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Probably a proof (if any exist) that calls upon Knuth's up-arrow notation or Busy Beaver.

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Once upon a time, this was Graham's number:'s_number. I have no idea what the answer is now. – Qiaochu Yuan Aug 15 '10 at 4:44
This should be community wiki. Perhaps closed – Casebash Aug 15 '10 at 6:34
Infinity $\infty$ is used in lots of proofs :) Anything bigger? – Pratik Deoghare Aug 16 '10 at 13:05
I once saw a programming contest along the following lines: Write a C program, under 5K bytes, that outputs the biggest number possible. Assume (contrary to fact) that C can handle arbitrarily large integers and that your program has unlimited computational resources (i.e. memory). The winning entries were amazing. – Frank Thorne Aug 10 '12 at 18:51
@Casebash: There is a unique answer to the question, so I do not see why it should be a community wiki. – user1729 Sep 19 '12 at 11:06
up vote 10 down vote accepted

The mathematician Harvey Friedman observed a special finite form of Kurskal's Tree Theorem. Regarding this form, Friedman discusses the existence of a rapidly growing function he calls $TREE(n)$.

The $TREE$ sequence begins $TREE(1)=1$ and $TREE(2)=3$, but $TREE(3)$ is a number so extremely large that its weak lower bound is $(A(...A(1)...))$, where the number of A's is $A(187196)$, and $A()$ is a version of Ackermann's function: $A(x) = 2↑↑...↑x$ with $x-1 ↑s$ (Knuth up-arrows).

Whereas Graham's Number is $A^{64}(4)$, the above mentioned lower bound is $A^{A(187196)}$. As you can imagine, the $TREE$ function keeps on growing rather quickly. For a discussion on the hierarchy of fast growing functions see here:

There are other examples of numbers greater than Graham's Number, as can be seen here:, although I'm not sure if this number is larger than Friedman's $TREE(3)$

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In one of Friedman's posts on the FOM mailing list, he mentions a number called SCG(13) that is far larger than TREE(3):

I couldn't find a lot of other information about it, though.

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TREE[3] is much larger than the numbers derived from Goodstein sequences for any reasonable input. See:

The Goodstein function is upper bounded by ε₀, whereas the TREE function is lower bounded by the small Veblen ordinal.

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BB(n) can surpass any recursive number, it is not computable. Perhaps, BB(1000) already inexpressible in any existing notation. You can also: BB(BB(n)), BB(BB(BB(...(BB(n))...)) (with "n" nested functions).

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Correct, but the question asked for a specific number, used in a proof. Do you know of a proof where BB(1000) appears? – Rick Decker Sep 21 '12 at 18:30

protected by Community May 6 '14 at 9:18

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