# Extending Knuth up-arrow/hyperoperations to non-positive values [duplicate]

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.