An equivalent problem was originally asked on MSE as Does every number base have at least one “Baseless number”?, but did not receive any answers that would help answer the main question about "existence for every $b$."
Recursive form of the equality $(x_d=y_d)$
Let $(c_n)=(c_1,c_2,\dots,c_d)\subseteq\mathbb Z,d\ge 2$ be a finite sequence such that $0\le c_i\lt b,\forall i$ and $c_1\ne 0$, where $b\ge4$ is an integer.
Define the following iterative sequences $x_0=y_0=0$ and for $n\ge 1$ as follows:
$$\begin{align} x_n&=\begin{cases} (x_{n-1}+c_{n})\cdot b, & n\lt d \\ (x_{n-1}+c_{n})\cdot 1, & n=d \end{cases} \\ y_n&=\begin{cases} (y_{n-1}+c_{n})\cdot c_{n}, & n\le d \end{cases} \end{align}$$
That is, the last terms will be:
$$\begin{align} x_d&=(\dots(((c_1)\cdot b +c_2)\cdot b+c_3)\cdot b\dots)\cdot b + c_d)\cdot 1 \\ y_d&=(\dots(((c_1)\cdot c_1 +c_2)\cdot c_2+c_3)\cdot c_3\dots)\cdot c_{d-1} + c_d)\cdot c_d \\ \end{align}$$
We call $(c_n)$ a solution for $b$ if $x_d = y_d$ and $x_d,y_d\gt 1$.
I call the number $x_d=y_d$ a "Baseless number".
Notice that $x_d=O(b^d)$ but $y_d=O((b-1)^d)$.
We can show $\exists\space d_0$ such that $x_d\gt y_d$ for all $d\ge d_0$.
That is, there are at most finitely many solutions $(c_n)$ for any given $b$.
It is also clear that $y_d$ is divisible by $c_d$, so the $x_d$ and $x_{d-1}$ must also be divisible by $c_d$.
Context, examples and questions
Notice that $(c_n)$ is equivalent to digits of $x_d$ in number base $b$, of some $d$ digit number.
For example, if $b=10$, it is known we have a unique solution $(c_n)=(8,3,8,5)$, which is a solution because:
$$ x_d=8385=((((8)\color{red}{10}+3)\color{red}{10}+8)\color{red}{10}+5)\color{red}{1}=((((8)\color{blue}{8}+3)\color{blue}{3}+8)\color{blue}{8}+5)\color{blue}{5}=8385=y_d $$
For another example, if $b=9$ then we have exactly $6$ solutions: $$ (b=9) \implies (c_n)\in\{(1,3),(2,3),(2,3,7),(2,7,5,5),(2,8,7,3),(4,4,8,6,7)\}$$
By using an exhaustive search I gave on MSE, I found all solutions for small bases $b\le 13$.
Question $a)$ Is it possible to find all solutions $(c_n)$ for larger bases $b$, efficiently?
Looking at $x_d=y_d$ for some $d$, we can rewrite the problem as the following equality in $(c_n)$:
$$ x_d=\sum_{i=1}^{d} c_{i}b^{d-i} = \sum_{i=1}^{d}c_i\prod_{j=i}^d c_j = y_d $$
For example, if we observe $d=2$ and $(c_n)=(1,c_2)$ then we get $b=c_2^2\implies c_2=\sqrt{b}$.
That is, we obtained that if $b$ is a perfect square, then $(c_n)=(1,\sqrt{b})$ is one solution.
This is the smallest solution, but not necessarily the only solution. Revisit the $b=9$ example.
I can find families of solutions, but I do not know how to determine if for some $b$, some $d$ case does not having any solutions, other than by an exhaustive search. Is there anything useful that can can be said about this problem in general?
Question $b)$ Is it true that for every $b\ge 4$, there is at least one solution?
Or can we find a counter-example?
The smallest $b$ for which I do not know any solutions, is $b=107$.
If a solution for $b=107$ exists, it has $x_d \gt 107^{6}\gt 1.5\cdot 10^{12}$. That is, has $d\gt 6$.
