Well, it's probably not the intended answer, but I believe it is a valid one.
Let's call them a,b,c=d, then I first check what factors can be found in d.
They are (2,2), (2,2,2), (2,2,3), (2,2,2,2,2) for the first line to the last line resp. Now I notice that if you do the same for b you'll get (2), (2,2), (2,3), (2,2,2) which is fairly close already. For the first 3 equations you could write d=2b and for the 4th it is d=4b. So how can we separate the fourth series from the other 3? If we take b-a, we get 0, 1, 1 and 2, which is almost what I want except for the first line. This can be solved by taking the factorial since 0! = 1! = 1.
So if I take (b-a)! I get 1, 1, 1, 2, multiplying that by 2 gives 2, 2, 2, 4 which are the numbers that I need to multiply b with to get d. So our formula becomes 2b(b-a)!=d, which works for all 4 examples but doesn't use the value of c.
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