Let \(d\geq 2\). Then \[ T_d(x)=\left\{\begin{array}{ll} \frac{x}{d} &\mbox{if \(x\equiv 0 \pmod{d}\),}\\ \frac{(d+1)x-j}{d} &\mbox{if \(x\equiv j \pmod{d}, 1\leq j\leq d-2\),}\\ \frac{(d+1)x+1}{d} &\mbox{if \(x\equiv -1 \pmod{d}\).} \end{array} \right. \] For example, \(d=2\) gives the \(3x+1\) mapping: \[ T_2(x)=\left\{\begin{array}{ll} \frac{x}{2} &\mbox{if \(x\equiv 0 \pmod{2}\),}\\ \frac{3x+1}{2} &\mbox{if \(x\equiv 1 \pmod{2}\).} \end{array} \right. \] \(T_d\) is an example of a relatively prime mapping, in the language of Matthews and Watts, where \(m_0=1, m_j=d+1, 1\leq j\leq d-1\) and where we have the inequality \[ \prod_{i=0}^{d-1}\frac{m_i}{d}=\frac{1}{d}\left(1+\frac{1}{d}\right)^{d-1}<1. \] So it seems certain that the sequence of iterates \(n, T_d(n), T_d^2(n),\ldots\) always eventually enters a cycle, and that there are only finitely many such cycles.
We have \(T_d(j)=j\) for \(-1\leq j\leq d-2, d>2\) and Wiggin conjectured that every trajectory starting at a positive integer will eventually reach some \(j\) in the range \(1\leq j\leq d-2\).
The conjectural picture for negative input is less clear. For example, with \(d=3, 6, 8, 9, 10\) it seems that all negative trajectories will end up at \(-1\). However, for \(d=4\), negative trajectories may end up in the cycle \(-9, -11, -14, -18, -23, -29, -36\) as well as \(-1\).
Here is a table of d for which there are exceptional cycles.
Last modified 13th March 2023
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