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The tetration is denoted $^n a$, where $a$ is called the base and $n$ is called the height, and is defined for $n\in\mathbb N\cup\{-1,\,0\}$ by the recurrence $$ {^{-1} a} = 0, \quad {^{n+1} a} = a^{\left({^n a}\right)},\tag1$$ so that $${^0 a}=1, \quad {^1 a} = a, \quad {^2 a} = a^a, \quad {^3 a} = a^{a^a}, \, \dots \quad {^n a} = \underbrace{a^{a^{{.^{.^{.^a}}}}}}_{n\,\text{levels}}.\tag2$$ Let $a$ be a real number in the interval $e^{-e} < a < e^{1/e}$. It is known that the following limit exists $$L(a) = \lim_{n\to\infty} {^n a},\tag3$$ where $L(a)$ satisfies $a^{L(a)}=L(a)$. For example, $L\left(\!\sqrt2\right)=2$.

It is also known that $$\lim_{n\to\infty} \, \frac{L(a) - {^{n+1} a}}{L(a) - {^n a}} = \ln L(a).\tag4$$ Finally, it is known that the following limit exists $$C(a) = \lim_{n\to\infty} \, \frac{L(a) - {^n a}}{\left(\ln L(a)\right)^n}.\tag5$$ Apparently, no closed form for the function $C(a)$ is known. But the numerical evidence suggests the following conjecture (basically, this is the coefficient of the linear term in the Taylor series expansion of $C(a)$ near $a=1$):

$$C'(1) = \lim_{a\to1} \, \frac{C(a)}{a-1} \stackrel?= 1.\tag{$\diamond$}$$

How can we prove it? Can we find values of some higher-order derivatives? Are they all integers? Is there a general formula, recurrence or an efficient algorithm to compute them?

Related questions: [1][2][3].

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