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

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Claim. Theorem 3 of Kolpakov and Talambutsa: if fi(x)=aix+bif_i(x)=a_ix+b_i for i=1,…,ni=1,\ldots,n with positive integers aia_i and rational bib_i, and ∑i1/ai>1\sum_i 1/a_i>1, then the semigroup $\langle f_1,\ldots,f_n\rangle$ is not free with that basis. The problem's maps, with ai,bi∈Na_i,b_i\in\mathbb N, satisfy the hypothesis, and a relation between two distinct words of equal length evaluated at 11 is a repeated entry of some AkA_k, as the page for Klarner explains. The authors state that their proof follows the scheme of Klarner's Theorem 1.1, reducing to the case of equal multipliers a1=⋯=an=ma_1=\cdots=a_n=m with n>mn>m, and they supply the second part, which the 1982 paper omits. The paper's main subject is the free direction: ping-pong criteria for freeness without arithmetic hypotheses.

Postings. arXiv:2105.09387, first posted 19 May 2021 (revised 15 September 2021); Proc. Amer. Math. Soc. 150 (2022), 2301--2307, published online 16 March 2022. The paper is digested on its library card. The site's commentary of 3 December 2025 names it as a generalization of Klarner's result.

Acceptance. Refereed publication in the Proceedings of the American Mathematical Society, and the acceptance of the site's curator, Thomas Bloom, in the commentary, which names the paper as a generalization of Klarner's result. Nothing was reviewed here.

Depends on. No page of this wiki.