In paper which deals with ordered factorizations, I found approximation formulae for the number of ordered factorizations of integer $n$ into $m$ distinct parts greater than one. In the paper it is stated:
If $1 \leq m = o(\sqrt{log(x)})$, then: \begin{equation} \label{eq:smallm} A(x, m) = x \frac{(\log(x))^{m - 1}}{(m - 1)!} e^{-\frac{((1 - \gamma)m + 1)(m - 1)}{\log(x)}} \left(1 + O\left(\frac{m^3}{(\log(x))^2}\right)\right) \end{equation} uniformly in $m$.
I am not sure how to interpret $1 \leq m = o(\sqrt{log(x)})$ expression, and I do not understand what uniformly in $m$ means.
Since the formula gives an approximation for the number of $m$-factorizations for all $n \leq x$, I compared the results given with the formula to the actual number $m$-factorizations obtained with enumeration. I assumed that $m$ is in interval $[1, \lfloor\sqrt{log(x)}\rfloor]$. This seemed to yield good results. In other words, I assumed that $1 \leq m = o(\sqrt{log(x)}) \implies [1, \lfloor\sqrt{log(x)}\rfloor]$. Is this true, and if not for
which $m$ does the expression for $A(x, m)$ hold?
Any comments or suggestions appreciated.