In my previous question on the properties of $\sum_{k=0}^nk^a\binom{n}{k}$, a comment directed me to the answers on this related question with formula $$ \newcommand{\stirtwo}[2]{\left\{{#1}\atop{#2}\right\}} \sum_{k=0}^n\binom{n}{k}k^m=2^{n-m}\sum_{j=0}^m\binom{n}{j}2^{m-j}\stirtwo{m}{j}j!\tag{1} $$ Unfortunately, this is also going to be a pain to calculate as I know of no recurrence relation of $\stirtwo{m}{j}$ on $j$ such that
$$\stirtwo{m}{j} = \sum_{i=0}^{j-1}F(\stirtwo{m}{i})\tag{2}$$
for some polynomial $F$, or something to that effect. I think it would be nicer than calculating $\stirtwo{x}{y}$ for every $0\lt x\lt m$ using the usual recurrence relation$\stirtwo{n}{k}=k\stirtwo{n-1}{k}+\stirtwo{n-1}{k-1}$. Or maybe an even simpler formula exists specific to the function I'm actually trying to calculate.
Any advice and suggestions welcome.
