Let $t \in \mathbb{N}$ and consider the function $f: \mathbb{N} \rightarrow \mathbb{N}$, defined by $f_t (m)= 2 \uparrow^{m} t$,
where "$\uparrow$" is Knuth's up-arrow notation (which can be recursively defined - as I can't seem to come up quickly with an online reference for this recursive definition - by $x\uparrow^{0}t=x\cdot t,\ x\uparrow^{m}0=1$ (if $m \neq 0$), $\ x\uparrow^{m+1}\left(t+1\right)=x\uparrow^{m}\left(x\uparrow^{m+1}t\right)$)
My question is, whether this function is primitive recursive for all $t \in \mathbb{N}$. For the values $t=0,1 \dots$ we get $f_0 (m)=$ sgn$(x)$, $f_1 (m)=x\uparrow^{m} 1,\ \dots $ so my guess is, since these are primitive recursive functions, that it be primitive recursive for all $t \in \mathbb{N}$.
(It seems obvious that a proof by induction is required, but I can't seem to figure out the induction step:
Supposing that $f_t (m)$ is primitive recursive, then we have to show that $f_{t+1} (m) $ is primitive recursive as well. My idea was to show this by using the definition of primitive recursive functions, i.e. by using primitive recursion:
$f_{t+1} (0)= 1 $ which is primitive recursive.
$f_{t+1} (m+1)= 2 \uparrow^{m+1} (t+1)=2 \uparrow^{m} (2 \uparrow^{m+1} t)$, but the problem is that I can't use the induction hypothesis to show that $2 \uparrow^{m} (2 \uparrow^{m+1} t)$ is a primitive recursive function, since $ (2 \uparrow^{m+1} t)>t$.)
Any help would be much appreciated.
EDIT (In response to chandok's answer; because this wouldn't have fitted in the comments):
Sorry, that I still haven't got it, but I am not sure, if I can modify the proof that the Ackermann function is not primitive recursive, to suit the needs for my function $f_3$. If I want to prove the closedness under composition - that is, given functions $h:\mathbb{N}^l \rightarrow \mathbb{N}$ and $g:\mathbb{N}^k \rightarrow \mathbb{N}$, which have the property, that:
$\exists N_1 \ \forall n_1,\ldots,n_{l}>N_1$ such that $h(n_1 ,\ldots ,n_{l})<2 \uparrow^{n_1 + \ldots + n_{l}} 3$
and
$\forall i \in \left\{1,\ldots,l \right\} \ \exists N_{i+1} \ \forall n_1,\ldots,n_{k}>N_{i+1} $ such that $g_i(n_1, \ldots ,n_{k})<2 \uparrow^{n_1 + \ldots + n_{k}} 3$
then there should exists an $N$ such that $\forall n_1,\ldots,n_{k}>N$ we should also have, that $t(n_1,\ldots,n_k):=f(g_1(n_1,\ldots ,n_k),\ldots ,g_l(n_1,\ldots,n_k))<2 \uparrow^{n_1 + \ldots + n_{k}} 3$
i run into problems: In trying to prove the above statement, I somehow have to make use of the above properties. But I can't use the "boudedness" of $h$ because, because I can't control how much the functions $g_i$ grow - if there aren't values $n_1,\ldots ,n_k$ such that each of the $g_i$'s grow above $N_1$, then I can't make use of $h$'s properties. Maybe there is some other way to prove the closedness under composition, but I can't see it. .
