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The Initial Parity Pattern

Abstract

Finiteness of the alphabet forces an apwenian substitution fixed point to begin with one, an even letter and one.

Theorem 1.1 (The forced initial pattern).

Lean statement: D5/S3/Combinatorics/Apwenian/GuoHanPrefix.initial_prefix

Proof. Machine-checked in Lean as D5/S3/Combinatorics/Apwenian/GuoHanPrefix.initial_prefix (✓ std3). ∎

Source. Repository-derived.

Acknowledgement. Ying-Jun Guo, Guo-Niu Han (2025). On a family of automatic apwenian sequences. DOI: 10.1016/j.disc.2025.114399. URL: https://irma.math.unistra.fr/~guoniu/papers/p120autoapw.pdf.

Commentary.

Let Sigma be a finite alphabet of nonnegative integers in which every odd letter equals one. Let p be an integer at least two, let sigma assign a word of length p to each nonnegative integer, and let a take values in Sigma and satisfy a(np + r) = sigma(a(n), r) for all nonnegative n and all r from zero through p minus one. If a is apwenian, then a(1) has image zero modulo two and a(2) = 1. Since a(0) = 1, the initial parity pattern is 101. The conclusion does not require every word assigned by sigma to every letter of Sigma to remain in Sigma.

References