Wiki
Wiki

Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

Updated


Kaneko, Suzuki and Tachiya, arXiv:2601.20743v1, Theorem 1, printed/PDF p. 3. Definitions are on pp. 2–3; the source record identifies the selected artifact and reading scope.

Definitions

Let q>1q>1 be a Pisot or Salem number, of degree dd over Q\mathbb Q. The source includes rational integers q≥2q\ge2 among Pisot numbers. For an algebraic number α\alpha, write h(α)\mathrm h(\alpha) for the maximum absolute value of its conjugates over Q\mathbb Q. This is the source's boxed coefficient size, not logarithmic height.

For a sequence c=(c(n))n≥1c=(c(n))_{n\ge1}, put

Nc={n≥1:c(n)≠0},Nc(x)=Nc∩[1,x),Sc(x)=∑1≤n<xh(c(n)).\mathcal N_c=\{n\ge1:c(n)\ne0\},\qquad \mathcal N_c(x)=\mathcal N_c\cap[1,x),\qquad S_c(x)=\sum_{1\le n<x}\mathrm h(c(n)).

For the distinguished real embedding and real x>1x>1, z≥0z\ge0, set

Rc(q,x,z)=∑n∈Z1≤n<x ∑j∈Zj≥z∣c(n+j)∣q−j.R_c(q,x,z)= \sum_{\substack{n\in\mathbb Z\\1\le n<x}} \ \sum_{\substack{j\in\mathbb Z\\j\ge z}}|c(n+j)|q^{-j}.

The inner sum is infinite. For rational integers, h(c(n))\mathrm h(c(n)) equals ∣c(n)∣|c(n)|.

Statement

Let a,ba,b be sequences of algebraic integers of Q(q)\mathbb Q(q) with a(n)≥0a(n)\ge0 for all n≥1n\ge1 and Na\mathcal N_a infinite. Suppose real sequences xj,yj,zj≥1x_j,y_j,z_j\ge1 and a fixed η∈(0,1]\eta\in(0,1] satisfy, as j→∞j\to\infty,

xj→∞,Sa(xj),Sb(xj)=O(yj),x_j\to\infty,\qquad S_a(x_j),S_b(x_j)=O(y_j), #Na(xj),#Nb(xj)=o(xj/zj),\#\mathcal N_a(x_j),\#\mathcal N_b(x_j)=o(x_j/z_j), Ra(q,ηxj,zj),Rb(q,ηxj,zj)=o(xj/yjd−1).R_a(q,\eta x_j,z_j),R_b(q,\eta x_j,z_j) =o(x_j/y_j^{d-1}).

If Nb\mathcal N_b is infinite, require constants Δ,L>1\Delta,L>1 such that for every two consecutive elements m<m+m<m_+ of Nb\mathcal N_b and every real μ≥L\mu\ge L,

m+Δμ<m+⟹Na∩[m+μ,m+Δμ)≠∅.(G)m+\Delta\mu<m_+ \quad\Longrightarrow\quad \mathcal N_a\cap[m+\mu,m+\Delta\mu)\ne\varnothing. \tag{G}

Then the convergent series ∑n≥1(a(n)+b(n))q−n\sum_{n\ge1}(a(n)+b(n))q^{-n} does not belong to Q(q)\mathbb Q(q). When Nb\mathcal N_b is finite, condition (G) is absent. The label (G) is this page's; the paper numbers the hypotheses (i)–(v), and (G) is its condition (v).

Proof pointer, standing and use

The proof on p. 11 combines Lemmas 1–3, pp. 6–11: rationality over the base field supplies a nonzero algebraic-integer tail with a lower norm bound; sparsity and the averaged estimate make many tails small; (G) supplies nonvanishing on enough of those indices.

The statement, its definitions and the full-tail convention were checked against the source. The proof route has been read, but no complete source-proof reconstruction or independent acceptance is recorded. The quantitative replacement of the tail hypothesis is Theorem 2.

Bears on. Problem 249 as a possible transformation criterion. Its dense numerator sequence does not meet the sparsity hypothesis.