Let $f(z)$ be a transcendental entire function. Hence $f$ is not a polynomial.
Assume $f(z)$ has infinitely many zero's $z_n$
$$f(z_n) = 0$$
Lets say that $f(z)$ is given by a taylor series.
Im interested in the distance between those zero's
$$d = D(f(z)) = \inf_{i \neq j} |(z_i - z_j)| s.t. f(z_n) = 0 $$
What is know about this ? What theorems do we have ? What open conjectures are studied ?
I assume
$d = 0$ occurs most often ?
In particular I am intrested in non-periodic functions.
A comment that is too long; not sure about the intent of the question as the comment regarding universality doesn't really apply to entire functions per se, as it is related to some kind of conditional convergence and the Taylor series of entire functions are absolutely convergent - now if one looks at say $L$ functions which are universal in $1/2<\Re z <1$ and entire, one can quibble with the above, but the universality of the $L$ functions come from their representations as conditionally convergent Dirichlet series in the critical strip and has little to do with the fact that they can be extended to be entire.
So coming back to the OP question, we know that for any sequence $z_k$ st $\lim |z_k| =\infty$ there are entire functions with zeroes at $z_k$ so the infimum above can be anything. In particular, we can realize the infimum to be any $d>0$ with a function of order $2$ (or lower) by picking the zeroes $d$ spaced apart on lines parallel with either coordinate axis (and of course we can wiggle them a little to avoid regularity so instead of $dk+dNi, dN+dki$ we can push them a little apart so the infimum distance is still $d$)
The only thing we can say is that if the exponent of convergence (the infimum of $a>0$ st $\sum_{z_k \ne 0}1/|z_k|^a< \infty$ if a finite $a$ with that property exists or $\infty$ otherwise) of the sequence $z_k$ is $2+\delta, \delta>0$ (or infinity of course), then indeed $d=0$ by the pigeonhole principle since then in any $N \times N$ square centered at the origin there are at least $cN^{2+\delta/2}$ zeroes for large $N$ and some fixed $c>0$ and then one can split it into $N^2$ interiorly disjoint squares $1 \times 1$, so there is at least one such containing $cN^{\delta/2}$ elements of the sequence, hence at least a pair with distance $o(1)$
In particular, if the order of a Hadamard product of $f$ is at least $2+\delta$ (eg $f$ has finite nonintegral order greater than $2$), then $d=0$, but one can easily find functions of order up to $2$ with any $d$ we want and then, of course, $e^{P(z)}f$ has integral order $\deg P$ for any polynomial with degree at least $3$ and $e^{e^z}f$ has infinite order and these still have any $d$ we want, so we cannot really say that "most" $f$ have $d$ in a really meaningful way