Looking at the table of $\frac{\left\{\begin{array}{c}k+m\\k\end{array}\right\}}{\binom{k+m}k}$, $1\leqslant k,m\leqslant10$, where $\left\{\begin{array}{c}k+m\\k\end{array}\right\}$ is the Stirling number of the second kind and $\binom{k+m}k$ the binomial coefficient, \begin{array}{cccccccccc} \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8} & \frac{1}{9} & \frac{1}{10} & \frac{1}{11} \\ 1 & \frac{7}{6} & \frac{3}{2} & \frac{31}{15} & 3 & \frac{127}{28} & \frac{85}{12} & \frac{511}{45} & \frac{93}{5} & \frac{2047}{66} \\ \frac{3}{2} & \frac{5}{2} & \frac{9}{2} & \frac{43}{5} & \frac{69}{4} & \frac{3025}{84} & \frac{311}{4} & \frac{2591}{15} & \frac{3933}{10} & \frac{20125}{22} \\ 2 & \frac{13}{3} & 10 & \frac{243}{10} & \frac{185}{3} & \frac{6821}{42} & \frac{1325}{3} & \frac{55591}{45} & 3542 & \frac{114195}{11} \\ \frac{5}{2} & \frac{20}{3} & \frac{75}{4} & \frac{331}{6} & \frac{675}{4} & \frac{11215}{21} & \frac{5225}{3} & \frac{52507}{9} & \frac{40035}{2} & \frac{772040}{11} \\ 3 & \frac{19}{2} & \frac{63}{2} & \frac{1087}{10} & \frac{777}{2} & \frac{30083}{21} & 5432 & \frac{63373}{3} & \frac{420273}{5} & \frac{15027069}{44} \\ \frac{7}{2} & \frac{77}{6} & 49 & \frac{1939}{10} & \frac{4753}{6} & \frac{9992}{3} & \frac{43120}{3} & \frac{952777}{15} & \frac{5737557}{20} & \frac{58163133}{44} \\ 4 & \frac{50}{3} & 72 & \frac{4819}{15} & 1476 & \frac{146240}{21} & 33664 & \frac{4992457}{30} & \frac{4199094}{5} & \frac{142561135}{33} \\ \frac{9}{2} & 21 & \frac{405}{4} & \frac{2514}{5} & 2565 & \frac{93886}{7} & \frac{573975}{8} & \frac{3919713}{10} & \frac{8735121}{4} & \frac{272592195}{22} \\ 5 & \frac{155}{6} & \frac{275}{2} & 752 & \frac{12650}{3} & \frac{677465}{28} & \frac{567325}{4} & \frac{15266213}{18} & \frac{10333565}{2} & \frac{1409284345}{44} \end{array} I noticed that the denominators of the $m$th column are divisible by $m+1$. However in the table of $(m+1)\frac{\left\{\begin{array}{c}k+m\\k\end{array}\right\}}{\binom{k+m} k}$, \begin{array}{cccccccccc} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 2 & \frac{7}{2} & 6 & \frac{31}{3} & 18 & \frac{127}{4} & \frac{170}{3} & \frac{511}{5} & 186 & \frac{2047}{6} \\ 3 & \frac{15}{2} & 18 & 43 & \frac{207}{2} & \frac{3025}{12} & 622 & \frac{7773}{5} & 3933 & \frac{20125}{2} \\ 4 & 13 & 40 & \frac{243}{2} & 370 & \frac{6821}{6} & \frac{10600}{3} & \frac{55591}{5} & 35420 & 114195 \\ 5 & 20 & 75 & \frac{1655}{6} & \frac{2025}{2} & \frac{11215}{3} & \frac{41800}{3} & 52507 & 200175 & 772040 \\ 6 & \frac{57}{2} & 126 & \frac{1087}{2} & 2331 & \frac{30083}{3} & 43456 & 190119 & 840546 & \frac{15027069}{4} \\ 7 & \frac{77}{2} & 196 & \frac{1939}{2} & 4753 & \frac{69944}{3} & \frac{344960}{3} & \frac{2858331}{5} & \frac{5737557}{2} & \frac{58163133}{4} \\ 8 & 50 & 288 & \frac{4819}{3} & 8856 & \frac{146240}{3} & 269312 & \frac{14977371}{10} & 8398188 & \frac{142561135}{3} \\ 9 & 63 & 405 & 2514 & 15390 & 93886 & 573975 & \frac{35277417}{10} & \frac{43675605}{2} & \frac{272592195}{2} \\ 10 & \frac{155}{2} & 550 & 3760 & 25300 & \frac{677465}{4} & 1134650 & \frac{15266213}{2} & 51667825 & \frac{1409284345}{4} \end{array} one still sees denominators, which still tend only to depend on $m$. Least common multiples of these denominators in the $m$th column seem to go like \begin{array}{cc} m \\ 1 & 1 \\ 2 & 2 \\ 3 & 1 \\ 4 & 6 \\ 5 & 2 \\ 6 & 12 \\ 7 & 3 \\ 8 & 10 \\ 9 & 2 \\ 10 & 12 \\ 11 & 2 \\ 12 & 420 \\ 13 & 60 \\ 14 & 24 \\ 15 & 3 \\ 16 & 90 \\ 17 & 10 \\ 18 & 420 \\ 19 & 42 \\ 20 & 660 \end{array} but here I am not sure enough since I only looked into the columns as deep as up to $k=20$.
What is this sequence? Can one give an explicit formula for the numerator and denominator of $\frac{\left\{\begin{array}{c}k+m\\k\end{array}\right\}}{\binom{k+m} k}$ after cancellation?