Yeah. This is a plot point used in a few stories, eg Carl Sagan's "Contact"
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Yes
Can you prove this? Or link a proof?
I don't know of one but the proof is simple. Let me try (badly) to make one up:
If it doesn't go into a loop of some kind, then it necessarily must include all finite strings (that's a theoretical compsci term).
Basically, take a string of any finite length, and then view pi in inrements of this length. Calculate it out to double the amount of substrings of length of your target string's interval you have [or intervals]. Check if your string one of those intervals. If not, do it again until it is, doubling how long you calculate each time.
Because pi is non-repeating, each doubling in intervals must necessarily include at least one new interval from all other previous ones. And because your target string length is finite, you have a finite upper limit to how many of these doublings you have to search. I think it's n in the length of your target string.
Someone please check my work I'm bad at these things, but that's the general idea. It's also wildly inefficient This doesn't work with Infinite strings because of diagnonalization.
Let me give another counterexample. Let x be the binary expansion of pi i.e. the infinite string representing pi in base 2.
Now you will not find 2 in this sequence by definition but it's still a non-repeating number.
Now one can validly say that we restricted our alphabet and we should look only for finite strings with digits that actually occure in the number. The answer is the string "23456789" concatenated with x.