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Variants of the \(3N+1\) conjecture and multiplicative semigroups. (English) Zbl 1120.11014

Entov, Michael (ed.) et al., Geometry, spectral theory, groups, and dynamics. Proceedings in memory of Robert Brooks, Haifa, Israel, December 29, 2003–January 2, 2004, January 5–9, 2004. Providence, RI: American Mathematical Society (AMS); Ramat Gan: Bar-Ilan University (ISBN 0-8218-3710-9/pbk). Contemporary Mathematics 387. Israel Mathematical Conference Proceedings, 121-127 (2005).
The still open “Collatz”-problem (or “\(3N+1\)”- or “Hasse”- or “Syracuse”- or “Kakutani”-problem) is to prove that for every \(n \in \mathbb{N}\) there exists a \(k\) with \(t^{(k)}(n)=1\) where the function \(t(n)\) takes odd numbers \(n\) to \((3n+1)/2\) and even numbers \(n\) to \(n/2\). In the paper under review the author considers some variants of this famous problem. Let \(n\) be an odd integer and define \(\delta(n)=1\) if \(n\equiv1\pmod 4\), resp. \(\delta(n)=3\) otherwise. Then he conjectures that the sequence \((F^{(k)}(n))_{k\in\mathbb{N}}\) will finally end in a periodic sequence just like the original sequence \((t^{(k)}(n))_{k\in\mathbb N}\) where \(F(n):=\frac{\delta(n)n+1}{2^{e(n)}}\) such that \(F(n)\) is odd again. While the author is not able to prove his conjecture he succeeds in showing that if one additionally divides out the powers of \(3\) in \(F(n)\) the sequence will terminate with \(1\) for any starting odd number \(n\). Furthermore, he studies some combinatorial problems and proves that every odd integer can be written as a product of some rational numbers of the form \(\frac{4n+1}{2n+1}\). All proofs are elementary and use induction.
For the entire collection see [Zbl 1075.58001].

MSC:

11B83 Special sequences and polynomials
20M25 Semigroup rings, multiplicative semigroups of rings