Little Picard Theorem:
If a function $f:\mathbb{C}\to\mathbb{C}$ is entire and non-constant, then the set of values that $f(z)$ assumes is either the whole complex plane or the plane minus a single point.
Define "Picard-0 functions" to be the functions which are entire, non-constant and their image is $\mathbb{C}$. Define "Picard-1 functions" to be the functions which are entire, non-constant and their image is $\mathbb{C}$ without a single point.
Some elementary examples of Picard-0 functions are $\sin$ and $\cos$. However, $\exp$ is a well known example of a Picard-1 function.
Just by looking at the power series expansions at the origin,
$$\sin z=\sum_{n=0}^\infty\operatorname{Im}(i^n)\frac{z^n}{n!},\quad z\in\mathbb{C},$$
$$\cos z=\sum_{n=0}^\infty\operatorname{Re}(i^n)\frac{z^n}{n!},\quad z\in\mathbb{C},$$
$$\exp z=\sum_{n=0}^\infty \frac{z^n}{n!},\quad z\in\mathbb{C},$$
it is not immediately clear whether they are Picard-0 functions or Picard-1 functions, given that they are entire and non-constant (this is just an illustrative example). (In the series, $0^0=1$ was assumed.)
If $f(z)=\sum_{n=0}^\infty a_n z^n$ and $f$ is entire and non-constant, is it possible to state some explicit conditions for $a_n$ which would determine whether $f$ is a Picard-0 function or a Picard-1 function?
Best Answer
What you call “Picard-1-functions” are exactly the functions of the form $$ \tag{1} f(z) = c + e^{g(z)} $$ with a complex constant $c$ and an entire function $g$.
I am not aware of a condition on the Taylor coefficients $(a_n)$ of $f$ which is necessary and sufficient for $f$ to be surjective or not. This might be difficult because small changes in a coefficient change the behavior: $e^z$ is not surjective, but $$ f(z) = e^z + a z^n $$ is surjective for any positive integer $n$ and arbitrary non-zero complex number $a$.
There are some necessary conditions which are related to the order and type of entire functions (compare Surjective entire functions on Math Overflow).
The order of an entire function is defined as $$ \tag{2} \rho = \limsup_{r \to \infty} \frac{\log \log M(r, f)}{\log r} $$ where $M(r, f) = \sup \{ |f(z)| : |z| \le r \}$. The order can be computed from the Taylor coefficients as $$ \tag{3} \rho = \limsup_{n \to \infty} \frac{n \log n}{- \log |a_n|} \, . $$
For entire functions of positive and finite order $\rho$, the type is defined as $$ \tag{4} \sigma = \limsup \frac{M(r, f)}{r^\rho} $$ which is related to the coefficients via $$ \tag{5} \sigma = \frac{1}{e \rho} \limsup_{n \to \infty} n |a_n|^{\rho/n} \, . $$
Now, if $f$ is not surjective and of finite order, then the function $g$ in $(1)$ is necessarily a polynomial, and that implies that the order $\rho$ in $(2)$ is an integer, and the type $\sigma$ in $(4)$ is positive.
Using the identities $(3)$ and $(5)$, we get necessary conditions on the (growth of the) coefficients $(a_n)$ for $f$ to be not surjective.