Let $n,r\in N$ and let $S(n,m)$ represent Stirling's number of the second kind. It is known that $\sum_{m=0}^n S(n,m)m!=F_n$ is a Fubini number. Is it possible to represent (or estimate from above) through Fubini numbers the following
$$ \sum_{m=0}^n S(n,m)\frac{\Gamma(r+m)}{\Gamma(r)}. $$
Denoting these numbers by $F_n^{(r)}$, we can find the exponential generating function (EGF) \begin{align} F_r(z):=\sum_{n=0}^{\infty}F_n^{(r)}\frac{z^n}{n!} &=\frac{1}{\Gamma(r)}\sum_{n=0}^\infty\frac{z^n}{n!}\sum_{m=0}^{n}S(n,m)\Gamma(m+r) \\&=\frac{1}{\Gamma(r)}\sum_{m=0}^\infty\Gamma(m+r)\sum_{n=m}^{\infty}S(n,m)\frac{z^n}{n!} \\&=\sum_{m=0}^\infty\binom{m+r-1}{m}(e^z-1)^m\color{blue}{=(2-e^z)^{-r}} \end{align} from the EGF of $n\mapsto S(n,m)$ and the binomial series. Thus, $F_r(z)=F_1(z)^r$, which gives \begin{align} \color{gray}{F_{r+s}(z)=F_r(z)F_s(z)}&\color{gray}{\implies} F_n^{(r+s)}=\sum_{m=0}^n\binom{n}{m}F_m^{(r)}F_{n-m}^{(s)}, \\ \color{gray}{F_1(z)F_r'(z)=rF_r(z)F_1'(z)}&\color{gray}{\implies} F_{n+1}^{(r)}=rF_{n+1}^{(1)}+\sum_{m=0}^{n-1}\binom{n}{m}(rF_{m+1}^{(1)}F_{n-m}^{(r)}-F_{m+1}^{(r)}F_{n-m}^{(1)}), \end{align} and so on (say, $\color{gray}{(2-e^z)F_r(z)=F_{r-1}(z)}$ gives yet another connection formula). Similarly to the case $r=1$, if $r$ is a positive integer, the contour integration approach gives a sum-of-residues formula useful for asymptotic analysis (for $r\notin\mathbb{Z}$, the idea of contour integration is still good, but the analysis itself is harder).