# Are hyperoperators primitive recursive?

I apologize if this question is too basic. I have read that the Ackerman function is the first example of a computable but NOT primitive recursive function. Hyperoperators seem to be closely related to these functions, but I am not sure if they still keep the property of being NOT primitive recursive. My intuition is that they are, but I am not sure. Any textbook or references to read to get a better understanding of this will be very appreciated (and a straight response too!).

NOTE: I read this related question, but it doesnt help me.

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Each given hyperoperation can be implemented by a bounded loop of the previous hyperoperation. –  ex0du5 May 11 '13 at 6:45
straight answer welcomed!!! (so, if I got it right, any of them can be computed using a do-loop?) –  julian fernandez May 11 '13 at 6:59

If I have understood well yes.

Because if follow from the definiton of hyperoperators definition.

$H_0(a,b):=S(b)$

is the successor function, and every hyperoperation is defined recursively from the successor function:

(recursive definition of $H_{n+1}$ using $H_n$)

$i)$ $H_{n+1}(a,0):=a_{n+1}$

$ii)$ $H_{n+1}(a,b+1):=H_n(a,H_{n+1}(a,b))$

Here $a_{n+1}$ is the initial value of the function when the argument is $0$ and in the case of hyporoperators we have that

$a_ {n+1}:= \begin{cases} a, & \text{if$n=0$} \\ 0, & \text{if$n=1$} \\ 1, & \text{if$n\gt 1$} \\ \end{cases}$

Since $H_0(a,b)=S(b)$ is the successor funtion and is basic primitive rucursive funtion, we have that all the hyperoperations (that are defined from $H_0(a,b)=S(b)$) are Primitve recursive functions.

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