# Tag Info

## New answers tagged tetration

0

The number of digits of $n$ is $\lfloor \log_{10}n\rfloor+1$ If we ignore the floor and the $+1$, we get $\log_{10}7^{7^7}=7^7 \log 7\approx 695974$ as shown by Alpha. This is fine for three layer tetrations, but is inadequate for taller ones. For example, if you ask Alpha for $7^{7^7}\log_{10}7$ you get back $10^{10^{5.842593328962333}}$ which is still ...

0

This answer is just to reproduce my derivation for my modified notation in my other answer. The original equation: $$1 + \sum_{n=0}^{\infty}\sum_{k=0}^{n}\bigg(\frac{x^{n+1} \log^{n+1+k}(x)C_{kn}}{\Gamma(n+2)}\bigg)$$ My notational modification: (indeed I had an error with the n at the inner loop): $$\displaystyle{ \begin{array} {lll} ^4 x ... 2 The way to do this, is to calculate the coefficients of the iterates {^2}(e^x), {^3}(e^x), {^4}(e^x), etc. first. The coefficients of the corresponding Piseaux for {^n}z are going to be the same. The problem then becomes: Let a series s with coefficients a_n be given, as s=\sum\limits_{n=0}^\infty a_n. Determine the coefficients b_n of ... 1 Note: this is not yet an answer, but only an ansatz to find a promising key for the pattern-detection To find patterns I would seriously try to simplify formulae, such that constants move out of loops/sums/products and repeatedly occuring functions with same parameter get a shorter symbol. So for instance$$ \; ^4 x= \sum_{n=0}^{\infty} \frac{(x ...

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You have it wrong. $$2\uparrow\uparrow\uparrow\uparrow2=2\uparrow\uparrow\uparrow2=2\uparrow\uparrow2=4$$ In fact, $2 \uparrow^n 2= 4$ for any $n$. However $2\uparrow\uparrow3=16$ and $2\uparrow\uparrow\uparrow3=65536$. $2\uparrow\uparrow\uparrow\uparrow3=2\uparrow\uparrow\uparrow4=2\uparrow\uparrow65536$. This is an stack of exponentation with height ...

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In this paper, it is described that $$\sum_{k = 1}^n k^k = 1 + \sum_{k = 2}^n | \prod_{1 \leq i < j \leq k} (r_i - r_j)^2 |$$ where $r_1, \ldots, r_k$ are the roots of $z^k - 1 = 0$.

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