Which sequences can be extended to analytic functions? (e. g., Ackermann's function)

It's a standard theorem in complex analysis that if $z_n$ is a sequence that goes to infinity, there is an entire function taking any prescribed values at the $z_n$. There is a function $f$ vanishing to order 1 at each $z_n$ (for $z_n=n$, you could take $f(z)=\sin \pi z$), and then consider $\sum_n a_nf(z)/(f'(z_n)(z-z_n))$. This may not converge, but you can tweak it by multiplying each term by something that is 1 at $z_n$ (eg, $\exp(c_n(z-z_n))$ for $c_n$ chosen appropriately) to make it converge.

(I don't know off the top of my head how to choose the $c_n$; this is copied from Exercise 1 on page 197 of Ahlfors's Complex Analysis.)

EDIT: It's easy to show that such $c_n$ exist. If you write $b_n=a_n/(z_n f'(z_n))$, then for any fixed $z$, the terms of the sum will be approximately $b_n \exp(c_n z_n)$ for $n$ large. You can obviously pick $c_n$ so that this converges.

Actually, there is an even stronger result, often called the interpolation theorem, which follows from a well-known theorem of Mittag-Leffler:

Let $(z_n)$ be a sequence of complex numbers with no limit point. For each $n$, let $l(n)$ be any integer greater or equal to $1$ and for $0 \leq k \leq l(n)$, let $(a_{n, k})$ be complex numbers. Then there exists an entire function $g$ such that

$$g^{(k)}(z_n) = a_{n,k}$$

for every $n \geq 1$ and every $0 \leq k \leq l(n)$.

That is, you can fix values for the derivative at the $z_j$'s.

If I remember correctly you can get even a stronger result. Let $a_n$, $b_n$ and $c_n$ be sequences of complex numbers and $i_n$ be a sequence of natural numbers. Then I believe there exists a meromorphic function that takes the value $b_n$ on $a_n$ for all $n$, and has a pole of order $i_n$ at $c_n$ for all $n$. Assuming that the $a_n$ and $c_n$ are disjoint and have no accumulation point. It's been a while since I thought about complex analysis but I seem to remember learning this.