User blog comment:SuperSpruce/The T array function: Part 2: Dimensional Arrays/@comment-38080588-20190222205602/@comment-38080588-20190223213125

But \(f_{2n}(n)\) is not equal to \(f_{\omega2}(n)\) since, while \(f_n(n)=f_\omega(n)\) because \(n\) replaces \(\omega\), \(f_{\omega2}(n)=f_{\omega+\omega}(n)=f_{\omega+n}(n)\), and when that is finally reduced down to many stacked \(f_\omega(n)\), it can be seen that \(f_\omega(f_\omega(n))\) is actually the same as \(f_{f_\omega(n)}(f_\omega(n))\) because \(f_\omega(n)\) becomes the new \(n)\ for the outside function to replace its \(\omega\), which makes it already much larger than \(f_{2n}(n)\), but still much smaller than \(f_{\omega2}(n)\).