Set of Sequence Codes is Primitive Recursive

Theorem
Let $\operatorname{Seq}$ be the set of all code numbers of finite sequences in $\N$.

Then $\operatorname{Seq}$ is primitive recursive.

Proof
By the definition of a primitive recursive set, it is sufficient to show that the characteristic function $\chi_{\operatorname{Seq}}$ of $\operatorname{Seq}$ is primitive recursive.

Let $p: \N \to \N$ be the prime enumeration function.

Let $\operatorname{len} \left({n}\right)$ be the length of $n$.

We note that $\chi_{\operatorname{Seq}} \left({n}\right) = 1$ iff $p \left({y}\right)$ divides $n$ for $1 \le y \le \operatorname{len} \left({n}\right)$.

That is, iff $\operatorname{div} \left({n, p \left({y}\right)}\right) = 1$ for $1 \le y \le \operatorname{len} \left({n}\right)$, where $\operatorname{div}$ is the divisor relation.

We then see that $\operatorname{div} \left({n, p \left({y}\right)}\right) = 1$ for $1 \le y \le \operatorname{len} \left({n}\right)$ iff their product equals $1$.

So we can define $\chi_{\operatorname{Seq}}$ by:
 * $\displaystyle \chi_{\operatorname{Seq}} \left({n}\right) = \begin{cases}

\prod_{y \mathop = 1}^{\operatorname{len} \left({n}\right)} \operatorname{div} \left({n, p \left({y}\right)}\right) & : n > 1 \\ 0 & : \text{otherwise} \end{cases}$

Then we define $g: \N^2 \to \N$ as:
 * $\displaystyle g \left({n, z}\right) = \begin{cases}

1 & : z = 0 \\ \prod_{y \mathop = 1}^z \operatorname{div} \left({n, p \left({y}\right)}\right) & : z \ne 0 \end{cases}$

We then apply Bounded Product is Primitive Recursive to the primitive recursive function $\operatorname{div} \left({n, p \left({y}\right)}\right)$, and see that $g$ is primitive recursive.

Finally, we have that:
 * $\chi_{\operatorname{Seq}} \left({n}\right) = \begin{cases}

g \left({n, \operatorname{len} \left({n}\right)}\right) & : n > 1 \\ 0 & : \text{otherwise} \end{cases}$ is obtained by substitution from: So $\chi_{\operatorname{Seq}}$ is primitive recursive.
 * the primitive recursive function $\operatorname{len}$
 * the primitive recursive function $g$
 * the primitive recursive relation $>$
 * the constant $1$

Hence the result.