# Arrange the output by 3-adic valuation

I have a number that has the form $$x=\sum_{k=1}^n c_k3^{q_k}$$, where $$q_k\in\mathbb{Q}$$ and $$0\neq c_k$$ is in $$\mathbb{Z}$$ or $$i\mathbb{Z}$$. We may define the $$3$$-adic valuation of $$c_k3^{q_k}$$ as $$v_3(c_k3^{q_k})=v_3(c_k)+q_k$$, where $$v_3(c_k)=\begin{cases}v_3(c_k),& if c_k\in\mathbb{Z}\\v_3(c_k/i),& if c_k\in i\mathbb{Z}\end{cases}$$. Now, how can I output $$x$$ in the form $$d_13^{r_1}+d_23^{r_2}+\cdots+d_n3^{r_n},$$ where $$d_k\in\mathbb{Z}$$ or $$i\mathbb{Z}$$ with $$v_3(d_k)=0$$, $$r_k\in\mathbb{Q}$$ with $$r_1\leq r_2\leq\cdots\leq r_n$$?

For example, if $$x=5i\cdot3^{1/2}-18\cdot3^{4/7}+2\cdot3^{1/2}+3\cdot3^{1/2}$$, the expected output should be something like $$2\ i\ 3^{1/2}+5\ 3^{1/2}+3^{3/2}-2\ 3^{18/7}.$$

• How would the number $27$ be decomposed: $1\times 3^3$, or something else entirely? May 17, 2020 at 9:27
• @J.M. You are right. It should be $1\cdot 3^3$, or simply $3^3$. May 17, 2020 at 9:31
• This will be easier to canonicalize if you work with a list of pairs of the form {d,e} where the oair denoted d*3^e. If the output formatting is critical maybe make the outer wrapper a symbol and use Format to display it. May 17, 2020 at 22:26