Length Function is Primitive Recursive

Theorem
Let $$n \in \N$$ and let $$\operatorname{len} \left({n}\right)$$ be the length of $$n$$.

Then the function $$\operatorname{len}: \N \to \N$$ is primitive recursive.

Proof
Clearly $$\operatorname{len} \left({n}\right) = 0$$.

For $$n > 0$$, we have:
 * $$\operatorname{len} \left({n}\right) = \sum_{y=1}^n \operatorname{div} \left({n, p \left({y}\right)}\right)$$

where:
 * $$\operatorname{div} \left({n, m}\right)$$ is defined as:
 * $$\operatorname{div} \left({n, y}\right) = \begin{cases}

1 & : y \backslash n \\ 0 & : y \nmid n \end{cases}$$
 * $$p \left({y}\right)$$ is the $$y$$th prime number.

Let $$g: \N^2 \to \N$$ be the function defined by:
 * $$g \left({n, z}\right) = \begin{cases}

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

We have that:
 * $\operatorname{div}$ is primitive recursive;
 * $p: \N \to \N$ is primitive recursive;
 * Bounded Summation is Primitive Recursive.

So it follows that $$g$$ is also primitive recursive.

Finally, as $$\operatorname{len} \left({n}\right) = g \left({n, n}\right)$$ it follows that $$\operatorname{len}$$ is primitive recursive.