A set $S$ of entire functions $\mathbb{C}\to \mathbb{C}$ is called interpolating if for any sequence $a_1<a_2<\dots$ tending to infinity and any sequence $b_1, b_2, \dots$ there exists $f\in S$ such that $f(a_i)=b_i$.
Is the closure of polynomial and exponential functions under composition an interpolating set?
The answer is "no". Gerald Edgar has already suggested to look at sequences that grow "too slowly". But in fact, one can also construct sequences that grow "too fast".
Lemma. Let $C$ be the closure of complex polynomials and $\exp$ under composition. Then for any $f\in C$ there is a natural number $n$ such that $$|f(z)|\leq \exp^n(|z|)$$ holds for all $z$ with sufficiently large absolute value. Here $\exp^n=\exp(\exp(\dotso))$ denotes the $n$-fold iteration of $\exp$.
Proof. Let $C'\subseteq C$ be the subset where this holds. $C'$ clearly contains all polynomials and the exponential function. Furthermore, if $f\in C'$, then also $p(f)\in C'$ for all polynomials $p$, and $\exp(f)\in C'$. Indeed, $|f(z)|\leq \exp^n(|z|)$ for all sufficiently large $|z|$ implies that $|p(f(z))|\leq \exp^{n+1}(|z|)$ and $|\exp(f(z))|\leq \exp^{n+1}(|z|)$ for all sufficiently large $|z|$. Thus $C'=C$. $\square$
To get an explicit counterexample, consider the sequence $a_n=n$ and take $b_n=\exp^n(n)$. By the lemma, there can't be any $f\in C$ with $f(a_n)=b_n$.
