What is the inverse to hyperoperation for positive integers?

Question

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 $.)

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

Answer 1

Examples:

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

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

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

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 $n=1$.