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Kaneko, Suzuki and Tachiya, arXiv:2601.20743v1, Theorem 3, printed/PDF p. 5. The inherited gap condition is Theorem 1(v), p. 3.

Statement

Fix an integer t≥2t\ge2. Let a,ba,b be integer sequences indexed by positive integers, with a(n)≥0a(n)\ge0 for all nn and infinitely many nonzero a(n)a(n). Write

Nc={n≥1:c(n)≠0},Nc(x)=Nc∩[1,x),Sc(x)=∑1≤n<x∣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}|c(n)|.

Assume

lim sup⁡n→∞max⁡{a(n),∣b(n)∣}1/n<t.\limsup_{n\to\infty}\max\{a(n),|b(n)|\}^{1/n}<t.

Suppose there are real sequences xj,zj≥1x_j,z_j\ge1 such that

xj→∞,Sa(xj),Sb(xj)=o(tzjxj),#Na(xj),#Nb(xj)=o(xj/zj).x_j\to\infty,\qquad S_a(x_j),S_b(x_j)=o(t^{z_j}x_j),\qquad \#\mathcal N_a(x_j),\#\mathcal N_b(x_j)=o(x_j/z_j).

If Nb\mathcal N_b is infinite, assume fixed constants Δ,L>1\Delta,L>1 such that for every consecutive pair m<m+m<m_+ in Nb\mathcal N_b and every real μ≥L\mu\ge L,

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

Then ∑n≥1(a(n)+b(n))t−n\sum_{n\ge1}(a(n)+b(n))t^{-n} is irrational. For finite Nb\mathcal N_b the gap hypothesis is absent. Both support bounds are required separately, even when bb has signed coefficients.

Proof and application limits

The paragraph preceding the theorem specializes Theorem 2 to q=tq=t of degree one. One can choose yjy_j to dominate both coefficient masses; the powers involving d−1d-1 disappear. This gives exactly the displayed integer hypotheses.

For a(n)=φ(n)a(n)=\varphi(n) and b=0b=0, every index is in Na\mathcal N_a. Since zj≥1z_j\ge1, its support count is not o(xj/zj)o(x_j/z_j). Any coefficientwise splitting a(n)+b(n)=φ(n)a(n)+b(n)=\varphi(n) also fails: the union of the two supports must contain every positive integer. This is a failure of direct application, not a ban on other series with the same value.

The theorem and inherited condition were compared with the page images. The proof route and integer specialization were read. No independent review, native tier or complete source-proof reconstruction is claimed.

Bears on. Problem 249 as a conditional method; the target's irrationality remains unresolved.