Real irrational algebraic numbers "never repeat" An oft-used phrase describing irrational numbers is that their (decimal) expansions "never repeat".
The sense of "never repeating" intended is, of course, that their expansions don't repeat forever.  And it's straight-forward to show that rational numbers do repeat forever.
It is also easy to show that irrational numbers never repeat forever because that would make them rational.
Now, the question:   Suppose I define "never repeating" as simply meaning that the first N digits of the expansion (in whatever base you like) are not repeated.   I.e. positions 1..N are not the same as positions (N+1)..2N.    Clearly I can construct transcendental numbers with that property.   But if we restrict ourselves to irrational algebraic numbers, can it be shown that there are any numbers in that set that "never repeat" in this sense?
It seems to me that even for a randomly chosen irrational algebraic number, the probably that it fulfils this property very very quickly becomes infinitessimally small.   I.e. the first billions digits will not match the next billion digits.    However, having looked at the first N digits, the next N digits could still be anything, and even though the probability shrinks as p^-N, there are sill infinitely many opportunities as N grows.
So, my question is: can it be shown that there exists a (real) irrational algebraic number for which it is never true that digits 1..N are the same as digits (N+1)..2N for any N?

Related:How to know that irrational numbers never repeat?
 A: Still not sure if we're talking about the same thing, but this is what I found.

Let $r$ be a non-negative real. If $r \ne 1, 2$ then there is a number base in which the first two digits of $r$  are the same. The proof below is partly by computer program.

If $r \ge 3$ the result is clear. Suppose $2 < r < 3$ and let $f$ be the fractional part of $r$. Then for a number base $b \ge 3$, the representation of $r$ begins $2.2$ iff $f \in [2/b, 3/b)$. Since $b > 2$ we have $3/(b + 1) > 2/b$, so the intervals for $b = 3, 4, \ldots$ cover the whole of $(0, 1).$

If $1 < r < 2$ the proof is similar, using bases $b \ge 2$.

The case $0 \le r < 1$ is the most interesting. If $b \ge 2$ is a number base and $d$ is a digit such that $0 \le d < b$, then the representation of $r$ in base $b$ begins $.dd$ iff
$$r \in [(bd + d)/b^2, (bd + d + 1)/b^2).$$
So we have intervals $[0, 1/4)$, $[3/4, 1)$, then $[0, 1/9)$, $[4/9, 5/9)$, $[8/9, 1)$, etc.
The question is, do these cover the interval $[0, 1)$? The answer isn't obvious, so I wrote a computer program to find out (using integer arithmetic throughout, to avoid rounding errors). The answer is that the intervals for $2 \le b \le 50$ cover $(0, 1)$, and $50$ is the least upper limit that will work.

In case this unexpected result was due a bug, I wrote programs to generate random reals and random rationals in $(0, 1)$ and look for a base in which the first two digits were the same. Such a base could always be found, and the maximum base required over $10^8$ trials was $50$.
