The Collatz map analogue in polynomial rings and in completions (Q6635116)

Let \(R\) be a commutative ring. The authors study Collatz maps on \(R[X]\), on \(R[[x]]\) (the \(x\)-completion of the polynomial ring \(R[x]\)) and on \(\mathbb Z_{2}\) (the \(2\)-completion of \(\mathbb Z\)). The Collatz map on \(\mathbb Z_{2}\) (\(T:Z_{2}\to \mathbb Z_{2}\)) is defined similarly to the classical Collatz map:\N\[\NT(s)=\begin{cases} \frac {s} {2} &\text{if \(2\) divides \(s\) in \(\mathbb Z_{2}\)},\\\N3s+1 &\text{otherwise} \end{cases}\N\]\Nfor each \(s\in \mathbb Z_{2}\).\N\NFor \(R[x]\), the map \(T:R[x]\to R[x]\) is defined as follows.\N\[\NT(f)=\begin{cases} \frac {f} {x} &\text{ if \(f(0)=0\)},\\\N(x+1)f -f(0) &\text{ if \(f(0)\ne0\)}. \end{cases}\N\]\NUse a similar definition for \(R[[x]]\).\N\NIf the characteristic of \(R\) is zero, then the only (minimal) \(T\)-cycles in \(R[x]\) are \((0)\) and the cycles \( (a, ax)\), for all \(a\ne0 \in R\). If char\(R\) is finite, then each polynomial \(f\in R[x]\) is eventually periodic. The authors determine the lengts of the corresponding cycles under suitable assumptions on \(f\).\N\NAssume that \(R\) is of of finite cardinality \(q\). The eventually \(T\)-periodic power series in \(R[[x]]\) are precisely the series of the form \(u(1 + xv)^{-1}\), where \(u\) and \(v\) are polynomials in\( R[x]\). Moreover, for \(n\ge1\), the number \(Z_{n}\) of the cycles of length \(n\) in \(R[[x]]\) is finite, and it is explicitly computed. Asymptotically, \(Z_{n}\sim \frac {\left(\frac {\sqrt {4q-3}} {2}\right)^{n}} {n}\) as \(n\to\infty\).\N\NAnalogously, the number \(Z_{n}\) of the \(T\)-cycles of length \(n\) in \(Z_{2}\) is finite. Asymptotically, \(Z_{n}\sim \frac {\phi^{n}} {n}\) as \(n\to\infty\), where \(\phi=\frac {1+\sqrt 5} {2}\) is the golden number.