Subcubic graph number

The subcubic graph numbers are the outputs of an extremely fast-growing al function. They were devised by Harvey Friedman.

A subcubic graph is a graph in which each vertex has a valence of at most three. (For the sake of this article, subcubic graphs are allowed to be s or s, and are not required to be connected.) Suppose we have a sequence of subcubic graphs G1, G2, ... such that each graph Gi has at most i + k vertices (for some integer k) and for no i < j is Gi into Gj.

Friedman showed that such a sequence cannot be infinite, and for each value of k, there is a sequence with maximal length. We denote this maximal length using SCG(k).

Specific values
It is possible to show that SCG(0) = 6. The first graph is one vertex with a loop, the second is two vertices  connected by a single edge, and the next four graphs consist of 3, 2, 1, and 0 unconnected vertices. All maximal sequences will peak and decline this way. SCG(1) is very large, it exceeds \(f_{\varepsilon_0}(n)\) for any reasonable n; precise bounds for it are still in search.

Friedman proved that SCG(13) is greater than the halting time of any such that it can be proven to halt in at most 20002 (exponential tower 22 2 ... 2 2 2     with 2000 2's) symbols in \(\Pi^1_1\)-\(\text{CA}_0\). Not much is known about it, although it is far larger than TREE(3), which is already believed to surpass many of Jonathan Bowers' works. In particular, SCG(13) is almost certainly greater than meameamealokkapoowa oompa &mdash; although since SCG(n) is a computable function, SCG(13) is still no match for Rayo's number.

The growth rate of SCG(n) is about \(\psi_{\Omega_1}(\Omega_\omega)\) in the fast-growing hierarchy. It is upper bounded by the Takeuti-Feferman-Buchholz ordinal.

Simple subcubic graph numbers
If we require the subcubic graphs to be simple (i.e. not pseudographs or multigraphs), we get the simple subcubic graph numbers, denoted SSCG. Adam P. Goucher has shown that SSCG(2) << TREE(3) << SSCG(3). Moreover, he has shown that even TREEn(3) for even very large n (for example n=TREE(3)) does not compete at all with SSCG(3).

SCG(n) and SSCG(n) have comparable growth rates: Goucher proved that \(SSCG(4n+3) \geq SCG(n)\). Further, it can be proven that \(SSCG(3n+4) \geq 3 SCG(n)\).

Values and bounds

 * SSCG(0) = 2
 * SSCG(1) = 5
 * SSCG(2) \(\geq 3*2^{3*2^{95}}-8 \approx 10^{3.5775*10^{28}}\)