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So... I had the silly idea to extend Knuth's up-arrow notation so that it included zero and negative arrows. It is normally defined as $$\begin{align*} a \uparrow b & = a^b \\ a \uparrow^n b & = \underbrace{a \uparrow^{n - 1} (a \uparrow^{n - 1} (\dots(a \uparrow^{n - 1} a) \dots ))}_{b\text{ copies of } a} \end{align*}$$ so, basically the hyperoperation sequence starting from exponentiation. For now, I will only consider $a,b > 0$.

If we try to go backwards from $a \uparrow b$, the "trivial" extension (letting down arrows represent negative up arrows, because why the heck not) is: $$\begin{align*} a \;b & = a \cdot b \\ a \downarrow b & = a + b \\ a \downarrow \downarrow b & = \text{see below} \end{align*} \\ \vdots$$ But I had trouble coming up with an expression for $a \downarrow \downarrow \downarrow b$.

Maybe it doesn't exist. Alternatively, maybe there is a way of defining $a \; b$ (zero arrows) such that it does exist. So my question is: Is there an extension of Knuth's up-arrow notation such that $a \downarrow^n b$ exists for all $n \geq 3$?


Edit: Welp, I messed this question up. I initially thought $a \downarrow \downarrow b = a + 1$ was correct, but it is actually $b + 1$. So I thought I had an example of an extension when I did not. I have modified the question accordingly.

An extension would define $a \uparrow^n b$ for each $n \leq 0$ which satisfies the recursive definition of the notation.


Edit 2: Okay, turns out $a \downarrow \downarrow b = b + 1$ isn't correct either, as this would imply $a \downarrow b = a + b - 1$. For example, $4 \downarrow 3 = 4 \downarrow \downarrow (4 \downarrow \downarrow 4) = 4 \downarrow \downarrow (4 + 1) = (4 + 1) + 1 = 6 = 4 + 3 - 1$. But it is really close; perhaps we need an exception, such as $$\begin{align*} a \downarrow \downarrow b = \begin{cases}b + 1 & \text{if } a < b \\ b + 2 & \text{if } a = b \end{cases}\end{align*}.$$ The case $a > b$ does not show up when evaluating $a \downarrow b$, but it will be need to be defined if we try to extend further to $a \downarrow \downarrow \downarrow b$. For instance, we could abuse the fact that the case $a > b$ is allowed to be anything, and let $$a \downarrow \downarrow b = b + 1 + \left\lfloor \frac{a}{b} \right\rfloor,$$ but finding $a \downarrow \downarrow \downarrow b$ may be intractable as a result.

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marked as duplicate by Simply Beautiful Art, José Carlos Santos, YuiTo Cheng, воитель, Brahadeesh Jul 3 at 4:46

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