Suppose $PA$ is Peano arithmetic. For $m \in \mathbb{N}$ define $\overline{m}$ as a term in the language of $PA$ using the following recurrence.
$$\overline{0} = 0$$
$$\overline{m + 1} = S(\overline{m})$$
Now let’s say that $n \in \mathbb{N}$ is described by a formula $\phi(x)$ with a single free variable $x$ iff $PA \vdash \forall x (\phi(x) \to (x = \overline{n}))$ and let’s say that a function $f: \mathbb{N} \to \mathbb{N}$ is described by a formula $\psi(x, y)$ with exactly two free variables $x$ and $y$ iff $\forall m \in \mathbb{N}$ $PA \vdash \forall x \forall y (((x = \overline{m}) \& \psi(x, y)) \to (y = \overline{f(x)}))$.
Any number is described by some formula (take for example $\phi(x) = ‘x = \overline{n}’$). However, not every function is described by a formula because there are continuum many functions but only countably many formulas.
Let’s define a function $L: \mathbb{N} \to \mathbb{N}$ that maps any natural number $n$ to the minimal possible length (number of symbols in the word) of a formula that describes $n$.
My question is:
Can $L$ be described by a formula?
No, it can not.
Define $\Gamma(n) = \min\{x| L(x) \geq 4n\}$. Then if $L(x)$ is described by a formula $\psi(x, y)$, then $\Gamma$ is described by a formula $$\xi(x, y) = "\forall t ((t < x\times\overline{4}) \to \neg \psi(y, t)) \& \forall z ((z < y) \to \exists d ((d \leq x\times\overline{4}) \& \psi(y, d)))"$$
Now suppose $\lambda$ is the length of $\xi$. From the fact, that $\xi(x, y)$ describes $\Gamma$ it follows, that $\phi(y) = "\forall x ((x = \overline{n}) \to \xi(x, y))"$ describes $\Gamma(n)$. From that we can conclude, that $L(\Gamma(n)) \leq 10 + \lambda + 3n$. But on the other hand, $L(\Gamma(n)) \geq 4n$ by definition. And here we get the contradiction.