Minimal growth of entire functions with given sequence of zeros (in Ukrainian)

Let $(z_j)$ be a sequence of complex numbers satisfying $\lim_{j\to\infty}z_j=\infty$ and $n(r)$ be its counting function such that $$ \varliminf_{r\to\infty}\frac{\ln n(r)}{\ln\ln r}=\alpha>0. $$ It is shown that there exists an entire function $f$ with zeros in the points $z_j$ and only in them, such that for any $\varepsilon>0$ there exists subset $E\subset [1,\infty)$ of finite logarithmic measure such that $$ \ln\ln \mathcal M (r,f)=o((n(r))^{\frac1\alpha+\varepsilon}),\;\; r\to\infty,\;\; r\notin E. $$ For $\varepsilon=0$ this assertion is not valid for some sequence $(z_j)$.