OliForum Matematico

Sia $ \displaystyle{n=\sum_{i=0}^r 2^i \alpha_i=\sum_{j=0}^s 3^j \beta_j} $; poniamo $ \displaystyle{f(n)=\sum_{i=0}^r 3^{i+1} \alpha_i} $ e $ \displaystyle{g(n)=\sum_{j=0}^s 2^j \beta_j} $

$n=2^r a = 3^s b $ e $f(n) = 3^{r+1} a $ e $g(n) = 2^s b $.
Infatti
\begin{equation}
f(g(n)) = f(g(3^s b)) = f(2^s b) = 3^{s+1} b = 3n
\end{equation}
e
\begin{equation}
g(f(n)) = g(f(2^r a)) = g(3^{r+1} a) = 2^{r+1} a = 2n
\end{equation}