# Hyperoperation: Why does $H_n(0,b) = 0$ for $n\ge4$, $b$ odd ($\ge -1$)

On the Wikipedia for hyperoperations is says that $H_n(0,b) = 0$ if $n\ge4$, $b$ odd ($\ge -1$).

From what I gether, the basic jist of what this is saying is that $0^{0^0}=0$ and from that we can gather that when $b$ is odd and $\ge1$, $H_n(0,b) = 0$ if $n\ge4$ and when $b$ is even and $\ge1$ that $H_n(0,b)$ is undefined if $n\ge4$.

My question is why is $0^{0^0}=0$ and why is $H_n(0,-1)=0$ if $n\ge4$?

It's mostly a matter of how the base cases are intended to be handled. In particular, for all $n\ge2$, it is intended that we have $H_n(a,1)=a$. Likewise, we also have $a=H_3(a,1)=H_2(a,H_3(a,0))=a\cdot H_3(a,0)$. For this to consistently hold, we need either $a=0$ or $H_3(a,0)=1$. Since the latter case holds for any $a$, choosing it for the definition makes it simpler. Hence, we reach the convention of $0^0=H_3(0,0)=1$.
From here, one may note that $H_4(0,b+1)=H_3(0,H_4(0,b))$, hence if $H_4(0,b)=1$, then $H_4(0,b+1)=0^1=0$, and if $H_4(0,b)=0$, then $H_4(0,b+1)=0^0=1$. And since $H_4(0,0)=1$, we deduce $H_4(0,2n)=1$ and $H_4(0,2n+1)=0$. In general, one may deduce by induction, that since $0$ and $1$ are even and odd respectively, then this extends to the higher hyperoperations.