According to Wikipedia hyperoperation for positive integers is defined as $$ H_{n}(a,b)=H_{n-1}(a,H_{n}(a,b-1)) $$ with some base conditions. (We take $ n \geqslant 1 $.)

Recursivly define a sequence of binary operators that inverts hyperoperation for all positive integers.
Meaning we are looking for $I_{n}$ such that $$ \forall n,a,b \in \mathbb{N}_{+}: I_{n}(H_{n}(a,b),a)=b $$

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    $\begingroup$ This is at most possible for $n \geq 1$, because $H_0$ ignores its first argument. $\endgroup$ – A.P. Oct 31 '15 at 15:09
  • $\begingroup$ $I_{0}$ I'd simply define as counting down (by $1$). $\endgroup$ – MrFrety Nov 5 '15 at 7:02
  • $\begingroup$ That wouldn't work. The problem is that you are asking for an inverse of $H_n(\cdot, b)$, but $H_0$ is not injective in it's first component. Indeed, $H_0(a, b) = b+1$ for every $a \in \Bbb{Z}_{\geq 0}$, so with your definition we would have $I_0(H_0(a,b), b) = b$ for every $a \in \Bbb{Z}_{\geq 0}$... $\endgroup$ – A.P. Nov 5 '15 at 11:06
  • $\begingroup$ I'd rather define $H_{0}(a,b)=a+1$. Mentioned it on the discussion page... Then it would work, wouldn't it? $\endgroup$ – MrFrety Nov 7 '15 at 20:48
  • $\begingroup$ No, it wouldn't, because you'd be messing with the base case of the definition of $H_n$. Indeed, consider $H_o'(a,b) = a + 1$. Then $H_1(1,b) = H_0'(1, H_1(1, b-1)) = 2$ for every $b > 0$... $\endgroup$ – A.P. Nov 7 '15 at 22:46


  • $9/3=1+(9-3)/3=1+(1+(6-3)/3)=1+(1+1)=3$

  • $log_{2}(8)=1+log_{2}(8/2) =1+(1+log_{2}(4/2))=1+(1+1)=3$

  • $log_{3}(27)=1+log_{3}(27/3) =1+(1+log_{3}(9/3))=1+(1+1)=3$

So in general: $$ I_{n}(c,a)=1+I_{n}(I_{n-1}(c,a),a) $$ whereby $I_{n}(a,a)=1$ for $n>1$ and $I_{1}(c,a)=c-a$ for $n=1$.

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    $\begingroup$ This definition will terminate only when the first argument of $I_n(\cdot, b)$ is of the form $H_n(a,b)$. Can the definition be meaningfully extended to other values? For example, we might say something like $I_n(c,d) = 0$ if $d > c$, to give the "quotient" of the (pseudo)inverse, discarding the "remainder" (I didn't actually test if this is sufficient). $\endgroup$ – A.P. Oct 31 '15 at 15:15
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    $\begingroup$ I have some ideas on how to expand it to other number sets or even construct new number sets from it. But it was too hard, so I restricted myself to positive integers for now. $\endgroup$ – MrFrety Nov 5 '15 at 6:58
  • $\begingroup$ This definition doesn't always hold true. It only occurs here due to log rules, mainly $1+\log_b(a)=\log_b(ab)$ $\endgroup$ – Simply Beautiful Art Jun 8 '17 at 22:23
  • $\begingroup$ I claim that it holds true for all n purely from the definition of the Hyperoperator. The inverse basically counts the number of repetitions... or it should... I'm always thankful for corrections... $\endgroup$ – MrFrety Apr 30 '18 at 3:47
  • $\begingroup$ After some adjustments the variable names are hopefully more consistent, now. $\endgroup$ – MrFrety Apr 30 '18 at 5:04

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