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I apologize in advance if this question is too ill-posed, but here we go.

As an additive identity, $x+0=x$.

As a multiplicative identity, $x\times 1=x$.

$2$ feels similar in a way I can't define as well, but $2+2=2\times 2=2^2={^2}{2}=H_{a}(2,2)$, where $H_{a}$ is an $a$th-level hyperoperation.

Are there any other integers that are special in a similar way? Some sort of hyper-hyperoperation where $3$ serves as an identity somehow, or some other integer? I know that $e$ is somewhat analogous in exponentiation, and that $i$ is a unit, and those aren't really what I'm after.

Trevor
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    It ultimately depends on the set $S$ and the operation $\ast$ you define. No doubt there exists some $S$ such that $2 \in S$ and $S \subseteq \Bbb R$, such that $s \ast 2 = 2 \ast s = s$ for all $s \in S$. It's mostly a matter of how pathological you allow $S$ and $\ast$ to be. (Of course I imagine the problem is much harder when $\ast$ is some hyperoperation in the sense you mention, and $S = \Bbb R$...) – PrincessEev Feb 29 '20 at 12:51
  • The $\forall x$ in the property is what makes an identity useful. There is no $\forall$ in your examples for $2$. –  Feb 29 '20 at 12:51
  • Well, I suppose $\forall x : H_x (2,2)=4$, if that counts. – Trevor Feb 29 '20 at 16:55

1 Answers1

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From Wikipedia, we have the special cases under the convention that $0^0=1$:

$H_n(0,b)=\begin{cases}0,&(2\le n\le3\land b\ge1)\lor(n\ge4\land b\text{ odd})\\1,&(n=3\land b=0)\lor(n\ge4\land b\text{ even})\\b,&n=1\\b+1,&n=0\end{cases}$

$H_n(1,b)=1,~n\ge3$

$H_n(a,0)=\begin{cases}a,&n=1\\0,&n=2\\1,&\text{else}\end{cases}$

$H_n(a,1)=a,~n\ge2$

$H_n(a,a)=H_{n+1}(a,2),~n\ge1$

$H_n(a,-1)=\begin{cases}a-1,&n=1\\-a,&n=2\\1/a,&n=3\\0,\text{else}\end{cases}$

$H_n(2,2)=\begin{cases}3,&n=0\\4,&\text{else}\end{cases}$

but aside from these there isn't much nice beyond expanding with the definition.

Simply Beautiful Art
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