Probably a proof (if any exist) that calls upon Knuth's up-arrow notation or Busy Beaver.
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7$\begingroup$ Once upon a time, this was Graham's number: en.wikipedia.org/wiki/Graham's_number. I have no idea what the answer is now. $\endgroup$– Qiaochu YuanCommented Aug 15, 2010 at 4:44
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2$\begingroup$ This should be community wiki. Perhaps closed $\endgroup$– CasebashCommented Aug 15, 2010 at 6:34
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2$\begingroup$ Infinity $\infty$ is used in lots of proofs :) Anything bigger? $\endgroup$– Pratik DeoghareCommented Aug 16, 2010 at 13:05
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1$\begingroup$ 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. $\endgroup$– academicCommented Aug 10, 2012 at 18:51
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1$\begingroup$ @Casebash: There is a unique answer to the question, so I do not see why it should be a community wiki. $\endgroup$– user1729Commented Sep 19, 2012 at 11:06
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3 Answers
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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: http://en.wikipedia.org/wiki/Fast-growing_hierarchy
There are other examples of numbers greater than Graham's Number, as can be seen here: http://en.wikipedia.org/wiki/Goodstein_function#Sequence_length_as_a_function_of_the_starting_value, although I'm not sure if this number is larger than Friedman's $TREE(3)$
answered Aug 15, 2010 at 6:04
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1$\begingroup$ In terms of fast growing hierarchy:$$\operatorname{TREE}(n)\gg f_{\theta(\Omega^\omega \omega,0)}(n)$$ math.stackexchange.com/a/1979838/272831 $\endgroup$ Commented Mar 7, 2017 at 14:41
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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): http://www.cs.nyu.edu/pipermail/fom/2006-April/010362.html
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: http://www.cs.nyu.edu/pipermail/fom/2006-March/010279.html
The Goodstein function is upper bounded by ε₀, whereas the TREE function is lower bounded by the small Veblen ordinal.