有理数全体の集合

\begin{align*} f\left(x\right) & =\frac{1}{\left\lfloor x\right\rfloor +1-\left\{ x\right\} }\\ & =\frac{1}{\left\lfloor x\right\rfloor +1-\left(x-\left\lfloor x\right\rfloor \right)}\\ & =\frac{1}{2\left\lfloor x\right\rfloor +1-x} \end{align*} となるので、
\[ f\left(0\right)=1 \] \begin{align*} f^{2\circ}\left(0\right) & =f\left(f\left(0\right)\right)\\ & =f\left(1\right)\\ & =\frac{1}{2\left\lfloor 1\right\rfloor +1-1}\\ & =\frac{1}{2} \end{align*} \begin{align*} f^{3\circ}\left(0\right) & =f\left(f^{2\circ}\left(0\right)\right)\\ & =f\left(\frac{1}{2}\right)\\ & =\frac{1}{2\left\lfloor \frac{1}{2}\right\rfloor +1-\frac{1}{2}}\\ & =2 \end{align*} \begin{align*} f^{4\circ}\left(0\right) & =f\left(f^{3\circ}\left(0\right)\right)\\ & =f\left(2\right)\\ & =\frac{1}{2\left\lfloor 2\right\rfloor +1-2}\\ & =\frac{1}{3} \end{align*} \begin{align*} f^{5\circ}\left(0\right) & =f\left(f^{4\circ}\left(0\right)\right)\\ & =f\left(\frac{1}{3}\right)\\ & =\frac{1}{2\left\lfloor \frac{1}{3}\right\rfloor +1-\frac{1}{3}}\\ & =\frac{3}{2} \end{align*} となっていき、更に計算をすると、
\[ \left(f^{i\circ}\left(0\right)\right)_{i\in\mathbb{N}_{0}}=\left(0,1,\frac{1}{2},2,\frac{1}{3},\frac{3}{2},\frac{2}{3},3,\frac{1}{4},\frac{4}{3},\frac{3}{5},\frac{5}{2},\frac{2}{5},\frac{5}{3},\frac{3}{4},4,\frac{1}{5},\frac{5}{4},\frac{4}{7},\frac{7}{3},\frac{3}{8},\frac{8}{5},\frac{5}{7},\frac{7}{2},\frac{2}{7},\frac{7}{5},\frac{5}{8},\frac{8}{3},\frac{3}{7},\frac{7}{4},\frac{4}{5},5\right) \] となる。