Prime Exponent Function is Primitive Recursive

Theorem
Let $$n \in \N$$ be a natural number.

Let $$\left({n, j}\right): \N^2 \to \N$$ be defined as:
 * $$\left({n, j}\right) = \left({n}\right)_j$$

where $$\left({n}\right)_j$$ is the prime exponent function.

Then $$\left({n, j}\right)$$ is primitive recursive.

Proof
Let $$p \left({j}\right)$$ be the prime enumeration function.

For $$n \ne 0$$ and $$j \ne 0$$, we see that $$\left({n}\right)_j$$ is the largest value of $$k$$ for which $$p \left({j}\right)^k$$ is a divisor of $$n$$.

Thus $$\left({n}\right)_j$$ is the smallest value of $$k$$ for which $$p \left({j}\right)^{k+1}$$ is not a divisor of $$n$$.

We note that if $$r \ge n$$ and $$j \ne 0$$, we have $$p \left({j}\right)^r \ge 2^r \ge 2^n> n$$.

Thus $$n$$ is a (generous) upper bound of $$\left({n}\right)_j$$ is $$n$$.

The condition that $$p \left({j}\right)^{k+1}$$ is not a divisor of $$n$$ can be expressed as:
 * $$\operatorname{div} \left({n, p \left({j}\right)^{k+1}}\right) = 0$$

where:
 * $\operatorname{div}$ is primitive recursive;
 * The Equality Relation is Primitive Recursive;
 * $p \left({j}\right)$ is primitive recursive;
 * Exponentiation is Primitive Recursive;
 * Addition is Primitive Recursive.

So we see that the relation:
 * $$\mathcal{R} \left({n, j, k}\right) \iff \operatorname{div} \left({n, p \left({j}\right)^{k+1}}\right) = 0$$

is primitive recursive.

From Bounded Minimization is Primitive Recursive, we also see that:
 * $$\left({n}\right)_j = \begin{cases}

\mu k \le n \mathcal{R} \left({n, j, k}\right) & : n \ne 0 \land j \ne 0 \\ 0 & : \text{otherwise} \end{cases}$$ is primitive recursive.

The result follows.