Polyominoes families filling a square

Here I consider the problem of covering a square by full series of polyominoes (possibly disregarding those with holes). The solutions I found are the following:

the 33 square, which can be covered by the monomino, the domino and the triominoes (1,2,3)-ominoes for brevity; the 99 square tiled by the (1,4,5)-ominoes; the 3131 square obtained with the (2,6,7)-ominoes and the largest one, the 61 61 square tiles by octominoes, heptaminoes, pentaminoes, triominoes and the domino.

According to math, the (2,4,6,8)-ominoes (without holes) can cover the 5656 square, since 12 + 54 + 356 + 3638 = 5656. Unfortunately, by means of parity considerations it is easy to see that this is not possible, since a checkerboard 5656 has an equal number of black and white squares, but the (2,4,6,8)-ominoes could not be positioned on a checkerboard in such a way to cover the same number of black and white squares.

The next square in this series should be the 107107, to be tiled with the monomino, the triominoes, the heptaminoes and the nonominoes, but currently it is beyond the capabilities of my programs.

Instead, I used all the pieces from the monomino to the octominoes, plus a octomino with a hole, to cover a 6363 square with the exception of two symmetric holes.