User blog comment:Luckyluxiuz/Question/@comment-37212455-20191219030832

It's omega plus one \(\omega + 1\).

\(n\uparrow n\) has a similar growth rate to \(f_2(n)\), while \(f_m(n)\) very generally has a growth rate comparable to \(n\uparrow^{m-1}n\). Now since \(f_{\omega}(n) = f_n(n)\) (under normal fundamental sequences), it means that \(f_{\omega}(n)\) has a similar growthrate to \(n\uparrow^{n}n\). The cool thing is now that we know this, we know that \(f_{\omega}(f_{\omega}(n))\) would grow at a comparable rate to \(n\uparrow^{n}n\uparrow^{n\uparrow^{n}n}n\uparrow^{n}n\). However, since it's really the \(n\uparrow^{n\uparrow^{n}n}n\) that makes a difference, we can safely simplify this down to \(n\uparrow^{n\uparrow^{n}n}n\). We can keep iterating things so that \(f_{\omega}(f_{\omega}(f_{\omega}(n)))\) would be about the same growth rate as \(n\uparrow^{n\uparrow^{n\uparrow^{n}n}n}n\), and so on. This is similar to successive terms of your \(S(n)\), which has growth rate similar to \(f_{\omega + 1}(n)\) on account of the fact that if you do the math each successive value of \(S(n)\) is like iterating \(f_{\omega}(n)\) another time.

Suggested reading: Graham's number and Arrow notation.