Comparison between two tetrations

Question

For a given natural number n, what ist the least number m, such that

$$e \uparrow \uparrow m > \pi \uparrow \uparrow n$$

It seems that m = n + 1 is the desired number. Is this true for all n ?

Answer 1

Use the equation $\ln(\ln(b^{b^z}))=z\ln(b)+\ln(\ln(b))$ to help calculate the limit of the iterated logarithms of the iterated exponentiation of $\pi$. For $k_6$, $z=\pi\uparrow\uparrow 4$, which has more than a billion digits and one can ignore the $\ln(\ln(b))$ term. Then the approximation for $k_6$ without this term, is exactly the same as the equation for $k_5$ with the $\ln(\ln(b))$ term.

$$k_n = \ln^{[on]}(\pi \uparrow\uparrow n) \\ \lim_{n=6 \to \infty} k_n = k_5 + O\frac{1}{\pi \uparrow \uparrow 4} \\ k_2 \approx 1.2798985874699298018433854796 \\ k_3 \approx 1.3167952100708763660472735425 \\ k_4 \approx 1.3176613010072958359625235748 \\ k_5 \approx 1.3176613010072958359630869692 $$

For each value of n, $k_n$ can be thought of the value for the "n+1" term concatenated on top of the $e\uparrow\uparrow n$ tetration stack to get equivalence . For n=3, $e\uparrow e \uparrow e \uparrow k_3 = \pi \uparrow \pi \uparrow \pi$. For $\pi$, since $k_n$ is < e, as n gets arbitrarily large, $$\pi \uparrow\uparrow n < e \uparrow \uparrow (n+1)$$

As a side note, there is an interesting base $b\approx 7.28550781987618684208203148323$. For this base b, the limit of the iterated logarithms is exactly e. $$\lim_{n \to \infty}\ln^{[on]}(b \uparrow\uparrow n)=e \\ \therefore \lim_{n \to \infty} b \uparrow\uparrow n = e \uparrow \uparrow (n+1)$$ $\pi$ falls into the same category as all bases$>$e, and $<$ b, where the limit of the iterated logarithms is greater than 1, and less than e.