Primitive Recursion and course-of-values recursion - examples?

I ran into examples that I not trivially understand on course-of-values recursion,


In defining a function by primitive recursion, the value of the next
argument $f(n+1)$ depends only on the value of the current argument
$f⁢(n)$. Definition of functions by course-of-values recursion,
$f⁢(n+1)$ depends on value(s) of some or all of the preceding
arguments $f⁢(n),…,f⁢(0)$. very basic examples of definition by
course-of-values recursion are Fibonacci numbers...

My examples, from the computation course, wrote following $f_1$ and $f_2$ is Primitive recursive. I reproduce them here:

Let $g$ be a primitive recursive function,

-
$f_1(0)=c_1, f_1(1)=c_2, f_1(x+2)=g(x,f_1(x),f_1(x+1))$, and

-
$f_2(x)=c, f_2(x+1)=g(x,[f_2(0),...,f_2(x)])$ are primitive recursive.

I couldn't say how $f_1$ and $f_2$ are Primitive Recursive? any idea
to finding these are P.R.?

Edit 1: after the Yuval Hint the $f_1$ is easy, but for $f_2$ there is a problem, I try more than three days, but not capable to create an easy encode for $f_2$. any hint or idea?

The idea is to use some appropriate coding. A primitive recursive encoding of pairs is an encoding $\mathbb{N}^2 \to \mathbb{N}$, denoted by $\langle x,y \rangle$, such that the following functions are recursive: $e(x,y) = \langle x,y \rangle$, $p_1(\langle x,y \rangle) = x$, $p_2(\langle x,y \rangle) = y$. Coming up with a primitive recursive encoding of pairs is tedious but possible. Using such an encoding you can implement the first type of recursion in your question. The second type uses a more elaborate encoding scheme, which you can figure out on your own.

More explicitly, here is how you would implement a recursion of the first type:
$$
G(x,y) = \langle p_2(y), g(x,p_1(y),p_2(y)) \rangle \\
h(0) = \langle c_1,c_2 \rangle, \quad h(x+1) = G(x,h(x)) \\
f_1(x) = p_1(h(x))
$$
The trick in this case is that instead of computing the function $f_1$ itself, we compute the function $h(x) = \langle f_1(x), f_1(x+1) \rangle$; the value of $h$ at the point $x+1$ only depends on its value at the point $x$.

For the second type, similarly we compute the function $k(x) = [f_2(0),\ldots,f_2(x)]$ instead of $f_2$ itself; the value of $k$ at the point $x+1$ only depends on its value at the point $x$.

StackExchange Computer Science Q#40768, answer score: 6