We know that for the by the following equations recursively defined operations $\uparrow_n\colon\mathbb{N}\times\mathbb{N}\to\mathbb{N}$:

 - $m \uparrow_1 k = m+k$
 - $m \uparrow_{n+1} 1 = m$
 - $m \uparrow_{n+1} (k+1) = m\uparrow_n (m \uparrow_{n+1} k)$

the function $n\mapsto n \uparrow_n n$ is not primitive recursive. (the next 2 operations here are as expected $m\uparrow_2 k = mk$, $m\uparrow_3 k = m^k$)

What do we know about the left-braced operations?:

 - $m \downarrow_1 k = m+k$
 - $m \downarrow_{n+1} 1 = m$
 - $m \downarrow_{n+1} (k+1) = (m\downarrow_{n+1} k) \downarrow_n m$

Here again $m\downarrow_2 k=mk$, $m\downarrow_3 k = m^k$ and additionally $m\downarrow_4 k = m^{m^{k-1}}$.
Is the function $n\mapsto n\downarrow_n n$ again not primitive recursive?