*An equivalent problem was originally asked on MSE as [Does every number base have at least one “Baseless number”?](https://math.stackexchange.com/q/3658214/318073), 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 $0\le c_i\lt b,\forall i$ where $b\ge4$ is an integer.

Define the following iterative sequences $x_0=y_0=0$ and for $n\ge 1$ as follows:

$$\begin{align}
x_n&=\begin{cases}
(x_{n-1}+c_{n-1})\cdot b, & n\lt d \\
(x_{n-1}+c_{n-1})\cdot 1, & n=d
\end{cases}
\\
y_n&=\begin{cases}
(y_{n-1}+c_{n-1})\cdot c_{n-1}, & n\le d
\end{cases}
\end{align}$$

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 the equality holds $x_d = y_d$.
> 
>  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 $c_d$.

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**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 $d\gt 6$.