Can we always find an analytic function if we know countable points? I hope to find an analytic function such that $f(n)=b_{n}$, $n\in \mathbb{N}$?
Can we always take an analytic function $f$?
 A: Yes, you can find such a function.
We can do it inductively. Start with $f_0(x) = 0$, the constant function. For convenience, let $h_n$ be a polynomial with $h_n(m) = 0$ for $m < n$ and $h_n(n) = 1$.
Suppose you have $f_n$ constructed for some $n$ so that $f_n(m) = b_m$ for $m < n$. We want to construct $f_{n+1}$. Let $N$ be a (huge) integer, which we will determine shortly. Consider the map $g_n(x) = (b_n - f_n(n)) (x/n)^N h_n(x)$. It is clear that $g_n(m) = 0$ and $g_n(n) = b_n - f_n(n)$. Define $f_{n+1} := f_n + g_n$; then $f_n(m) = b_m$ for $m < n+1$. Now, let use determine how large $N$ is supposed to be. We would like $|g_n(x)| < 2^{-n}$ for $|x| < n-1$. This will hold if $N$ is so large that $\left(\frac{n-1}{n}\right)^N \cdot \max_{|x| < n-1} (b_n - f_n(n)) h_n(x) < 2^{-n}$, but we don't want to describe $N$ in any more explicit way.
Now, $f_n$ is uniformly convergent on compact sets, and all $f_n$ are analytic. It follows that their limit, call it $f$, is analytic as well. And clearly $f(m) = b_m$ for all $m$. 
A: Yes. As a consequence of Weierstrass's and Mittag-Leffler's theorems, for any
sequences $a_n$ and $b_n$ with $a_n$ distinct and $|a_n| \to \infty$ as $n \to \infty$ there is an entire function $f$ with $f(a_n) = b_n$.
