are not in C$\mathcal C$. Removing the horizontal edges from C$\mathcal C$ and replacing them by the non-horizontal edges yields a new cycle cover. However, if C′$\mathcal C'$ is a cycle cover other than C$\mathcal C$, then the symmetric difference of the respective sets of edges in C$\mathcal C$ and C′$\mathcal C'$ can be partitioned into alternating paths of the above form.