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$$n=a^{(a+1)^{(a+2)^{(a+3)\cdots}}}$$

How would one go about solving in this equation? I am more used to solving equations in this form:

$$n=a^{a^{a^{a\cdots}}}$$

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Which you solve in this form:

$$a^n=n$$

But how would you solve that equation at the top of the page though? If you were curious, and I know that SE likes what I have tried, so I will show steps that I have attempted.

$$f(a)=a^{(a+1)^{(a+2)^{(a+3)\cdots}}}$$

$$f(a)=a^{f(a+1)}$$

But from here I am not sure what to do from here. Can someone please help me evaluate this equation for $a$ in therms of $n$? Can the proof of this also be somewhat rigorous please?

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We need additional resources to handle “general” sequences of infinite exponentials. In particular:

$\mathbf{Definition:}$ Suppose $A_k=\{a_1,a_2,\ldots,a_k\}$, $k\in\mathbb{N}$, with $a_k>0$ and $n\le |A_k|=k$. A general infinite exponential is:

$$e_n(A_k) =

\begin{cases}

a_k, & \text{if $n=1$} \\

a_{k-n+1}^{e_{n-1}(A_k)}, & \text{if $n>1$}

\end{cases}$$

Now, if one sets $b_n=e_n(A_n)$, then the sequence $b_n$, $n\in\mathbb{N}$ expresses the sequence of ascending exponentials:

$$a_1,\,a_1^{a_2},\,a_1^{a_2^{a_3}},\,\ldots$$

Now you need the following theorem:

$\mathbf{Theorem\,\,(Barrow):}$ The sequence $b_n$, $n\in\mathbb{N}$ converges iff:

- $a_n$ converges
- $\exists n_0:\forall n>n_0:b_n\in [e^{-e},e^{1/e}]$

From the above theorem it is clear that your example is unsolvable, since your sequence is $a_n=a_{n-1}+1$, with $a_1=a$ and this sequence diverges, therefore it fails bullet one.

Suppose instead that you are called to solve, using an $a_n$ which satisfies the theorem’s assumptions:

$$y=a_1^{a_2^{a_3^{\cdots}}}\Rightarrow$$

$$\ln(y)=\ln(a_1)\cdot a_2^{a_3^{a_4^{\cdots}}}\Rightarrow$$

$$\frac{\ln(y)}{\ln(a_1)}=a_2^{a_3^{a_4^{\cdots}}}$$

Now, $a_n$ converges, so if we fix $\epsilon>0$ we are guaranteed a $k>0$ such that for all $n>k$, we have $a_n \sim a$, where $a=\lim\limits_{n\to\infty}a_n$.

We can therefore continue the iteration with logarithms as above, all the way to $k$:

$$\frac{(\cdots)}{\ln(a_k)}=a_{k+1}^{a_{k+2}^{a_{k+3}^\cdots}}$$

which is equivalent within $\epsilon$, to:

$$Y_{\epsilon}=\frac{(\cdots)}{\ln(a_k)}\sim a^{a^{a^{\cdots}}}$$

the latter being now solvable using the trick you mention in your question (or any other valid method) either for $a$ or for $Y_{\epsilon}$. In particular:

$$a\sim Y_\epsilon^{\frac{1}{Y_{\epsilon}}}\Leftrightarrow Y_{\epsilon}=\frac{W(-\ln(a))}{-\ln(a)}$$

where $W$ is the Lambert function.

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