9^9^9.

[Andy_L]

@DPRK

You probably mean (some variant of) Ackermann’s function, or some other fast-growing, yet computable, function. Graham’s number is just a number, remarkable as an explicit huge number that arises “in the wild.”

I think I was conflating Ackermann’s function and Knuth’s up-arrow notation Knuth's up-arrow notation - Wikipedia.

[DPRK]

@Chronos

I’m not sure what the standard notation would be for it, but the fastest-growing function I’ve ever encountered “in the wild”, so to speak, is defined by the recurrence relation D(n) = D(n-1)+2^(D(n-1)) .

Doesn’t that just grow like tetration?

[Senegoid]

I knew a guy at Berkeley, 40-some years ago, whose hobby was devising ways to write the largest possible numbers with the fewest possible symbols. (Note, fewest symbols, not merely fewest digits.)

Most of his schemes were various massively recursive functions based on the symbol 0 (zero) and the successor function S (written without parentheses, as those are superfluous). Thus:

0
S0 ( = 1 )
SS0 ( = 2 )
etc. and messy recursive functions on top of those.

So I suggested something naive like 9^9^9 and variations on that. His immediate response was: First you have to define 9 and you’ve already wasted too many symbols doing that ( e.g., SSSSSSSSS0 ).

[Monocracy]

@Dr.Strangelove

And a somewhat more accessible video explanation.

OK, either this guy isn’t explaining it correctly, or I’m calling bullshit on the whole TREE thing. This is how he gets to TREE(2) = 3. He draws a green node, two red nodes, then a red node for 3. Why can’t he draw two green nodes, a green node, two red nodes, then a red node for 4? Or for TREE(1), draw ten green nodes, nine green nodes, eight green nodes, seven green nodes, etc, for TREE(1) = 10?

What am I missing?

[Ludovic]

Isn’t the maximum number of nodes determined by the tree’s order in the sequence, i.e the first tree can only have 1 node (or up to 3, I’m not really clear on that.)

[minor7flat5]

@Pantastic

Or TREE(TREE(TREE(9))) if you want to stop messing around, since the lower case tree function isn’t quite as insanely fast growing. For people who aren’t familiar, the TREE function is a function that grows absurdly fast; TREE(1) is 1, TREE(2) is 3, and TREE(3) is so vastly incredibly huge that you need special notation to even vaguely talk about how big it is. TREE sequence | Googology Wiki | Fandom

I am stunned by the opacity of that page.

And I’m certain that to some subset of the population that is absolutely clear and logical. I am not a member of that subset.

[Left_Hand_of_Dorkness]

@Jim_B

Interesting thing that I read a child, in a children’s book, naturally.

The largest 3-digit number isn’t 999. It’s 9^9^9. Or, nine taken to the ninth power taken to the ninth power.

This number is apparently gargantuan. In ordinary type, it would reach from New York to Cleveland. Also, some mathematicians have speculated on what digits the behemoth number begins with, and ends with. I forget the exact details.

Believe it or not, I may even still have the book in my collection (it was a very interesting book, well worth holding on to). I will check.

In the meantime, does anyone else know what I am talking about? What else is known about this number, and what more has been learned in the 30+ years since I got the book?

Please divulge.

FWIW I read the same book when I was a kid–or at least the same factoid. I don’t remember which book it was; perhaps an Isaac Asimov Book of Awesome Crap for Kids? It looks like tetration was defined in 1947, so unless the book had some parameter I’m forgetting, it was probably wrong. It may have said not that this was the largest number, but that it was far larger than 999, or something like that.

[DPRK]

I suppose tetration is uncommon enough to not have a truly standard notation. Is [sup][sup]9[/sup]9[/sup]9 really more familiar to you than 9↑↑↑3 ? On the other hand, 9[sup]9[/sup] is quite standard notation.

[Johanna]

@Half_Man_Half_Wit

But 9^9^9 can be written using only the symbol ‘9’, in the standard exponentiation notation as 9[sup]9[sup]9[/sup][/sup]; so you don’t actually need any other symbols—in that sense, it’s truly a three-digit number, not a three-digit-plus-any-other-symbols-you-care-to-add number.

T.I.L.:
You can stack superscripts in this site’s code!

[DPRK]

@minor7flat5

I am stunned by the opacity of that page.

And I’m certain that to some subset of the population that is absolutely clear and logical. I am not a member of that subset.

It looks like there are no examples or proofs (or definitions of terminology) on that page, so no wonder it seems opaque. Nevertheless, near the top there is a definition of the function TREE(n) in question. No proof that TREE(2) = 3 (though that is an easy exercise) or that TREE(3) is huge; that is hardly obvious from the bare definition— it’s not just you.

[Velocity]

That seems like a sort of…cheating definition. Surely “digits” means just the placeholders for numbers from 0-9? 456, 724, etc.?

[begbert2]

Can I take an 8, and tip it on its side?

[Dr.Strangelove]

@Ludovic

Isn’t the maximum number of nodes determined by the tree’s order in the sequence, i.e the first tree can only have 1 node

That’s right. The first tree in the sequence can only have one node; the second two nodes, etc. Otherwise, the whole exercise wouldn’t make much sense–you could make any sequence as long as you liked just by making a humongous initial tree and shrinking it from there.

[k9bfriender]

@Monocracy

OK, either this guy isn’t explaining it correctly, or I’m calling bullshit on the whole TREE thing. This is how he gets to TREE(2) = 3. He draws a green node, two red nodes, then a red node for 3. Why can’t he draw two green nodes, a green node, two red nodes, then a red node for 4? Or for TREE(1), draw ten green nodes, nine green nodes, eight green nodes, seven green nodes, etc, for TREE(1) = 10?

What am I missing?

@Ludovic

Isn’t the maximum number of nodes determined by the tree’s order in the sequence, i.e the first tree can only have 1 node (or up to 3, I’m not really clear on that.)

It’s the number of colors allowed, the number of seeds. If you duplicate a previous tree, then you end. So if you draw one green node, the only think you can do is draw another green node, and that is a duplicate of the previous tree.

With 2, if he drew 2 green nodes, then the second one would be a duplicate of the first.

The point of it is how many unique tree structures can be made, copies and duplicates do not count.

In other words, the rules that define the function specifically say that he cannot do this.

If you poke around on numberphile, you will find that most of the people on there are maths professors or better, if it seems wrong, if it seems as though it is time to call bullshit, it is actually more likely that you misunderstood something.

[iamthewalrus_3]

Oh, you guys with your Trees and your Ackermans and your Arrow notations.

Behold Busy Beaver and weep. Busy Beaver grows so quickly its value is not computable.

A fun essay on big numbers.

[Pantastic]

@Monocracy

OK, either this guy isn’t explaining it correctly, or I’m calling bullshit on the whole TREE thing. This is how he gets to TREE(2) = 3. He draws a green node, two red nodes, then a red node for 3. Why can’t he draw two green nodes, a green node, two red nodes, then a red node for 4? Or for TREE(1), draw ten green nodes, nine green nodes, eight green nodes, seven green nodes, etc, for TREE(1) = 10?

Watch the video from the start instead of where you linked, in the first minute he explains it. The first tree can have at most one node, the second two, third three, and so on. So drawing two nodes in the first tree is invalid, much less ten.

[BigT]

@Riemann

Yes, expressed that way the constraint makes more sense. (Although the expression is ambiguous, as has been noted.)

It is my understanding that the order of operations deals with that ambiguity. Within the same level, you go from left to right. So it inherently means (9^9)^9.

That said, I only get a 78 digit number: 196 627 050 475 552 913 618 075 908 526 912 116 283 103 450 944 214 766 927 315 415 537 966 391 196 809

So clearly the book means 9^(9^9), which I can’t find anything that will even begin to calculate.

[BigT]

@iamthewalrus_3

Oh, you guys with your Trees and your Ackermans and your Arrow notations.

Behold Busy Beaver and weep. Busy Beaver grows so quickly its value is not computable.

A fun essay on big numbers.

Busy Beaver has been discussed on Computerphile, a sister channel to Numberphile.

I could have sworn Busy Beaver was mentioned in the Numberphile videos, but I don’t want to watch them both all the way again to see.

[msmith537]

@Jim_B

The largest 3-digit number isn’t 999. It’s 9^9^9. Or, nine taken to the ninth power taken to the ninth power.

This number is apparently gargantuan. In ordinary type, it would reach from New York to Cleveland. Also, some mathematicians have speculated on what digits the behemoth number begins with, and ends with. I forget the exact details.

Am I missing something? I plugged that into Excel and came up with (obvious rounding):
196,627,050,475,553,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000

I don’t think that would reach to Cleveland unless you wrote it in letters 6 miles tall.

@Jim_B

So clearly the book means 9^(9^9), which I can’t find anything that will even begin to calculate.

That’s the same thing. It’s just 9^81.

[Andy_L]

@msmith537

That’s the same thing. It’s just 9^81.

9^9^9 is either (9^9)^9 (or 387420489^9) or the vastly larger 9^(387420489) (which has 369,693,100 digits or so http://mathforum.org/library/drmath/view/59172.html ); 9^81 doesn’t enter into it (that would be 9^(9^2) )