Let $\{a\_n\}$ be a sequence of complex numbers indexed by the positive integers. Does there always exist an analytic function $f$ such that $f(n) = \{a\_n\}$ for $n=1,2,...$? If not, are there any simple necessary or sufficient conditions for the existence of such $f$? This analytic function should be defined on some connected domain in the complex plane containing the positive integers. 


To make this concrete, consider Ackermann's function, which is defined recursively: first define the sequence of functions $A\_k$, $k=1,2,...$, as 


$A\_1(n) = 2n$,  

$A\_k(1) = 2$,
$A\_k(n) = A\_{k-1}(A\_k(n-1))$,


and then define Ackermann's function as the diagonal $A(n) = A\_n(n)$ for $n \geq 1$. Does there exist an analytic function $f$ such that $f(n) = A(n)$ for $n = 1,2,...$?


Actually, the individual functions $A\_k$ are interesting as well. $A\_1(n) = 2n$, as given above; $A\_2(n) = 2^n$, and $A\_3(n) = 2^{2^{\ldots^{2^2}}}$ (with $n$ twos in the expression). Obviously, $A\_1$ and $A\_2$ have analytic extensions. According to [Wikipedia](http://en.wikipedia.org/wiki/Ackermann_function) (which uses a slightly different definition and notation), analytic extensions of $A\_3$ or $A\_k$ for any other $k$ aren't known, but from the language, it isn't clear whether the existence of an extension is itself in question, or whether one simply hasn't been found yet. Also, it doesn't say anything about the diagonal $A(n)$ (unless I missed it). 


There are many other obvious sequences that don't seem to have obvious analytic extensions, like the prime-counting function (just to name one!). As far as my knowledge is concerned, this seems more like the rule than the exception. My knowledge here is admittedly very limited, though, so anything at all that you can share will probably teach me something. 



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 $z\_n$](http://en.wikipedia.org/wiki/Weierstrass_factorization#Existence_of_entire_function_with_specified_zeroes) (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.


