I observed a strange anomaly in the sequence of bases of a number system.
For every $n$-base, $n \geq 2$, it holds that its number symbols are written with numerals which are all less than $n$. For example, a binary ($2$-ary) base has number symbols which consist only of $1$s and $0$s.
The anomaly manifests when we investigate $n = 1$. By the above pattern, its number symbols should consist only of $0$s. But if this were so, the only number we could express would be $0$. Hence, we have to break the pattern and allow $1$ to appear in our number symbols. We have to break another pattern, namely that the number symbols of a $n$-base consist precisely of $n$ different types of numerals, if we want $0$ to be representable in our system. That is, if we allow only $1$s in our number symbols, we cannot represent $0$.
For clarity, here is a sample definition of the value of number symbols, and an example from my imagined unary system. $$ (k_m k_{m-1} \dots k_1 k_0)_n = \sum_{i=0}^m n^i k_i $$ Then, the number symbols of the unary system are the same as that of the binary, with the difference that in the unary system, they denote a natural number equal to the number of $1$s which occur in them. For example, $(1101)_1$ is $3$.
I find it surprising that the elegant pattern described above completely breaks down at $n = 1$. Do we have analogous examples, or a deeper reason that this happens? Thank you for your input! :)
(If we are happy only being able to represent natural numbers) the empty string can represent 0 and the only allowed digit can be 1.
In a way it will be more elegant as the number 1 will correspond to the "successor" operator in the Peano arithmetical sense.
But as Jean Marie states, this will not be a positional number system. Every "1" symbol will have the same value which it adds.