After some thinking of how to incorporate c in our formula, I came up with this:
Look at the differences between the numbers:
b-a gives 0, 1, 1 and 2 resp.
c-b gives 1, 1, 2 and 0 resp.
I want to get a sequence of 2, 2, 2, 4 so I can multiply with b to get d. The above series don't seem to help much unless I subtract 1 more, then I get:
b-a-1 gives -1, 0, 0 and 1 resp.
c-b-1 gives 0, 0, 1 and -1 resp.
By multiplying these I can single out the last equation:
(b-a-1)(c-b-1) gives 0, 0, 0, -1
subtracting from 1 gives
1 - (b-a-1)(c-b-1) gives 1, 1, 1, 2
this is half of what I want so I multiply by 2:
2(1 - (b-a-1)(c-b-1)) gives 2, 2, 2, 4
This can be multiplied by b to get d:
2b(1 - (b-a-1)(c-b-1)) gives 4, 8, 12, 32
So there is another solution. ;)