Solving tricky Knuth Up Arrow Notations How would I solve something like
$2\uparrow\uparrow n$? when n ≤1?
Or
$2\uparrow^{-2}2$?  
Thanks!
 A: I'm not aware of an interpretation with a negative number of arrows, so I'll just address the case where $n \geq 1$.
With up arrow notation, you can "strip away" an arrow and then write $n$ copies of 2, each separated by one fewer arrows than you had previously. In your case, we'd have
$$
2 \uparrow \uparrow n = \underbrace{2 \uparrow 2 \uparrow \cdots \uparrow 2}_{n \text{ copies of } 2}.
$$
A single uparrow is just regular exponentiation, so we'll get an exponent tower of 2's having height $n$.
For example, when $n = 3$, this all works out to
$$
\begin{align*}
2 \uparrow \uparrow 3 &= 2 \uparrow 2 \uparrow 2\\
&= (2 \uparrow 2) \uparrow 2\\
&= (2^2) \uparrow 2\\
&= 2^{2^2}.
\end{align*}
$$
A: Here is a possible answer:
We know that for every positive integer n, m↑↑n = logm[m↑↑(n+1)]. If we assume the same equation to hold for nonpositive n, we get:
m↑↑0 = logm(m↑↑1) = logmm = 1
m↑↑-1 = logm(m↑↑0) = logm1 = 0
m↑↑-2 = logm(m↑↑-1) = logm0 = -infinity (in other words, it is undefined) 
This seems to be the only logical extension of the double-arrow operator to nonpositive integer "exponents". 
As for the extension to a "nonpositive number of arrows", it is even simpler:
m↑0n = m*n (ordinary multiplication)
m↑-1n = m+n (ordinary addition)
But the next one is somewhat surprising:
m↑-2n = (m+n+4)/2 (you might want to try and verify on your own that this is correct, by proving that m+n is indeed equal to this function iterated n times on m)
So the expression in your original question, 2↑-22, evaluates to (2+2+4)/2 = 4. Of-course, this is hardly surprising, since 2[any-number-of-arrows]2 always evaluates to 4.
