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I hope someone can answer this question. If you can answer it, then you already know what SCG(13) is.

SCG(13) is a very, very large number which is part of a theorem about graphs.

It's at least one of the largest (if not the largest) numbers whose lower-bound is given in a proved mathematical theorem. If you haven't heard about it before, you wouldn't believe how big it is. It isn't something that you could write in ordinary scientific notation. A special, much more compact, large-number notation is used for such numbers.

I'll refrain from going into detail, because it's something that people can look up if they want to.

Anyway, can anyone tell me what is the largest proven lower-bound for SCG(13)?

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Michael Ossipoff
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    Did you already check the references at this question? Certainly TREE(3) is a lower-bound. – Dietrich Burde Dec 30 '17 at 20:48
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    Yes, I read that question and its answers, and followed the link to Prof. Friedman's post. He mentions a theorem relating SCG(13) to the halting-time of a Turning machine. But I'm asking for an answer expressed (in some manner) in terms of the g(n) notation for large numbers (the notation in which Graham's number is g(64) ), (Paragraph) .Friedman says that, compared to SCG(13), TREE(3) is unnoticeable. So, sure, TREE(3) is a lower bound. But is there a larger lower-bound? – Michael Ossipoff Dec 30 '17 at 21:03
  • That’s actually not what you asked for. – Randall Dec 30 '17 at 21:08
  • Yes,, and so I wanted to clarify that I'd like an answer expressed in terms of the g(n) large-number notation, which is defined in terms of familiar operations like exponentiation. For example, some at the googology websites say that SCG(13) is greater than TREE(TREE(TREE(...(TREE(3)...))), with the nesting iterated TREE(3) times. ...and that TREE(3) is greater than g(g(g(...(g(64)...))) ...with the nesting iterated g(64) times. I've read those things at the googology websites, but I'm asking for verification. Of course g(n) is defined in terms of the familiar exponentiation. – Michael Ossipoff Dec 30 '17 at 21:22
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    $TREE(3)$ is already so big that it is not even known where iit is located n the fast-growing-hierarchy. If $SCG(13)$ is so large that $TREE(3)$ is "unnoticable" , it will be very difficult to find a "good" lower bound. What we can say that the nested function $g(g(g(\cdots (g(64)\cdots ))))$ will be nowhere near even to $TREE(3)$, unless the number of $g's$ is not much smaller than $TREE(3)$. – Peter Dec 30 '17 at 22:03
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    $TREE(3)$ is known to exceed the $\Gamma_0$-level, which is much higher than the $\epsilon_0$-level, which itself is not reachable with the nested $g$-function with a "reasonable" number of $g's$. – Peter Dec 30 '17 at 22:07
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    By the way, what does "SCG" stand for ? – Peter Dec 30 '17 at 22:09
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    SCG stands for sub-cubic graph. I don't know graphs. Maybe it's something to do with the fact that a cube's vertices have 3 connections. Thanks for the answer. Evidently I was asking for information that isn't known, and you gave the best answer based on what can be said. The number is from a theorem about drawing a succession of graphs according to certain rules. SCG(13) is that sequence's stopping-point when 13 extra vertices are allowed in each of the graphs. – Michael Ossipoff Dec 30 '17 at 22:15
  • @MichaelOssipoff Is this function related to the $TREE$-function , or something completely different ? – Peter Dec 30 '17 at 22:23
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    @MichaelOssipoff I wonder whether it is certain that $SCG(n)\ge TREE(n)$ for all positive integers $n$. In this case, we would have the enormous lower bound $TREE(13)$ – Peter Dec 30 '17 at 22:40
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    [Fixing a typo] SCG(n) is related to TREE(n) -- they're both from theorems about graphs. SCN(n) is about planar graphs, whose points can be multiply-connected, but needn't all be connected to others. Limited to 3 line-end connections to any point. Single looped line doubly connected to a point is allowed. First graph can have 1 point. 2nd graph can have up to 2 points...etc. Graphs that can be reduced to a previous graph (by deletion of points or lines, or merging of points) aren't permitted. SCG(13) is the maximum length the sequence can reach if 13 extra points are allowed for each graph. – Michael Ossipoff Dec 30 '17 at 23:23
  • Wikipedia says that SSCG(3) >>TREE(3). Wikipedia says: "SSCG(3) is not only larger than TREE(3), it is much, much larger than TREE(TREE(…TREE(3)…))[3][citation needed] where the total nesting depth of the formula is TREE(3) levels of the TREE function." Reliable sources say SCG(N) is equal to or greater than SSCG(n). – Michael Ossipoff Dec 30 '17 at 23:44
  • @MichaelOssipoff Here : http://googology.wikia.com/wiki/Subcubic_graph_number you can also see how large those numbers are. – Peter Dec 31 '17 at 09:55

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According to this link and this link, we most likely have

$$f_{\psi(\Omega_\omega)}(n)<^\star\operatorname{scg}(n)<^\star f_{\psi(\varepsilon_{\Omega_\omega+1})}(n)$$

though I'm not entirely sure what fundamental sequences are being used here nor of a particular proof.

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