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EXAMPLE
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For n <= 4, all polyominoes with n-1 cells are needed to obtain all polyominoes with n cells by adding one cell, so a(n) = A000105(n-1).
For n = 5, all but the square tetromino are needed to obtain all pentominoes, so a(5) = A000105(4)-1 = 4.
For n = 6, there are 5 different sets of a(6) = 8 pentominoes that are sufficient to obtain all hexominoes. One of these sets consists of the I, L, N, P, U, V, W, and Y pentominoes. The X pentomino is the only pentomino that does not appear in any of these sets. The I, L, N, and W pentominoes are needed in all such sets.
For n = 7, there are 8 different sets of a(7) = 19 hexominoes that are sufficient to obtain all heptominoes. 14
hexominoes appear in all these sets, 10 appear in none of them.
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