$\frac{1}{2}$
$\frac{1}{4}$ $\frac{1}{2}$
$\frac{1}{6}$ $\frac{1}{4}$ $\frac{11}{24}$
$\frac{1}{8}$ $\frac{1}{6}$ $\frac{11}{48}$ $\frac{5}{12}$
$\frac{1}{10}$ $\frac{1}{8}$ $\frac{11}{72}$ $\frac{5}{24}$ $\frac{137}{360}$
$\frac{1}{12}$ $\frac{1}{10}$ $\frac{11}{96}$ $\frac{5}{36}$ $\frac{137}{720}$ $\frac{7}{20}$
$\frac{1}{14}$ $\frac{1}{12}$ $\frac{11}{120}$ $\frac{5}{48}$ $\frac{137}{1080}$ $\frac{7}{40}$ $\frac{363}{1120}$
...
Does anyone know what's the generate term expression for this number sequence
They are $$\frac{|S_{m,2}|}{m!n}$$ where $S_{m,2}$ are the Stirling numbers of the first kind, $m$ is the column (starting from $m=2$), and $n$ is the row counting from the diagonal (i.e. row-col+1).
The Stirling numbers $|S_{m,k}|=\begin{bmatrix}m\\k\end{bmatrix}$ are defined here.
For $k=2$, they are $0, 1, 3, 11, 50, 274, 1764$, etc. So starting with $m=2$ and $n=1$, we get $$\begin{array}{l|lllll} n=1&\frac{1}{2!}=\frac{1}{2}&\frac{3}{3!}=\frac{1}{2}& \frac{11}{4!}=\frac{11}{24}& \frac{50}{5!}=\frac{5}{12}& \frac{274}{6!}=\frac{137}{360}\\ n=2&\frac{1}{2!2}&\frac{3}{3!2}& \frac{11}{4!2}& \frac{50}{5!2}& \frac{274}{6!2}\\ n=3&\frac{1}{2!3}&\frac{3}{3!3}& \frac{11}{4!3}& \frac{50}{5!3}& \frac{274}{6!3} \end{array}$$