For all $(n_1, \cdots, n_k) \in \mathbb{N}^k$ let's define $<n_1, \dots, n_k>:=p_1^{n_1+1} \dotsc p_k^{n_k+1}$, whereby $(p_1, p_2, p_3, \cdots)=(2,3,5,7,\cdots)$ are the
prime numbers.
For every function $f: \mathbb{N}^k \times \mathbb{N} \rightarrow \mathbb{N}$ we define $\hat{f}: \mathbb{N}^k \times \mathbb{N} \rightarrow \mathbb{N}$ as follows:
$\hat{f}(\vec{x}, y)=<f(\vec{x}, 0), \cdots, f(\vec{x}, y-1)>$,
i.e. $f(\vec{x}, 0) = <>=1$ and $f(\vec{x},1) = <f(\vec{x}, 0)> = 2^{f(\vec{x}, 0) + 1}$.
Now we want to proof that f is primitive recursive if and only if $\hat{f}$ is primitive recursive. (Moreover: If f is total, then f is computable if and only if $\hat{f}$ is computable.)
In order to clarify the definition of primitve recursive functions:
We define a set $\mathcal{P}$ of operations on $\mathbb{N}$, the so-called "primitive recursive functions":
For every $k \geq 0$ the constant function from $\mathbb{N}^k$ to $\mathbb{N}$ with value 0 is primitive recursive.
The function $n \mapsto n+1$ is primitive recursive.
For every $n \geq 1$ and every k so that $1 \leq k \leq n$ the projection function $\Pi_k^n: \mathbb{N}^n \rightarrow \mathbb{N}, \pi_k^n(x_1, ..., x_n) = x_k$ is primitive recursive.
If g: $\mathbb{N}^k \rightarrow \mathbb{N}$ and $f_1, \cdots, f_k: \mathbb{N}^n \rightarrow \mathbb{N}$ all are primitive recursive, then also the function $g(f_1, \cdots, f_k): \mathbb{N}^n \rightarrow \mathbb{N}$, defined by:
$\forall \vec{x}=(x_1, \cdots, x_n): g(f_1, \cdots f_k)(\vec{x})):=g(f_1(\vec{x}), \cdots, f_k(\vec{x}))$.
For all $k \geq 0$, all primitive recursive funtions h: $\mathbb{N}^k \rightarrow \mathbb{N}$ and all primitive recursive functions g, the function defined by
$\forall \vec{x} \in \mathbb{N}: f(\vec{x}, 0) = h(\vec{x}), \quad \forall y \in \mathbb{N} \forall \vec{x} \in \mathbb{N}^k: f(\vec{x}, y+1) = g(f(\vec{x}, y), \vec{x}, y)$ is primitive recursive. This f is called $PR(g,h)$.
This are all primitive recursive funtions.
I'd appreciate any idea on how to proof the statements mentioned above.