I've come across a two-parameter sequence $a_{nk}$ with $n=1,2,\ldots$ and $k=1,2,\ldots,n$, and I would like to identify a closed expression for it. So far I have the elements
$a_{1k}=\{1\}\\ a_{2k}=\{1,3\}\\ a_{3k}=\{1,8,12\}\\ a_{4k}=\{1,15,60,60\}\\ a_{5k}=\{1,24,180,480,360\}\\ a_{6k}=\{1,35,420,2100,4200,2520\}\\ a_{7k}=\{1,48,840,6720,25200,40320,20160\}\ ,$
with $k$ increasing to the right.
There seems to be a lot of order in the elements, and I think there's probably a not too complicated expression for it - but I haven't been able to find one, despite trying the integer sequence encyclopedia on various subsets and other ways to look for a system.
So far I've identified the following:
Looking at the second $k$ for each $n>1$ they can be described with a simple formula giving $\{3,8,15,24,35,48\}=\{1\times3,2\times4,3\times5,4\times6,5\times7,6\times8\}$.
Also each row $n$ is divisible by $n-1$ except for $n=1$ and $k=1$. Performing this division gives (neglecting all $a_{nk}$ for $k=1$)
$a_{2k}=\{3\}\\ a_{3k}=\{4,6\}\\ a_{4k}=\{5,20,20\}\\ a_{5k}=\{6,45,120,90\}\\ a_{6k}=\{7,84,420,840,504\}\\ a_{7k}=\{8,140,1120,4200,6720,3360\}\ .$
Now the previously mentioned order in $k=2$ is obvious.
But there I'm stuck. Any ideas?
UPDATE: I've realised the elements have a factor ${n+1 \choose k-1}$. Dividing out this factor leaves the elements
1
1 1
1 2 2
1 3 6 6
1 4 12 24 24
1 5 20 60 120 120
1 6 30 120 360 720 720 ,
which in itself looks pretty binomial-ish..
Hint: You can find the numbers in your updated section as A008279 in OEIS.