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2

The number of times you have to take the logarithm to make the number handy is about $3^{27}$, smaller than the number Ross Millikan mentions, but still very large. To be exact, we have $$10\uparrow\uparrow (3^{27}-1)<3\uparrow\uparrow\uparrow 3=3\uparrow\uparrow (3^{27})<10\uparrow\uparrow (3^{27})$$ If we take $log_{10}$ $3^{27}-2$ times, we ...

0

I think , I have proven that for natural $a,b\ge 2$, we have $a^{a^a}>b^b\implies a\uparrow\uparrow(n+1)>b\uparrow\uparrow n$ for all $n\ge 2$ Could someone check the proof ? It works as follows. The basic idea is that $S=kT$ implies $a^S=b^T\cdot (\frac{a^k}{b})^T$ If $(\frac{a^k}{b})^T\ge k$ holds for $S=a^a$ and $T=b$, then we can apply ...

1

Since $\log(a↑↑b)=a↑↑(b-1)\cdot\log a$, we have \begin{align} m ↑↑ (n+1) &= n ↑↑ n \\ m ↑↑ n \cdot \log m &= n ↑↑ (n-1) \cdot \log n \\ m ↑↑ (n-1) \cdot \log m + \log \log m &= n ↑↑ (n-2) \cdot \log n + \log\log n \end{align} If we consider $m ↑↑ x \cdot \log m \gg \log\log m$, we could ignore double-log component and collapse the equation into ...

2

Not a complete answer, but too cumbersome for a comment. For nearly every value of $n$, the answer will be the largest value of $m$ such that $\frac{m^m}{m-1}$ is less than $n$. With a little work one can show that if $n \le m^{m-1}$ then $n\uparrow\uparrow n < m \uparrow\uparrow (n+1)$, whereas it is easy to see that if $n \ge \frac{m^m}{m-1}$ then ...

1

One can prove your initial claim by proving the following more general statement: Theorem. $a^{a^b} = c^{c^{c^d}}$ with $a > c \ge 2$ and $b,d \ge 1$ has no solutions. Proof. Write $a^z = c^y$ with $\text{gcd}(z,y) = 1$. Then $a^{\frac{1}{y}} = c^{\frac{1}{z}} = x$ is an integer, and $a = x^y, c = x^z$. So $$x^{y x^{yb}} = x^{z x^{z x^{zd}}}$$ $$y x ... 2 In short: I don't see any possibility to introduce associativity between iteration-height (here h=7/2) and base (here b=3) such that ^7(^{1/2}3) and  (^{7 \cdot {1/2}})3 (or even  ^{1/2}(^73)) relate easily. But there is an ansatz to assume the notation$$ z_h = \exp^{°h}_b(z_0)  as an iteration with fractional iteration-"height" $h$ of the ...

0

A Puiseux series is formal Laurent power series in $T^{\frac1n}$ for some $n$ (the $n$ may vary with the series). The set of Puiseux series is a field, denoted $k{\ll} T{\gg}$, and it is the algebraic closure of the field of formal power series $k[[T]]$.

4

This is an extended response to Yiannis Galidakis's question. I have been experimenting with the value $c$ they gave in the previous post. It is in case 3b since $|t| = |W(-\ln c)| = 1$ and there is no $n \in \mathbb{N}$ such that $t^n = 1$. I have evaluated the sequence $a_n$ for $0 \leq n \leq 10^8$. As in Gottfried Helm's analysis, I have found no ...

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This is not a new answer, but only intended to give some illustration for the cases, that $|b|=1$ and the two subcases, that 1) $b$ is a rational-order complex root of the unit $b = \exp( 2 \pi î /q)$ where $q \in \mathbb Q$ and 2) $b$ is an irrational-order complex root of the unit. Remark: the examples are computed using Pari/GP with ...

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