It is generally believed that all irrational algebraic numbers $\alpha$ are normal, in all bases. In base $2$ that implies that there are arbitrary large $n$ such that for the binary expansion
$$\alpha = \ldots .a_1a_2a_2 \ldots a_{2n}\ldots$$
there are equal numbers of $0$, and $1$ in the sequence $(a_1, \ldots, a_{2n})$.
Indeed, if instead of $0$, $1$, we have $-1$, $1$, then we get a random walk on the line and it will return to the origin infinitely often.
I considered $1$ million binary digits of $\sqrt{2}$ after the dot, converted $0$ to $-1$, and took the partial sums, then ploted them on the vertical
Same thing, but for $\sqrt{3}$.
Same thing, but for $\sqrt{6}$. There seems to be an excess of $1$'s over $0$' that will never be corrected. However:
Same thing for $\sqrt{6}$, $5$ million digits. There are corrections, and again we are hitting even
The above seems to be related to large deviations.
However, I would like to see examples in larger bases $b$ ($3$, $4$, $\ldots$) where there would be an equal amount of digits $k$, $0\le k\le b-1$, in the chunk $a_1\ldots a_{bn}$. That I could not find, just by searching ternary expansions of some radicals. It does not mean that they should not exist, but perhaps the theory of probability says that it is improbable to happen. Still, I wonder if anybody has some concrete examples of algebraic numbers with large "balanced chunks" ( in base $10$ that looks quite improbable...)
Thank you for your interest. Any feedback would be appreciated!
Notes:
@Greg Martin rightly pointed out that, since the algebraic numbers ( even quadratic numbers) are dense, there will be example of them with arbitrary initial chunks. Sill, if somebody says: take this fairly simple algebraic number ( small definition), and check that in the first 3 million ternary digits there are equal numbers of $0$, $1$, $2$, that would be a pleasant surprise. If another simple example of a number would have equal number of $0$, $1$, $\ldots$, $9$'s in the first $1$ billion decimal digits, surely that would be impressive. ( to quantify simple, the sum of absolute values of the coefficients of the defining equation is small).
@Dave L. Renfro correctly pointed out that normal numbers have a clear definition that does not imply "random" binary digits. However, lots of people believe that irrational algebraic do have random digits, whatever random may mean. So that would settle the binary expansion of algebraic irrational, modulo some (unattainable) proofs.
The equation
$$ 9091x+x^n=114 $$
has a solution close to
$$ \frac{114}{9091}=0.\overline{0125398746} $$
that gets closer as $n$ increases, so you can get algebraic numbers with arbitarily long balanced digit sequences, all with the same moderately sized coefficients.
Similar equations for bases $2$ to $9$:
\begin{eqnarray*} 11_2x + x^n&=&1_2\\ 222_3x + x^n&=&12_3\\ 101_4x + x^n&=&2_4\\ 1032_5x + x^n&=&14_5\\ 51_6x + x^n&=&2_6\\ 36412_7x + x^n&=&524_7\\ 335_8x + x^n&=&5_8\\ 608_9x + x^n&=&25_9\\ \end{eqnarray*}
Here’s the Java code I used to find these.
Perhaps $51_6x + x^n=2_6$ is the one that’s most in the spirit of your question; in base $10$ it’s
$$ 31x+x^n=2 $$
with coefficients that could reasonably be called “small”. The solutions are near $\frac2{31}$, which is $0.\overline{021534}_6$.