Ackermann function and primitive recursiveness

Question

If we define $b_n(m) := a(n,m)$ for all $n$ and $m \in \mathbb{N}$. For which $n$ is the function $b_n$ primitive recursive and for which $n$ it is not a primitive recursive function? Can anyone please help me out with this?

Answer 1

I assume that (as the title suggests) $$ a(n,m) = \begin{cases} m +1 & n= 0\\ a(n-1, 1) & n > 0, m=0\\ a\bigl(n-1, a(n,m-1)\bigr) \end{cases} $$ denotes the ackermann function. Rewriting this in terms of $b_n$, we have $b_0 = \cdot + 1$ is the successor function and $$ b_n(m) = \begin{cases} b_{n-1}(1) & m = 0 \\ b_{n-1}\bigl(b_n(m-1)\bigr)\end{cases} $$ If $b_{n-1}$ is primitive recursive, this defines a primitive recursive function $b_n$. As $\cdot + 1$ is primitive recursive, by induction, all $b_n$ are.