I would like to represent $\pi$ in an arbitary mixed base. Is this possible to do formulaically and uniquely, before knowing the base of the $n^{th}$ digit?
For example, say the number is to be represented by$$A_5B_3C_5D_7E_3\dots$$ where the letters are digits and the numbers represent the base of the associated digit. My goal is to formulate a set of rules that puts $\mathbb{R}$ into a bijective correspondence with all the possible number representations in the given mixed base. Is this possible?
My first thought is that the $n^{th}$ digit need not represent some $n^{th}$ power of $\frac 1{b}$ else there will be unrepresentable numbers.
Edit: the purpose of expressing a number this way is to enable a digit walk on a non-uniform tessellation graph. For example, expressing $\pi$ in base four allows for a digit walk on a square lattice. Well, what if you want to do a digit walk on the edge graph of an arbitrary tiling of the plane (without uniform degree)?
Yes you can. Let us consider representing $\pi$ as $3.A_5B_3C_5\ldots$ We can use the leading $3$ as long as the base in the ones place is at least $4$. Any digit in the place after the radix point is worth $\frac 15$ and $\pi \lt 3\frac 15$, so $A=0$. Another way to see that is we multiply the fractional part by $5$ and take the integer part of the result. As $5 \cdot 0.14159265 =0.7796325$, which is less than $1$ we again see $A=0$. Now since the next base is $3$ we multiply by $3$ and take the integer part. $3 \cdot 0.7796325=2.3388675$ so $B=2$ and we strip that off to continue. Now $5 \cdot 0.3388675=1.6943375$, so $C=1$ and we have $\pi \approx 3.0_52_31_5\ldots$ You have an unambiguous representation as long as you use digits less than the base in each place unless the expression would terminate. The $1$ in the third place represents $\frac 1{5\cdot 3 \cdot 5}=\frac 1{75}$, so we have $\pi \approx 3+\frac 2{15}+\frac 1{75}+\ldots$. You can do this, but I don't see why you would want to. The point of numeric representations is to understand numbers or to compute with them. This makes both hard.