Smallest region of the plane that contains all free n-ominoes
In the interest of getting the process started, here's a quick (but not very optimal) answer.

Pattern:

n = 8:
########
 ######
  #####
   ####
    ###
     ##


Take a triangle with length n - 1, stick an extra square onto the corner, and cut off the bottom square.

Proof that all n-ominos fit:

Note that any n-omino can fit in a rectangle with length + width at most n + 1.

If an n-omino fits in a rectangle with length + width at most n, it fits nicely in the triangle (which is the union of all such rectangles). If it happens to use the cut-off square, transposing it will fit in the triangle.

Otherwise, we have a chain with at most one branch.  We can always fit one end of the chain into the extra square (prove this with casework), and the rest fits into a rectangle with length + width at most n, reducing to the case above.

The only case where the above doesn't work is the case where we use both the extra square and the cut-off square.  There's only one such n-omino (the long L), and that one fits inside the triangle transposed.

Code (Python 2):

def f(n):
    if n &lt; 7:
        return [0, 1, 2, 4, 6, 9, 12][n]
    return n * (n - 1) / 2


Table:

 1:   1
 2:   2
 3:   4
 4:   6
 5:   9
 6:  12
 7:  21
 8:  28
 9:  36
10:  45
11:  55
12:  66
13:  78
14:  91
15: 105
16: 120
17: 136
18: 153
19: 171
20: 190
... more cases can be generated if necessary.