Perhaps such numbers might be referred to as “Liouville Sluggers”?
If pi extends into infinity (does it? I think so) then isn’t the answer that zeros will at some point appear an infinite number of times consecutively?
In base pi, pi is 1 followed by an infinite number of zeros after the “decimal” point 
Brian
No because then pi would become a rational number.
For context there are 10^82 atoms in the observable universe. There are perhaps something under 10^186 Planck lengths in the volume of the observable universe.
Oh pish posh. Let’s consider the dinner party problem. For N guests, how many ways are there to seat them? That would be N!.
2! = 2
10! = 3,628,800
61! = 5.076e83, more than the number of atoms in the visible universe
120! = 6.69e198 , far more than a billion times the number of Planck units in the volume of the observable universe.
500! = 1.22e1134 , TILT, though this falls way short of Graham’s number, never mind TREE(3).
Yeah, factorial growth is practically a flat line compared to some functions out there.
Does infinity actually exist? It’s a very useful intangible abstraction and we should not discard it but I’m not sure it actually exists. Any answer to this question is dependent on what one means by the words “actually” and “exist.”
It would be possible for a number to meet your criterion, while still not being simply normal. For instance, you could have a number with a hundred zeroes after the decimal point, and all other digits limited to a hundred in a row, but then no more zeroes ever after the 101st decimal place.
Most mathematics doesn’t actually concern itself directly with the concept of “infinity”, but simply dances around the edges of it. You can discuss all of the concepts in this thread without ever caring about the “actual existence” of infinity.
If you took a number N, and had a rule that says “add a number of factorial symbols equal to the number itself”, and then applied that rule N times, it would still be a flat line compared to TREE(N), etc. And any “normal” number, whether 500 or a googolplex with that rule applied would be infinitesimal compared to TREE(3).
But remember, TREE(3) is a very small number. Almost all positive integers are much, much larger than it.
Is the TREE(n) function monotonic in n? Has that been proven? This seems like a highly relevant question for applications where n>3.
No number actually exists–they’re all very useful intangible abstractions.
That’s reassuring, somehow.
Well of course in-finite literally means without end; properly one (usually) cannot talk about having “arrived at” infinity.
Though some infinities are bigger than others:
That’s a controversial statement and false in practice. Most mathematicians are perfectly happy working with completed infinities, particularly sets with infinitely many elements, like \mathbb Z, the set of all integers. Even collections that are too large to be called sets, like the class of all sets. (Russell’s Paradox tells us that we aren’t allowed to call that a set.) Cantor showed us the way.
What does “completed infinities” mean? Is that a term of art in mathematics?
The subheading of that article… “…has been a mystery”, “Now, it turns out,…”, makes it seem like it’s talking about some very recent discovery (say, since the previous issue of Scientific American), not something that’d been settled for nearly 60 years at the latest.
Quite true. Though almost all those numbers are inaccessible. There is a function, called the Busy Beaver function or BB(N), which tells you the most number of steps an N-state Turing machine can run for (among Turing machines that halt eventually). There is a 745-state Turing machine that halts if and only if ZFC is inconsistent. So if we knew BB(745), we could run the experiment of running that Turing machine BB(745) steps. If it halted in that time, then ZFC must be inconsistent and our entire basis for computing BB(745) was wrong. But if it didn’t halt, then we’ve proved the consistency of ZFC using only its own axioms, which we know from Godel is impossible.
Of course we knew already that BB was incomputable, since otherwise we could solve the halting problem. But we might think that in any particular case, you could maybe just enumerate the possibilities and figure it out. And we can do so up to BB(5). Just not BB(745)! And probably not for much lower numbers as well. That entire space of integers is just inaccessible to mathematics, even in principle.
As opposed to the potential infinities @Lumpy was referencing. Consider the integers: Someone who believes only in potential infinities would say that you can’t take the whole set of integers; to say that there are infinitely many integers is only to say that, if you have finitely many in hand, there is always another you don’t have. Someone who allows completed infinities (like most working mathematicians) is perfectly happy talking about the set of all integers as a complete thing in itself.
[I may not have done this justice; I’m a working mathematician, not a philosopher.]