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The first nine terms of the MacLaurin series of the following function is:

$$(1+x)^{(1+x)^{(1+x)^{...}}}= 1+x+x^2+\frac{3}{2}x^3+\frac{7}{3}x^4+4x^5+\frac{283}{40}x^6+\frac{4681}{360}x^7+\frac{123101}{5040}x^8+...$$

This can be verified by evaluating the series of large tetrations of $(1+x)$.

However, I seem to have some difficulty trying to come up with the general coefficient for powers of $x$.

Here are the values for $f^{n}(0)$:

$f(0)=1$

$f^{1}(0)=1$

$f^{2}(0)=2$

$f^{3}(0)=9$

$f^{4}(0)=56$

$f^{5}(0)=480$

$f^{6}(0)=5094$

$f^{7}(0)=65534$

$f^{8}(0)=984808$

One thing I noted is that they seem to be divisible by $n$; however I do not know where to go from there.

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    http://oeis.org/search?q=1%2C1%2C2%2C9%2C56%2C480&sort=&language=english&go=Search – 2017-02-02

2 Answers 2

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If we define the power tower $f(x):=(1+x)^{(1+x)^{(1+x)^{...}}}$ with $\sum\limits_{k=0}^\infty x^k a_k$ then we get by using it's derivative $\displaystyle\enspace f'(x)=\frac{f^2(x)}{(1+x)(1-f(x)\ln(1+x))}\enspace$ the recursion

$$\sum\limits_{v=0}^k ( (k-v)a_{k-v} +(k+1-v)a_{k+1-v})\sum\limits_{j=1}^v\frac{(-1)^j}{j}a_{v-j}=\sum\limits_{v=0}^k a_v a_{k-v}$$

where $\enspace a_0=1\enspace $ and $\displaystyle\enspace (\sum\limits_{j=1}^v\frac{(-1)^j}{j}a_{v-j})|_{v=0}:=1\enspace $ .

Example: $\enspace a_0=1$ , $\displaystyle a_2=1=\frac{1}{1!}$ , $\displaystyle a_2=1=\frac{2}{2!}$ , $\displaystyle a_3=\frac{3}{2}=\frac{9}{3!}$ , ...

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    I'm sorry but could you explain to me how you derived the formula using the derivative? Also it's not very clear what the general term is. – 2017-02-02
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    @LinusChoy : I have added the derivate above. :-) By comparing the coefficients of $x^k$ the recursion follows. You can explicitly resolve for $a_{k+1}=...(a_0,...,a_k)$ but it doesn't look nice. :-) *Note:* With that you don't get the explicit solution but the evaluation is easier. – 2017-02-02
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    How do you derive the higher general terms with your formula? Such as $a_{4}$, $a_{5}$, $a_{6}$. I'm sorry but I'm still quite confused as to how to use your formula. – 2017-02-02
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    $\displaystyle\sum\limits_{v=0}^k ( (k-v)a_{k-v} +(k+1-v)a_{k+1-v})\sum\limits_{j=1}^v\frac{(-1)^j}{j}a_{v-j}=\sum\limits_{v=0}^k a_v a_{k-v}$ can be changed to $\displaystyle a_{k+1}=\frac{1}{k+1}(-ka_k-\sum\limits_{v=1}^k ( (k-v)a_{k-v} +(k+1-v)a_{k+1-v})\sum\limits_{j=1}^v\frac{(-1)^j}{j}a_{v-j}+\sum\limits_{v=0}^k a_v a_{k-v})$ . Start with $k=0$ . – 2017-02-02
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    Ah I see now, thank you for your patience. – 2017-02-02
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    You are welcome! – 2017-02-02
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This is an infinite power tower and we have

$$(1+x)^{(1+x)^{\dots}}=\frac{W(-\ln(1+x))}{-\ln(1+x)}$$

Where $W(x)$ is the Lambert W function and has known series expansion. I'm not sure about any closed form for the $n$th term, but this gives a more direct approach to this problem.

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    Yes, I'm aware it is related to the Lambert W function, however it appears that its series consists of alternating positive and negative coefficients in contrast with this series which are all positive, thus I'm at a loss as to how to derive it from there. – 2017-02-02
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    Well, the logarithm is there too. – 2017-02-02