Is there an upper bound on the growth rate of analytic functions? This problem comes from a solution tactic used in Is there a rational surjection $\Bbb N\to\Bbb Q$?, where I discovered that there is an analytic function $f(z)$ that takes the values $f(n)=a_n$ for all $n\in\Bbb N$, as long as $a_n$ has at most polynomial growth; here I am interested in seeing how far I can relax the "polynomial growth" constraint.
Let us call an analytic function a kernel if it satisfies $k(0)=1$ and $k(n)=0$ for all $0\ne n\in\Bbb Z$. The main kernel used in the above question/answer was the function ${\rm sincz}(z)=\frac\pi z\sin(\frac z\pi)$, which has growth rate $O(z^{-1})$ (in the positive and negative real direction). Then ${\rm sincz}^m(z)$ is also a kernel, with growth rate $O(z^{-m})$, and any kernel yields an analytic function via $f(z)=\sum_{n\in\Bbb N}a_nk(n)$, which works for all sequences whose growth rate is no more than $O(\frac1{k(n)n^2})$ (or substitute some other summable series for $n^{-2}$).
But if $k$ is a kernel and $g$ is any analytic function with $g(0)=1$, then $g(z)k(z)$ is also a kernel, which allows for much faster-decaying kernels, such as $e^{-z^2}{\rm sincz}(z)$. In fact, given any eventually monotonic analytic function $g(z)$ with $g(0)=1$, the function $\frac{{\rm sincz}^2(z)}{g(z^2)}$ is a kernel with growth rate $O(\frac1{g(z^2)z^2})$, which can create analytic functions for any sequence of growth rate less than $O(g(n^2))\supseteq O(g(n))$.
So the problem is reduced to the question in the title:

Is there any upper bound on the growth rate of analytic functions? That is, is there a definable sequence $a_n$ which grows so fast that it eventually outpaces any analytic function $f(z)$ sampled at the natural numbers?

The examples given still fall far short of such fast-growing functions as the Ackermann function or Graham's sequence, but it is not obvious to me that there are not similar techniques for producing extremely fast-growing analytic functions.
 A: No, there is no upper bound. In fact, we can say something much stronger.

Theorem Suppose that $a_n \in \mathbb{C}$ satisfy $a_n \to \infty$, and that $A_n$ are arbitrary complex numbers. Then there exists an entire function $f(z)$ satisfying $f(a_n) = A_n$.

In sketch, I'll just tell you such a function (more or less). Let $g(z)$ be an entire function with simple zeros at the $a_n$. That such a function exists is a theorem of Hadamard. Then for a sufficiently clever choice of $\gamma_n$, the following function
$$ \sum_{n \geq 1} g(z) \frac{e^{\gamma_n(z - a_n)}}{z - a_n} \frac{A_n}{g'(a_n)}$$
converges everywhere, is entire, and satisfies $f(a_n) = A_n$.
This also appears as Exercise 1 in section 5.2.3 of my copy of Ahlfor's Complex Analysis. It has been asked and answered on this site as well.
I happen to remember this from a complex analysis graduate exam question I faced some years ago. In fact, the actual question posed of me was more astounding, and I had to prove the following.

Theorem Suppose that $a_n \in \mathbb{C}$ satisfy $a_n \to \infty$ and that $\mathcal{A}_n = \{ A_n^{(0)}, A_n^{(1)}, \ldots, A_n^{(j)}\}$ are arbitrary finite lists of complex numbers. Then there exists an entire function $f(z)$ satisfying $f^{(j)}(a_n) = A_n^{(j)}$ for all $a_n$ and all prescribed $A_n^{j}$.

In short, it is possible to specify infinitely many points with arbitrarily (but finitely many) derivatives at each point, as long as the points tend to infinity. I do not actually remember how to prove this anymore, but I'm sure that it's built off of the function above in a clever way.
Returning to your question: you can choose $A_n$ that grow as arbitrarily fast as you want, and yet you can still find a complex analytic function that takes those values at the integers.
