Is an anomaly in base-n arithmetic discoverable in base-m arithmetic?

Question

I have always been fascinated by the book "Contact" by Carl Sagan. The final chapter of the book (not the film!) reports about an anomaly in the n-millionth decimal of pi, optimally visible when pi is calculated in base-11 arithmetic: A series of only ones and zeroes, composing a digital image of a circle.

My question: If there really exists such an anomaly in some base-n arithmetic for the decimals of an irrational constant, would it be detectable at all in base-m arithmetic? Would it stand out to a mathematics professional? Would it be visible to the untrained eye like mine?

In short: Assuming such an anomaly exists in base-17, and I start calculating pi in base-13 arithmetic, would I be able to detect the anomaly? How?

EDIT: The anomaly presented by Carl Sagan (I didn't reread the details) was something like this: Hidden in the representation of pi in base-11 is a section of the digits only zeroes and ones. This section has the size of a prime squared (to make it likely to try a two-dimensional approach), and if laid-out as a true square, the zeroes and ones look like a circle. I imagine something like this:

00000100000
00011011000
00110001100
00100000100
01000000010
01000000010
01000000010
00100000100
00110001100
00011011000
00000100000

My question is a little more generic, if patterns like this can be easily found, even if not the correct base is used for the analysis.

Answer 1

There is an objective (base-independent) sense in which long 0-1 strings are rare in base $b$; the probability of a length $n$ string selected from the set of all base $b$ strings being entirely composed of $0$ and $1$ is $(\frac2b)^n$, which is a very small number for any $n$ large enough to give a pattern of the sort you are referring to.

Nevertheless, it is very difficult to discern this pattern in bases other than powers or roots of $b$. There are some results suggesting that the digits of $2^n$ in base $10$ are normal (although I don't think this is proven, and I'm sure this can be stated more precisely), even though $2^n$ is composed entirely of $1$'s and $0$'s (mostly $0$'s) in base $4$, $8$ and so on. This basically means that the digits appear essentially random in base $10$, even though they happen to fall in a very narrow range of possible numbers. Proving normality is really hard, from what I understand, so there aren't too many sure results about this sort of thing, but most results seem to suggest that there is some "strong mixing" when you transition from one base to another that is not a power of some common base.

But there is something more to be said. If instead of considering random strings of $0$'s and $1$'s, we consider long repeating patterns (like the big open spaces of $0$'s in the middle of your circle drawing), it's another matter. For concreteness, I'll suppose it is a long string of zeros. Any such number will be exceptionally close to the rational number formed if you make the pattern of zeros go on forever, meaning that the continued fraction expansion will have an enormous term there, on the order $b^n$ if there was a length $n$ run of zeros in the base $b$ expansion. Moreover, this rationality will be visible in other bases, because rational numbers are periodic in every base, so there will be a repeating pattern of the same length as the original pattern. The period of repetition may be very long, though; I'm not sure under what conditions it will be short enough to be visible before the end of the pattern.

Answer 2

If a strange pattern appears in base $b$, you should be able to see something in bases of the form $b^k$, because the digits are just grouped together by blocks of $k$. For instance if it is a circle in base $2$ as in the book, you will likely get a circle again in base $4$ or $8$, if you write the digits in a certain way. It probably won't work for base that are not powers of $b$, because you will have a shifting that disrupt most patterns.