Yes, Block 'g' also fits the same pattern beginning with element No. 121. Only it would make ADOMAH PT 18 elements (or 9 units) taller.
You are absolutely right. There is the relationship of the sizes of the sub shells to the size of the tetrahedron. The tetrahedron shown on http://www.perfectperiodictable.com site has edge equal to 9 units. Therefore, there could be only 4 rectangles possible (if only prime numbers are used): 1x8 (s-block), 3x6 (p-block), 5x4(d-block) and 7x2(f-block). There could be no other rectangles possible, given that only prime numbers are used. (green slices also assume the same dimensions if the tetrahedron edge is odd number, only in opposite order).
Now, try to make edge of the tetrahedron E=10 units long. We would have following combinations, for "red" slices (spaced 2 units appart as measured along the edges):
1x9 (s-block), 3x7 (p-block), 5x5(d-block), 7x3(f-block) and 9x1(g-block), These are the "red layers", representing the blocks.
The green layers would be 8x2, 6x4, 4x6 and 2x8 (just spacers, or matrix, or, perhaps the neutrons within the nucleus).
You can see that g-block is 18 elements long (where 1 unit is a pair of elements).
If edge is E=11 units, then red layers would be:
1x10 (s-block), 3x8(p), 5x6(d), 7x4(f) and 9x2(block-g, 18 element tall, total of 36 elemets).
Therefore, if tetrahedron has edge "E" and rectangles have sides "a" and "b", where side "a" is always odd and "b" can be either even or odd, the dimensions of the subshell blocks follow rule: a + b = E (using only prime numbers). Number of slices, or sub shells: NOS = E/2 with no reminder.
All other blocks, such as 'h', 'i', 'j' ... can also be derived in the same manner by making tetrahedron edge E = 12, 13, 14, 15, 16, 17 ... units.
This is not a coincidence: The Periodic Law can be completely understood in terms of the tetrahedral geometry.