Indexed metadata

Powers of the Thue--Morse Series: 2-Adic Valuations and Automatic Odd Parts

Zhao Shen

Source record

Source: arXiv

Published: Sep 15, 2026

arXiv: 2609.16966

Open original source ↗

Source abstract

Let T(x)=j0(1x2j)=n0t(n)xnT(x)=\prod_{j\ge 0}(1-x^{2^j})=\sum_{n\ge 0}t(n)x^n be the Thue--Morse generating function, and write T(x)m=n0tm(n)xnT(x)^m=\sum_{n\ge 0}t_m(n)x^n for a positive integer mm. For a nonzero integer aa, let ν2(a)ν_2(a) be the exponent of 22 in aa and put odd(a)=a/2ν2(a)\operatorname{odd}(a)=a/2^{ν_2(a)}. For every s1s\ge 1, we prove that the sequence (odd(tm(n))mod2s)n0(\operatorname{odd}(t_m(n))\bmod 2^s)_{n\ge 0} is 22-automatic when m=2rm=2^r, r1r\ge 1, or m=32rm=3\cdot 2^r, r2r\ge 2. In both families, ν2(tm(n))=ν2(n+m1m1)ν_2(t_m(n))=ν_2\binom{n+m-1}{m-1}. The valuation identity for m=2rm=2^r is known; our proof recovers it and also establishes the automaticity assertion. For m=6m=6 we prove ν2(t6(n))=ν2(n+55)+1n3(mod4)ν_2(t_6(n))=ν_2\binom{n+5}{5}+\mathbf{1}_{n\equiv 3\pmod 4} and show that its odd parts modulo every 2s2^s are 22-automatic.

Evidence graph

No public relationships recorded yet.

Integrity note: This page is a factual metadata record created by deterministic ingestion. It is not a claim that the work moves a mathematical frontier or has been independently verified.